3.516 \(\int (d \csc (e+f x))^{3/2} \sin ^4(e+f x) \, dx\)

Optimal. Leaf size=77 \[ \frac{6 d^2 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right )}{5 f \sqrt{\sin (e+f x)} \sqrt{d \csc (e+f x)}}-\frac{2 d^3 \cos (e+f x)}{5 f (d \csc (e+f x))^{3/2}} \]

[Out]

(-2*d^3*Cos[e + f*x])/(5*f*(d*Csc[e + f*x])^(3/2)) + (6*d^2*EllipticE[(e - Pi/2 + f*x)/2, 2])/(5*f*Sqrt[d*Csc[
e + f*x]]*Sqrt[Sin[e + f*x]])

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Rubi [A]  time = 0.0556424, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {16, 3769, 3771, 2639} \[ \frac{6 d^2 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right )}{5 f \sqrt{\sin (e+f x)} \sqrt{d \csc (e+f x)}}-\frac{2 d^3 \cos (e+f x)}{5 f (d \csc (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Csc[e + f*x])^(3/2)*Sin[e + f*x]^4,x]

[Out]

(-2*d^3*Cos[e + f*x])/(5*f*(d*Csc[e + f*x])^(3/2)) + (6*d^2*EllipticE[(e - Pi/2 + f*x)/2, 2])/(5*f*Sqrt[d*Csc[
e + f*x]]*Sqrt[Sin[e + f*x]])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

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

Rubi steps

\begin{align*} \int (d \csc (e+f x))^{3/2} \sin ^4(e+f x) \, dx &=d^4 \int \frac{1}{(d \csc (e+f x))^{5/2}} \, dx\\ &=-\frac{2 d^3 \cos (e+f x)}{5 f (d \csc (e+f x))^{3/2}}+\frac{1}{5} \left (3 d^2\right ) \int \frac{1}{\sqrt{d \csc (e+f x)}} \, dx\\ &=-\frac{2 d^3 \cos (e+f x)}{5 f (d \csc (e+f x))^{3/2}}+\frac{\left (3 d^2\right ) \int \sqrt{\sin (e+f x)} \, dx}{5 \sqrt{d \csc (e+f x)} \sqrt{\sin (e+f x)}}\\ &=-\frac{2 d^3 \cos (e+f x)}{5 f (d \csc (e+f x))^{3/2}}+\frac{6 d^2 E\left (\left .\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )\right |2\right )}{5 f \sqrt{d \csc (e+f x)} \sqrt{\sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.188771, size = 62, normalized size = 0.81 \[ -\frac{2 (d \csc (e+f x))^{3/2} \left (\sin ^3(e+f x) \cos (e+f x)+3 \sin ^{\frac{3}{2}}(e+f x) E\left (\left .\frac{1}{4} (-2 e-2 f x+\pi )\right |2\right )\right )}{5 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Csc[e + f*x])^(3/2)*Sin[e + f*x]^4,x]

[Out]

(-2*(d*Csc[e + f*x])^(3/2)*(3*EllipticE[(-2*e + Pi - 2*f*x)/4, 2]*Sin[e + f*x]^(3/2) + Cos[e + f*x]*Sin[e + f*
x]^3))/(5*f)

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Maple [C]  time = 0.151, size = 545, normalized size = 7.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*csc(f*x+e))^(3/2)*sin(f*x+e)^4,x)

[Out]

-1/5/f*2^(1/2)*(6*cos(f*x+e)*(-I*(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((I*cos(f*x+e)+sin(f*x+e)-I)/sin(f*x+e))^(1
/2)*(-(I*cos(f*x+e)-sin(f*x+e)-I)/sin(f*x+e))^(1/2)*EllipticE(((I*cos(f*x+e)+sin(f*x+e)-I)/sin(f*x+e))^(1/2),1
/2*2^(1/2))-3*cos(f*x+e)*(-I*(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((I*cos(f*x+e)+sin(f*x+e)-I)/sin(f*x+e))^(1/2)*
(-(I*cos(f*x+e)-sin(f*x+e)-I)/sin(f*x+e))^(1/2)*EllipticF(((I*cos(f*x+e)+sin(f*x+e)-I)/sin(f*x+e))^(1/2),1/2*2
^(1/2))+6*(-I*(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((I*cos(f*x+e)+sin(f*x+e)-I)/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)
-sin(f*x+e)-I)/sin(f*x+e))^(1/2)*EllipticE(((I*cos(f*x+e)+sin(f*x+e)-I)/sin(f*x+e))^(1/2),1/2*2^(1/2))-3*(-I*(
-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((I*cos(f*x+e)+sin(f*x+e)-I)/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-sin(f*x+e)-I)/
sin(f*x+e))^(1/2)*EllipticF(((I*cos(f*x+e)+sin(f*x+e)-I)/sin(f*x+e))^(1/2),1/2*2^(1/2))-2^(1/2)*cos(f*x+e)^3+4
*2^(1/2)*cos(f*x+e)-3*2^(1/2))*(d/sin(f*x+e))^(3/2)*sin(f*x+e)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \csc \left (f x + e\right )\right )^{\frac{3}{2}} \sin \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(f*x+e))^(3/2)*sin(f*x+e)^4,x, algorithm="maxima")

[Out]

integrate((d*csc(f*x + e))^(3/2)*sin(f*x + e)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (d \cos \left (f x + e\right )^{4} - 2 \, d \cos \left (f x + e\right )^{2} + d\right )} \sqrt{d \csc \left (f x + e\right )} \csc \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(f*x+e))^(3/2)*sin(f*x+e)^4,x, algorithm="fricas")

[Out]

integral((d*cos(f*x + e)^4 - 2*d*cos(f*x + e)^2 + d)*sqrt(d*csc(f*x + e))*csc(f*x + e), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(f*x+e))**(3/2)*sin(f*x+e)**4,x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(f*x+e))^(3/2)*sin(f*x+e)^4,x, algorithm="giac")

[Out]

Timed out