\(\int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \, dx\) [114]

Optimal result

Integrand size = 21, antiderivative size = 116 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\frac {64 a^3 \sin (c+d x)}{21 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{7 d}+\frac {2 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 d} \]

[Out]

2/7*a*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/d+64/21*a^3*sin(d*x+c)/d/(a+a*
cos(d*x+c))^(1/2)+16/21*a^2*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2830, 2726, 2725} \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\frac {64 a^3 \sin (c+d x)}{21 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{21 d}+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d} \]

[In]

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

[Out]

(64*a^3*Sin[c + d*x])/(21*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(21*d)
+ (2*a*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(7*d) + (2*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*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)]

Rubi steps \begin{align*} \text {integral}& = \frac {2 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {5}{7} \int (a+a \cos (c+d x))^{5/2} \, dx \\ & = \frac {2 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{7 d}+\frac {2 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{7} (8 a) \int (a+a \cos (c+d x))^{3/2} \, dx \\ & = \frac {16 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{7 d}+\frac {2 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{21} \left (32 a^2\right ) \int \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {64 a^3 \sin (c+d x)}{21 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{7 d}+\frac {2 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.72 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (315 \sin \left (\frac {1}{2} (c+d x)\right )+77 \sin \left (\frac {3}{2} (c+d x)\right )+3 \left (7 \sin \left (\frac {5}{2} (c+d x)\right )+\sin \left (\frac {7}{2} (c+d x)\right )\right )\right )}{84 d} \]

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(315*Sin[(c + d*x)/2] + 77*Sin[(3*(c + d*x))/2] + 3*(7*Sin[(5
*(c + d*x))/2] + Sin[(7*(c + d*x))/2])))/(84*d)

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (6 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8\right ) \sqrt {2}}{21 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(86\)

[In]

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

[Out]

8/21*cos(1/2*d*x+1/2*c)*a^3*sin(1/2*d*x+1/2*c)*(6*cos(1/2*d*x+1/2*c)^6+3*cos(1/2*d*x+1/2*c)^4+4*cos(1/2*d*x+1/
2*c)^2+8)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.65 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\frac {2 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{3} + 12 \, a^{2} \cos \left (d x + c\right )^{2} + 23 \, a^{2} \cos \left (d x + c\right ) + 46 \, a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{21 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

2/21*(3*a^2*cos(d*x + c)^3 + 12*a^2*cos(d*x + c)^2 + 23*a^2*cos(d*x + c) + 46*a^2)*sqrt(a*cos(d*x + c) + a)*si
n(d*x + c)/(d*cos(d*x + c) + d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.66 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\frac {{\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 77 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 315 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{84 \, d} \]

[In]

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

[Out]

1/84*(3*sqrt(2)*a^2*sin(7/2*d*x + 7/2*c) + 21*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 77*sqrt(2)*a^2*sin(3/2*d*x +
3/2*c) + 315*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93 \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\frac {\sqrt {2} {\left (3 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 77 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 315 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{84 \, d} \]

[In]

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

[Out]

1/84*sqrt(2)*(3*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c) + 21*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*
d*x + 5/2*c) + 77*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(3/2*d*x + 3/2*c) + 315*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. \[ \int \cos (c+d x) (a+a \cos (c+d x))^{5/2} \, dx=\int \cos \left (c+d\,x\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]

[In]

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

[Out]

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