3.9.2 \(\int \frac {\sec ^2(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^2} \, dx\) [802]

Optimal. Leaf size=164 \[ \frac {(b B-2 a C) \tanh ^{-1}(\sin (c+d x))}{b^3 d}-\frac {2 a \left (a^2 b B-2 b^3 B-2 a^3 C+3 a b^2 C\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^3 (a+b)^{3/2} d}+\frac {C \tan (c+d x)}{b^2 d}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]

[Out]

(B*b-2*C*a)*arctanh(sin(d*x+c))/b^3/d-2*a*(B*a^2*b-2*B*b^3-2*C*a^3+3*C*a*b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+
1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/b^3/(a+b)^(3/2)/d+C*tan(d*x+c)/b^2/d-a^2*(B*b-C*a)*tan(d*x+c)/b^2/(a^2-b^2)/d/
(a+b*sec(d*x+c))

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Rubi [A]
time = 0.46, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4157, 4113, 4167, 4083, 3855, 3916, 2738, 214} \begin {gather*} -\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {2 a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {(b B-2 a C) \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac {C \tan (c+d x)}{b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B - 2*a*C)*ArcTanh[Sin[c + d*x]])/(b^3*d) - (2*a*(a^2*b*B - 2*b^3*B - 2*a^3*C + 3*a*b^2*C)*ArcTanh[(Sqrt[a
- b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*b^3*(a + b)^(3/2)*d) + (C*Tan[c + d*x])/(b^2*d) - (a^2*(b
*B - a*C)*Tan[c + d*x])/(b^2*(a^2 - b^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 4113

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> Simp[(-a^2)*(A*b - a*B)*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*Csc[e + f*x])^(m + 1)*Simp[a*b*(A*b - a*B)*
(m + 1) - (A*b - a*B)*(a^2 + b^2*(m + 1))*Csc[e + f*x] + b*B*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 4157

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 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]

Rubi steps

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

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Mathematica [A]
time = 2.34, size = 240, normalized size = 1.46 \begin {gather*} \frac {-\frac {2 a \left (-a^2 b B+2 b^3 B+2 a^3 C-3 a b^2 C\right ) \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-b B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+b B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 a C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2 b (-b B+a C) \sin (c+d x)}{(a-b) (a+b) (b+a \cos (c+d x))}+b C \tan (c+d x)}{b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a*(-(a^2*b*B) + 2*b^3*B + 2*a^3*C - 3*a*b^2*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2
- b^2)^(3/2) - b*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*a*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]
+ b*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 2*a*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^2*b*(-(b*
B) + a*C)*Sin[c + d*x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])) + b*C*Tan[c + d*x])/(b^3*d)

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Maple [A]
time = 0.41, size = 245, normalized size = 1.49

method	result	size
derivativedivides	\(\frac {-\frac {C}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-b B +2 a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}-\frac {C}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (b B -2 a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}+\frac {2 a \left (\frac {a b \left (b B -a C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}-\frac {\left (a^{2} b B -2 b^{3} B -2 a^{3} C +3 C \,b^{2} a \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3}}}{d}\)	\(245\)
default	\(\frac {-\frac {C}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-b B +2 a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}-\frac {C}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (b B -2 a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}+\frac {2 a \left (\frac {a b \left (b B -a C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}-\frac {\left (a^{2} b B -2 b^{3} B -2 a^{3} C +3 C \,b^{2} a \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3}}}{d}\)	\(245\)
risch	\(\frac {2 i \left (B a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-C \,a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+B \,a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 C \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+C a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+B a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-3 C \,a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+2 b^{3} C \,{\mathrm e}^{i \left (d x +c \right )}+a^{2} b B -2 a^{3} C +C \,b^{2} a \right )}{d \,b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left (-a^{2}+b^{2}\right ) \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{b^{2} d}-\frac {2 a C \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b^{3} d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {2 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) a^{4} C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C \,a^{2}}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{b^{2} d}+\frac {2 a C \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b^{3} d}\)	\(977\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-C/b^2/(tan(1/2*d*x+1/2*c)-1)+1/b^3*(-B*b+2*C*a)*ln(tan(1/2*d*x+1/2*c)-1)-C/b^2/(tan(1/2*d*x+1/2*c)+1)+(B
*b-2*C*a)/b^3*ln(tan(1/2*d*x+1/2*c)+1)+2*a/b^3*(a*b*(B*b-C*a)/(a^2-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)-(B*a^2*b-2*B*b^3-2*C*a^3+3*C*a*b^2)/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((
a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))))

