3.136 \(\int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=175 \[ \frac {64 a^3 (15 A+13 B) \tan (c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (15 A+13 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {2 (9 A-2 B) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d}+\frac {2 a (15 A+13 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d} \]

[Out]

2/105*a*(15*A+13*B)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/63*(9*A-2*B)*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d+2/9
*B*(a+a*sec(d*x+c))^(7/2)*tan(d*x+c)/a/d+64/315*a^3*(15*A+13*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/315*a^2
*(15*A+13*B)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.35, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4010, 4001, 3793, 3792} \[ \frac {16 a^2 (15 A+13 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {64 a^3 (15 A+13 B) \tan (c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (9 A-2 B) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d}+\frac {2 a (15 A+13 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(64*a^3*(15*A + 13*B)*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(15*A + 13*B)*Sqrt[a + a*Sec[c
+ d*x]]*Tan[c + d*x])/(315*d) + (2*a*(15*A + 13*B)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(105*d) + (2*(9*A
- 2*B)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(63*d) + (2*B*(a + a*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*a*d)

Rule 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*Sqrt[a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3793

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b*Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[(a*(2*m - 1))/m, Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x
], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && IntegerQ[2*m]

Rule 4001

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

Rule 4010

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

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx &=\frac {2 B (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {2 \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {7 a B}{2}+\frac {1}{2} a (9 A-2 B) \sec (c+d x)\right ) \, dx}{9 a}\\ &=\frac {2 (9 A-2 B) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 B (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{21} (15 A+13 B) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\\ &=\frac {2 a (15 A+13 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {2 (9 A-2 B) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 B (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{105} (8 a (15 A+13 B)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac {16 a^2 (15 A+13 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a (15 A+13 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {2 (9 A-2 B) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 B (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{315} \left (32 a^2 (15 A+13 B)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {64 a^3 (15 A+13 B) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (15 A+13 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a (15 A+13 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {2 (9 A-2 B) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 B (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}\\ \end {align*}

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Mathematica [A]  time = 0.64, size = 96, normalized size = 0.55 \[ \frac {2 a^3 \tan (c+d x) \left (5 (9 A+26 B) \sec ^3(c+d x)+3 (60 A+73 B) \sec ^2(c+d x)+(345 A+292 B) \sec (c+d x)+690 A+35 B \sec ^4(c+d x)+584 B\right )}{315 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(2*a^3*(690*A + 584*B + (345*A + 292*B)*Sec[c + d*x] + 3*(60*A + 73*B)*Sec[c + d*x]^2 + 5*(9*A + 26*B)*Sec[c +
d*x]^3 + 35*B*Sec[c + d*x]^4)*Tan[c + d*x])/(315*d*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 0.43, size = 136, normalized size = 0.78 \[ \frac {2 \, {\left (2 \, {\left (345 \, A + 292 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (345 \, A + 292 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (60 \, A + 73 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (9 \, A + 26 \, B\right )} a^{2} \cos \left (d x + c\right ) + 35 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

2/315*(2*(345*A + 292*B)*a^2*cos(d*x + c)^4 + (345*A + 292*B)*a^2*cos(d*x + c)^3 + 3*(60*A + 73*B)*a^2*cos(d*x
+ c)^2 + 5*(9*A + 26*B)*a^2*cos(d*x + c) + 35*B*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*
cos(d*x + c)^5 + d*cos(d*x + c)^4)

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giac [A]  time = 2.06, size = 261, normalized size = 1.49 \[ \frac {8 \, {\left ({\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (15 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 13 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \sqrt {2} {\left (15 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 13 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 63 \, \sqrt {2} {\left (15 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 13 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 210 \, \sqrt {2} {\left (4 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 315 \, \sqrt {2} {\left (A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

8/315*(((4*(2*sqrt(2)*(15*A*a^7*sgn(cos(d*x + c)) + 13*B*a^7*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 9*sqr
t(2)*(15*A*a^7*sgn(cos(d*x + c)) + 13*B*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 63*sqrt(2)*(15*A*a^7*
sgn(cos(d*x + c)) + 13*B*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 210*sqrt(2)*(4*A*a^7*sgn(cos(d*x + c
)) + 3*B*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 315*sqrt(2)*(A*a^7*sgn(cos(d*x + c)) + B*a^7*sgn(cos
(d*x + c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)

