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

Optimal. Leaf size=343 \[ \frac{4 a^3 (143 A+121 B+105 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{231 d}+\frac{4 a^3 (264 A+253 B+210 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{1155 d}+\frac{4 a^3 (143 A+121 B+105 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{231 d}+\frac{2 (99 A+143 B+105 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{693 d}+\frac{4 a^3 (21 A+17 B+15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}-\frac{4 a^3 (21 A+17 B+15 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 (11 B+6 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{99 a d}+\frac{2 C \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^3}{11 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.699487, 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 = {4088, 4018, 3997, 3787, 3768, 3771, 2639, 2641} \[ \frac{4 a^3 (264 A+253 B+210 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{1155 d}+\frac{4 a^3 (143 A+121 B+105 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{231 d}+\frac{2 (99 A+143 B+105 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{693 d}+\frac{4 a^3 (21 A+17 B+15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}+\frac{4 a^3 (143 A+121 B+105 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}-\frac{4 a^3 (21 A+17 B+15 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 (11 B+6 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{99 a d}+\frac{2 C \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^3}{11 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4088

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[(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 + b*B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A,
B, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] &&  !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 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]

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 \int \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (\frac{1}{2} a (11 A+3 C)+\frac{1}{2} a (11 B+6 C) \sec (c+d x)\right ) \, dx}{11 a}\\ &=\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (11 B+6 C) \sec ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{4 \int \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (\frac{3}{4} a^2 (33 A+11 B+15 C)+\frac{1}{4} a^2 (99 A+143 B+105 C) \sec (c+d x)\right ) \, dx}{99 a}\\ &=\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (11 B+6 C) \sec ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{2 (99 A+143 B+105 C) \sec ^{\frac{5}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d}+\frac{8 \int \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x)) \left (\frac{15}{4} a^3 (33 A+22 B+21 C)+\frac{3}{4} a^3 (264 A+253 B+210 C) \sec (c+d x)\right ) \, dx}{693 a}\\ &=\frac{4 a^3 (264 A+253 B+210 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (11 B+6 C) \sec ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{2 (99 A+143 B+105 C) \sec ^{\frac{5}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d}+\frac{16 \int \sec ^{\frac{3}{2}}(c+d x) \left (\frac{231}{8} a^4 (21 A+17 B+15 C)+\frac{45}{8} a^4 (143 A+121 B+105 C) \sec (c+d x)\right ) \, dx}{3465 a}\\ &=\frac{4 a^3 (264 A+253 B+210 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (11 B+6 C) \sec ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{2 (99 A+143 B+105 C) \sec ^{\frac{5}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d}+\frac{1}{15} \left (2 a^3 (21 A+17 B+15 C)\right ) \int \sec ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{77} \left (2 a^3 (143 A+121 B+105 C)\right ) \int \sec ^{\frac{5}{2}}(c+d x) \, dx\\ &=\frac{4 a^3 (21 A+17 B+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{4 a^3 (143 A+121 B+105 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{231 d}+\frac{4 a^3 (264 A+253 B+210 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (11 B+6 C) \sec ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{2 (99 A+143 B+105 C) \sec ^{\frac{5}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d}-\frac{1}{15} \left (2 a^3 (21 A+17 B+15 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{231} \left (2 a^3 (143 A+121 B+105 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{4 a^3 (21 A+17 B+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{4 a^3 (143 A+121 B+105 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{231 d}+\frac{4 a^3 (264 A+253 B+210 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (11 B+6 C) \sec ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{2 (99 A+143 B+105 C) \sec ^{\frac{5}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d}-\frac{1}{15} \left (2 a^3 (21 A+17 B+15 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{231} \left (2 a^3 (143 A+121 B+105 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 (21 A+17 B+15 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 (143 A+121 B+105 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 (21 A+17 B+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{4 a^3 (143 A+121 B+105 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{231 d}+\frac{4 a^3 (264 A+253 B+210 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{1155 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (11 B+6 C) \sec ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{2 (99 A+143 B+105 C) \sec ^{\frac{5}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{693 d}\\ \end{align*}

