\(\int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx\) [200]

Optimal result

Integrand size = 33, antiderivative size = 277 \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {4 a^3 (15 A+17 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^3 (105 A+121 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{231 d}+\frac {20 a^3 (21 A+22 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {4 a^3 (15 A+17 B) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (105 A+121 B) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (15 A+11 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)} \]

[Out]

20/693*a^3*(21*A+22*B)*sin(d*x+c)/d/sec(d*x+c)^(5/2)+4/45*a^3*(15*A+17*B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/11*a
*A*(a+a*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(9/2)+2/99*(15*A+11*B)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/d/sec(d*x
+c)^(7/2)+4/231*a^3*(105*A+121*B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+4/15*a^3*(15*A+17*B)*(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*(10
5*A+121*B)*(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 = 0.56 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4102, 4081, 3872, 3854, 3856, 2719, 2720} \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {4 a^3 (15 A+17 B) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {20 a^3 (21 A+22 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 (15 A+11 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {4 a^3 (105 A+121 B) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {4 a^3 (105 A+121 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{231 d}+\frac {4 a^3 (15 A+17 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^2}{11 d \sec ^{\frac {9}{2}}(c+d x)} \]

[In]

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

[Out]

(4*a^3*(15*A + 17*B)*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)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (20*a^3*(21*A + 22*B)*Sin[c
+ d*x])/(693*d*Sec[c + d*x]^(5/2)) + (4*a^3*(15*A + 17*B)*Sin[c + d*x])/(45*d*Sec[c + d*x]^(3/2)) + (4*a^3*(1
05*A + 121*B)*Sin[c + d*x])/(231*d*Sqrt[Sec[c + d*x]]) + (2*a*A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(11*d*Sec
[c + d*x]^(9/2)) + (2*(15*A + 11*B)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(99*d*Sec[c + d*x]^(7/2))

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2}{11} \int \frac {(a+a \sec (c+d x))^2 \left (\frac {1}{2} a (15 A+11 B)+\frac {1}{2} a (5 A+11 B) \sec (c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx \\ & = \frac {2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (15 A+11 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {4}{99} \int \frac {(a+a \sec (c+d x)) \left (\frac {5}{2} a^2 (21 A+22 B)+\frac {1}{2} a^2 (60 A+77 B) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {20 a^3 (21 A+22 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (15 A+11 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}-\frac {8}{693} \int \frac {-\frac {77}{4} a^3 (15 A+17 B)-\frac {9}{4} a^3 (105 A+121 B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {20 a^3 (21 A+22 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (15 A+11 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{9} \left (2 a^3 (15 A+17 B)\right ) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx+\frac {1}{77} \left (2 a^3 (105 A+121 B)\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {20 a^3 (21 A+22 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {4 a^3 (15 A+17 B) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (105 A+121 B) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (15 A+11 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{15} \left (2 a^3 (15 A+17 B)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{231} \left (2 a^3 (105 A+121 B)\right ) \int \sqrt {\sec (c+d x)} \, dx \\ & = \frac {20 a^3 (21 A+22 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {4 a^3 (15 A+17 B) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (105 A+121 B) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (15 A+11 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{15} \left (2 a^3 (15 A+17 B) \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 (105 A+121 B) \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) \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) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{231 d}+\frac {20 a^3 (21 A+22 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {4 a^3 (15 A+17 B) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (105 A+121 B) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \sec (c+d x))^2 \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 (15 A+11 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{99 d \sec ^{\frac {7}{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 = 4.18 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.86 \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {a^3 e^{-i d x} \sqrt {\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (480 (105 A+121 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-2464 i (15 A+17 B) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+\cos (c+d x) (110880 i A+125664 i B+30 (1953 A+2134 B) \sin (c+d x)+308 (75 A+73 B) \sin (2 (c+d x))+8505 A \sin (3 (c+d x))+5940 B \sin (3 (c+d x))+2310 A \sin (4 (c+d x))+770 B \sin (4 (c+d x))+315 A \sin (5 (c+d x)))\right )}{27720 d} \]

[In]

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

[Out]

(a^3*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(480*(105*A + 121*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2,
2] - (2464*I)*(15*A + 17*B)*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 + 30*(1953*A + 2134*B)*Sin[c + d*x] + 308*(75*
A + 73*B)*Sin[2*(c + d*x)] + 8505*A*Sin[3*(c + d*x)] + 5940*B*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*E^(I*d*x))

Maple [A] (verified)

Time = 30.20 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.59

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 (10080 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (-43680 A -6160 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{10} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (77280 A +24200 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-72240 A -37532 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (39270 A +29722 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-8820 A -8118 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1575 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3465 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+1815 B \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3927 B \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3465 \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}\)	\(441\)
parts	\(\text {Expression too large to display}\)	\(1059\)

[In]

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

[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)*sin(1/2*d*x+1/2*c)^8*cos(
1/2*d*x+1/2*c)+(-72240*A-37532*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(39270*A+29722*B)*sin(1/2*d*x+1/2*c)
^4*cos(1/2*d*x+1/2*c)+(-8820*A-8118*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+1575*A*(2*sin(1/2*d*x+1/2*c)^2-
1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3465*A*(2*sin(1/2*d*x+1/2*c)^2-1)^
(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+1815*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/
2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3927*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*
(sin(1/2*d*x+1/2*c)^2)^(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 = 251, normalized size of antiderivative = 0.91 \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (15 i \, \sqrt {2} {\left (105 \, A + 121 \, B\right )} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 i \, \sqrt {2} {\left (105 \, A + 121 \, B\right )} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 231 i \, \sqrt {2} {\left (15 \, A + 17 \, B\right )} a^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 231 i \, \sqrt {2} {\left (15 \, A + 17 \, B\right )} a^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (315 \, A a^{3} \cos \left (d x + c\right )^{5} + 385 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{4} + 135 \, {\left (14 \, A + 11 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 154 \, {\left (15 \, A + 17 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 30 \, {\left (105 \, A + 121 \, B\right )} a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{3465 \, d} \]

[In]

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

[Out]

-2/3465*(15*I*sqrt(2)*(105*A + 121*B)*a^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 15*I*sqr
t(2)*(105*A + 121*B)*a^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 231*I*sqrt(2)*(15*A + 17*
B)*a^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 231*I*sqrt(2)*(15*A
+ 17*B)*a^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (315*A*a^3*co
s(d*x + c)^5 + 385*(3*A + B)*a^3*cos(d*x + c)^4 + 135*(14*A + 11*B)*a^3*cos(d*x + c)^3 + 154*(15*A + 17*B)*a^3
*cos(d*x + c)^2 + 30*(105*A + 121*B)*a^3*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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