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

Optimal. Leaf size=174 \[ \frac {8 a^2 (63 A+47 C) \tan (c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (63 A+22 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{315 d}+\frac {2 a (63 A+47 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{21 a d} \]

[Out]

2/315*(63*A+22*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/9*C*sec(d*x+c)^2*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2
/21*C*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/a/d+8/315*a^2*(63*A+47*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/315*a*
(63*A+47*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.48, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4089, 4010, 4001, 3793, 3792} \[ \frac {8 a^2 (63 A+47 C) \tan (c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (63 A+22 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{315 d}+\frac {2 a (63 A+47 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{21 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*(63*A + 47*C)*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(63*A + 47*C)*Sqrt[a + a*Sec[c + d*
x]]*Tan[c + d*x])/(315*d) + (2*(63*A + 22*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(315*d) + (2*C*Sec[c + d
*x]^2*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(9*d) + (2*C*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(21*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]

Rule 4089

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

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{9 d}+\frac {2 \int \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{2} a (9 A+4 C)+\frac {3}{2} a C \sec (c+d x)\right ) \, dx}{9 a}\\ &=\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{9 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{21 a d}+\frac {4 \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {15 a^2 C}{4}+\frac {1}{4} a^2 (63 A+22 C) \sec (c+d x)\right ) \, dx}{63 a^2}\\ &=\frac {2 (63 A+22 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{315 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{9 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{21 a d}+\frac {1}{105} (63 A+47 C) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac {2 a (63 A+47 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 (63 A+22 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{315 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{9 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{21 a d}+\frac {1}{315} (4 a (63 A+47 C)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {8 a^2 (63 A+47 C) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (63 A+47 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 (63 A+22 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{315 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{9 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{21 a d}\\ \end {align*}

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Mathematica [A]  time = 1.31, size = 121, normalized size = 0.70 \[ \frac {a \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \sqrt {a (\sec (c+d x)+1)} ((567 A+748 C) \cos (c+d x)+(882 A+748 C) \cos (2 (c+d x))+189 A \cos (3 (c+d x))+189 A \cos (4 (c+d x))+693 A+136 C \cos (3 (c+d x))+136 C \cos (4 (c+d x))+752 C)}{630 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(693*A + 752*C + (567*A + 748*C)*Cos[c + d*x] + (882*A + 748*C)*Cos[2*(c + d*x)] + 189*A*Cos[3*(c + d*x)] +
136*C*Cos[3*(c + d*x)] + 189*A*Cos[4*(c + d*x)] + 136*C*Cos[4*(c + d*x)])*Sec[c + d*x]^4*Sqrt[a*(1 + Sec[c +
d*x])]*Tan[(c + d*x)/2])/(630*d)

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fricas [A]  time = 0.47, size = 120, normalized size = 0.69 \[ \frac {2 \, {\left (2 \, {\left (189 \, A + 136 \, C\right )} a \cos \left (d x + c\right )^{4} + {\left (189 \, A + 136 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 34 \, C\right )} a \cos \left (d x + c\right )^{2} + 85 \, C a \cos \left (d x + c\right ) + 35 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(2*(189*A + 136*C)*a*cos(d*x + c)^4 + (189*A + 136*C)*a*cos(d*x + c)^3 + 3*(21*A + 34*C)*a*cos(d*x + c)^
2 + 85*C*a*cos(d*x + c) + 35*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^5 + d*c
os(d*x + c)^4)