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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)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,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. 529 vs. \(2 (157) = 314\).
time = 19.92, size = 1114, normalized size = 6.79 \begin {gather*} \left [\frac {{\left ({\left (2 \, C a^{5} - B a^{4} b - 3 \, C a^{3} b^{2} + 2 \, B a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, C a^{4} b - B a^{3} b^{2} - 3 \, C a^{2} b^{3} + 2 \, B a b^{4}\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 ) - {\left ({\left (2 \, C a^{6} - B a^{5} b - 4 \, C a^{4} b^{2} + 2 \, B a^{3} b^{3} + 2 \, C a^{2} b^{4} - B a b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, C a^{5} b - B a^{4} b^{2} - 4 \, C a^{3} b^{3} + 2 \, B a^{2} b^{4} + 2 \, C a b^{5} - B b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (2 \, C a^{6} - B a^{5} b - 4 \, C a^{4} b^{2} + 2 \, B a^{3} b^{3} + 2 \, C a^{2} b^{4} - B a b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, C a^{5} b - B a^{4} b^{2} - 4 \, C a^{3} b^{3} + 2 \, B a^{2} b^{4} + 2 \, C a b^{5} - B b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{4} b^{2} - 2 \, C a^{2} b^{4} + C b^{6} + {\left (2 \, C a^{5} b - B a^{4} b^{2} - 3 \, C a^{3} b^{3} + B a^{2} b^{4} + C a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )\right )}}, \frac {2 \, {\left ({\left (2 \, C a^{5} - B a^{4} b - 3 \, C a^{3} b^{2} + 2 \, B a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, C a^{4} b - B a^{3} b^{2} - 3 \, C a^{2} b^{3} + 2 \, B a b^{4}\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 ) - {\left ({\left (2 \, C a^{6} - B a^{5} b - 4 \, C a^{4} b^{2} + 2 \, B a^{3} b^{3} + 2 \, C a^{2} b^{4} - B a b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, C a^{5} b - B a^{4} b^{2} - 4 \, C a^{3} b^{3} + 2 \, B a^{2} b^{4} + 2 \, C a b^{5} - B b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (2 \, C a^{6} - B a^{5} b - 4 \, C a^{4} b^{2} + 2 \, B a^{3} b^{3} + 2 \, C a^{2} b^{4} - B a b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, C a^{5} b - B a^{4} b^{2} - 4 \, C a^{3} b^{3} + 2 \, B a^{2} b^{4} + 2 \, C a b^{5} - B b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{4} b^{2} - 2 \, C a^{2} b^{4} + C b^{6} + {\left (2 \, C a^{5} b - B a^{4} b^{2} - 3 \, C a^{3} b^{3} + B a^{2} b^{4} + C a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\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)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/2*(((2*C*a^5 - B*a^4*b - 3*C*a^3*b^2 + 2*B*a^2*b^3)*cos(d*x + c)^2 + (2*C*a^4*b - B*a^3*b^2 - 3*C*a^2*b^3 +
2*B*a*b^4)*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)) - ((2
*C*a^6 - B*a^5*b - 4*C*a^4*b^2 + 2*B*a^3*b^3 + 2*C*a^2*b^4 - B*a*b^5)*cos(d*x + c)^2 + (2*C*a^5*b - B*a^4*b^2
