3.514 \(\int \csc ^3(e+f x) \sqrt {d \csc (e+f x)} \, dx\)

Optimal. Leaf size=100 \[ -\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d^2 f}-\frac {6 \cos (e+f x) \sqrt {d \csc (e+f x)}}{5 f}-\frac {6 d E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right )}{5 f \sqrt {\sin (e+f x)} \sqrt {d \csc (e+f x)}} \]

[Out]

-2/5*cos(f*x+e)*(d*csc(f*x+e))^(5/2)/d^2/f-6/5*cos(f*x+e)*(d*csc(f*x+e))^(1/2)/f+6/5*d*(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))/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 = 100, 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} \[ -\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d^2 f}-\frac {6 \cos (e+f x) \sqrt {d \csc (e+f x)}}{5 f}-\frac {6 d E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right )}{5 f \sqrt {\sin (e+f x)} \sqrt {d \csc (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^3*Sqrt[d*Csc[e + f*x]],x]

[Out]

(-6*Cos[e + f*x]*Sqrt[d*Csc[e + f*x]])/(5*f) - (2*Cos[e + f*x]*(d*Csc[e + f*x])^(5/2))/(5*d^2*f) - (6*d*Ellipt
icE[(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 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 \csc ^3(e+f x) \sqrt {d \csc (e+f x)} \, dx &=\frac {\int (d \csc (e+f x))^{7/2} \, dx}{d^3}\\ &=-\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d^2 f}+\frac {3 \int (d \csc (e+f x))^{3/2} \, dx}{5 d}\\ &=-\frac {6 \cos (e+f x) \sqrt {d \csc (e+f x)}}{5 f}-\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d^2 f}-\frac {1}{5} (3 d) \int \frac {1}{\sqrt {d \csc (e+f x)}} \, dx\\ &=-\frac {6 \cos (e+f x) \sqrt {d \csc (e+f x)}}{5 f}-\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d^2 f}-\frac {(3 d) \int \sqrt {\sin (e+f x)} \, dx}{5 \sqrt {d \csc (e+f x)} \sqrt {\sin (e+f x)}}\\ &=-\frac {6 \cos (e+f x) \sqrt {d \csc (e+f x)}}{5 f}-\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d^2 f}-\frac {6 d 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*}

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Mathematica [A]  time = 0.17, size = 68, normalized size = 0.68 \[ -\frac {2 \sqrt {d \csc (e+f x)} \left (3 \cos (e+f x)+\cot (e+f x) \csc (e+f x)-3 \sqrt {\sin (e+f x)} E\left (\left .\frac {1}{4} (-2 e-2 f x+\pi )\right |2\right )\right )}{5 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^3*Sqrt[d*Csc[e + f*x]],x]

[Out]

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

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \csc \left (f x + e\right )} \csc \left (f x + e\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(d*csc(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(d*csc(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(d*csc(f*x + e))*csc(f*x + e)^3, x)

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maple [C]  time = 0.21, size = 1054, normalized size = 10.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(d*csc(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*csc(f*x + e))*csc(f*x + e)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**3*(d*csc(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(d*csc(e + f*x))*csc(e + f*x)**3, x)

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