3.7.92 \(\int \frac {\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^3} \, dx\) [692]

Optimal. Leaf size=271 \[ -\frac {3 a C \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac {\left (2 A b^6+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} b^4 (a+b)^{5/2} d}+\frac {\left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \]

[Out]

-3*a*C*arctanh(sin(d*x+c))/b^4/d+(2*A*b^6+6*a^6*C-15*a^4*b^2*C+a^2*b^4*(A+12*C))*arctanh((a-b)^(1/2)*tan(1/2*d
*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/b^4/(a+b)^(5/2)/d+1/2*(A*b^2+3*C*a^2-2*C*b^2)*tan(d*x+c)/b^3/(a^2-b^2)/d-1/
2*(A*b^2+C*a^2)*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^2-1/2*a*(2*A*b^4-3*a^4*C+a^2*b^2*(A+6*C
))*tan(d*x+c)/b^3/(a^2-b^2)^2/d/(a+b*sec(d*x+c))

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Rubi [A]
time = 0.73, antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4184, 4175, 4167, 4083, 3855, 3916, 2738, 214} \begin {gather*} -\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac {\left (3 a^2 C+A b^2-2 b^2 C\right ) \tan (c+d x)}{2 b^3 d \left (a^2-b^2\right )}-\frac {a \left (-3 a^4 C+a^2 b^2 (A+6 C)+2 A b^4\right ) \tan (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac {\left (6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)+2 A b^6\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac {3 a C \tanh ^{-1}(\sin (c+d x))}{b^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^3,x]

[Out]

(-3*a*C*ArcTanh[Sin[c + d*x]])/(b^4*d) + ((2*A*b^6 + 6*a^6*C - 15*a^4*b^2*C + a^2*b^4*(A + 12*C))*ArcTanh[(Sqr
t[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^4*(a + b)^(5/2)*d) + ((A*b^2 + 3*a^2*C - 2*b^2*C)*Ta
n[c + d*x])/(2*b^3*(a^2 - b^2)*d) - ((A*b^2 + a^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Se
c[c + d*x])^2) - (a*(2*A*b^4 - 3*a^4*C + a^2*b^2*(A + 6*C))*Tan[c + d*x])/(2*b^3*(a^2 - b^2)^2*d*(a + b*Sec[c
+ d*x]))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4083

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 4167

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m
+ 2))), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*
B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 4175

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e
+ f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*C
sc[e + f*x])^(m + 1)*Simp[b*(m + 1)*((-a)*(b*B - a*C) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) +
C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 4184

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(-d)*(A*b^2 + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e +
f*x])^(n - 1)/(b*f*(a^2 - b^2)*(m + 1))), x] + Dist[d/(b*(a^2 - b^2)*(m + 1)), Int[(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^(n - 1)*Simp[A*b^2*(n - 1) + a^2*C*(n - 1) + a*b*(A + C)*(m + 1)*Csc[e + f*x] - (A*b^2*(m
+ n + 1) + C*(a^2*n + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2 -
b^2, 0] && LtQ[m, -1] && GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx &=-\frac {\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\int \frac {\sec ^2(c+d x) \left (2 \left (A b^2+a^2 C\right )-2 a b (A+C) \sec (c+d x)-\left (A b^2+3 a^2 C-2 b^2 C\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac {\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\int \frac {\sec (c+d x) \left (-b \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right )-a \left (a^2-b^2\right ) \left (A b^2-3 a^2 C+4 b^2 C\right ) \sec (c+d x)-b \left (a^2-b^2\right ) \left (A b^2+3 a^2 C-2 b^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\int \frac {\sec (c+d x) \left (-b^2 \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right )+6 a b \left (a^2-b^2\right )^2 C \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=\frac {\left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {(3 a C) \int \sec (c+d x) \, dx}{b^4}+\frac {\left (2 A b^6+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac {3 a C \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac {\left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\left (2 A b^6+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{2 b^5 \left (a^2-b^2\right )^2}\\ &=-\frac {3 a C \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac {\left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\left (2 A b^6+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^5 \left (a^2-b^2\right )^2 d}\\ &=-\frac {3 a C \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac {\left (a^2 A b^4+2 A b^6+6 a^6 C-15 a^4 b^2 C+12 a^2 b^4 C\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} b^4 (a+b)^{5/2} d}+\frac {\left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 2.52, size = 421, normalized size = 1.55 \begin {gather*} \frac {(b+a \cos (c+d x)) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {2 \left (2 A b^6+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (c+d x))^2}{\left (a^2-b^2\right )^{5/2}}+6 a C (b+a \cos (c+d x))^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 a C (b+a \cos (c+d x))^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 b C (b+a \cos (c+d x))^2 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {2 b C (b+a \cos (c+d x))^2 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {a b^2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{(a-b) (a+b)}+\frac {a b \left (-3 A b^4+4 a^4 C-7 a^2 b^2 C\right ) (b+a \cos (c+d x)) \sin (c+d x)}{(a-b)^2 (a+b)^2}\right )}{b^4 d (A+2 C+A \cos (2 (c+d x))) (a+b \sec (c+d x))^3} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^3,x]