- 4*C*a^3*b^3 + 2*B*a^2*b^4 + 2*C*a*b^5 - B*b^6)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((2*C*a^6 - B*a^5*b - 4
*C*a^4*b^2 + 2*B*a^3*b^3 + 2*C*a^2*b^4 - B*a*b^5)*cos(d*x + c)^2 + (2*C*a^5*b - B*a^4*b^2 - 4*C*a^3*b^3 + 2*B*
a^2*b^4 + 2*C*a*b^5 - B*b^6)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(C*a^4*b^2 - 2*C*a^2*b^4 + C*b^6 + (2*C*
a^5*b - B*a^4*b^2 - 3*C*a^3*b^3 + B*a^2*b^4 + C*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^5*b^3 - 2*a^3*b^5 + a*b
^7)*d*cos(d*x + c)^2 + (a^4*b^4 - 2*a^2*b^6 + b^8)*d*cos(d*x + c)), 1/2*(2*((2*C*a^5 - B*a^4*b - 3*C*a^3*b^2 +
2*B*a^2*b^3)*cos(d*x + c)^2 + (2*C*a^4*b - B*a^3*b^2 - 3*C*a^2*b^3 + 2*B*a*b^4)*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))) - ((2*C*a^6 - B*a^5*b - 4*C*a^4*b^
2 + 2*B*a^3*b^3 + 2*C*a^2*b^4 - B*a*b^5)*cos(d*x + c)^2 + (2*C*a^5*b - B*a^4*b^2 - 4*C*a^3*b^3 + 2*B*a^2*b^4 +
2*C*a*b^5 - B*b^6)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((2*C*a^6 - B*a^5*b - 4*C*a^4*b^2 + 2*B*a^3*b^3 + 2*
C*a^2*b^4 - B*a*b^5)*cos(d*x + c)^2 + (2*C*a^5*b - B*a^4*b^2 - 4*C*a^3*b^3 + 2*B*a^2*b^4 + 2*C*a*b^5 - B*b^6)*
cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(C*a^4*b^2 - 2*C*a^2*b^4 + C*b^6 + (2*C*a^5*b - B*a^4*b^2 - 3*C*a^3*b
^3 + B*a^2*b^4 + C*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^5*b^3 - 2*a^3*b^5 + a*b^7)*d*cos(d*x + c)^2 + (a^4*b
^4 - 2*a^2*b^6 + b^8)*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 (B + C \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 404 vs. \(2 (157) = 314\).
time = 0.49, size = 404, normalized size = 2.46 \begin {gather*} \frac {\frac {2 \, {\left (2 \, C a^{4} - B a^{3} b - 3 \, C a^{2} b^{2} + 2 \, B a b^{3}\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^{2} b^{3} - b^{5}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, {\left (2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} {\left (a^{2} b^{2} - b^{4}\right )}} - \frac {{\left (2 \, C a - B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac {{\left (2 \, C a - B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(2*C*a^4 - B*a^3*b - 3*C*a^2*b^2 + 2*B*a*b^3)*(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^2*b^3 - b^5)*sqrt(-a^2 + b^2)) - 2*(2*
C*a^3*tan(1/2*d*x + 1/2*c)^3 - B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - C*a*b^2*tan(1
/2*d*x + 1/2*c)^3 + C*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*C*a^3*tan(1/2*d*x + 1/2*c) + B*a^2*b*tan(1/2*d*x + 1/2*c)
- C*a^2*b*tan(1/2*d*x + 1/2*c) + C*a*b^2*tan(1/2*d*x + 1/2*c) + C*b^3*tan(1/2*d*x + 1/2*c))/((a*tan(1/2*d*x +
1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b)*(a^2*b^2 - b^4)) - (2*C*a - B*b)*lo
g(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 + (2*C*a - B*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^3)/d

