3.10.25 \(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^4} \, dx\) [925]

Optimal. Leaf size=314 \[ \frac {\left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 C)\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \]

[Out]

(a^3*B+4*a*b^2*B-b^3*(A+2*C)-a^2*b*(4*A+3*C))*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(7/2)/
(a+b)^(7/2)/d+1/3*a*(A*b^2-a*(B*b-C*a))*tan(d*x+c)/b^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^3+1/6*(3*A*b^4+a^3*b*B-6*a
*b^3*B-4*a^4*C+a^2*b^2*(2*A+9*C))*tan(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^2+1/6*(a^4*b*B-10*a^2*b^3*B-6*
b^5*B+a^3*b^2*(2*A-5*C)+2*a^5*C+a*b^4*(13*A+18*C))*tan(d*x+c)/b^2/(a^2-b^2)^3/d/(a+b*sec(d*x+c))

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Rubi [A]
time = 0.75, antiderivative size = 314, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4175, 4165, 4088, 12, 3916, 2738, 214} \begin {gather*} \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}+\frac {\left (a^3 B-a^2 b (4 A+3 C)+4 a b^2 B-b^3 (A+2 C)\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {\tan (c+d x) \left (-4 a^4 C+a^3 b B+a^2 b^2 (2 A+9 C)-6 a b^3 B+3 A b^4\right )}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac {\tan (c+d x) \left (2 a^5 C+a^4 b B+a^3 b^2 (2 A-5 C)-10 a^2 b^3 B+a b^4 (13 A+18 C)-6 b^5 B\right )}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^4,x]

[Out]

((a^3*B + 4*a*b^2*B - b^3*(A + 2*C) - a^2*b*(4*A + 3*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/
((a - b)^(7/2)*(a + b)^(7/2)*d) + (a*(A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Sec[c +
d*x])^3) + ((3*A*b^4 + a^3*b*B - 6*a*b^3*B - 4*a^4*C + a^2*b^2*(2*A + 9*C))*Tan[c + d*x])/(6*b^2*(a^2 - b^2)^
2*d*(a + b*Sec[c + d*x])^2) + ((a^4*b*B - 10*a^2*b^3*B - 6*b^5*B + a^3*b^2*(2*A - 5*C) + 2*a^5*C + a*b^4*(13*A
+ 18*C))*Tan[c + d*x])/(6*b^2*(a^2 - b^2)^3*d*(a + b*Sec[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 4088

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

Rule 4165

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[(-(A*b^2 - a*b*B + a^2*C))*Cot[e + f*x]*((a + b*Csc[e +
f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e
+ f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

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]

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx &=\frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\int \frac {\sec (c+d x) \left (-3 b \left (A b^2-a (b B-a C)\right )+\left (a^2 b B-3 b^3 B-a^3 C+a b^2 (2 A+3 C)\right ) \sec (c+d x)+3 b \left (a^2-b^2\right ) C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac {\int \frac {\sec (c+d x) \left (-2 b^2 \left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right )-b \left (a^3 b B-6 a b^3 B+a^2 b^2 (2 A-3 C)+2 a^4 C+3 b^4 (A+2 C)\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b^3 \left (a^2-b^2\right )^2}\\ &=\frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\int \frac {3 b^3 \left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )^3}\\ &=\frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 C)\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^3}\\ &=\frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 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 \left (a^2-b^2\right )^3 d}\\ &=-\frac {\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B+3 a^2 b C+2 b^3 C\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 1.62, size = 299, normalized size = 0.95 \begin {gather*} \frac {\frac {24 \left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 C)\right ) \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {2 \left (6 a^5 A+14 a^3 A b^2+25 a A b^4-11 a^4 b B-22 a^2 b^3 B-12 b^5 B+8 a^5 C+a^3 b^2 C+36 a b^4 C+6 \left (-A b^5+a^5 B-9 a^3 b^2 B-2 a b^4 B+9 a^2 b^3 (A+C)+a^4 b (2 A+C)\right ) \cos (c+d x)+a \left (-A b^4-13 a^3 b B-2 a b^3 B+a^4 (6 A+4 C)+a^2 b^2 (10 A+11 C)\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{(b+a \cos (c+d x))^3}}{24 \left (-a^2+b^2\right )^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^4,x]

[Out]