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*((-2*(2*A*b^6 + 6*a^6*C - 15*a^4*b^2*C + a^2*b^4*(A
+ 12*C))*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x])^2)/(a^2 - b^2)^(5/2) + 6*a*
C*(b + a*Cos[c + d*x])^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 6*a*C*(b + a*Cos[c + d*x])^2*Log[Cos[(c +
d*x)/2] + Sin[(c + d*x)/2]] + (2*b*C*(b + a*Cos[c + d*x])^2*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x
)/2]) + (2*b*C*(b + a*Cos[c + d*x])^2*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + (a*b^2*(A*b^2
+ a^2*C)*Sin[c + d*x])/((a - b)*(a + b)) + (a*b*(-3*A*b^4 + 4*a^4*C - 7*a^2*b^2*C)*(b + a*Cos[c + d*x])*Sin[c
+ d*x])/((a - b)^2*(a + b)^2)))/(b^4*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^3)

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Maple [A]
time = 0.74, size = 356, normalized size = 1.31

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (\frac {-\frac {\left (a A \,b^{3}+4 A \,b^{4}-4 a^{4} C +a^{3} b C +8 C \,a^{2} b^{2}\right ) a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (a A \,b^{3}-4 A \,b^{4}+4 a^{4} C +a^{3} b C -8 C \,a^{2} b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )^{2}}-\frac {\left (a^{2} A \,b^{4}+2 A \,b^{6}+6 a^{6} C -15 a^{4} b^{2} C +12 C \,a^{2} b^{4}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}-\frac {C}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 a C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{4}}-\frac {C}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 a C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{4}}}{d}\)	\(356\)
default	\(\frac {-\frac {2 \left (\frac {-\frac {\left (a A \,b^{3}+4 A \,b^{4}-4 a^{4} C +a^{3} b C +8 C \,a^{2} b^{2}\right ) a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (a A \,b^{3}-4 A \,b^{4}+4 a^{4} C +a^{3} b C -8 C \,a^{2} b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )^{2}}-\frac {\left (a^{2} A \,b^{4}+2 A \,b^{6}+6 a^{6} C -15 a^{4} b^{2} C +12 C \,a^{2} b^{4}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}-\frac {C}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 a C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{4}}-\frac {C}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 a C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{4}}}{d}\)	\(356\)
risch	\(\text {Expression too large to display}\)	\(1393\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-2/b^4*((-1/2*(A*a*b^3+4*A*b^4-4*C*a^4+C*a^3*b+8*C*a^2*b^2)*a*b/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^
3-1/2*(A*a*b^3-4*A*b^4+4*C*a^4+C*a^3*b-8*C*a^2*b^2)*a*b/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c))/(a*tan(1/2*d
*x+1/2*c)^2-b*tan(1/2*d*x+1/2*c)^2-a-b)^2-1/2*(A*a^2*b^4+2*A*b^6+6*C*a^6-15*C*a^4*b^2+12*C*a^2*b^4)/(a^4-2*a^2
*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))-C/b^3/(tan(1/2*d*x+1/2*c)
-1)+3*a*C/b^4*ln(tan(1/2*d*x+1/2*c)-1)-C/b^3/(tan(1/2*d*x+1/2*c)+1)-3*a*C/b^4*ln(tan(1/2*d*x+1/2*c)+1))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 747 vs. \(2 (257) = 514\).
time = 26.67, size = 1552, normalized size = 5.73 \begin {gather*} \left [\frac {{\left ({\left (6 \, C a^{8} - 15 \, C a^{6} b^{2} + {\left (A + 12 \, C\right )} a^{4} b^{4} + 2 \, A a^{2} b^{6}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (6 \, C a^{7} b - 15 \, C a^{5} b^{3} + {\left (A + 12 \, C\right )} a^{3} b^{5} + 2 \, A a b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (6 \, C a^{6} b^{2} - 15 \, C a^{4} b^{4} + {\left (A + 12 \, C\right )} a^{2} b^{6} + 2 \, A b^{8}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 6 \, {\left ({\left (C a^{9} - 3 \, C a^{7} b^{2} + 3 \, C