\(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx\) [510]

Optimal result

Integrand size = 35, antiderivative size = 322 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\frac {32 a^3 (4184 A+4615 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{45045 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^3 (4184 A+4615 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45045 d \sqrt {a+a \cos (c+d x)}}+\frac {4 a^3 (4184 A+4615 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15015 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (4184 A+4615 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (280 A+299 B) \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{1287 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (16 A+13 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {13}{2}}(c+d x) \sin (c+d x)}{13 d} \]

[Out]

2/13*a*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(13/2)*sin(d*x+c)/d+16/45045*a^3*(4184*A+4615*B)*sec(d*x+c)^(3/2)*s
in(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+4/15015*a^3*(4184*A+4615*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^
(1/2)+2/9009*a^3*(4184*A+4615*B)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/1287*a^3*(280*A+299*B)
*sec(d*x+c)^(9/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/143*a^2*(16*A+13*B)*sec(d*x+c)^(11/2)*sin(d*x+c)*(a+a*
cos(d*x+c))^(1/2)/d+32/45045*a^3*(4184*A+4615*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.36 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3040, 3054, 3059, 2851, 2850} \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\frac {2 a^3 (280 A+299 B) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{1287 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^3 (4184 A+4615 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{9009 d \sqrt {a \cos (c+d x)+a}}+\frac {4 a^3 (4184 A+4615 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{15015 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a^3 (4184 A+4615 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{45045 d \sqrt {a \cos (c+d x)+a}}+\frac {32 a^3 (4184 A+4615 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{45045 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (16 A+13 B) \sin (c+d x) \sec ^{\frac {11}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{143 d}+\frac {2 a A \sin (c+d x) \sec ^{\frac {13}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d} \]

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(15/2),x]

[Out]

(32*a^3*(4184*A + 4615*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(45045*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^3*(4184*
A + 4615*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(45045*d*Sqrt[a + a*Cos[c + d*x]]) + (4*a^3*(4184*A + 4615*B)*Sec
[c + d*x]^(5/2)*Sin[c + d*x])/(15015*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(4184*A + 4615*B)*Sec[c + d*x]^(7/2)
*Sin[c + d*x])/(9009*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(280*A + 299*B)*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(12
87*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(16*A + 13*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(11/2)*Sin[c + d*x
])/(143*d) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(13/2)*Sin[c + d*x])/(13*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 {15}{2}}(c+d x)} \, dx \\ & = \frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {13}{2}}(c+d x) \sin (c+d x)}{13 d}+\frac {1}{13} \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 (16 A+13 B)+\frac {1}{2} a (8 A+13 B) \cos (c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (16 A+13 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {13}{2}}(c+d x) \sin (c+d x)}{13 d}+\frac {1}{143} \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 (280 A+299 B)+\frac {1}{4} a^2 (216 A+247 B) \cos (c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx \\ & = \frac {2 a^3 (280 A+299 B) \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{1287 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (16 A+13 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {13}{2}}(c+d x) \sin (c+d x)}{13 d}+\frac {\left (a^2 (4184 A+4615 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx}{1287} \\ & = \frac {2 a^3 (4184 A+4615 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (280 A+299 B) \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{1287 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (16 A+13 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {13}{2}}(c+d x) \sin (c+d x)}{13 d}+\frac {\left (2 a^2 (4184 A+4615 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}{3003} \\ & = \frac {4 a^3 (4184 A+4615 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15015 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (4184 A+4615 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (280 A+299 B) \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{1287 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (16 A+13 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {13}{2}}(c+d x) \sin (c+d x)}{13 d}+\frac {\left (8 a^2 (4184 A+4615 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}{15015} \\ & = \frac {16 a^3 (4184 A+4615 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45045 d \sqrt {a+a \cos (c+d x)}}+\frac {4 a^3 (4184 A+4615 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15015 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (4184 A+4615 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (280 A+299 B) \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{1287 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (16 A+13 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {13}{2}}(c+d x) \sin (c+d x)}{13 d}+\frac {\left (16 a^2 (4184 A+4615 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}{45045} \\ & = \frac {32 a^3 (4184 A+4615 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{45045 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^3 (4184 A+4615 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45045 d \sqrt {a+a \cos (c+d x)}}+\frac {4 a^3 (4184 A+4615 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15015 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (4184 A+4615 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (280 A+299 B) \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{1287 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (16 A+13 B) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {13}{2}}(c+d x) \sin (c+d x)}{13 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.53 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (171806 A+162955 B+35 (5552 A+5083 B) \cos (c+d x)+14 (15167 A+15925 B) \cos (2 (c+d x))+62760 A \cos (3 (c+d x))+69225 B \cos (3 (c+d x))+62760 A \cos (4 (c+d x))+69225 B \cos (4 (c+d x))+8368 A \cos (5 (c+d x))+9230 B \cos (5 (c+d x))+8368 A \cos (6 (c+d x))+9230 B \cos (6 (c+d x))) \sec ^{\frac {13}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{90090 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(15/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(171806*A + 162955*B + 35*(5552*A + 5083*B)*Cos[c + d*x] + 14*(15167*A + 15925
*B)*Cos[2*(c + d*x)] + 62760*A*Cos[3*(c + d*x)] + 69225*B*Cos[3*(c + d*x)] + 62760*A*Cos[4*(c + d*x)] + 69225*
B*Cos[4*(c + d*x)] + 8368*A*Cos[5*(c + d*x)] + 9230*B*Cos[5*(c + d*x)] + 8368*A*Cos[6*(c + d*x)] + 9230*B*Cos[
6*(c + d*x)])*Sec[c + d*x]^(13/2)*Tan[(c + d*x)/2])/(90090*d)

