3.327 \(\int \frac{x^m}{\csc ^{\frac{3}{2}}(a+b \log (c x^n))} \, dx\)

Optimal. Leaf size=130 \[ \frac{2 x^{m+1} \text{Hypergeometric2F1}\left (-\frac{3}{2},-\frac{3 b n+2 i m+2 i}{4 b n},-\frac{-b n+2 i m+2 i}{4 b n},e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(-3 i b n+2 m+2) \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \csc ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

[Out]

(2*x^(1 + m)*Hypergeometric2F1[-3/2, -(2*I + (2*I)*m + 3*b*n)/(4*b*n), -(2*I + (2*I)*m - b*n)/(4*b*n), E^((2*I
)*a)*(c*x^n)^((2*I)*b)])/((2 + 2*m - (3*I)*b*n)*(1 - E^((2*I)*a)*(c*x^n)^((2*I)*b))^(3/2)*Csc[a + b*Log[c*x^n]
]^(3/2))

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Rubi [A]  time = 0.0943862, antiderivative size = 126, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4510, 4508, 364} \[ \frac{2 x^{m+1} \, _2F_1\left (-\frac{3}{2},\frac{1}{4} \left (-\frac{2 i (m+1)}{b n}-3\right );-\frac{2 i m-b n+2 i}{4 b n};e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(-3 i b n+2 m+2) \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \csc ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^m/Csc[a + b*Log[c*x^n]]^(3/2),x]

[Out]

(2*x^(1 + m)*Hypergeometric2F1[-3/2, (-3 - ((2*I)*(1 + m))/(b*n))/4, -(2*I + (2*I)*m - b*n)/(4*b*n), E^((2*I)*
a)*(c*x^n)^((2*I)*b)])/((2 + 2*m - (3*I)*b*n)*(1 - E^((2*I)*a)*(c*x^n)^((2*I)*b))^(3/2)*Csc[a + b*Log[c*x^n]]^
(3/2))

Rule 4510

Int[Csc[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(e*x)^(m + 1)
/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Csc[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a
, b, c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])

Rule 4508

Int[Csc[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(Csc[d*(a + b*Log[x])]^p*(1
- E^(2*I*a*d)*x^(2*I*b*d))^p)/x^(I*b*d*p), Int[((e*x)^m*x^(I*b*d*p))/(1 - E^(2*I*a*d)*x^(2*I*b*d))^p, x], x]
/; FreeQ[{a, b, d, e, m, p}, x] &&  !IntegerQ[p]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{x^m}{\csc ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx &=\frac{\left (x^{1+m} \left (c x^n\right )^{-\frac{1+m}{n}}\right ) \operatorname{Subst}\left (\int \frac{x^{-1+\frac{1+m}{n}}}{\csc ^{\frac{3}{2}}(a+b \log (x))} \, dx,x,c x^n\right )}{n}\\ &=\frac{\left (x^{1+m} \left (c x^n\right )^{\frac{3 i b}{2}-\frac{1+m}{n}}\right ) \operatorname{Subst}\left (\int x^{-1-\frac{3 i b}{2}+\frac{1+m}{n}} \left (1-e^{2 i a} x^{2 i b}\right )^{3/2} \, dx,x,c x^n\right )}{n \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \csc ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}\\ &=\frac{2 x^{1+m} \, _2F_1\left (-\frac{3}{2},\frac{1}{4} \left (-3-\frac{2 i (1+m)}{b n}\right );-\frac{2 i+2 i m-b n}{4 b n};e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2+2 m-3 i b n) \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \csc ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}\\ \end{align*}

Mathematica [A]  time = 2.40706, size = 218, normalized size = 1.68 \[ \frac{2 x^{m+1} \left (3 e^{-2 i a} b^2 n^2 \left (c x^n\right )^{-2 i b} \left (-1+e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \csc ^2\left (a+b \log \left (c x^n\right )\right ) \text{Hypergeometric2F1}\left (1,\frac{3 b n+2 i m+2 i}{4 b n},\frac{5 b n+2 i m+2 i}{4 b n},e^{-2 i \left (a+b \log \left (c x^n\right )\right )}\right )+(-i b n+2 m+2) \left (-3 b n \cot \left (a+b \log \left (c x^n\right )\right )+2 m+2\right )\right )}{(-i b n+2 m+2) (-3 i b n+2 m+2) (3 i b n+2 m+2) \csc ^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^m/Csc[a + b*Log[c*x^n]]^(3/2),x]

[Out]

(2*x^(1 + m)*((2 + 2*m - I*b*n)*(2 + 2*m - 3*b*n*Cot[a + b*Log[c*x^n]]) + (3*b^2*n^2*(-1 + E^((2*I)*a)*(c*x^n)
^((2*I)*b))*Csc[a + b*Log[c*x^n]]^2*Hypergeometric2F1[1, (2*I + (2*I)*m + 3*b*n)/(4*b*n), (2*I + (2*I)*m + 5*b
*n)/(4*b*n), E^((-2*I)*(a + b*Log[c*x^n]))])/(E^((2*I)*a)*(c*x^n)^((2*I)*b))))/((2 + 2*m - I*b*n)*(2 + 2*m - (
3*I)*b*n)*(2 + 2*m + (3*I)*b*n)*Csc[a + b*Log[c*x^n]]^(3/2))

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Maple [F]  time = 0.307, size = 0, normalized size = 0. \begin{align*} \int{{x}^{m} \left ( \csc \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/csc(a+b*ln(c*x^n))^(3/2),x)

[Out]

int(x^m/csc(a+b*ln(c*x^n))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{\csc \left (b \log \left (c x^{n}\right ) + a\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/csc(a+b*log(c*x^n))^(3/2),x, algorithm="maxima")

[Out]

integrate(x^m/csc(b*log(c*x^n) + a)^(3/2), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/csc(a+b*log(c*x^n))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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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(x**m/csc(a+b*ln(c*x**n))**(3/2),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(x^m/csc(a+b*log(c*x^n))^(3/2),x, algorithm="giac")

[Out]

Timed out