3.537 \(\int \frac {\csc ^5(e+f x)}{(d \csc (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=105 \[ -\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d^4 f}-\frac {6 \cos (e+f x) \sqrt {d \csc (e+f x)}}{5 d^2 f}-\frac {6 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right )}{5 d f \sqrt {\sin (e+f x)} \sqrt {d \csc (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 3768, 3771, 2639} \[ -\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d^4 f}-\frac {6 \cos (e+f x) \sqrt {d \csc (e+f x)}}{5 d^2 f}-\frac {6 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right )}{5 d f \sqrt {\sin (e+f x)} \sqrt {d \csc (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^5/(d*Csc[e + f*x])^(3/2),x]

[Out]

(-6*Cos[e + f*x]*Sqrt[d*Csc[e + f*x]])/(5*d^2*f) - (2*Cos[e + f*x]*(d*Csc[e + f*x])^(5/2))/(5*d^4*f) - (6*Elli
pticE[(e - Pi/2 + f*x)/2, 2])/(5*d*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 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]

Rule 3768

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

Rubi steps

\begin {align*} \int \frac {\csc ^5(e+f x)}{(d \csc (e+f x))^{3/2}} \, dx &=\frac {\int (d \csc (e+f x))^{7/2} \, dx}{d^5}\\ &=-\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d^4 f}+\frac {3 \int (d \csc (e+f x))^{3/2} \, dx}{5 d^3}\\ &=-\frac {6 \cos (e+f x) \sqrt {d \csc (e+f x)}}{5 d^2 f}-\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d^4 f}-\frac {3 \int \frac {1}{\sqrt {d \csc (e+f x)}} \, dx}{5 d}\\ &=-\frac {6 \cos (e+f x) \sqrt {d \csc (e+f x)}}{5 d^2 f}-\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d^4 f}-\frac {3 \int \sqrt {\sin (e+f x)} \, dx}{5 d \sqrt {d \csc (e+f x)} \sqrt {\sin (e+f x)}}\\ &=-\frac {6 \cos (e+f x) \sqrt {d \csc (e+f x)}}{5 d^2 f}-\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d^4 f}-\frac {6 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right )}{5 d f \sqrt {d \csc (e+f x)} \sqrt {\sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 73, normalized size = 0.70 \[ \frac {\csc ^4(e+f x) \left (-7 \cos (e+f x)+3 \cos (3 (e+f x))+12 \sin ^{\frac {5}{2}}(e+f x) E\left (\left .\frac {1}{4} (-2 e-2 f x+\pi )\right |2\right )\right )}{10 f (d \csc (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^5/(d*Csc[e + f*x])^(3/2),x]

[Out]

(Csc[e + f*x]^4*(-7*Cos[e + f*x] + 3*Cos[3*(e + f*x)] + 12*EllipticE[(-2*e + Pi - 2*f*x)/4, 2]*Sin[e + f*x]^(5
/2)))/(10*f*(d*Csc[e + f*x])^(3/2))

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \csc \left (f x + e\right )} \csc \left (f x + e\right )^{3}}{d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(d*csc(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*csc(f*x + e))*csc(f*x + e)^3/d^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )^{5}}{\left (d \csc \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(d*csc(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^5/(d*csc(f*x + e))^(3/2), x)

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maple [C]  time = 0.18, size = 1054, normalized size = 10.04 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^5/(d*csc(f*x+e))^(3/2),x)

[Out]

-1/5/f*(6*cos(f*x+e)^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)*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)^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))+6*cos(f*x+e)^2*(-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*c
os(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)^2*(-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*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*co
s(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)/s
in(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))/s
in(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))-3*cos(f*x+e)^2*2^(1/2)+cos(f*x+e)*2^(1
/2)+3*2^(1/2))/sin(f*x+e)^4/(d/sin(f*x+e))^(3/2)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )^{5}}{\left (d \csc \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(d*csc(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate(csc(f*x + e)^5/(d*csc(f*x + e))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\sin \left (e+f\,x\right )}^5\,{\left (\frac {d}{\sin \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^5*(d/sin(e + f*x))^(3/2)),x)

[Out]

int(1/(sin(e + f*x)^5*(d/sin(e + f*x))^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{5}{\left (e + f x \right )}}{\left (d \csc {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**5/(d*csc(f*x+e))**(3/2),x)

[Out]

Integral(csc(e + f*x)**5/(d*csc(e + f*x))**(3/2), x)

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