3.127 \(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=189 \[ \frac {2 a^2 (9 A+10 B) \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (39 A+34 B) \tan (c+d x)}{45 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (39 A+34 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}-\frac {4 a (39 A+34 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d} \]

[Out]

2/105*(39*A+34*B)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/45*a^2*(39*A+34*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
+2/63*a^2*(9*A+10*B)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/315*a*(39*A+34*B)*(a+a*sec(d*x+c))^(1/
2)*tan(d*x+c)/d+2/9*a*B*sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.46, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4018, 4016, 3800, 4001, 3792} \[ \frac {2 a^2 (9 A+10 B) \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (39 A+34 B) \tan (c+d x)}{45 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (39 A+34 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}-\frac {4 a (39 A+34 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(39*A + 34*B)*Tan[c + d*x])/(45*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(9*A + 10*B)*Sec[c + d*x]^3*Tan[c
+ d*x])/(63*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*(39*A + 34*B)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(315*d) +
(2*a*B*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(9*d) + (2*(39*A + 34*B)*(a + a*Sec[c + d*x])^(3/
2)*Tan[c + d*x])/(105*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 3800

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

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 4016

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

Rule 4018

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

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx &=\frac {2 a B \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac {2}{9} \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {3}{2} a (3 A+2 B)+\frac {1}{2} a (9 A+10 B) \sec (c+d x)\right ) \, dx\\ &=\frac {2 a^2 (9 A+10 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a B \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac {1}{21} (a (39 A+34 B)) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^2 (9 A+10 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a B \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac {2 (39 A+34 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {1}{105} (2 (39 A+34 B)) \int \sec (c+d x) \left (\frac {3 a}{2}-a \sec (c+d x)\right ) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^2 (9 A+10 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a (39 A+34 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a B \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac {2 (39 A+34 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {1}{45} (a (39 A+34 B)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^2 (39 A+34 B) \tan (c+d x)}{45 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (9 A+10 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a (39 A+34 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a B \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac {2 (39 A+34 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}\\ \end {align*}

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Mathematica [A]  time = 0.79, size = 100, normalized size = 0.53 \[ \frac {2 a^2 \tan (c+d x) \left (5 (9 A+17 B) \sec ^3(c+d x)+3 (39 A+34 B) \sec ^2(c+d x)+4 (39 A+34 B) \sec (c+d x)+8 (39 A+34 B)+35 B \sec ^4(c+d x)\right )}{315 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(8*(39*A + 34*B) + 4*(39*A + 34*B)*Sec[c + d*x] + 3*(39*A + 34*B)*Sec[c + d*x]^2 + 5*(9*A + 17*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 = 0.42, size = 127, normalized size = 0.67 \[ \frac {2 \, {\left (8 \, {\left (39 \, A + 34 \, B\right )} a \cos \left (d x + c\right )^{4} + 4 \, {\left (39 \, A + 34 \, B\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (39 \, A + 34 \, B\right )} a \cos \left (d x + c\right )^{2} + 5 \, {\left (9 \, A + 17 \, B\right )} a \cos \left (d x + c\right ) + 35 \, B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(8*(39*A + 34*B)*a*cos(d*x + c)^4 + 4*(39*A + 34*B)*a*cos(d*x + c)^3 + 3*(39*A + 34*B)*a*cos(d*x + c)^2
+ 5*(9*A + 17*B)*a*cos(d*x + c) + 35*B*a)*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 = 2.09, size = 258, normalized size = 1.37 \[ \frac {4 \, {\left ({\left ({\left ({\left (2 \, \sqrt {2} {\left (57 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 47 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \sqrt {2} {\left (57 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 47 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 819 \, \sqrt {2} {\left (A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, \sqrt {2} {\left (7 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 315 \, \sqrt {2} {\left (A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/315*((((2*sqrt(2)*(57*A*a^6*sgn(cos(d*x + c)) + 47*B*a^6*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 9*sqrt(
2)*(57*A*a^6*sgn(cos(d*x + c)) + 47*B*a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 819*sqrt(2)*(A*a^6*sgn(
cos(d*x + c)) + B*a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 105*sqrt(2)*(7*A*a^6*sgn(cos(d*x + c)) + 5*
B*a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 315*sqrt(2)*(A*a^6*sgn(cos(d*x + c)) + B*a^6*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 = 1.60, size = 139, normalized size = 0.74 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (312 A \left (\cos ^{4}\left (d x +c \right )\right )+272 B \left (\cos ^{4}\left (d x +c \right )\right )+156 A \left (\cos ^{3}\left (d x +c \right )\right )+136 B \left (\cos ^{3}\left (d x +c \right )\right )+117 A \left (\cos ^{2}\left (d x +c \right )\right )+102 B \left (\cos ^{2}\left (d x +c \right )\right )+45 A \cos \left (d x +c \right )+85 B \cos \left (d x +c \right )+35 B \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{315 d \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x)

