3.556 \(\int \frac{(a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac{11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=307 \[ \frac{4 a^3 (105 A+121 B+143 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{231 d}+\frac{4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (105 A+143 B+99 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (105 A+121 B+143 C) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (15 A+17 B+21 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 (6 A+11 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{99 a d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{11 d \sec ^{\frac{9}{2}}(c+d x)} \]

[Out]

(4*a^3*(15*A + 17*B + 21*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^3*(
105*A + 121*B + 143*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (4*a^3*(210*
A + 253*B + 264*C)*Sin[c + d*x])/(1155*d*Sec[c + d*x]^(3/2)) + (4*a^3*(105*A + 121*B + 143*C)*Sin[c + d*x])/(2
31*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(11*d*Sec[c + d*x]^(9/2)) + (2*(6*A + 11*
B)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(99*a*d*Sec[c + d*x]^(7/2)) + (2*(105*A + 143*B + 99*C)*(a^3 + a^3
*Sec[c + d*x])*Sin[c + d*x])/(693*d*Sec[c + d*x]^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.675096, antiderivative size = 307, normalized size of antiderivative = 1., number of steps used = 10, 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, 2641, 2639} \[ \frac{4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (105 A+143 B+99 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (105 A+121 B+143 C) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}+\frac{4 a^3 (105 A+121 B+143 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{4 a^3 (15 A+17 B+21 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 (6 A+11 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{99 a d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{11 d \sec ^{\frac{9}{2}}(c+d x)} \]

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]^(11/2),x]

[Out]

(4*a^3*(15*A + 17*B + 21*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^3*(
105*A + 121*B + 143*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (4*a^3*(210*
A + 253*B + 264*C)*Sin[c + d*x])/(1155*d*Sec[c + d*x]^(3/2)) + (4*a^3*(105*A + 121*B + 143*C)*Sin[c + d*x])/(2
31*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(11*d*Sec[c + d*x]^(9/2)) + (2*(6*A + 11*
B)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(99*a*d*Sec[c + d*x]^(7/2)) + (2*(105*A + 143*B + 99*C)*(a^3 + a^3
*Sec[c + d*x])*Sin[c + d*x])/(693*d*Sec[c + d*x]^(5/2))

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 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]

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]

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{11}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \sec (c+d x))^3 \left (\frac{1}{2} a (6 A+11 B)+\frac{1}{2} a (3 A+11 C) \sec (c+d x)\right )}{\sec ^{\frac{9}{2}}(c+d x)} \, dx}{11 a}\\ &=\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{7}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \sec (c+d x))^2 \left (\frac{1}{4} a^2 (105 A+143 B+99 C)+\frac{3}{4} a^2 (15 A+11 B+33 C) \sec (c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx}{99 a}\\ &=\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (105 A+143 B+99 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{8 \int \frac{(a+a \sec (c+d x)) \left (\frac{3}{4} a^3 (210 A+253 B+264 C)+\frac{15}{4} a^3 (21 A+22 B+33 C) \sec (c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{693 a}\\ &=\frac{4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (105 A+143 B+99 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{16 \int \frac{-\frac{45}{8} a^4 (105 A+121 B+143 C)-\frac{231}{8} a^4 (15 A+17 B+21 C) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{3465 a}\\ &=\frac{4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (105 A+143 B+99 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{15} \left (2 a^3 (15 A+17 B+21 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{77} \left (2 a^3 (105 A+121 B+143 C)\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (105 A+121 B+143 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)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (105 A+143 B+99 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{231} \left (2 a^3 (105 A+121 B+143 C)\right ) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{15} \left (2 a^3 (15 A+17 B+21 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{4 a^3 (15 A+17 B+21 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{15 d}+\frac{4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (105 A+121 B+143 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)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (105 A+143 B+99 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{231} \left (2 a^3 (105 A+121 B+143 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{4 a^3 (15 A+17 B+21 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{15 d}+\frac{4 a^3 (105 A+121 B+143 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{4 a^3 (210 A+253 B+264 C) \sin (c+d x)}{1155 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (105 A+121 B+143 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)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 (6 A+11 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (105 A+143 B+99 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [C]  time = 5.1811, size = 246, normalized size = 0.8 \[ \frac{a^3 \sqrt{\sec (c+d x)} \left (-2464 i (15 A+17 B+21 C) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+480 (105 A+121 B+143 C) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\cos (c+d x) (30 (1953 A+2134 B+2354 C) \sin (c+d x)+308 (75 A+73 B+54 C) \sin (2 (c+d x))+8505 A \sin (3 (c+d x))+2310 A \sin (4 (c+d x))+315 A \sin (5 (c+d x))+110880 i A+5940 B \sin (3 (c+d x))+770 B \sin (4 (c+d x))+125664 i B+1980 C \sin (3 (c+d x))+155232 i C)\right )}{27720 d} \]

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]^(11/2),x]

[Out]

(a^3*Sqrt[Sec[c + d*x]]*(480*(105*A + 121*B + 143*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - (2464*I)*(
15*A + 17*B + 21*C)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(
c + d*x))] + Cos[c + d*x]*((110880*I)*A + (125664*I)*B + (155232*I)*C + 30*(1953*A + 2134*B + 2354*C)*Sin[c +
d*x] + 308*(75*A + 73*B + 54*C)*Sin[2*(c + d*x)] + 8505*A*Sin[3*(c + d*x)] + 5940*B*Sin[3*(c + d*x)] + 1980*C*
Sin[3*(c + d*x)] + 2310*A*Sin[4*(c + d*x)] + 770*B*Sin[4*(c + d*x)] + 315*A*Sin[5*(c + d*x)])))/(27720*d)

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Maple [A]  time = 2.188, size = 545, normalized size = 1.8 \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)^(11/2),x)

[Out]

-4/3465*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(10080*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-43680*A-6160*B)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(77280*A+24200*B+3960*C)*sin(1/2*d*x+1/2*c)
^8*cos(1/2*d*x+1/2*c)+(-72240*A-37532*B-14256*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(39270*A+29722*B+1986
6*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-8820*A-8118*B-6864*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+1
575*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))-3465
*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))+1815*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))-3927*B*(si
n(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))+2145*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))-4851*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+sin(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)^(11/2),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{11}{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)^(11/2),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)^(11/2), 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)**(11/2),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{11}{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)^(11/2),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)^(11/2), x)