a^{5} b^{4} - C a^{3} b^{6}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (C a^{8} b - 3 \, C a^{6} b^{3} + 3 \, C a^{4} b^{5} - C a^{2} b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (C a^{7} b^{2} - 3 \, C a^{5} b^{4} + 3 \, C a^{3} b^{6} - C a b^{8}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 6 \, {\left ({\left (C a^{9} - 3 \, C a^{7} b^{2} + 3 \, C a^{5} b^{4} - C a^{3} b^{6}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (C a^{8} b - 3 \, C a^{6} b^{3} + 3 \, C a^{4} b^{5} - C a^{2} b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (C a^{7} b^{2} - 3 \, C a^{5} b^{4} + 3 \, C a^{3} b^{6} - C a b^{8}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C a^{6} b^{3} - 6 \, C a^{4} b^{5} + 6 \, C a^{2} b^{7} - 2 \, C b^{9} + {\left (6 \, C a^{8} b - 17 \, C a^{6} b^{3} - {\left (3 \, A - 13 \, C\right )} a^{4} b^{5} + {\left (3 \, A - 2 \, C\right )} a^{2} b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (9 \, C a^{7} b^{2} + {\left (A - 25 \, C\right )} a^{5} b^{4} - 5 \, {\left (A - 4 \, C\right )} a^{3} b^{6} + 4 \, {\left (A - C\right )} a b^{8}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} b^{4} - 3 \, a^{6} b^{6} + 3 \, a^{4} b^{8} - a^{2} b^{10}\right )} d \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{7} b^{5} - 3 \, a^{5} b^{7} + 3 \, a^{3} b^{9} - a b^{11}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} b^{6} - 3 \, a^{4} b^{8} + 3 \, a^{2} b^{10} - b^{12}\right )} d \cos \left (d x + c\right )\right )}}, \frac {{\left ({\left (6 \, C a^{8} - 15 \, C a^{6} b^{2} + {\left (A + 12 \, C\right )} a^{4} b^{4} + 2 \, A a^{2} b^{6}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (6 \, C a^{7} b - 15 \, C a^{5} b^{3} + {\left (A + 12 \, C\right )} a^{3} b^{5} + 2 \, A a b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (6 \, C a^{6} b^{2} - 15 \, C a^{4} b^{4} + {\left (A + 12 \, C\right )} a^{2} b^{6} + 2 \, A b^{8}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - 3 \, {\left ({\left (C a^{9} - 3 \, C a^{7} b^{2} + 3 \, C a^{5} b^{4} - C a^{3} b^{6}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (C a^{8} b - 3 \, C a^{6} b^{3} + 3 \, C a^{4} b^{5} - C a^{2} b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (C a^{7} b^{2} - 3 \, C a^{5} b^{4} + 3 \, C a^{3} b^{6} - C a b^{8}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (C a^{9} - 3 \, C a^{7} b^{2} + 3 \, C a^{5} b^{4} - C a^{3} b^{6}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (C a^{8} b - 3 \, C a^{6} b^{3} + 3 \, C a^{4} b^{5} - C a^{2} b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (C a^{7} b^{2} - 3 \, C a^{5} b^{4} + 3 \, C a^{3} b^{6} - C a b^{8}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{6} b^{3} - 6 \, C a^{4} b^{5} + 6 \, C a^{2} b^{7} - 2 \, C b^{9} + {\left (6 \, C a^{8} b - 17 \, C a^{6} b^{3} - {\left (3 \, A - 13 \, C\right )} a^{4} b^{5} + {\left (3 \, A - 2 \, C\right )} a^{2} b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (9 \, C a^{7} b^{2} + {\left (A - 25 \, C\right )} a^{5} b^{4} - 5 \, {\left (A - 4 \, C\right )} a^{3} b^{6} + 4 \, {\left (A - C\right )} a b^{8}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{8} b^{4} - 3 \, a^{6} b^{6} + 3 \, a^{4} b^{8} - a^{2} b^{10}\right )} d \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{7} b^{5} - 3 \, a^{5} b^{7} + 3 \, a^{3} b^{9} - a b^{11}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} b^{6} - 3 \, a^{4} b^{8} + 3 \, a^{2} b^{10} - b^{12}\right )} d \cos \left (d x + c\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(((6*C*a^8 - 15*C*a^6*b^2 + (A + 12*C)*a^4*b^4 + 2*A*a^2*b^6)*cos(d*x + c)^3 + 2*(6*C*a^7*b - 15*C*a^5*b^
