\(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {13}{2}}(c+d x) \, dx\) [511]

Optimal result

Integrand size = 35, antiderivative size = 275 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {13}{2}}(c+d x) \, dx=\frac {16 a^3 (710 A+803 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \cos (c+d x)}}+\frac {8 a^3 (710 A+803 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3465 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (710 A+803 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (194 A+209 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (14 A+11 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d} \]

[Out]

2/11*a*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(11/2)*sin(d*x+c)/d+8/3465*a^3*(710*A+803*B)*sec(d*x+c)^(3/2)*sin(d
*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/1155*a^3*(710*A+803*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2
/693*a^3*(194*A+209*B)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/99*a^2*(14*A+11*B)*sec(d*x+c)^(9
/2)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+16/3465*a^3*(710*A+803*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c
))^(1/2)

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3040, 3054, 3059, 2851, 2850} \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {13}{2}}(c+d x) \, dx=\frac {2 a^3 (194 A+209 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{693 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^3 (710 A+803 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{1155 d \sqrt {a \cos (c+d x)+a}}+\frac {8 a^3 (710 A+803 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3465 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a^3 (710 A+803 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3465 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (14 A+11 B) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{99 d}+\frac {2 a A \sin (c+d x) \sec ^{\frac {11}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d} \]

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(13/2),x]

[Out]

(16*a^3*(710*A + 803*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (8*a^3*(710*A + 8
03*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3465*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(710*A + 803*B)*Sec[c + d*x]
^(5/2)*Sin[c + d*x])/(1155*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(194*A + 209*B)*Sec[c + d*x]^(7/2)*Sin[c + d*x
])/(693*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(14*A + 11*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(9/2)*Sin[c +
d*x])/(99*d) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(11/2)*Sin[c + d*x])/(11*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 2851

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

Rule 3040

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[(a + b*Sin[e + f*x])^m*((
c + d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

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]

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} (A+B \cos (c+d x))}{\cos ^{\frac {13}{2}}(c+d x)} \, dx \\ & = \frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac {1}{11} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{3/2} \left (\frac {1}{2} a (14 A+11 B)+\frac {1}{2} a (6 A+11 B) \cos (c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (14 A+11 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac {1}{99} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {1}{4} a^2 (194 A+209 B)+\frac {3}{4} a^2 (46 A+55 B) \cos (c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx \\ & = \frac {2 a^3 (194 A+209 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (14 A+11 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac {1}{231} \left (a^2 (710 A+803 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 a^3 (710 A+803 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (194 A+209 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (14 A+11 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac {\left (4 a^2 (710 A+803 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{1155} \\ & = \frac {8 a^3 (710 A+803 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3465 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (710 A+803 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (194 A+209 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (14 A+11 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac {\left (8 a^2 (710 A+803 B) \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}{3465} \\ & = \frac {16 a^3 (710 A+803 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \cos (c+d x)}}+\frac {8 a^3 (710 A+803 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3465 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (710 A+803 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (194 A+209 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (14 A+11 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{11 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.53 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {13}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (9070 A+7678 B+(25070 A+24827 B) \cos (c+d x)+(9230 A+9284 B) \cos (2 (c+d x))+9230 A \cos (3 (c+d x))+10439 B \cos (3 (c+d x))+1420 A \cos (4 (c+d x))+1606 B \cos (4 (c+d x))+1420 A \cos (5 (c+d x))+1606 B \cos (5 (c+d x))) \sec ^{\frac {11}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{6930 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(13/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(9070*A + 7678*B + (25070*A + 24827*B)*Cos[c + d*x] + (9230*A + 9284*B)*Cos[2*
(c + d*x)] + 9230*A*Cos[3*(c + d*x)] + 10439*B*Cos[3*(c + d*x)] + 1420*A*Cos[4*(c + d*x)] + 1606*B*Cos[4*(c +
d*x)] + 1420*A*Cos[5*(c + d*x)] + 1606*B*Cos[5*(c + d*x)])*Sec[c + d*x]^(11/2)*Tan[(c + d*x)/2])/(6930*d)

Maple [A] (verified)

Time = 2.21 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.52

\[-\frac {2 a^{2} \cot \left (d x +c \right ) \left (\cos \left (d x +c \right )-1\right ) \left (\left (5680 \left (\cos ^{5}\left (d x +c \right )\right )+2840 \left (\cos ^{4}\left (d x +c \right )\right )+2130 \left (\cos ^{3}\left (d x +c \right )\right )+1775 \left (\cos ^{2}\left (d x +c \right )\right )+1120 \cos \left (d x +c \right )+315\right ) A +\cos \left (d x +c \right ) \left (6424 \left (\cos ^{4}\left (d x +c \right )\right )+3212 \left (\cos ^{3}\left (d x +c \right )\right )+2409 \left (\cos ^{2}\left (d x +c \right )\right )+1430 \cos \left (d x +c \right )+385\right ) B \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sec ^{\frac {13}{2}}\left (d x +c \right )\right )}{3465 d}\]