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Mupad [B]
time = 12.68, size = 2500, normalized size = 15.24 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*tan(c/2 + (d*x)/2)^3*(B*a^2*b - C*b^3 - 2*C*a^3 + C*a*b^2 + C*a^2*b))/(b^2*(a + b)*(a - b)) - (2*tan(c/2 +
(d*x)/2)*(C*b^3 - 2*C*a^3 + B*a^2*b + C*a*b^2 - C*a^2*b))/(b^2*(a + b)*(a - b)))/(d*(a + b + tan(c/2 + (d*x)/
2)^4*(a - b) - 2*a*tan(c/2 + (d*x)/2)^2)) + (atan((((B*b - 2*C*a)*((32*tan(c/2 + (d*x)/2)*(B^2*b^8 + 8*C^2*a^8
- 2*B^2*a*b^7 - 8*C^2*a^7*b + 3*B^2*a^2*b^6 + 4*B^2*a^3*b^5 - 5*B^2*a^4*b^4 - 2*B^2*a^5*b^3 + 2*B^2*a^6*b^2 +
4*C^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*C^2*a^4*b^4 + 16*C^2*a^5*b^3 - 16*C^2*a^6*b^2 - 4*B*C*a*b^7 - 8*B*C*a^7*b +
8*B*C*a^2*b^6 - 8*B*C*a^3*b^5 - 16*B*C*a^4*b^4 + 18*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^
3*b^4) + (((32*(B*a^2*b^10 - B*b^12 - 3*B*a^3*b^9 + B*a^5*b^7 - 3*C*a^2*b^10 - 3*C*a^3*b^9 + 5*C*a^4*b^8 + C*a
^5*b^7 - 2*C*a^6*b^6 + 2*B*a*b^11 + 2*C*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (32*tan(c/2 + (d*x)/2)*(B
*b - 2*C*a)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/(b^3*(a*b^6 + b^7 - a^2*b
^5 - a^3*b^4)))*(B*b - 2*C*a))/b^3)*1i)/b^3 + ((B*b - 2*C*a)*((32*tan(c/2 + (d*x)/2)*(B^2*b^8 + 8*C^2*a^8 - 2*
B^2*a*b^7 - 8*C^2*a^7*b + 3*B^2*a^2*b^6 + 4*B^2*a^3*b^5 - 5*B^2*a^4*b^4 - 2*B^2*a^5*b^3 + 2*B^2*a^6*b^2 + 4*C^
2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*C^2*a^4*b^4 + 16*C^2*a^5*b^3 - 16*C^2*a^6*b^2 - 4*B*C*a*b^7 - 8*B*C*a^7*b + 8*B*
C*a^2*b^6 - 8*B*C*a^3*b^5 - 16*B*C*a^4*b^4 + 18*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4
) - (((32*(B*a^2*b^10 - B*b^12 - 3*B*a^3*b^9 + B*a^5*b^7 - 3*C*a^2*b^10 - 3*C*a^3*b^9 + 5*C*a^4*b^8 + C*a^5*b^
7 - 2*C*a^6*b^6 + 2*B*a*b^11 + 2*C*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (32*tan(c/2 + (d*x)/2)*(B*b -
2*C*a)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/(b^3*(a*b^6 + b^7 - a^2*b^5 -
a^3*b^4)))*(B*b - 2*C*a))/b^3)*1i)/b^3)/((64*(8*C^3*a^8 - 2*B^3*a*b^7 - 4*C^3*a^7*b - 2*B^3*a^2*b^6 + 3*B^3*a^
3*b^5 + B^3*a^4*b^4 - B^3*a^5*b^3 + 12*C^3*a^4*b^4 + 6*C^3*a^5*b^3 - 20*C^3*a^6*b^2 - 12*B*C^2*a^7*b - 20*B*C^
2*a^3*b^5 - 13*B*C^2*a^4*b^4 + 32*B*C^2*a^5*b^3 + 8*B*C^2*a^6*b^2 + 11*B^2*C*a^2*b^6 + 9*B^2*C*a^3*b^5 - 17*B^
2*C*a^4*b^4 - 5*B^2*C*a^5*b^3 + 6*B^2*C*a^6*b^2))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + ((B*b - 2*C*a)*((32*tan(
c/2 + (d*x)/2)*(B^2*b^8 + 8*C^2*a^8 - 2*B^2*a*b^7 - 8*C^2*a^7*b + 3*B^2*a^2*b^6 + 4*B^2*a^3*b^5 - 5*B^2*a^4*b^
4 - 2*B^2*a^5*b^3 + 2*B^2*a^6*b^2 + 4*C^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*C^2*a^4*b^4 + 16*C^2*a^5*b^3 - 16*C^2*a^
6*b^2 - 4*B*C*a*b^7 - 8*B*C*a^7*b + 8*B*C*a^2*b^6 - 8*B*C*a^3*b^5 - 16*B*C*a^4*b^4 + 18*B*C*a^5*b^3 + 8*B*C*a^