3 + (A + 12*C)*a^3*b^5 + 2*A*a*b^7)*cos(d*x + c)^2 + (6*C*a^6*b^2 - 15*C*a^4*b^4 + (A + 12*C)*a^2*b^6 + 2*A*b^
8)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b
*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 6*((C*a^9 -
3*C*a^7*b^2 + 3*C*a^5*b^4 - C*a^3*b^6)*cos(d*x + c)^3 + 2*(C*a^8*b - 3*C*a^6*b^3 + 3*C*a^4*b^5 - C*a^2*b^7)*co
s(d*x + c)^2 + (C*a^7*b^2 - 3*C*a^5*b^4 + 3*C*a^3*b^6 - C*a*b^8)*cos(d*x + c))*log(sin(d*x + c) + 1) + 6*((C*a
^9 - 3*C*a^7*b^2 + 3*C*a^5*b^4 - C*a^3*b^6)*cos(d*x + c)^3 + 2*(C*a^8*b - 3*C*a^6*b^3 + 3*C*a^4*b^5 - C*a^2*b^
7)*cos(d*x + c)^2 + (C*a^7*b^2 - 3*C*a^5*b^4 + 3*C*a^3*b^6 - C*a*b^8)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2
*(2*C*a^6*b^3 - 6*C*a^4*b^5 + 6*C*a^2*b^7 - 2*C*b^9 + (6*C*a^8*b - 17*C*a^6*b^3 - (3*A - 13*C)*a^4*b^5 + (3*A
- 2*C)*a^2*b^7)*cos(d*x + c)^2 + (9*C*a^7*b^2 + (A - 25*C)*a^5*b^4 - 5*(A - 4*C)*a^3*b^6 + 4*(A - C)*a*b^8)*co
s(d*x + c))*sin(d*x + c))/((a^8*b^4 - 3*a^6*b^6 + 3*a^4*b^8 - a^2*b^10)*d*cos(d*x + c)^3 + 2*(a^7*b^5 - 3*a^5*
b^7 + 3*a^3*b^9 - a*b^11)*d*cos(d*x + c)^2 + (a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*d*cos(d*x + c)), 1/2*((
(6*C*a^8 - 15*C*a^6*b^2 + (A + 12*C)*a^4*b^4 + 2*A*a^2*b^6)*cos(d*x + c)^3 + 2*(6*C*a^7*b - 15*C*a^5*b^3 + (A
+ 12*C)*a^3*b^5 + 2*A*a*b^7)*cos(d*x + c)^2 + (6*C*a^6*b^2 - 15*C*a^4*b^4 + (A + 12*C)*a^2*b^6 + 2*A*b^8)*cos(
d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - 3*((C*a
^9 - 3*C*a^7*b^2 + 3*C*a^5*b^4 - C*a^3*b^6)*cos(d*x + c)^3 + 2*(C*a^8*b - 3*C*a^6*b^3 + 3*C*a^4*b^5 - C*a^2*b^
7)*cos(d*x + c)^2 + (C*a^7*b^2 - 3*C*a^5*b^4 + 3*C*a^3*b^6 - C*a*b^8)*cos(d*x + c))*log(sin(d*x + c) + 1) + 3*
((C*a^9 - 3*C*a^7*b^2 + 3*C*a^5*b^4 - C*a^3*b^6)*cos(d*x + c)^3 + 2*(C*a^8*b - 3*C*a^6*b^3 + 3*C*a^4*b^5 - C*a
^2*b^7)*cos(d*x + c)^2 + (C*a^7*b^2 - 3*C*a^5*b^4 + 3*C*a^3*b^6 - C*a*b^8)*cos(d*x + c))*log(-sin(d*x + c) + 1
) + (2*C*a^6*b^3 - 6*C*a^4*b^5 + 6*C*a^2*b^7 - 2*C*b^9 + (6*C*a^8*b - 17*C*a^6*b^3 - (3*A - 13*C)*a^4*b^5 + (3
*A - 2*C)*a^2*b^7)*cos(d*x + c)^2 + (9*C*a^7*b^2 + (A - 25*C)*a^5*b^4 - 5*(A - 4*C)*a^3*b^6 + 4*(A - C)*a*b^8)
*cos(d*x + c))*sin(d*x + c))/((a^8*b^4 - 3*a^6*b^6 + 3*a^4*b^8 - a^2*b^10)*d*cos(d*x + c)^3 + 2*(a^7*b^5 - 3*a
^5*b^7 + 3*a^3*b^9 - a*b^11)*d*cos(d*x + c)^2 + (a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*d*cos(d*x + c))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**3,x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a + b*sec(c + d*x))**3, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 521 vs. \(2 (257) = 514\).
time = 0.57, size = 521, normalized size = 1.92 \begin {gather*} \frac {\frac {{\left (6 \, C a^{6} - 15 \, C a^{4} b^{2} + A a^{2} b^{4} + 12 \, C a^{2} b^{4} + 2 \, A b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {3 \, C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} + \frac {3 \, C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} - \frac {4 \, C a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, C a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, C a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, C a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, A a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, A a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, C a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, C a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, C a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, A a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}^{2}} - \frac {2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} b^{3}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