[In]

int((a+cos(d*x+c)*a)^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(13/2),x)

[Out]

-2/3465*a^2/d*cot(d*x+c)*(cos(d*x+c)-1)*((5680*cos(d*x+c)^5+2840*cos(d*x+c)^4+2130*cos(d*x+c)^3+1775*cos(d*x+c
)^2+1120*cos(d*x+c)+315)*A+cos(d*x+c)*(6424*cos(d*x+c)^4+3212*cos(d*x+c)^3+2409*cos(d*x+c)^2+1430*cos(d*x+c)+3
85)*B)*(a*(1+cos(d*x+c)))^(1/2)*sec(d*x+c)^(13/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.57 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {13}{2}}(c+d x) \, dx=\frac {2 \, {\left (8 \, {\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 4 \, {\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (355 \, A + 286 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 35 \, {\left (32 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right ) + 315 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )} \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(13/2),x, algorithm="fricas")

[Out]

2/3465*(8*(710*A + 803*B)*a^2*cos(d*x + c)^5 + 4*(710*A + 803*B)*a^2*cos(d*x + c)^4 + 3*(710*A + 803*B)*a^2*co
s(d*x + c)^3 + 5*(355*A + 286*B)*a^2*cos(d*x + c)^2 + 35*(32*A + 11*B)*a^2*cos(d*x + c) + 315*A*a^2)*sqrt(a*co
s(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^6 + d*cos(d*x + c)^5)*sqrt(cos(d*x + c)))

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {13}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**(13/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 672 vs. \(2 (239) = 478\).

Time = 0.37 (sec) , antiderivative size = 672, normalized size of antiderivative = 2.44 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {13}{2}}(c+d x) \, dx=\text {Too large to display} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(13/2),x, algorithm="maxima")

[Out]

8/3465*(5*(693*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 2310*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 4620*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5478*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(
cos(d*x + c) + 1)^7 + 3575*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 1300*sqrt(2)*a^(5/2)*sin(d*x
+ c)^11/(cos(d*x + c) + 1)^11 + 200*sqrt(2)*a^(5/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13)*A*(sin(d*x + c)^2/(
cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(13/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1
)^(13/2)*(4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x + c)^6/(co
s(d*x + c) + 1)^6 + sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 1)) + 11*(315*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x
+ c) + 1) - 1260*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2394*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 - 2736*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 1859*sqrt(2)*a^(5/2)*sin(d*x +
c)^9/(cos(d*x + c) + 1)^9 - 676*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 104*sqrt(2)*a^(5/2)*si
n(d*x + c)^13/(cos(d*x + c) + 1)^13)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos(d*x + c
) + 1) + 1)^(13/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(13/2)*(4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*si
n(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + sin(d*x + c)^8/(cos(d*x + c) + 1)^
8 + 1)))/d

Giac [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {13}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(13/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 5.44 (sec) , antiderivative size = 751, normalized size of antiderivative = 2.73 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {13}{2}}(c+d x) \, dx=\text {Too large to display} \]

[In]

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(13/2)*(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)*(710*A + 803*B)*16i)/(3465*d) - (B*a^2*exp(c*3i + d*x*3i)*(a + a*(exp(- c*1i - d*x*1i)/2 + ex
p(c*1i + d*x*1i)/2))^(1/2)*8i)/(3*d) + (B*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)*8i)/(3*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)*(30*A + 41*B)*8i)/(15*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)*(30*A + 41*B)*8i)/(15*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)*(160*A + 157*B)*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)*(160*A + 157*B)*8i)/(35*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)*(710*A + 803*B)*8i)/(315*d) - (a^2*exp(c*9i + d*x*9i)*(a + a*(exp(- c*1i - d*x*1i)/2 + ex
p(c*1i + d*x*1i)/2))^(1/2)*(710*A + 803*B)*8i)/(315*d) - (a^2*exp(c*11i + d*x*11i)*(a + a*(exp(- c*1i - d*x*1i
)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(710*A + 803*B)*16i)/(3465*d)))/(exp(c*1i + d*x*1i) + 5*exp(c*2i + d*x*2i)
+ 5*exp(c*3i + d*x*3i) + 10*exp(c*4i + d*x*4i) + 10*exp(c*5i + d*x*5i) + 10*exp(c*6i + d*x*6i) + 10*exp(c*7i +
d*x*7i) + 5*exp(c*8i + d*x*8i) + 5*exp(c*9i + d*x*9i) + exp(c*10i + d*x*10i) + exp(c*11i + d*x*11i) + 1)