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giac [A]  time = 18.85, size = 268, normalized size = 1.54 \[ \frac {4 \, {\left (315 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 315 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (945 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 525 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (1071 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 819 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (567 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 423 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 2 \, {\left (63 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 47 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/315*(315*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 315*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (945*sqrt(2)*A*a^6*sgn(cos(
d*x + c)) + 525*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (1071*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 819*sqrt(2)*C*a^6*sg
n(cos(d*x + c)) - (567*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 423*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - 2*(63*sqrt(2)*A
*a^6*sgn(cos(d*x + c)) + 47*sqrt(2)*C*a^6*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*t
an(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*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 = 1.97, size = 130, normalized size = 0.75 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (378 A \left (\cos ^{4}\left (d x +c \right )\right )+272 C \left (\cos ^{4}\left (d x +c \right )\right )+189 A \left (\cos ^{3}\left (d x +c \right )\right )+136 C \left (\cos ^{3}\left (d x +c \right )\right )+63 A \left (\cos ^{2}\left (d x +c \right )\right )+102 C \left (\cos ^{2}\left (d x +c \right )\right )+85 C \cos \left (d x +c \right )+35 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{315 d \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315/d*(-1+cos(d*x+c))*(378*A*cos(d*x+c)^4+272*C*cos(d*x+c)^4+189*A*cos(d*x+c)^3+136*C*cos(d*x+c)^3+63*A*cos
(d*x+c)^2+102*C*cos(d*x+c)^2+85*C*cos(d*x+c)+35*C)*(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)^2*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 10.68, size = 621, normalized size = 3.57 \[ \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 (5\,A+4\,C\right )\,4{}\mathrm {i}}{9\,d}-\frac {a\,\left (7\,A+12\,C\right )\,4{}\mathrm {i}}{9\,d}+\frac {A\,a\,8{}\mathrm {i}}{9\,d}\right )-\frac {a\,\left (5\,A+4\,C\right )\,4{}\mathrm {i}}{9\,d}+\frac {a\,\left (7\,A+12\,C\right )\,4{}\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 {\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+4\,C\right )\,4{}\mathrm {i}}{5\,d}-\frac {A\,a\,4{}\mathrm {i}}{5\,d}+\frac {C\,a\,16{}\mathrm {i}}{105\,d}\right )-\frac {A\,a\,12{}\mathrm {i}}{5\,d}+\frac {a\,\left (A+12\,C\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\,\left (3\,A+8\,C\right )\,4{}\mathrm {i}}{7\,d}-\frac {a\,\left (A+C\right )\,16{}\mathrm {i}}{7\,d}+\frac {A\,a\,4{}\mathrm {i}}{7\,d}+\frac {C\,a\,32{}\mathrm {i}}{63\,d}\right )+\frac {A\,a\,12{}\mathrm {i}}{7\,d}-\frac {a\,\left (A+3\,C\right )\,16{}\mathrm {i}}{7\,d}+\frac {a\,\left (A-8\,C\right )\,4{}\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 (21\,A+34\,C\right )\,8{}\mathrm {i}}{315\,d}-\frac {A\,a\,4{}\mathrm {i}}{3\,d}\right )-\frac {A\,a\,4{}\mathrm {i}}{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 )}-\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 (189\,A+136\,C\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 + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(3/2))/cos(c + d*x)^2,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*(5*A + 4*C)*4i)/(9*d) -
(a*(7*A + 12*C)*4i)/(9*d) + (A*a*8i)/(9*d)) - (a*(5*A + 4*C)*4i)/(9*d) + (a*(7*A + 12*C)*4i)/(9*d) - (A*a*8i)
/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) + ((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 + 4*C)*4i)/(5*d) - (A*a*4i)/(5*d) + (C*a*16i)/(105*d)) - (A*a*1
2i)/(5*d) + (a*(A + 12*C)*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*(3*A + 8*C)*4i)/(7*d) - (a*(A + C)*16i)
/(7*d) + (A*a*4i)/(7*d) + (C*a*32i)/(63*d)) + (A*a*12i)/(7*d) - (a*(A + 3*C)*16i)/(7*d) + (a*(A - 8*C)*4i)/(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*(21*A + 34*C)*8i)/(315*d) - (A*a*4i)/(3*d)) - (A*a*4i)/d))/((exp(c*1i +
d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) - (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)*(189*A + 136*C)*4i)/(315*d*(exp(c*1i + d*x*1i) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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