((6*C*a^6 - 15*C*a^4*b^2 + A*a^2*b^4 + 12*C*a^2*b^4 + 2*A*b^6)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*
b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^4*b^4 - 2*a^2*b^6 + b^8)
*sqrt(-a^2 + b^2)) - 3*C*a*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 + 3*C*a*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b
^4 - (4*C*a^6*tan(1/2*d*x + 1/2*c)^3 - 5*C*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 7*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 -
A*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 8*C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*A*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 +
4*A*a*b^5*tan(1/2*d*x + 1/2*c)^3 - 4*C*a^6*tan(1/2*d*x + 1/2*c) - 5*C*a^5*b*tan(1/2*d*x + 1/2*c) + 7*C*a^4*b^2
*tan(1/2*d*x + 1/2*c) - A*a^3*b^3*tan(1/2*d*x + 1/2*c) + 8*C*a^3*b^3*tan(1/2*d*x + 1/2*c) + 3*A*a^2*b^4*tan(1/
2*d*x + 1/2*c) + 4*A*a*b^5*tan(1/2*d*x + 1/2*c))/((a^4*b^3 - 2*a^2*b^5 + b^7)*(a*tan(1/2*d*x + 1/2*c)^2 - b*ta
n(1/2*d*x + 1/2*c)^2 - a - b)^2) - 2*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*b^3))/d

