\(\int \frac {(a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx\) [229]

Optimal result

Integrand size = 35, antiderivative size = 286 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {4 a^3 (5 A+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (105 A+143 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{231 d}+\frac {8 a^3 (35 A+44 C) \sin (c+d x)}{385 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (105 A+143 C) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{33 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (35 A+33 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)} \]

[Out]

8/385*a^3*(35*A+44*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/11*A*(a+a*sec(d*x+c))^3*sin(d*x+c)/d/sec(d*x+c)^(9/2)+4/
33*A*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/a/d/sec(d*x+c)^(7/2)+2/231*(35*A+33*C)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/
d/sec(d*x+c)^(5/2)+4/231*a^3*(105*A+143*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+4/5*a^3*(5*A+7*C)*(cos(1/2*d*x+1/2*c)
^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+4/231*a
^3*(105*A+143*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x
+c)^(1/2)*sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4172, 4102, 4081, 3872, 3854, 3856, 2720, 2719} \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {8 a^3 (35 A+44 C) \sin (c+d x)}{385 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (35 A+33 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {4 a^3 (105 A+143 C) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {4 a^3 (105 A+143 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{231 d}+\frac {4 a^3 (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{33 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{11 d \sec ^{\frac {9}{2}}(c+d x)} \]

[In]

Int[((a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(4*a^3*(5*A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(105*A + 14
3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (8*a^3*(35*A + 44*C)*Sin[c + d
*x])/(385*d*Sec[c + d*x]^(3/2)) + (4*a^3*(105*A + 143*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])/(11*d*Sec[c + d*x]^(9/2)) + (4*A*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(33*
a*d*Sec[c + d*x]^(7/2)) + (2*(35*A + 33*C)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(231*d*Sec[c + d*x]^(5/2))

Rule 2719

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

Rule 2720

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

Rule 3854

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 3856

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 3872

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 4081

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 4102

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 4172

Int[((A_.) + 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] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps \begin{align*} \text {integral}& = \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 (3 a A+\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 {4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{33 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \sec (c+d x))^2 \left (\frac {3}{4} a^2 (35 A+33 C)+\frac {9}{4} a^2 (5 A+11 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 {4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{33 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (35 A+33 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8 \int \frac {(a+a \sec (c+d x)) \left (\frac {9}{2} a^3 (35 A+44 C)+\frac {45}{4} a^3 (7 A+11 C) \sec (c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{693 a} \\ & = \frac {8 a^3 (35 A+44 C) \sin (c+d x)}{385 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 {4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{33 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (35 A+33 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {16 \int \frac {-\frac {45}{8} a^4 (105 A+143 C)-\frac {693}{8} a^4 (5 A+7 C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{3465 a} \\ & = \frac {8 a^3 (35 A+44 C) \sin (c+d x)}{385 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 {4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{33 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (35 A+33 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (2 a^3 (5 A+7 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{77} \left (2 a^3 (105 A+143 C)\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {8 a^3 (35 A+44 C) \sin (c+d x)}{385 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (105 A+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 {4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{33 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (35 A+33 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{231} \left (2 a^3 (105 A+143 C)\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{5} \left (2 a^3 (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = \frac {4 a^3 (5 A+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {8 a^3 (35 A+44 C) \sin (c+d x)}{385 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (105 A+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 {4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{33 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (35 A+33 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{231} \left (2 a^3 (105 A+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 (5 A+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (105 A+143 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{231 d}+\frac {8 a^3 (35 A+44 C) \sin (c+d x)}{385 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (105 A+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 {4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{33 a d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (35 A+33 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 6.59 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.80 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {a^3 e^{-i d x} \sqrt {\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (160 (105 A+143 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-2464 i (5 A+7 C) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+\cos (c+d x) (36960 i A+51744 i C+10 (1953 A+2354 C) \sin (c+d x)+308 (25 A+18 C) \sin (2 (c+d x))+2835 A \sin (3 (c+d x))+660 C \sin (3 (c+d x))+770 A \sin (4 (c+d x))+105 A \sin (5 (c+d x)))\right )}{9240 d} \]

[In]

Integrate[((a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(a^3*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(160*(105*A + 143*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2,
2] - (2464*I)*(5*A + 7*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]*((36960*I)*A + (51744*I)*C + 10*(1953*A + 2354*C)*Sin[c + d*x] + 308*(25*A +
18*C)*Sin[2*(c + d*x)] + 2835*A*Sin[3*(c + d*x)] + 660*C*Sin[3*(c + d*x)] + 770*A*Sin[4*(c + d*x)] + 105*A*Sin
[5*(c + d*x)])))/(9240*d*E^(I*d*x))

Maple [A] (verified)

Time = 9.73 (sec) , antiderivative size = 436, normalized size of antiderivative = 1.52

method	result	size
default	\(-\frac {4 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a^{3} \left (3360 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-14560 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (25760 A +1320 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-24080 A -4752 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (13090 A +6622 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-2940 A -2288 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+525 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-1155 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+715 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-1617 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{1155 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(436\)
parts	\(\text {Expression too large to display}\)	\(1186\)

[In]

int((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x,method=_RETURNVERBOSE)

[Out]

-4/1155*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(3360*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^12-14560*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10+(25760*A+1320*C)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*
c)+(-24080*A-4752*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(13090*A+6622*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x
+1/2*c)+(-2940*A-2288*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+525*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))-1155*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))+715*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))-1617*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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.86 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (105 \, A + 143 \, C\right )} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (105 \, A + 143 \, C\right )} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 231 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} a^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 231 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} a^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (105 \, A a^{3} \cos \left (d x + c\right )^{5} + 385 \, A a^{3} \cos \left (d x + c\right )^{4} + 15 \, {\left (42 \, A + 11 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 77 \, {\left (10 \, A + 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 10 \, {\left (105 \, A + 143 \, C\right )} a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{1155 \, d} \]

[In]

integrate((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

-2/1155*(5*I*sqrt(2)*(105*A + 143*C)*a^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(
2)*(105*A + 143*C)*a^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 231*I*sqrt(2)*(5*A + 7*C)*a
^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 231*I*sqrt(2)*(5*A + 7*
C)*a^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (105*A*a^3*cos(d*x
+ c)^5 + 385*A*a^3*cos(d*x + c)^4 + 15*(42*A + 11*C)*a^3*cos(d*x + c)^3 + 77*(10*A + 9*C)*a^3*cos(d*x + c)^2 +
10*(105*A + 143*C)*a^3*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))**3*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(11/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {11}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^3/sec(d*x + c)^(11/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{11/2}} \,d x \]

[In]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3)/(1/cos(c + d*x))^(11/2),x)

[Out]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3)/(1/cos(c + d*x))^(11/2), x)