Maple [A] (verified)

Time = 2.20 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.51

\[-\frac {2 a^{2} \cot \left (d x +c \right ) \left (\cos \left (d x +c \right )-1\right ) \left (\left (66944 \left (\cos ^{6}\left (d x +c \right )\right )+33472 \left (\cos ^{5}\left (d x +c \right )\right )+25104 \left (\cos ^{4}\left (d x +c \right )\right )+20920 \left (\cos ^{3}\left (d x +c \right )\right )+18305 \left (\cos ^{2}\left (d x +c \right )\right )+11970 \cos \left (d x +c \right )+3465\right ) A +\cos \left (d x +c \right ) \left (73840 \left (\cos ^{5}\left (d x +c \right )\right )+36920 \left (\cos ^{4}\left (d x +c \right )\right )+27690 \left (\cos ^{3}\left (d x +c \right )\right )+23075 \left (\cos ^{2}\left (d x +c \right )\right )+14560 \cos \left (d x +c \right )+4095\right ) B \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sec ^{\frac {15}{2}}\left (d x +c \right )\right )}{45045 d}\]

[In]

int((a+cos(d*x+c)*a)^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(15/2),x)

[Out]

-2/45045*a^2/d*cot(d*x+c)*(cos(d*x+c)-1)*((66944*cos(d*x+c)^6+33472*cos(d*x+c)^5+25104*cos(d*x+c)^4+20920*cos(
d*x+c)^3+18305*cos(d*x+c)^2+11970*cos(d*x+c)+3465)*A+cos(d*x+c)*(73840*cos(d*x+c)^5+36920*cos(d*x+c)^4+27690*c
os(d*x+c)^3+23075*cos(d*x+c)^2+14560*cos(d*x+c)+4095)*B)*(a*(1+cos(d*x+c)))^(1/2)*sec(d*x+c)^(15/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.55 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\frac {2 \, {\left (16 \, {\left (4184 \, A + 4615 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} + 8 \, {\left (4184 \, A + 4615 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 6 \, {\left (4184 \, A + 4615 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (4184 \, A + 4615 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 35 \, {\left (523 \, A + 416 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 315 \, {\left (38 \, A + 13 \, B\right )} a^{2} \cos \left (d x + c\right ) + 3465 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )} \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(15/2),x, algorithm="fricas")

