3.6.22 \(\int \csc ^2(e+f x) (d \csc (e+f x))^{3/2} \, dx\) [522]

Optimal. Leaf size=103 \[ -\frac {6 d \cos (e+f x) \sqrt {d \csc (e+f x)}}{5 f}-\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d f}-\frac {6 d^2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right )}{5 f \sqrt {d \csc (e+f x)} \sqrt {\sin (e+f x)}} \]

[Out]

-2/5*cos(f*x+e)*(d*csc(f*x+e))^(5/2)/d/f-6/5*d*cos(f*x+e)*(d*csc(f*x+e))^(1/2)/f+6/5*d^2*(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)/s
in(f*x+e)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 103, 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, 3853, 3856, 2719} \begin {gather*} -\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 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d f}-\frac {6 d \cos (e+f x) \sqrt {d \csc (e+f x)}}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*d*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*f) - (6*d^2*Elli
pticE[(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 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 3853

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 3856

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 ^2(e+f x) (d \csc (e+f x))^{3/2} \, dx &=\frac {\int (d \csc (e+f x))^{7/2} \, dx}{d^2}\\ &=-\frac {2 \cos (e+f x) (d \csc (e+f x))^{5/2}}{5 d f}+\frac {3}{5} \int (d \csc (e+f x))^{3/2} \, dx\\ &=-\frac {6 d \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 f}-\frac {1}{5} \left (3 d^2\right ) \int \frac {1}{\sqrt {d \csc (e+f x)}} \, dx\\ &=-\frac {6 d \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 f}-\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 {6 d \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 f}-\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*}

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Mathematica [A]
time = 0.18, size = 68, normalized size = 0.66 \begin {gather*} \frac {(d \csc (e+f x))^{5/2} \left (-7 \cos (e+f x)+3 \cos (3 (e+f x))+12 E\left (\left .\frac {1}{4} (-2 e+\pi -2 f x)\right |2\right ) \sin ^{\frac {5}{2}}(e+f x)\right )}{10 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d*Csc[e + f*x])^(5/2)*(-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*d*f)

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Maple [C] Result contains complex when optimal does not.
time = 0.15, size = 1055, normalized size = 10.24

method	result	size
default	\(\text {Expression too large to display}\)	\(1055\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^2*(d*csc(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 148, normalized size = 1.44 \begin {gather*} -\frac {3 \, {\left (d \cos \left (f x + e\right )^{2} - d\right )} \sqrt {2 i \, d} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, {\left (d \cos \left (f x + e\right )^{2} - d\right )} \sqrt {-2 i \, d} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, {\left (3 \, d \cos \left (f x + e\right )^{3} - 4 \, d \cos \left (f x + e\right )\right )} \sqrt {\frac {d}{\sin \left (f x + e\right )}}}{5 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(3*(d*cos(f*x + e)^2 - d)*sqrt(2*I*d)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(f*x + e) + I*si
n(f*x + e))) + 3*(d*cos(f*x + e)^2 - d)*sqrt(-2*I*d)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(f*x +
e) - I*sin(f*x + e))) + 2*(3*d*cos(f*x + e)^3 - 4*d*cos(f*x + e))*sqrt(d/sin(f*x + e)))/(f*cos(f*x + e)^2 - f
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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