3.557 \(\int \frac{(a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac{13}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=343 \[ \frac{4 a^3 (95 A+105 B+121 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{231 d}+\frac{4 a^3 (175 A+195 B+221 C) \sin (c+d x)}{585 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{20 a^3 (236 A+273 B+286 C) \sin (c+d x)}{9009 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (145 A+195 B+143 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{1287 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{4 a^3 (95 A+105 B+121 C) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (175 A+195 B+221 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{195 d}+\frac{2 (6 A+13 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{143 a d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{13 d \sec ^{\frac{11}{2}}(c+d x)} \]

[Out]

(4*a^3*(175*A + 195*B + 221*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(195*d) + (4*a
^3*(95*A + 105*B + 121*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (20*a^3*(
236*A + 273*B + 286*C)*Sin[c + d*x])/(9009*d*Sec[c + d*x]^(5/2)) + (4*a^3*(175*A + 195*B + 221*C)*Sin[c + d*x]
)/(585*d*Sec[c + d*x]^(3/2)) + (4*a^3*(95*A + 105*B + 121*C)*Sin[c + d*x])/(231*d*Sqrt[Sec[c + d*x]]) + (2*A*(
a + a*Sec[c + d*x])^3*Sin[c + d*x])/(13*d*Sec[c + d*x]^(11/2)) + (2*(6*A + 13*B)*(a^2 + a^2*Sec[c + d*x])^2*Si
n[c + d*x])/(143*a*d*Sec[c + d*x]^(9/2)) + (2*(145*A + 195*B + 143*C)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(
1287*d*Sec[c + d*x]^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.717735, antiderivative size = 343, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {4086, 4017, 3996, 3787, 3769, 3771, 2639, 2641} \[ \frac{4 a^3 (175 A+195 B+221 C) \sin (c+d x)}{585 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{20 a^3 (236 A+273 B+286 C) \sin (c+d x)}{9009 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (145 A+195 B+143 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{1287 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{4 a^3 (95 A+105 B+121 C) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (95 A+105 B+121 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{4 a^3 (175 A+195 B+221 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{195 d}+\frac{2 (6 A+13 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{143 a d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{13 d \sec ^{\frac{11}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(13/2),x]

[Out]

(4*a^3*(175*A + 195*B + 221*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(195*d) + (4*a
^3*(95*A + 105*B + 121*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (20*a^3*(
236*A + 273*B + 286*C)*Sin[c + d*x])/(9009*d*Sec[c + d*x]^(5/2)) + (4*a^3*(175*A + 195*B + 221*C)*Sin[c + d*x]
)/(585*d*Sec[c + d*x]^(3/2)) + (4*a^3*(95*A + 105*B + 121*C)*Sin[c + d*x])/(231*d*Sqrt[Sec[c + d*x]]) + (2*A*(
a + a*Sec[c + d*x])^3*Sin[c + d*x])/(13*d*Sec[c + d*x]^(11/2)) + (2*(6*A + 13*B)*(a^2 + a^2*Sec[c + d*x])^2*Si
n[c + d*x])/(143*a*d*Sec[c + d*x]^(9/2)) + (2*(145*A + 195*B + 143*C)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(
1287*d*Sec[c + d*x]^(7/2))

Rule 4086

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

Rule 4017

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

Rule 3996

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{13}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{13 d \sec ^{\frac{11}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \sec (c+d x))^3 \left (\frac{1}{2} a (6 A+13 B)+\frac{1}{2} a (5 A+13 C) \sec (c+d x)\right )}{\sec ^{\frac{11}{2}}(c+d x)} \, dx}{13 a}\\ &=\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{13 d \sec ^{\frac{11}{2}}(c+d x)}+\frac{2 (6 A+13 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{143 a d \sec ^{\frac{9}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \sec (c+d x))^2 \left (\frac{1}{4} a^2 (145 A+195 B+143 C)+\frac{1}{4} a^2 (85 A+65 B+143 C) \sec (c+d x)\right )}{\sec ^{\frac{9}{2}}(c+d x)} \, dx}{143 a}\\ &=\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{13 d \sec ^{\frac{11}{2}}(c+d x)}+\frac{2 (6 A+13 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{143 a d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 (145 A+195 B+143 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{1287 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{8 \int \frac{(a+a \sec (c+d x)) \left (\frac{5}{4} a^3 (236 A+273 B+286 C)+\frac{1}{4} a^3 (745 A+780 B+1001 C) \sec (c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx}{1287 a}\\ &=\frac{20 a^3 (236 A+273 B+286 C) \sin (c+d x)}{9009 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{13 d \sec ^{\frac{11}{2}}(c+d x)}+\frac{2 (6 A+13 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{143 a d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 (145 A+195 B+143 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{1287 d \sec ^{\frac{7}{2}}(c+d x)}-\frac{16 \int \frac{-\frac{77}{8} a^4 (175 A+195 B+221 C)-\frac{117}{8} a^4 (95 A+105 B+121 C) \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{9009 a}\\ &=\frac{20 a^3 (236 A+273 B+286 C) \sin (c+d x)}{9009 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{13 d \sec ^{\frac{11}{2}}(c+d x)}+\frac{2 (6 A+13 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{143 a d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 (145 A+195 B+143 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{1287 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{1}{77} \left (2 a^3 (95 A+105 B+121 C)\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx+\frac{1}{117} \left (2 a^3 (175 A+195 B+221 C)\right ) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{20 a^3 (236 A+273 B+286 C) \sin (c+d x)}{9009 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (175 A+195 B+221 C) \sin (c+d x)}{585 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (95 A+105 B+121 C) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{13 d \sec ^{\frac{11}{2}}(c+d x)}+\frac{2 (6 A+13 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{143 a d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 (145 A+195 B+143 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{1287 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{1}{231} \left (2 a^3 (95 A+105 B+121 C)\right ) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{195} \left (2 a^3 (175 A+195 B+221 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{20 a^3 (236 A+273 B+286 C) \sin (c+d x)}{9009 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (175 A+195 B+221 C) \sin (c+d x)}{585 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (95 A+105 B+121 C) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{13 d \sec ^{\frac{11}{2}}(c+d x)}+\frac{2 (6 A+13 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{143 a d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 (145 A+195 B+143 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{1287 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{1}{231} \left (2 a^3 (95 A+105 B+121 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{195} \left (2 a^3 (175 A+195 B+221 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{4 a^3 (175 A+195 B+221 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{195 d}+\frac{4 a^3 (95 A+105 B+121 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{231 d}+\frac{20 a^3 (236 A+273 B+286 C) \sin (c+d x)}{9009 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (175 A+195 B+221 C) \sin (c+d x)}{585 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (95 A+105 B+121 C) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{13 d \sec ^{\frac{11}{2}}(c+d x)}+\frac{2 (6 A+13 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{143 a d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 (145 A+195 B+143 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{1287 d \sec ^{\frac{7}{2}}(c+d x)}\\ \end{align*}

