\(\int (c \sin (a+b x))^{5/2} \, dx\) [26]

Optimal result

Integrand size = 12, antiderivative size = 75 \[ \int (c \sin (a+b x))^{5/2} \, dx=\frac {6 c^2 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b \sqrt {\sin (a+b x)}}-\frac {2 c \cos (a+b x) (c \sin (a+b x))^{3/2}}{5 b} \]

[Out]

-2/5*c*cos(b*x+a)*(c*sin(b*x+a))^(3/2)/b-6/5*c^2*(sin(1/2*a+1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/2*a+1/4*Pi+1/2*b*x)
*EllipticE(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))*(c*sin(b*x+a))^(1/2)/b/sin(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2715, 2721, 2719} \[ \int (c \sin (a+b x))^{5/2} \, dx=\frac {6 c^2 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b \sqrt {\sin (a+b x)}}-\frac {2 c \cos (a+b x) (c \sin (a+b x))^{3/2}}{5 b} \]

[In]

Int[(c*Sin[a + b*x])^(5/2),x]

[Out]

(6*c^2*EllipticE[(a - Pi/2 + b*x)/2, 2]*Sqrt[c*Sin[a + b*x]])/(5*b*Sqrt[Sin[a + b*x]]) - (2*c*Cos[a + b*x]*(c*
Sin[a + b*x])^(3/2))/(5*b)

Rule 2715

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

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 c \cos (a+b x) (c \sin (a+b x))^{3/2}}{5 b}+\frac {1}{5} \left (3 c^2\right ) \int \sqrt {c \sin (a+b x)} \, dx \\ & = -\frac {2 c \cos (a+b x) (c \sin (a+b x))^{3/2}}{5 b}+\frac {\left (3 c^2 \sqrt {c \sin (a+b x)}\right ) \int \sqrt {\sin (a+b x)} \, dx}{5 \sqrt {\sin (a+b x)}} \\ & = \frac {6 c^2 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b \sqrt {\sin (a+b x)}}-\frac {2 c \cos (a+b x) (c \sin (a+b x))^{3/2}}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88 \[ \int (c \sin (a+b x))^{5/2} \, dx=-\frac {(c \sin (a+b x))^{5/2} \left (6 E\left (\left .\frac {1}{4} (-2 a+\pi -2 b x)\right |2\right )+\sqrt {\sin (a+b x)} \sin (2 (a+b x))\right )}{5 b \sin ^{\frac {5}{2}}(a+b x)} \]

[In]

Integrate[(c*Sin[a + b*x])^(5/2),x]

[Out]

-1/5*((c*Sin[a + b*x])^(5/2)*(6*EllipticE[(-2*a + Pi - 2*b*x)/4, 2] + Sqrt[Sin[a + b*x]]*Sin[2*(a + b*x)]))/(b
*Sin[a + b*x]^(5/2))

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.03

method	result	size
default	\(-\frac {c^{3} \left (6 \sqrt {-\sin \left (b x +a \right )+1}\, \sqrt {2 \sin \left (b x +a \right )+2}\, \left (\sqrt {\sin }\left (b x +a \right )\right ) E\left (\sqrt {-\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (b x +a \right )+1}\, \sqrt {2 \sin \left (b x +a \right )+2}\, \left (\sqrt {\sin }\left (b x +a \right )\right ) F\left (\sqrt {-\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{4}\left (b x +a \right )\right )+2 \left (\sin ^{2}\left (b x +a \right )\right )\right )}{5 \cos \left (b x +a \right ) \sqrt {c \sin \left (b x +a \right )}\, b}\)	\(152\)

[In]

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

[Out]

-1/5*c^3*(6*(-sin(b*x+a)+1)^(1/2)*(2*sin(b*x+a)+2)^(1/2)*sin(b*x+a)^(1/2)*EllipticE((-sin(b*x+a)+1)^(1/2),1/2*
2^(1/2))-3*(-sin(b*x+a)+1)^(1/2)*(2*sin(b*x+a)+2)^(1/2)*sin(b*x+a)^(1/2)*EllipticF((-sin(b*x+a)+1)^(1/2),1/2*2
^(1/2))-2*sin(b*x+a)^4+2*sin(b*x+a)^2)/cos(b*x+a)/(c*sin(b*x+a))^(1/2)/b

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.35 \[ \int (c \sin (a+b x))^{5/2} \, dx=-\frac {2 \, \sqrt {c \sin \left (b x + a\right )} c^{2} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 3 i \, \sqrt {2} \sqrt {-i \, c} c^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 3 i \, \sqrt {2} \sqrt {i \, c} c^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )}{5 \, b} \]

[In]

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

[Out]

-1/5*(2*sqrt(c*sin(b*x + a))*c^2*cos(b*x + a)*sin(b*x + a) - 3*I*sqrt(2)*sqrt(-I*c)*c^2*weierstrassZeta(4, 0,
weierstrassPInverse(4, 0, cos(b*x + a) + I*sin(b*x + a))) + 3*I*sqrt(2)*sqrt(I*c)*c^2*weierstrassZeta(4, 0, we
ierstrassPInverse(4, 0, cos(b*x + a) - I*sin(b*x + a))))/b

Sympy [F]

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

[In]

integrate((c*sin(b*x+a))**(5/2),x)

[Out]

Integral((c*sin(a + b*x))**(5/2), x)

Maxima [F]

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

[In]

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

[Out]

integrate((c*sin(b*x + a))^(5/2), x)

Giac [F]

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

[In]

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

[Out]

integrate((c*sin(b*x + a))^(5/2), x)

Mupad [F(-1)]

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

[In]

int((c*sin(a + b*x))^(5/2),x)

[Out]

int((c*sin(a + b*x))^(5/2), x)