Mathematica [C]  time = 7.35393, size = 1324, normalized size = 3.86 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(7*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^5*Csc[c]*(-3*S
qrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c
+ d*x))])*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(15*Sqrt[2]*d*E
^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (17*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*
x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^5*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-
1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d
*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(45*Sqrt[2]*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*
c + 2*d*x])) + (C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^5
*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4,
-E^((2*I)*(c + d*x))])*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3
*Sqrt[2]*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (13*A*Sqrt[Cos[c + d*x]]*EllipticF[(
c + d*x)/2, 2]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(21*d*(A +
2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(9/2)) + (11*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*
x)/2, 2]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(21*d*(A + 2*C +
2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(9/2)) + (5*C*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2
]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(11*d*(A + 2*C + 2*B*Co
s[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(9/2)) + (Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec
[c + d*x] + C*Sec[c + d*x]^2)*(((21*A + 17*B + 15*C)*Cos[d*x]*Csc[c])/(15*d) + (C*Sec[c]*Sec[c + d*x]^5*Sin[d*
x])/(22*d) + (Sec[c]*Sec[c + d*x]^4*(9*C*Sin[c] + 11*B*Sin[d*x] + 33*C*Sin[d*x]))/(198*d) + (Sec[c]*Sec[c + d*
x]^3*(77*B*Sin[c] + 231*C*Sin[c] + 99*A*Sin[d*x] + 297*B*Sin[d*x] + 378*C*Sin[d*x]))/(1386*d) + (Sec[c]*Sec[c
+ d*x]^2*(495*A*Sin[c] + 1485*B*Sin[c] + 1890*C*Sin[c] + 2079*A*Sin[d*x] + 2618*B*Sin[d*x] + 2310*C*Sin[d*x]))
/(6930*d) + (Sec[c]*Sec[c + d*x]*(2079*A*Sin[c] + 2618*B*Sin[c] + 2310*C*Sin[c] + 4290*A*Sin[d*x] + 3630*B*Sin
[d*x] + 3150*C*Sin[d*x]))/(6930*d) + ((143*A + 121*B + 105*C)*Tan[c])/(231*d)))/((A + 2*C + 2*B*Cos[c + d*x] +
A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(9/2))

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Maple [B]  time = 12.427, size = 1427, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*(1/8*A+3/8*B+3/8*C)*(-1/56*cos(1/2*d*x+1/2*
c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2*d*x+1/2*c)*(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+5/21*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*
d*x+1/2*c),2^(1/2)))-16/5*(3/8*A+3/8*B+1/8*C)/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/
2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(cos(1/2*
d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(
1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(
sin(1/2*d*x+1/2*c)^2)^(1/2)-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/
2*c)^2)^(1/2)+16*(3/8*A+1/8*B)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(
cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x
+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*C*(-1/352*cos(1/2*d*x+1/2*c)*(-
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^6-9/616*cos(1/2*d*x+1/2*c)*(-2*s
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-15/154*cos(1/2*d*x+1/2*c)*(-2*sin
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+15/77*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1
/2*c),2^(1/2)))+16*(1/8*B+3/8*C)*(-1/144*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/
2)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-7/180*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/
(cos(1/2*d*x+1/2*c)^2-1/2)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1
/2*d*x+1/2*c)^2)^(1/2)+7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2
*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*
cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*
c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))+2*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*(-2*sin
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2
*sin(1/2*d*x+1/2*c)^2-1))/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(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^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 ({\left (C a^{3} \sec \left (d x + c\right )^{6} +{\left (B + 3 \, C\right )} a^{3} \sec \left (d x + c\right )^{5} +{\left (A + 3 \, B + 3 \, C\right )} a^{3} \sec \left (d x + c\right )^{4} +{\left (3 \, A + 3 \, B + C\right )} a^{3} \sec \left (d x + c\right )^{3} +{\left (3 \, A + B\right )} a^{3} \sec \left (d x + c\right )^{2} + A a^{3} \sec \left (d x + c\right )\right )} \sqrt{\sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*a^3*sec(d*x + c)^6 + (B + 3*C)*a^3*sec(d*x + c)^5 + (A + 3*B + 3*C)*a^3*sec(d*x + c)^4 + (3*A + 3*
B + C)*a^3*sec(d*x + c)^3 + (3*A + B)*a^3*sec(d*x + c)^2 + A*a^3*sec(d*x + c))*sqrt(sec(d*x + c)), 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(sec(d*x+c)**(3/2)*(a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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