((24*(a^3*B + 4*a*b^2*B - b^3*(A + 2*C) - a^2*b*(4*A + 3*C))*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^
2]])/Sqrt[a^2 - b^2] - (2*(6*a^5*A + 14*a^3*A*b^2 + 25*a*A*b^4 - 11*a^4*b*B - 22*a^2*b^3*B - 12*b^5*B + 8*a^5*
C + a^3*b^2*C + 36*a*b^4*C + 6*(-(A*b^5) + a^5*B - 9*a^3*b^2*B - 2*a*b^4*B + 9*a^2*b^3*(A + C) + a^4*b*(2*A +
C))*Cos[c + d*x] + a*(-(A*b^4) - 13*a^3*b*B - 2*a*b^3*B + a^4*(6*A + 4*C) + a^2*b^2*(10*A + 11*C))*Cos[2*(c +
d*x)])*Sin[c + d*x])/(b + a*Cos[c + d*x])^3)/(24*(-a^2 + b^2)^3*d)

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Maple [A]
time = 0.53, size = 453, normalized size = 1.44

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (2 A \,a^{3}+2 A \,a^{2} b +6 a A \,b^{2}+A \,b^{3}-a^{3} B -6 a^{2} b B -2 a \,b^{2} B -2 b^{3} B +2 a^{3} C +3 a^{2} b C +6 C \,b^{2} a \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {4 \left (3 A \,a^{3}+7 a A \,b^{2}-7 a^{2} b B -3 b^{3} B +a^{3} C +9 C \,b^{2} a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (2 A \,a^{3}-2 A \,a^{2} b +6 a A \,b^{2}-A \,b^{3}+a^{3} B -6 a^{2} b B +2 a \,b^{2} B -2 b^{3} B +2 a^{3} C -3 a^{2} b C +6 C \,b^{2} a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )^{3}}-\frac {\left (4 A \,a^{2} b +A \,b^{3}-a^{3} B -4 a \,b^{2} B +3 a^{2} b C +2 C \,b^{3}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(453\)
default	\(\frac {\frac {-\frac {\left (2 A \,a^{3}+2 A \,a^{2} b +6 a A \,b^{2}+A \,b^{3}-a^{3} B -6 a^{2} b B -2 a \,b^{2} B -2 b^{3} B +2 a^{3} C +3 a^{2} b C +6 C \,b^{2} a \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {4 \left (3 A \,a^{3}+7 a A \,b^{2}-7 a^{2} b B -3 b^{3} B +a^{3} C +9 C \,b^{2} a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (2 A \,a^{3}-2 A \,a^{2} b +6 a A \,b^{2}-A \,b^{3}+a^{3} B -6 a^{2} b B +2 a \,b^{2} B -2 b^{3} B +2 a^{3} C -3 a^{2} b C +6 C \,b^{2} a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )^{3}}-\frac {\left (4 A \,a^{2} b +A \,b^{3}-a^{3} B -4 a \,b^{2} B +3 a^{2} b C +2 C \,b^{3}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(453\)
risch	\(\text {Expression too large to display}\)	\(1833\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(2*(-1/2*(2*A*a^3+2*A*a^2*b+6*A*a*b^2+A*b^3-B*a^3-6*B*a^2*b-2*B*a*b^2-2*B*b^3+2*C*a^3+3*C*a^2*b+6*C*a*b^2)
/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5+2/3*(3*A*a^3+7*A*a*b^2-7*B*a^2*b-3*B*b^3+C*a^3+9*C*a*b^2
)/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1/2*(2*A*a^3-2*A*a^2*b+6*A*a*b^2-A*b^3+B*a^3-6*B*a^2*b+
2*B*a*b^2-2*B*b^3+2*C*a^3-3*C*a^2*b+6*C*a*b^2)/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*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)^3-(4*A*a^2*b+A*b^3-B*a^3-4*B*a*b^2+3*C*a^2*b+2*C*b^3)/(a^6-3*a^4*b^2+
3*a^2*b^4-b^6)/((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*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,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. 681 vs. \(2 (300) = 600\).
