\(\int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx\) [91]

Optimal result

Integrand size = 33, antiderivative size = 237 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {2 a^3 (803 A+710 B) \sin (c+d x)}{495 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (209 A+194 B) \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}-\frac {4 a^2 (803 A+710 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3465 d}+\frac {2 a^2 (11 A+14 B) \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{99 d}+\frac {2 a (803 A+710 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{1155 d}+\frac {2 a B \cos ^3(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{11 d} \]

[Out]

2/1155*a*(803*A+710*B)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/11*a*B*cos(d*x+c)^3*(a+a*cos(d*x+c))^(3/2)*sin(d*
x+c)/d+2/495*a^3*(803*A+710*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/693*a^3*(209*A+194*B)*cos(d*x+c)^3*sin(d*
x+c)/d/(a+a*cos(d*x+c))^(1/2)-4/3465*a^2*(803*A+710*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+2/99*a^2*(11*A+14*B
)*cos(d*x+c)^3*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3055, 3060, 2838, 2830, 2725} \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {2 a^3 (209 A+194 B) \sin (c+d x) \cos ^3(c+d x)}{693 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^3 (803 A+710 B) \sin (c+d x)}{495 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (11 A+14 B) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{99 d}-\frac {4 a^2 (803 A+710 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3465 d}+\frac {2 a (803 A+710 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{1155 d}+\frac {2 a B \sin (c+d x) \cos ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d} \]

[In]

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

[Out]

(2*a^3*(803*A + 710*B)*Sin[c + d*x])/(495*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(209*A + 194*B)*Cos[c + d*x]^3*
Sin[c + d*x])/(693*d*Sqrt[a + a*Cos[c + d*x]]) - (4*a^2*(803*A + 710*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])
/(3465*d) + (2*a^2*(11*A + 14*B)*Cos[c + d*x]^3*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(99*d) + (2*a*(803*A +
710*B)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(1155*d) + (2*a*B*Cos[c + d*x]^3*(a + a*Cos[c + d*x])^(3/2)*Si
n[c + d*x])/(11*d)

Rule 2725

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

Rule 2830

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

Rule 2838

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-Cos[e + f*x])*(
(a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) -
a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 3055

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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x
])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*
x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))
*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 3060

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt
[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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}& = \frac {2 a B \cos ^3(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac {2}{11} \int \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \left (\frac {1}{2} a (11 A+6 B)+\frac {1}{2} a (11 A+14 B) \cos (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (11 A+14 B) \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{99 d}+\frac {2 a B \cos ^3(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac {4}{99} \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \left (\frac {3}{4} a^2 (55 A+46 B)+\frac {1}{4} a^2 (209 A+194 B) \cos (c+d x)\right ) \, dx \\ & = \frac {2 a^3 (209 A+194 B) \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (11 A+14 B) \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{99 d}+\frac {2 a B \cos ^3(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac {1}{231} \left (a^2 (803 A+710 B)\right ) \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {2 a^3 (209 A+194 B) \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (11 A+14 B) \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{99 d}+\frac {2 a (803 A+710 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{1155 d}+\frac {2 a B \cos ^3(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac {(2 a (803 A+710 B)) \int \left (\frac {3 a}{2}-a \cos (c+d x)\right ) \sqrt {a+a \cos (c+d x)} \, dx}{1155} \\ & = \frac {2 a^3 (209 A+194 B) \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}-\frac {4 a^2 (803 A+710 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3465 d}+\frac {2 a^2 (11 A+14 B) \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{99 d}+\frac {2 a (803 A+710 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{1155 d}+\frac {2 a B \cos ^3(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac {1}{495} \left (a^2 (803 A+710 B)\right ) \int \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {2 a^3 (803 A+710 B) \sin (c+d x)}{495 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (209 A+194 B) \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \cos (c+d x)}}-\frac {4 a^2 (803 A+710 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3465 d}+\frac {2 a^2 (11 A+14 B) \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{99 d}+\frac {2 a (803 A+710 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{1155 d}+\frac {2 a B \cos ^3(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{11 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.54 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (124366 A+114640 B+(68552 A+69890 B) \cos (c+d x)+16 (1397 A+1625 B) \cos (2 (c+d x))+5720 A \cos (3 (c+d x))+8675 B \cos (3 (c+d x))+770 A \cos (4 (c+d x))+2240 B \cos (4 (c+d x))+315 B \cos (5 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{27720 d} \]

