\(\int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \, dx\) [113]

Optimal result

Integrand size = 23, antiderivative size = 146 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\frac {832 a^3 \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {208 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {26 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}-\frac {4 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d} \]

[Out]

26/105*a*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d-4/63*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/d+2/9*(a+a*cos(d*x+c))^(7/
2)*sin(d*x+c)/a/d+832/315*a^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+208/315*a^2*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2
)/d

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2838, 2830, 2726, 2725} \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\frac {832 a^3 \sin (c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {208 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{7/2}}{9 a d}-\frac {4 \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{63 d}+\frac {26 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 d} \]

[In]

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

[Out]

(832*a^3*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (208*a^2*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(315
*d) + (26*a*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(105*d) - (4*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(63
*d) + (2*(a + a*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(9*a*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 2726

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
- 1)/(d*n)), x] + Dist[a*((2*n - 1)/n), Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 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)]

Rubi steps \begin{align*} \text {integral}& = \frac {2 (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d}+\frac {2 \int \left (\frac {7 a}{2}-a \cos (c+d x)\right ) (a+a \cos (c+d x))^{5/2} \, dx}{9 a} \\ & = -\frac {4 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d}+\frac {13}{21} \int (a+a \cos (c+d x))^{5/2} \, dx \\ & = \frac {26 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}-\frac {4 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d}+\frac {1}{105} (104 a) \int (a+a \cos (c+d x))^{3/2} \, dx \\ & = \frac {208 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {26 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}-\frac {4 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d}+\frac {1}{315} \left (416 a^2\right ) \int \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {832 a^3 \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {208 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {26 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}-\frac {4 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 (a+a \cos (c+d x))^{7/2} \sin (c+d x)}{9 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.65 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (8190 \sin \left (\frac {1}{2} (c+d x)\right )+2100 \sin \left (\frac {3}{2} (c+d x)\right )+756 \sin \left (\frac {5}{2} (c+d x)\right )+225 \sin \left (\frac {7}{2} (c+d x)\right )+35 \sin \left (\frac {9}{2} (c+d x)\right )\right )}{2520 d} \]

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(8190*Sin[(c + d*x)/2] + 2100*Sin[(3*(c + d*x))/2] + 756*Sin[
(5*(c + d*x))/2] + 225*Sin[(7*(c + d*x))/2] + 35*Sin[(9*(c + d*x))/2]))/(2520*d)

Maple [A] (verified)

Time = 2.15 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.68

method	result	size
default	\(\frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (140 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+39 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+52 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+104\right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(99\)

[In]

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

[Out]

8/315*cos(1/2*d*x+1/2*c)*a^3*sin(1/2*d*x+1/2*c)*(140*cos(1/2*d*x+1/2*c)^8-20*cos(1/2*d*x+1/2*c)^6+39*cos(1/2*d
*x+1/2*c)^4+52*cos(1/2*d*x+1/2*c)^2+104)*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 = 88, normalized size of antiderivative = 0.60 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\frac {2 \, {\left (35 \, a^{2} \cos \left (d x + c\right )^{4} + 130 \, a^{2} \cos \left (d x + c\right )^{3} + 219 \, a^{2} \cos \left (d x + c\right )^{2} + 292 \, a^{2} \cos \left (d x + c\right ) + 584 \, a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

2/315*(35*a^2*cos(d*x + c)^4 + 130*a^2*cos(d*x + c)^3 + 219*a^2*cos(d*x + c)^2 + 292*a^2*cos(d*x + c) + 584*a^
2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.64 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\frac {{\left (35 \, \sqrt {2} a^{2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 225 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 756 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 2100 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 8190 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{2520 \, d} \]

[In]

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

[Out]

1/2520*(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))*sqrt(a)/d

Giac [A] (verification not implemented)

none

Time = 0.81 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.90 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\frac {\sqrt {2} {\left (35 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 225 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 756 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 2100 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 8190 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{2520 \, d} \]

[In]

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

[Out]

1/2520*sqrt(2)*(35*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(9/2*d*x + 9/2*c) + 225*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(
7/2*d*x + 7/2*c) + 756*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c) + 2100*a^2*sgn(cos(1/2*d*x + 1/2*c))
*sin(3/2*d*x + 3/2*c) + 8190*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. \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]

[In]

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

[Out]

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