time = 2.80, size = 1420, normalized size = 4.52 \begin {gather*} \left [\frac {3 \, {\left (B a^{3} b^{3} - {\left (4 \, A + 3 \, C\right )} a^{2} b^{4} + 4 \, B a b^{5} - {\left (A + 2 \, C\right )} b^{6} + {\left (B a^{6} - {\left (4 \, A + 3 \, C\right )} a^{5} b + 4 \, B a^{4} b^{2} - {\left (A + 2 \, C\right )} a^{3} b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (B a^{5} b - {\left (4 \, A + 3 \, C\right )} a^{4} b^{2} + 4 \, B a^{3} b^{3} - {\left (A + 2 \, C\right )} a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{4} b^{2} - {\left (4 \, A + 3 \, C\right )} a^{3} b^{3} + 4 \, B a^{2} b^{4} - {\left (A + 2 \, C\right )} a b^{5}\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 ) + 2 \, {\left (2 \, C a^{7} + B a^{6} b + {\left (2 \, A - 7 \, C\right )} a^{5} b^{2} - 11 \, B a^{4} b^{3} + {\left (11 \, A + 23 \, C\right )} a^{3} b^{4} + 4 \, B a^{2} b^{5} - {\left (13 \, A + 18 \, C\right )} a b^{6} + 6 \, B b^{7} + {\left (2 \, {\left (3 \, A + 2 \, C\right )} a^{7} - 13 \, B a^{6} b + {\left (4 \, A + 7 \, C\right )} a^{5} b^{2} + 11 \, B a^{4} b^{3} - 11 \, {\left (A + C\right )} a^{3} b^{4} + 2 \, B a^{2} b^{5} + A a b^{6}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{7} + {\left (2 \, A + C\right )} a^{6} b - 10 \, B a^{5} b^{2} + {\left (7 \, A + 8 \, C\right )} a^{4} b^{3} + 7 \, B a^{3} b^{4} - {\left (10 \, A + 9 \, C\right )} a^{2} b^{5} + 2 \, B a b^{6} + A b^{7}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left ({\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d\right )}}, \frac {3 \, {\left (B a^{3} b^{3} - {\left (4 \, A + 3 \, C\right )} a^{2} b^{4} + 4 \, B a b^{5} - {\left (A + 2 \, C\right )} b^{6} + {\left (B a^{6} - {\left (4 \, A + 3 \, C\right )} a^{5} b + 4 \, B a^{4} b^{2} - {\left (A + 2 \, C\right )} a^{3} b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (B a^{5} b - {\left (4 \, A + 3 \, C\right )} a^{4} b^{2} + 4 \, B a^{3} b^{3} - {\left (A + 2 \, C\right )} a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{4} b^{2} - {\left (4 \, A + 3 \, C\right )} a^{3} b^{3} + 4 \, B a^{2} b^{4} - {\left (A + 2 \, C\right )} a b^{5}\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 (2 \, C a^{7} + B a^{6} b + {\left (2 \, A - 7 \, C\right )} a^{5} b^{2} - 11 \, B a^{4} b^{3} + {\left (11 \, A + 23 \, C\right )} a^{3} b^{4} + 4 \, B a^{2} b^{5} - {\left (13 \, A + 18 \, C\right )} a b^{6} + 6 \, B b^{7} + {\left (2 \, {\left (3 \, A + 2 \, C\right )} a^{7} - 13 \, B a^{6} b + {\left (4 \, A + 7 \, C\right )} a^{5} b^{2} + 11 \, B a^{4} b^{3} - 11 \, {\left (A + C\right )} a^{3} b^{4} + 2 \, B a^{2} b^{5} + A a b^{6}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{7} + {\left (2 \, A + C\right )} a^{6} b - 10 \, B a^{5} b^{2} + {\left (7 \, A + 8 \, C\right )} a^{4} b^{3} + 7 \, B a^{3} b^{4} - {\left (10 \, A + 9 \, C\right )} a^{2} b^{5} + 2 \, B a b^{6} + A b^{7}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

