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

Optimal result

Integrand size = 14, antiderivative size = 89 \[ \int (a+a \cos (c+d x))^{5/2} \, dx=\frac {64 a^3 \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d} \]

[Out]

2/5*a*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+64/15*a^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+16/15*a^2*sin(d*x+c)*(
a+a*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2726, 2725} \[ \int (a+a \cos (c+d x))^{5/2} \, dx=\frac {64 a^3 \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{15 d}+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d} \]

[In]

Int[(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(64*a^3*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^2*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(15*d)
+ (2*a*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {1}{5} (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)}{15 d}+\frac {2 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {1}{15} \left (32 a^2\right ) \int \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {64 a^3 \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.80 \[ \int (a+a \cos (c+d x))^{5/2} \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (150 \sin \left (\frac {1}{2} (c+d x)\right )+25 \sin \left (\frac {3}{2} (c+d x)\right )+3 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{30 d} \]

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(150*Sin[(c + d*x)/2] + 25*Sin[(3*(c + d*x))/2] + 3*Sin[(5*(c
+ d*x))/2]))/(30*d)

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {8 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (3 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8\right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(73\)

[In]

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

[Out]

8/15*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)*(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*co
s(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.70 \[ \int (a+a \cos (c+d x))^{5/2} \, dx=\frac {2 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{2} + 14 \, a^{2} \cos \left (d x + c\right ) + 43 \, a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

2/15*(3*a^2*cos(d*x + c)^2 + 14*a^2*cos(d*x + c) + 43*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x +
c) + d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.67 \[ \int (a+a \cos (c+d x))^{5/2} \, dx=\frac {{\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 25 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 150 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \]

[In]

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

[Out]

1/30*(3*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 150*sqrt(2)*a^2*sin(1/2*d*x +
1/2*c))*sqrt(a)/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94 \[ \int (a+a \cos (c+d x))^{5/2} \, dx=\frac {\sqrt {2} {\left (3 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 25 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 150 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \]

[In]

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

[Out]

1/30*sqrt(2)*(3*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c) + 25*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(3/2*
d*x + 3/2*c) + 150*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. \[ \int (a+a \cos (c+d x))^{5/2} \, dx=\int {\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]

[In]

int((a + a*cos(c + d*x))^(5/2),x)

[Out]

int((a + a*cos(c + d*x))^(5/2), x)