[Out]

-2/315/d*(-1+cos(d*x+c))*(312*A*cos(d*x+c)^4+272*B*cos(d*x+c)^4+156*A*cos(d*x+c)^3+136*B*cos(d*x+c)^3+117*A*co
s(d*x+c)^2+102*B*cos(d*x+c)^2+45*A*cos(d*x+c)+85*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

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

[Out]

Timed out

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mupad [B]  time = 9.60, size = 596, normalized size = 3.15 \[ \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\,\left (3\,A+2\,B\right )\,8{}\mathrm {i}}{7\,d}+\frac {a\,\left (A+4\,B\right )\,8{}\mathrm {i}}{7\,d}+\frac {B\,a\,32{}\mathrm {i}}{63\,d}\right )+\frac {A\,a\,8{}\mathrm {i}}{7\,d}-\frac {a\,\left (A+2\,B\right )\,24{}\mathrm {i}}{7\,d}-\frac {B\,a\,32{}\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\,\left (2\,A+3\,B\right )\,16{}\mathrm {i}}{9\,d}+\frac {a\,\left (3\,A+2\,B\right )\,8{}\mathrm {i}}{9\,d}+\frac {A\,a\,8{}\mathrm {i}}{9\,d}\right )+\frac {a\,\left (2\,A+3\,B\right )\,16{}\mathrm {i}}{9\,d}-\frac {a\,\left (3\,A+2\,B\right )\,8{}\mathrm {i}}{9\,d}-\frac {A\,a\,8{}\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 {\left (\frac {A\,a\,8{}\mathrm {i}}{3\,d}-\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (39\,A+34\,B\right )\,8{}\mathrm {i}}{315\,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\,\left (3\,A+2\,B\right )\,8{}\mathrm {i}}{5\,d}+\frac {a\,\left (3\,A+B\right )\,16{}\mathrm {i}}{105\,d}\right )-\frac {A\,a\,8{}\mathrm {i}}{5\,d}+\frac {a\,\left (A+3\,B\right )\,16{}\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 {a\,{\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 (39\,A+34\,B\right )\,16{}\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))^(3/2))/cos(c + d*x)^3,x)

[Out]

((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 + 4*B)*8i)/(7*d) - (
a*(3*A + 2*B)*8i)/(7*d) + (B*a*32i)/(63*d)) + (A*a*8i)/(7*d) - (a*(A + 2*B)*24i)/(7*d) - (B*a*32i)/(7*d)))/((e
xp(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*(3*A + 2*B)*8i)/(9*d) - (a*(2*A + 3*B)*16i)/(9*d) + (A*a*8i)/(9*d)) + (a*(2*A +
3*B)*16i)/(9*d) - (a*(3*A + 2*B)*8i)/(9*d) - (A*a*8i)/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) +
1)^4) + (((A*a*8i)/(3*d) - (a*exp(c*1i + d*x*1i)*(39*A + 34*B)*8i)/(315*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + e
xp(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*1
i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a*(3*A + 2*B)*8i)/(5*d) + (a*(3*A + B)*16i)/(105*d))
- (A*a*8i)/(5*d) + (a*(A + 3*B)*16i)/(5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - (a*exp(c
*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(39*A + 34*B)*16i)/(315*d*(exp(c*1
i + d*x*1i) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sec(c + d*x) + 1))**(3/2)*(A + B*sec(c + d*x))*sec(c + d*x)**3, x)

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