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Mupad [B]
time = 13.52, size = 2500, normalized size = 9.23 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x))^3),x)

[Out]

((tan(c/2 + (d*x)/2)*(6*C*a^5 + 2*C*b^5 + A*a^2*b^3 - 4*C*a^2*b^3 - 12*C*a^3*b^2 - 4*A*a*b^4 + 2*C*a*b^4 + 3*C
*a^4*b))/((a + b)*(b^5 - 2*a*b^4 + a^2*b^3)) - (tan(c/2 + (d*x)/2)^5*(2*C*b^5 - 6*C*a^5 + A*a^2*b^3 - 4*C*a^2*
b^3 + 12*C*a^3*b^2 + 4*A*a*b^4 - 2*C*a*b^4 + 3*C*a^4*b))/((a*b^3 - b^4)*(a + b)^2) + (2*tan(c/2 + (d*x)/2)^3*(
2*C*b^6 - 6*C*a^6 + 3*A*a^2*b^4 - 6*C*a^2*b^4 + 13*C*a^4*b^2))/(b*(a*b^2 - b^3)*(a + b)^2*(a - b)))/(d*(2*a*b
- tan(c/2 + (d*x)/2)^2*(2*a*b + 3*a^2 - b^2) - tan(c/2 + (d*x)/2)^6*(a^2 - 2*a*b + b^2) + a^2 + b^2 - tan(c/2
+ (d*x)/2)^4*(2*a*b - 3*a^2 + b^2))) + (C*a*atan(((C*a*((8*tan(c/2 + (d*x)/2)*(4*A^2*b^12 + 72*C^2*a^12 - 72*C
^2*a^11*b + 4*A^2*a^2*b^10 + A^2*a^4*b^8 + 36*C^2*a^2*b^10 - 72*C^2*a^3*b^9 + 36*C^2*a^4*b^8 + 288*C^2*a^5*b^7
- 288*C^2*a^6*b^6 - 432*C^2*a^7*b^5 + 441*C^2*a^8*b^4 + 288*C^2*a^9*b^3 - 288*C^2*a^10*b^2 + 48*A*C*a^2*b^10
- 36*A*C*a^4*b^8 - 6*A*C*a^6*b^6 + 12*A*C*a^8*b^4))/(a*b^12 + b^13 - 3*a^2*b^11 - 3*a^3*b^10 + 3*a^4*b^9 + 3*a
^5*b^8 - a^6*b^7 - a^7*b^6) - (3*C*a*((8*(4*A*b^18 - 6*A*a^2*b^16 + 6*A*a^3*b^15 + 2*A*a^6*b^12 - 2*A*a^7*b^11
+ 24*C*a^2*b^16 + 36*C*a^3*b^15 - 78*C*a^4*b^14 - 42*C*a^5*b^13 + 96*C*a^6*b^12 + 24*C*a^7*b^11 - 54*C*a^8*b^
10 - 6*C*a^9*b^9 + 12*C*a^10*b^8 - 4*A*a*b^17 - 12*C*a*b^17))/(a*b^15 + b^16 - 3*a^2*b^14 - 3*a^3*b^13 + 3*a^4
*b^12 + 3*a^5*b^11 - a^6*b^10 - a^7*b^9) - (24*C*a*tan(c/2 + (d*x)/2)*(8*a*b^17 - 8*a^2*b^16 - 32*a^3*b^15 + 3
2*a^4*b^14 + 48*a^5*b^13 - 48*a^6*b^12 - 32*a^7*b^11 + 32*a^8*b^10 + 8*a^9*b^9 - 8*a^10*b^8))/(b^4*(a*b^12 + b