Mathematica [C]  time = 6.50075, size = 300, normalized size = 0.87 \[ \frac{a^3 e^{-i d x} \sqrt{\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (-4928 i (175 A+195 B+221 C) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+12480 (95 A+105 B+121 C) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\cos (c+d x) (780 (1811 A+1953 B+2134 C) \sin (c+d x)+77 (7825 A+7800 B+7592 C) \sin (2 (c+d x))+251550 A \sin (3 (c+d x))+90860 A \sin (4 (c+d x))+24570 A \sin (5 (c+d x))+3465 A \sin (6 (c+d x))+2587200 i A+221130 B \sin (3 (c+d x))+60060 B \sin (4 (c+d x))+8190 B \sin (5 (c+d x))+2882880 i B+154440 C \sin (3 (c+d x))+20020 C \sin (4 (c+d x))+3267264 i C)\right )}{720720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(13/2),x]

[Out]

(a^3*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(12480*(95*A + 105*B + 121*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c
+ d*x)/2, 2] - (4928*I)*(175*A + 195*B + 221*C)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F
1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + Cos[c + d*x]*((2587200*I)*A + (2882880*I)*B + (3267264*I)*C + 780*(18
11*A + 1953*B + 2134*C)*Sin[c + d*x] + 77*(7825*A + 7800*B + 7592*C)*Sin[2*(c + d*x)] + 251550*A*Sin[3*(c + d*
x)] + 221130*B*Sin[3*(c + d*x)] + 154440*C*Sin[3*(c + d*x)] + 90860*A*Sin[4*(c + d*x)] + 60060*B*Sin[4*(c + d*
x)] + 20020*C*Sin[4*(c + d*x)] + 24570*A*Sin[5*(c + d*x)] + 8190*B*Sin[5*(c + d*x)] + 3465*A*Sin[6*(c + d*x)])
))/(720720*d*E^(I*d*x))

________________________________________________________________________________________

Maple [A]  time = 2.474, size = 576, normalized size = 1.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(13/2),x)

[Out]

-4/45045*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(-221760*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^14+(1058400*A+131040*B)*sin(1/2*d*x+1/2*c)^12*cos(1/2*d*x+1/2*c)+(-2122400*A-567840*B-80080*C)*sin(1/2
*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(2331040*A+1004640*B+314600*C)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-153
5860*A-939120*B-487916*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(633710*A+510510*B+386386*C)*sin(1/2*d*x+1/2
*c)^4*cos(1/2*d*x+1/2*c)+(-121230*A-114660*B-105534*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+18525*A*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-40425*A*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+20475*B*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-45045*B*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+23595*C*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-51051*C*(sin(1/2*d*x+1/2*
c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+s
in(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(13/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C a^{3} \sec \left (d x + c\right )^{5} +{\left (B + 3 \, C\right )} a^{3} \sec \left (d x + c\right )^{4} +{\left (A + 3 \, B + 3 \, C\right )} a^{3} \sec \left (d x + c\right )^{3} +{\left (3 \, A + 3 \, B + C\right )} a^{3} \sec \left (d x + c\right )^{2} +{\left (3 \, A + B\right )} a^{3} \sec \left (d x + c\right ) + A a^{3}}{\sec \left (d x + c\right )^{\frac{13}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(13/2),x, algorithm="fricas")

[Out]

integral((C*a^3*sec(d*x + c)^5 + (B + 3*C)*a^3*sec(d*x + c)^4 + (A + 3*B + 3*C)*a^3*sec(d*x + c)^3 + (3*A + 3*
B + C)*a^3*sec(d*x + c)^2 + (3*A + B)*a^3*sec(d*x + c) + A*a^3)/sec(d*x + c)^(13/2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(13/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(13/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^3/sec(d*x + c)^(13/2), x)