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maple [A]  time = 1.52, size = 141, normalized size = 0.81 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (690 A \left (\cos ^{4}\left (d x +c \right )\right )+584 B \left (\cos ^{4}\left (d x +c \right )\right )+345 A \left (\cos ^{3}\left (d x +c \right )\right )+292 B \left (\cos ^{3}\left (d x +c \right )\right )+180 A \left (\cos ^{2}\left (d x +c \right )\right )+219 B \left (\cos ^{2}\left (d x +c \right )\right )+45 A \cos \left (d x +c \right )+130 B \cos \left (d x +c \right )+35 B \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{315 d \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)

[Out]

-2/315/d*(-1+cos(d*x+c))*(690*A*cos(d*x+c)^4+584*B*cos(d*x+c)^4+345*A*cos(d*x+c)^3+292*B*cos(d*x+c)^3+180*A*co
s(d*x+c)^2+219*B*cos(d*x+c)^2+45*A*cos(d*x+c)+130*B*cos(d*x+c)+35*B)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d
*x+c)^4/sin(d*x+c)*a^2

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 10.81, size = 723, normalized size = 4.13 \[ \frac {\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,a^2\,4{}\mathrm {i}}{3\,d}-\frac {a^2\,\left (60\,A+73\,B\right )\,8{}\mathrm {i}}{315\,d}\right )+\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\mathrm {i}}{3\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (-\frac {A\,a^2\,4{}\mathrm {i}}{5\,d}+\frac {a^2\,\left (3\,A+4\,B\right )\,16{}\mathrm {i}}{105\,d}+\frac {a^2\,\left (9\,A+10\,B\right )\,4{}\mathrm {i}}{5\,d}\right )-\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\mathrm {i}}{5\,d}+\frac {a^2\,\left (5\,A+16\,B\right )\,4{}\mathrm {i}}{5\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,a^2\,4{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (A+2\,B\right )\,20{}\mathrm {i}}{7\,d}+\frac {B\,a^2\,32{}\mathrm {i}}{63\,d}-\frac {a^2\,\left (A+B\right )\,40{}\mathrm {i}}{7\,d}\right )+\frac {a^2\,\left (A-8\,B\right )\,4{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (5\,A+9\,B\right )\,8{}\mathrm {i}}{7\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,a^2\,4{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (3\,A+4\,B\right )\,20{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (11\,A+10\,B\right )\,4{}\mathrm {i}}{9\,d}\right )-\frac {A\,a^2\,4{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (3\,A+4\,B\right )\,20{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (11\,A+10\,B\right )\,4{}\mathrm {i}}{9\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (345\,A+292\,B\right )\,4{}\mathrm {i}}{315\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2))/cos(c + d*x)^2,x)

[Out]

((exp(c*1i + d*x*1i)*((A*a^2*4i)/(3*d) - (a^2*(60*A + 73*B)*8i)/(315*d)) + (a^2*(5*A + 2*B)*4i)/(3*d))*(a + a/
(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) +
((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a^2*(3*A + 4*B)*16i)/(105
*d) - (A*a^2*4i)/(5*d) + (a^2*(9*A + 10*B)*4i)/(5*d)) - (a^2*(5*A + 2*B)*4i)/(5*d) + (a^2*(5*A + 16*B)*4i)/(5*
d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*
1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2*4i)/(7*d) + (a^2*(A + 2*B)*20i)/(7*d) + (B*a^2*32i)/(63*d) - (a^2*(A
+ B)*40i)/(7*d)) + (a^2*(A - 8*B)*4i)/(7*d) + (a^2*(5*A + 2*B)*4i)/(7*d) - (a^2*(5*A + 9*B)*8i)/(7*d)))/((exp
(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1
/2)*(exp(c*1i + d*x*1i)*((A*a^2*4i)/(9*d) + (a^2*(3*A + 4*B)*20i)/(9*d) - (a^2*(5*A + 2*B)*4i)/(9*d) - (a^2*(1
1*A + 10*B)*4i)/(9*d)) - (A*a^2*4i)/(9*d) - (a^2*(3*A + 4*B)*20i)/(9*d) + (a^2*(5*A + 2*B)*4i)/(9*d) + (a^2*(1
1*A + 10*B)*4i)/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) - (a^2*exp(c*1i + d*x*1i)*(a + a
/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(345*A + 292*B)*4i)/(315*d*(exp(c*1i + d*x*1i) + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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