^13 - 3*a^2*b^11 - 3*a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b^6))))/b^4)*3i)/b^4 + (C*a*((8*tan(c/2
+ (d*x)/2)*(4*A^2*b^12 + 72*C^2*a^12 - 72*C^2*a^11*b + 4*A^2*a^2*b^10 + A^2*a^4*b^8 + 36*C^2*a^2*b^10 - 72*C^2
*a^3*b^9 + 36*C^2*a^4*b^8 + 288*C^2*a^5*b^7 - 288*C^2*a^6*b^6 - 432*C^2*a^7*b^5 + 441*C^2*a^8*b^4 + 288*C^2*a^
9*b^3 - 288*C^2*a^10*b^2 + 48*A*C*a^2*b^10 - 36*A*C*a^4*b^8 - 6*A*C*a^6*b^6 + 12*A*C*a^8*b^4))/(a*b^12 + b^13
- 3*a^2*b^11 - 3*a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b^6) + (3*C*a*((8*(4*A*b^18 - 6*A*a^2*b^16 +
6*A*a^3*b^15 + 2*A*a^6*b^12 - 2*A*a^7*b^11 + 24*C*a^2*b^16 + 36*C*a^3*b^15 - 78*C*a^4*b^14 - 42*C*a^5*b^13 +
96*C*a^6*b^12 + 24*C*a^7*b^11 - 54*C*a^8*b^10 - 6*C*a^9*b^9 + 12*C*a^10*b^8 - 4*A*a*b^17 - 12*C*a*b^17))/(a*b^
15 + b^16 - 3*a^2*b^14 - 3*a^3*b^13 + 3*a^4*b^12 + 3*a^5*b^11 - a^6*b^10 - a^7*b^9) + (24*C*a*tan(c/2 + (d*x)/
2)*(8*a*b^17 - 8*a^2*b^16 - 32*a^3*b^15 + 32*a^4*b^14 + 48*a^5*b^13 - 48*a^6*b^12 - 32*a^7*b^11 + 32*a^8*b^10
+ 8*a^9*b^9 - 8*a^10*b^8))/(b^4*(a*b^12 + b^13 - 3*a^2*b^11 - 3*a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a
^7*b^6))))/b^4)*3i)/b^4)/((16*(108*C^3*a^12 - 54*C^3*a^11*b + 216*C^3*a^4*b^8 + 216*C^3*a^5*b^7 - 702*C^3*a^6*
b^6 - 378*C^3*a^7*b^5 + 864*C^3*a^8*b^4 + 243*C^3*a^9*b^3 - 486*C^3*a^10*b^2 + 12*A^2*C*a*b^11 + 36*A*C^2*a^2*
b^10 + 108*A*C^2*a^3*b^9 - 54*A*C^2*a^4*b^8 - 54*A*C^2*a^5*b^7 - 18*A*C^2*a^7*b^5 + 18*A*C^2*a^8*b^4 + 18*A*C^
2*a^9*b^3 + 12*A^2*C*a^3*b^9 + 3*A^2*C*a^5*b^7))/(a*b^15 + b^16 - 3*a^2*b^14 - 3*a^3*b^13 + 3*a^4*b^12 + 3*a^5
*b^11 - a^6*b^10 - a^7*b^9) + (3*C*a*((8*tan(c/2 + (d*x)/2)*(4*A^2*b^12 + 72*C^2*a^12 - 72*C^2*a^11*b + 4*A^2*
a^2*b^10 + A^2*a^4*b^8 + 36*C^2*a^2*b^10 - 72*C^2*a^3*b^9 + 36*C^2*a^4*b^8 + 288*C^2*a^5*b^7 - 288*C^2*a^6*b^6