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(124366*A + 114640*B + (68552*A + 69890*B)*Cos[c + d*x] + 16*(1397*A + 1625*B)
*Cos[2*(c + d*x)] + 5720*A*Cos[3*(c + d*x)] + 8675*B*Cos[3*(c + d*x)] + 770*A*Cos[4*(c + d*x)] + 2240*B*Cos[4*
(c + d*x)] + 315*B*Cos[5*(c + d*x)])*Tan[(c + d*x)/2])/(27720*d)

Maple [A] (verified)

Time = 11.39 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-2520 B \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (1540 A +10780 B \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-5940 A -18810 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (9009 A +17325 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-6930 A -9240 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3465 A +3465 B \right ) \sqrt {2}}{3465 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(142\)
parts	\(\frac {8 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (140 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+39 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+52 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+104\right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {8 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (504 \left (\cos ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-364 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+178 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+75 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+100 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+200\right ) \sqrt {2}}{693 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(213\)

[In]

int(cos(d*x+c)^2*(a+cos(d*x+c)*a)^(5/2)*(A+B*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

8/3465*cos(1/2*d*x+1/2*c)*a^3*sin(1/2*d*x+1/2*c)*(-2520*B*sin(1/2*d*x+1/2*c)^10+(1540*A+10780*B)*sin(1/2*d*x+1
/2*c)^8+(-5940*A-18810*B)*sin(1/2*d*x+1/2*c)^6+(9009*A+17325*B)*sin(1/2*d*x+1/2*c)^4+(-6930*A-9240*B)*sin(1/2*
d*x+1/2*c)^2+3465*A+3465*B)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.58 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {2 \, {\left (315 \, B a^{2} \cos \left (d x + c\right )^{5} + 35 \, {\left (11 \, A + 32 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (286 \, A + 355 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (803 \, A + 710 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 4 \, {\left (803 \, A + 710 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, {\left (803 \, A + 710 \, B\right )} a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.87 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {22 \, {\left (35 \, \sqrt {2} a^{2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 225 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 756 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 2100 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 8190 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + 5 \, {\left (63 \, \sqrt {2} a^{2} \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 385 \, \sqrt {2} a^{2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 1287 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 3465 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 8778 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 31878 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{55440 \, d} \]

[In]

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

[Out]

1/55440*(22*(35*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) + 225*sqrt(2)*a^2*sin(7/2*d*x + 7/2*c) + 756*sqrt(2)*a^2*sin(
5/2*d*x + 5/2*c) + 2100*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 8190*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*A*sqrt(a) +
5*(63*sqrt(2)*a^2*sin(11/2*d*x + 11/2*c) + 385*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) + 1287*sqrt(2)*a^2*sin(7/2*d*x
+ 7/2*c) + 3465*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 8778*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 31878*sqrt(2)*a^2*
sin(1/2*d*x + 1/2*c))*B*sqrt(a))/d

Giac [A] (verification not implemented)

none

Time = 3.28 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.08 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} {\left (315 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 385 \, {\left (2 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 495 \, {\left (10 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 13 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 693 \, {\left (24 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 25 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 2310 \, {\left (20 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 19 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 6930 \, {\left (26 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 23 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{55440 \, d} \]

[In]

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

[Out]

1/55440*sqrt(2)*(315*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(11/2*d*x + 11/2*c) + 385*(2*A*a^2*sgn(cos(1/2*d*x + 1
/2*c)) + 5*B*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(9/2*d*x + 9/2*c) + 495*(10*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 1
3*B*a^2*sgn(cos(1/2*d*x + 1/2*c)))*sin(7/2*d*x + 7/2*c) + 693*(24*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 25*B*a^2*s
gn(cos(1/2*d*x + 1/2*c)))*sin(5/2*d*x + 5/2*c) + 2310*(20*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 19*B*a^2*sgn(cos(1
/2*d*x + 1/2*c)))*sin(3/2*d*x + 3/2*c) + 6930*(26*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 23*B*a^2*sgn(cos(1/2*d*x +
1/2*c)))*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

Mupad [F(-1)]

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

[In]

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

[Out]

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