[Out]

2/45045*(16*(4184*A + 4615*B)*a^2*cos(d*x + c)^6 + 8*(4184*A + 4615*B)*a^2*cos(d*x + c)^5 + 6*(4184*A + 4615*B
)*a^2*cos(d*x + c)^4 + 5*(4184*A + 4615*B)*a^2*cos(d*x + c)^3 + 35*(523*A + 416*B)*a^2*cos(d*x + c)^2 + 315*(3
8*A + 13*B)*a^2*cos(d*x + c) + 3465*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^7 + d*cos(d*
x + c)^6)*sqrt(cos(d*x + c)))

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**(15/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 763 vs. \(2 (280) = 560\).

Time = 0.37 (sec) , antiderivative size = 763, normalized size of antiderivative = 2.37 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\text {Too large to display} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(15/2),x, algorithm="maxima")

[Out]

8/45045*((45045*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 165165*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 414414*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 604890*sqrt(2)*a^(5/2)*sin(d*x +
c)^7/(cos(d*x + c) + 1)^7 + 522665*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 289185*sqrt(2)*a^(5/
2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 88980*sqrt(2)*a^(5/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 11864
*sqrt(2)*a^(5/2)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15)*A*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^5/((sin(d*
x + c)/(cos(d*x + c) + 1) + 1)^(15/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(15/2)*(5*sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 + 10*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*sin(d*x + c
)^8/(cos(d*x + c) + 1)^8 + sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 1)) + 65*(693*sqrt(2)*a^(5/2)*sin(d*x + c)/
(cos(d*x + c) + 1) - 3003*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 6930*sqrt(2)*a^(5/2)*sin(d*x +
c)^5/(cos(d*x + c) + 1)^5 - 10098*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 9053*sqrt(2)*a^(5/2)*
sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 4875*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 1500*sqrt(2
)*a^(5/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 200*sqrt(2)*a^(5/2)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15)*B
*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^5/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(15/2)*(-sin(d*x + c)/(cos
(d*x + c) + 1) + 1)^(15/2)*(5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1
0*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + sin(d*x + c)^10/(cos(d*x + c)
+ 1)^10 + 1)))/d

Giac [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(15/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 6.74 (sec) , antiderivative size = 789, normalized size of antiderivative = 2.45 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx=\text {Too large to display} \]

[In]

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(15/2)*(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)*(4184*A + 4615*B)*32i)/(45045*d) - (a^2*exp(c*5i + d*x*5i)*(a + a*(exp(- c*1i - d*x*1i)/2 + e
xp(c*1i + d*x*1i)/2))^(1/2)*(2*A + 5*B)*16i)/(5*d) + (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)*(2*A + 5*B)*16i)/(5*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)*(116*A + 115*B)*16i)/(35*d) - (a^2*exp(c*7i + d*x*7i)*(a + a*(exp(- c*1i - d*x*1
i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(116*A + 115*B)*16i)/(35*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)*(1046*A + 1075*B)*16i)/(315*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)*(1046*A + 1075*B)*16i)/(315*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)*(4184*A + 4615*B)*16i)/(3465*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)*(4184*A + 4615*B)*16i)/(3465*d) - (a^2
*exp(c*13i + d*x*13i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(4184*A + 4615*B)*32i)/(45
045*d)))/(exp(c*1i + d*x*1i) + 6*exp(c*2i + d*x*2i) + 6*exp(c*3i + d*x*3i) + 15*exp(c*4i + d*x*4i) + 15*exp(c*
5i + d*x*5i) + 20*exp(c*6i + d*x*6i) + 20*exp(c*7i + d*x*7i) + 15*exp(c*8i + d*x*8i) + 15*exp(c*9i + d*x*9i) +
6*exp(c*10i + d*x*10i) + 6*exp(c*11i + d*x*11i) + exp(c*12i + d*x*12i) + exp(c*13i + d*x*13i) + 1)