[1/12*(3*(B*a^3*b^3 - (4*A + 3*C)*a^2*b^4 + 4*B*a*b^5 - (A + 2*C)*b^6 + (B*a^6 - (4*A + 3*C)*a^5*b + 4*B*a^4*b
^2 - (A + 2*C)*a^3*b^3)*cos(d*x + c)^3 + 3*(B*a^5*b - (4*A + 3*C)*a^4*b^2 + 4*B*a^3*b^3 - (A + 2*C)*a^2*b^4)*c
os(d*x + c)^2 + 3*(B*a^4*b^2 - (4*A + 3*C)*a^3*b^3 + 4*B*a^2*b^4 - (A + 2*C)*a*b^5)*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*(2*C*a^7 + B*a^6*b + (2*A - 7*C)*a^5*b^
2 - 11*B*a^4*b^3 + (11*A + 23*C)*a^3*b^4 + 4*B*a^2*b^5 - (13*A + 18*C)*a*b^6 + 6*B*b^7 + (2*(3*A + 2*C)*a^7 -
13*B*a^6*b + (4*A + 7*C)*a^5*b^2 + 11*B*a^4*b^3 - 11*(A + C)*a^3*b^4 + 2*B*a^2*b^5 + A*a*b^6)*cos(d*x + c)^2 +
3*(B*a^7 + (2*A + C)*a^6*b - 10*B*a^5*b^2 + (7*A + 8*C)*a^4*b^3 + 7*B*a^3*b^4 - (10*A + 9*C)*a^2*b^5 + 2*B*a*
b^6 + A*b^7)*cos(d*x + c))*sin(d*x + c))/((a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d*cos(d*x + c)^
3 + 3*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5
*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d), 1/6*(3*(B
*a^3*b^3 - (4*A + 3*C)*a^2*b^4 + 4*B*a*b^5 - (A + 2*C)*b^6 + (B*a^6 - (4*A + 3*C)*a^5*b + 4*B*a^4*b^2 - (A + 2
*C)*a^3*b^3)*cos(d*x + c)^3 + 3*(B*a^5*b - (4*A + 3*C)*a^4*b^2 + 4*B*a^3*b^3 - (A + 2*C)*a^2*b^4)*cos(d*x + c)
^2 + 3*(B*a^4*b^2 - (4*A + 3*C)*a^3*b^3 + 4*B*a^2*b^4 - (A + 2*C)*a*b^5)*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^7 + B*a^6*b + (2*A - 7*C)*a^5*b^2
- 11*B*a^4*b^3 + (11*A + 23*C)*a^3*b^4 + 4*B*a^2*b^5 - (13*A + 18*C)*a*b^6 + 6*B*b^7 + (2*(3*A + 2*C)*a^7 - 1
3*B*a^6*b + (4*A + 7*C)*a^5*b^2 + 11*B*a^4*b^3 - 11*(A + C)*a^3*b^4 + 2*B*a^2*b^5 + A*a*b^6)*cos(d*x + c)^2 +
3*(B*a^7 + (2*A + C)*a^6*b - 10*B*a^5*b^2 + (7*A + 8*C)*a^4*b^3 + 7*B*a^3*b^4 - (10*A + 9*C)*a^2*b^5 + 2*B*a*b
^6 + A*b^7)*cos(d*x + c))*sin(d*x + c))/((a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d*cos(d*x + c)^3
+ 3*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*
b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**4,x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**2/(a + b*sec(c + d*x))**4, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 970 vs. \(2 (300) = 600\).
time = 0.60, size = 970, normalized size = 3.09 \begin {gather*} \frac {\frac {3 \, {\left (B a^{3} - 4 \, A a^{2} b - 3 \, C a^{2} b + 4 \, B a b^{2} - A b^{3} - 2 \, C 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^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {6 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 27 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, C a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 28 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 32 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 28 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, C a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 27 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, C a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\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 )}^{3}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/3*(3*(B*a^3 - 4*A*a^2*b - 3*C*a^2*b + 4*B*a*b^2 - A*b^3 - 2*C*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^6 - 3*a^4*b^2 + 3*
a^2*b^4 - b^6)*sqrt(-a^2 + b^2)) - (6*A*a^5*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^5*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^5*
tan(1/2*d*x + 1/2*c)^5 - 6*A*a^4*b*tan(1/2*d*x + 1/2*c)^5 - 12*B*a^4*b*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^4*b*tan(
1/2*d*x + 1/2*c)^5 + 12*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 27*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^3*b^2*t
an(1/2*d*x + 1/2*c)^5 - 27*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 12*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 27*C*a^2*b
^3*tan(1/2*d*x + 1/2*c)^5 + 12*A*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*B*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 18*C*a*b^4*
tan(1/2*d*x + 1/2*c)^5 + 3*A*b^5*tan(1/2*d*x + 1/2*c)^5 - 6*B*b^5*tan(1/2*d*x + 1/2*c)^5 - 12*A*a^5*tan(1/2*d*
x + 1/2*c)^3 - 4*C*a^5*tan(1/2*d*x + 1/2*c)^3 + 28*B*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 16*A*a^3*b^2*tan(1/2*d*x +
1/2*c)^3 - 32*C*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 16*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 28*A*a*b^4*tan(1/2*d*x
+ 1/2*c)^3 + 36*C*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 12*B*b^5*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^5*tan(1/2*d*x + 1/2*
c) + 3*B*a^5*tan(1/2*d*x + 1/2*c) + 6*C*a^5*tan(1/2*d*x + 1/2*c) + 6*A*a^4*b*tan(1/2*d*x + 1/2*c) - 12*B*a^4*b
*tan(1/2*d*x + 1/2*c) + 3*C*a^4*b*tan(1/2*d*x + 1/2*c) + 12*A*a^3*b^2*tan(1/2*d*x + 1/2*c) - 27*B*a^3*b^2*tan(
1/2*d*x + 1/2*c) + 6*C*a^3*b^2*tan(1/2*d*x + 1/2*c) + 27*A*a^2*b^3*tan(1/2*d*x + 1/2*c) - 12*B*a^2*b^3*tan(1/2
*d*x + 1/2*c) + 27*C*a^2*b^3*tan(1/2*d*x + 1/2*c) + 12*A*a*b^4*tan(1/2*d*x + 1/2*c) - 6*B*a*b^4*tan(1/2*d*x +
1/2*c) + 18*C*a*b^4*tan(1/2*d*x + 1/2*c) - 3*A*b^5*tan(1/2*d*x + 1/2*c) - 6*B*b^5*tan(1/2*d*x + 1/2*c))/((a^6
- 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^3))/d

