\(\int (a+a \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \sec ^{\frac {11}{2}}(c+d x) \, dx\) [1223]

Optimal result

Integrand size = 37, antiderivative size = 219 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\frac {2 a^3 (584 A+903 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (8 A+11 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (64 A+63 C) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {10 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d} \]

[Out]

10/63*a*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(7/2)*sin(d*x+c)/d+2/9*A*(a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(9/2)*s
in(d*x+c)/d+2/15*a^3*(8*A+11*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/315*a^2*(64*A+63*C)*sec
(d*x+c)^(5/2)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+2/315*a^3*(584*A+903*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*c
os(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {4306, 3123, 3054, 3059, 2850} \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\frac {2 a^3 (8 A+11 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^3 (584 A+903 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (64 A+63 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d}+\frac {10 a A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{63 d} \]

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(11/2),x]

[Out]

(2*a^3*(584*A + 903*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(8*A + 11*C)
*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(64*A + 63*C)*Sqrt[a + a*Cos[c + d*
x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(315*d) + (10*a*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(7/2)*Sin[c + d
*x])/(63*d) + (2*A*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(9*d)

Rule 2850

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3054

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

Rule 3059

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n +
1)*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]

Rule 3123

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Si
n[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x]
)^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n
+ 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2,
0])

Rule 4306

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx \\ & = \frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{5/2} \left (\frac {5 a A}{2}+\frac {1}{2} a (2 A+9 C) \cos (c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx}{9 a} \\ & = \frac {10 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{3/2} \left (\frac {1}{4} a^2 (64 A+63 C)+\frac {3}{4} a^2 (8 A+21 C) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{63 a} \\ & = \frac {2 a^2 (64 A+63 C) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {10 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {63}{8} a^3 (8 A+11 C)+\frac {1}{8} a^3 (248 A+441 C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{315 a} \\ & = \frac {2 a^3 (8 A+11 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (64 A+63 C) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {10 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{315} \left (a^2 (584 A+903 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a^3 (584 A+903 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (8 A+11 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (64 A+63 C) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {10 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.58 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (2908 A+2961 C+4 (698 A+441 C) \cos (c+d x)+4 (803 A+966 C) \cos (2 (c+d x))+584 A \cos (3 (c+d x))+588 C \cos (3 (c+d x))+584 A \cos (4 (c+d x))+903 C \cos (4 (c+d x))) \sec ^{\frac {9}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{1260 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(11/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(2908*A + 2961*C + 4*(698*A + 441*C)*Cos[c + d*x] + 4*(803*A + 966*C)*Cos[2*(c
+ d*x)] + 584*A*Cos[3*(c + d*x)] + 588*C*Cos[3*(c + d*x)] + 584*A*Cos[4*(c + d*x)] + 903*C*Cos[4*(c + d*x)])*
Sec[c + d*x]^(9/2)*Tan[(c + d*x)/2])/(1260*d)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.53

\[-\frac {2 a^{2} \left (\cos \left (d x +c \right )-1\right ) \left (\left (584 \left (\cos ^{4}\left (d x +c \right )\right )+292 \left (\cos ^{3}\left (d x +c \right )\right )+219 \left (\cos ^{2}\left (d x +c \right )\right )+130 \cos \left (d x +c \right )+35\right ) A +\left (\cos ^{2}\left (d x +c \right )\right ) \left (903 \left (\cos ^{2}\left (d x +c \right )\right )+294 \cos \left (d x +c \right )+63\right ) C \right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sec ^{\frac {11}{2}}\left (d x +c \right )\right ) \cot \left (d x +c \right )}{315 d}\]

[In]

int((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x)

[Out]

-2/315*a^2/d*(cos(d*x+c)-1)*((584*cos(d*x+c)^4+292*cos(d*x+c)^3+219*cos(d*x+c)^2+130*cos(d*x+c)+35)*A+cos(d*x+
c)^2*(903*cos(d*x+c)^2+294*cos(d*x+c)+63)*C)*((1+cos(d*x+c))*a)^(1/2)*sec(d*x+c)^(11/2)*cot(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.59 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\frac {2 \, {\left ({\left (584 \, A + 903 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 2 \, {\left (146 \, A + 147 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (73 \, A + 21 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 130 \, A a^{2} \cos \left (d x + c\right ) + 35 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )} \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

2/315*((584*A + 903*C)*a^2*cos(d*x + c)^4 + 2*(146*A + 147*C)*a^2*cos(d*x + c)^3 + 3*(73*A + 21*C)*a^2*cos(d*x
+ c)^2 + 130*A*a^2*cos(d*x + c) + 35*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^5 + d*cos(
d*x + c)^4)*sqrt(cos(d*x + c)))

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(11/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 579 vs. \(2 (189) = 378\).

Time = 0.54 (sec) , antiderivative size = 579, normalized size of antiderivative = 2.64 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\frac {8 \, {\left (\frac {{\left (\frac {315 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {945 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {1449 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {1287 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {572 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {104 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}\right )} A {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}} {\left (\frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}} + \frac {21 \, {\left (\frac {15 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {65 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {113 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {99 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {44 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {8 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}\right )} C {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}} {\left (\frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}}\right )}}{315 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

8/315*((315*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 945*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 1449*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1287*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(
d*x + c) + 1)^7 + 572*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 104*sqrt(2)*a^(5/2)*sin(d*x + c)^1
1/(cos(d*x + c) + 1)^11)*A*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^
(11/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4
/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)) + 21*(15*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d
*x + c) + 1) - 65*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 113*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 - 99*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 44*sqrt(2)*a^(5/2)*sin(d*x + c)^9
/(cos(d*x + c) + 1)^9 - 8*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11)*C*(sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(
3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) +
1)^6 + 1)))/d

Giac [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 7.19 (sec) , antiderivative size = 721, normalized size of antiderivative = 3.29 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\text {Too large to display} \]

[In]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(11/2)*(a + a*cos(c + d*x))^(5/2),x)

[Out]

((1/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((a^2*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d
*x*1i)/2))^(1/2)*(584*A + 903*C)*2i)/(315*d) - (a^2*exp(c*3i + d*x*3i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*
1i + d*x*1i)/2))^(1/2)*(A + 5*C)*8i)/(3*d) + (a^2*exp(c*6i + d*x*6i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i
+ d*x*1i)/2))^(1/2)*(A + 5*C)*8i)/(3*d) - (C*a^2*exp(c*1i + d*x*1i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i
+ d*x*1i)/2))^(1/2)*2i)/d + (C*a^2*exp(c*8i + d*x*8i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))
^(1/2)*2i)/d + (a^2*exp(c*4i + d*x*4i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(8*A + 11
*C)*12i)/(5*d) - (a^2*exp(c*5i + d*x*5i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(8*A +
11*C)*12i)/(5*d) + (a^2*exp(c*2i + d*x*2i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(73*A
+ 91*C)*8i)/(35*d) - (a^2*exp(c*7i + d*x*7i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(7
3*A + 91*C)*8i)/(35*d) - (a^2*exp(c*9i + d*x*9i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)
*(584*A + 903*C)*2i)/(315*d)))/(exp(c*1i + d*x*1i) + 4*exp(c*2i + d*x*2i) + 4*exp(c*3i + d*x*3i) + 6*exp(c*4i
+ d*x*4i) + 6*exp(c*5i + d*x*5i) + 4*exp(c*6i + d*x*6i) + 4*exp(c*7i + d*x*7i) + exp(c*8i + d*x*8i) + exp(c*9i
+ d*x*9i) + 1)