- 432*C^2*a^7*b^5 + 441*C^2*a^8*b^4 + 288*C^2*a^9*b^3 - 288*C^2*a^10*b^2 + 48*A*C*a^2*b^10 - 36*A*C*a^4*b^8 -
6*A*C*a^6*b^6 + 12*A*C*a^8*b^4))/(a*b^12 + b^13 - 3*a^2*b^11 - 3*a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 -
a^7*b^6) - (3*C*a*((8*(4*A*b^18 - 6*A*a^2*b^16 + 6*A*a^3*b^15 + 2*A*a^6*b^12 - 2*A*a^7*b^11 + 24*C*a^2*b^16 +
36*C*a^3*b^15 - 78*C*a^4*b^14 - 42*C*a^5*b^13 + 96*C*a^6*b^12 + 24*C*a^7*b^11 - 54*C*a^8*b^10 - 6*C*a^9*b^9 +
12*C*a^10*b^8 - 4*A*a*b^17 - 12*C*a*b^17))/(a*b^15 + b^16 - 3*a^2*b^14 - 3*a^3*b^13 + 3*a^4*b^12 + 3*a^5*b^11
- a^6*b^10 - a^7*b^9) - (24*C*a*tan(c/2 + (d*x)/2)*(8*a*b^17 - 8*a^2*b^16 - 32*a^3*b^15 + 32*a^4*b^14 + 48*a^
5*b^13 - 48*a^6*b^12 - 32*a^7*b^11 + 32*a^8*b^10 + 8*a^9*b^9 - 8*a^10*b^8))/(b^4*(a*b^12 + b^13 - 3*a^2*b^11 -
3*a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b^6))))/b^4))/b^4 - (3*C*a*((8*tan(c/2 + (d*x)/2)*(4*A^2*b
^12 + 72*C^2*a^12 - 72*C^2*a^11*b + 4*A^2*a^2*b^10 + A^2*a^4*b^8 + 36*C^2*a^2*b^10 - 72*C^2*a^3*b^9 + 36*C^2*a
^4*b^8 + 288*C^2*a^5*b^7 - 288*C^2*a^6*b^6 - 432*C^2*a^7*b^5 + 441*C^2*a^8*b^4 + 288*C^2*a^9*b^3 - 288*C^2*a^1
0*b^2 + 48*A*C*a^2*b^10 - 36*A*C*a^4*b^8 - 6*A*C*a^6*b^6 + 12*A*C*a^8*b^4))/(a*b^12 + b^13 - 3*a^2*b^11 - 3*a^
3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b^6) + (3*C*a*((8*(4*A*b^18 - 6*A*a^2*b^16 + 6*A*a^3*b^15 + 2*A
*a^6*b^12 - 2*A*a^7*b^11 + 24*C*a^2*b^16 + 36*C*a^3*b^15 - 78*C*a^4*b^14 - 42*C*a^5*b^13 + 96*C*a^6*b^12 + 24*
C*a^7*b^11 - 54*C*a^8*b^10 - 6*C*a^9*b^9 + 12*C*a^10*b^8 - 4*A*a*b^17 - 12*C*a*b^17))/(a*b^15 + b^16 - 3*a^2*b
^14 - 3*a^3*b^13 + 3*a^4*b^12 + 3*a^5*b^11 - a^6*b^10 - a^7*b^9) + (24*C*a*tan(c/2 + (d*x)/2)*(8*a*b^17 - 8*a^
2*b^16 - 32*a^3*b^15 + 32*a^4*b^14 + 48*a^5*b^1...

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