6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) + (((32*(B*a^2*b^10 - B*b^12 - 3*B*a^3*b^9 + B*a^5*b^7 - 3*C*a^2*b^1
0 - 3*C*a^3*b^9 + 5*C*a^4*b^8 + C*a^5*b^7 - 2*C*a^6*b^6 + 2*B*a*b^11 + 2*C*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a
^3*b^6) + (32*tan(c/2 + (d*x)/2)*(B*b - 2*C*a)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*
a^6*b^6))/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4)))*(B*b - 2*C*a))/b^3))/b^3 - ((B*b - 2*C*a)*((32*tan(c/2 + (d
*x)/2)*(B^2*b^8 + 8*C^2*a^8 - 2*B^2*a*b^7 - 8*C^2*a^7*b + 3*B^2*a^2*b^6 + 4*B^2*a^3*b^5 - 5*B^2*a^4*b^4 - 2*B^
2*a^5*b^3 + 2*B^2*a^6*b^2 + 4*C^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*C^2*a^4*b^4 + 16*C^2*a^5*b^3 - 16*C^2*a^6*b^2 -
4*B*C*a*b^7 - 8*B*C*a^7*b + 8*B*C*a^2*b^6 - 8*B*C*a^3*b^5 - 16*B*C*a^4*b^4 + 18*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/
(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) - (((32*(B*a^2*b^10 - B*b^12 - 3*B*a^3*b^9 + B*a^5*b^7 - 3*C*a^2*b^10 - 3*C*
a^3*b^9 + 5*C*a^4*b^8 + C*a^5*b^7 - 2*C*a^6*b^6 + 2*B*a*b^11 + 2*C*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6)
- (32*tan(c/2 + (d*x)/2)*(B*b - 2*C*a)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6)
)/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4)))*(B*b - 2*C*a))/b^3))/b^3))*(B*b - 2*C*a)*2i)/(b^3*d) + (a*atan(((a*
((a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)*(B^2*b^8 + 8*C^2*a^8 - 2*B^2*a*b^7 - 8*C^2*a^7*b + 3*B^2*a
^2*b^6 + 4*B^2*a^3*b^5 - 5*B^2*a^4*b^4 - 2*B^2*a^5*b^3 + 2*B^2*a^6*b^2 + 4*C^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*C^2
*a^4*b^4 + 16*C^2*a^5*b^3 - 16*C^2*a^6*b^2 - 4*B*C*a*b^7 - 8*B*C*a^7*b + 8*B*C*a^2*b^6 - 8*B*C*a^3*b^5 - 16*B*
C*a^4*b^4 + 18*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) + (a*((32*(B*a^2*b^10 - B*b^12
- 3*B*a^3*b^9 + B*a^5*b^7 - 3*C*a^2*b^10 - 3*C*a^3*b^9 + 5*C*a^4*b^8 + C*a^5*b^7 - 2*C*a^6*b^6 + 2*B*a*b^11 +
2*C*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (32*a*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(2*B*b^3
+ 2*C*a^3 - B*a^2*b - 3*C*a*b^2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/((a
*b^6 + b^7 - a^2*b^5 - a^3*b^4)*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))*((a + b)^3*(a - b)^3)^(1/2)*(2*B*b^3
+ 2*C*a^3 - B*a^2*b - 3*C*a*b^2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*(2*B*b^3 + 2*C*a^3 - B*a^2*b - 3*C
*a*b^2)*1i)/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3) + (a*((a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)*(
B^2*b^8 + 8*C^2*a^8 - 2*B^2*a*b^7 - 8*C^2*a^7*b + 3*B^2*a^2*b^6 + 4*B^2*a^3*b^5 - 5*B^2*a^4*b^4 - 2*B^2*a^5*b^
3 + 2*B^2*a^6*b^2 + 4*C^2*a^2*b^6 - 8*C^2*a^3*b...

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