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Mupad [B]
time = 7.70, size = 516, normalized size = 1.64 \begin {gather*} \frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a^3-A\,b^3+B\,a^3-2\,B\,b^3+2\,C\,a^3+6\,A\,a\,b^2-2\,A\,a^2\,b+2\,B\,a\,b^2-6\,B\,a^2\,b+6\,C\,a\,b^2-3\,C\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A\,a^3-3\,B\,b^3+C\,a^3+7\,A\,a\,b^2-7\,B\,a^2\,b+9\,C\,a\,b^2\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,a^3+A\,b^3-B\,a^3-2\,B\,b^3+2\,C\,a^3+6\,A\,a\,b^2+2\,A\,a^2\,b-2\,B\,a\,b^2-6\,B\,a^2\,b+6\,C\,a\,b^2+3\,C\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (A\,b^3-B\,a^3+2\,C\,b^3+4\,A\,a^2\,b-4\,B\,a\,b^2+3\,C\,a^2\,b\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + b/cos(c + d*x))^4),x)

[Out]

((tan(c/2 + (d*x)/2)*(2*A*a^3 - A*b^3 + B*a^3 - 2*B*b^3 + 2*C*a^3 + 6*A*a*b^2 - 2*A*a^2*b + 2*B*a*b^2 - 6*B*a^
2*b + 6*C*a*b^2 - 3*C*a^2*b))/((a + b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) - (4*tan(c/2 + (d*x)/2)^3*(3*A*a^3 - 3
*B*b^3 + C*a^3 + 7*A*a*b^2 - 7*B*a^2*b + 9*C*a*b^2))/(3*(a + b)^2*(a^2 - 2*a*b + b^2)) + (tan(c/2 + (d*x)/2)^5
*(2*A*a^3 + A*b^3 - B*a^3 - 2*B*b^3 + 2*C*a^3 + 6*A*a*b^2 + 2*A*a^2*b - 2*B*a*b^2 - 6*B*a^2*b + 6*C*a*b^2 + 3*
C*a^2*b))/((a + b)^3*(a - b)))/(d*(tan(c/2 + (d*x)/2)^2*(3*a*b^2 - 3*a^2*b - 3*a^3 + 3*b^3) - tan(c/2 + (d*x)/
2)^4*(3*a*b^2 + 3*a^2*b - 3*a^3 - 3*b^3) + 3*a*b^2 + 3*a^2*b + a^3 + b^3 - tan(c/2 + (d*x)/2)^6*(3*a*b^2 - 3*a
^2*b + a^3 - b^3))) - (atanh((tan(c/2 + (d*x)/2)*(2*a - 2*b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3))/(2*(a + b)^(1/2)
*(a - b)^(7/2)))*(A*b^3 - B*a^3 + 2*C*b^3 + 4*A*a^2*b - 4*B*a*b^2 + 3*C*a^2*b))/(d*(a + b)^(7/2)*(a - b)^(7/2)
)

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