3.4.27 \(\int \frac {x^m}{\csc ^{\frac {3}{2}}(a+b \log (c x^n))} \, dx\) [327]

Optimal. Leaf size=130 \[ \frac {2 x^{1+m} \, _2F_1\left (-\frac {3}{2},-\frac {2 i+2 i m+3 b n}{4 b n};-\frac {2 i+2 i m-b n}{4 b n};e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2+2 m-3 i b n) \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2} \csc ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

[Out]

2*x^(1+m)*hypergeom([-3/2, 1/4*(-2*I-2*I*m-3*b*n)/b/n],[1/4*(-2*I-2*I*m+b*n)/b/n],exp(2*I*a)*(c*x^n)^(2*I*b))/
(2+2*m-3*I*b*n)/(1-exp(2*I*a)*(c*x^n)^(2*I*b))^(3/2)/csc(a+b*ln(c*x^n))^(3/2)

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Rubi [A]
time = 0.07, 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 = {4606, 4604, 371} \begin {gather*} \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 )} \end {gather*}

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, -1/4*(2*I + (2*I)*m - b*n)/(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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 4604

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 4606

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])

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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 6.98, size = 228, normalized size = 1.75 \begin {gather*} \frac {2 x^{1+m} \left (\frac {2+2 m-3 b n \cot \left (a+b \log \left (c x^n\right )\right )}{\csc ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac {3 b^2 n^2 \sqrt {2-2 e^{2 i a} \left (c x^n\right )^{2 i b}} \sqrt {\frac {i e^{i a} \left (c x^n\right )^{i b}}{-1+e^{2 i a} \left (c x^n\right )^{2 i b}}} \, _2F_1\left (\frac {1}{2},\frac {-2 i-2 i m+b n}{4 b n};-\frac {2 i+2 i m-5 b n}{4 b n};e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{2+2 m+i b n}\right )}{4+8 m+4 m^2+9 b^2 n^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(1 + m)*((2 + 2*m - 3*b*n*Cot[a + b*Log[c*x^n]])/Csc[a + b*Log[c*x^n]]^(3/2) + (3*b^2*n^2*Sqrt[2 - 2*E^((
2*I)*a)*(c*x^n)^((2*I)*b)]*Sqrt[(I*E^(I*a)*(c*x^n)^(I*b))/(-1 + E^((2*I)*a)*(c*x^n)^((2*I)*b))]*Hypergeometric
2F1[1/2, (-2*I - (2*I)*m + b*n)/(4*b*n), -1/4*(2*I + (2*I)*m - 5*b*n)/(b*n), E^((2*I)*a)*(c*x^n)^((2*I)*b)])/(
2 + 2*m + I*b*n)))/(4 + 8*m + 4*m^2 + 9*b^2*n^2)

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \frac {x^{m}}{\csc \left (a +b \ln \left (c \,x^{n}\right )\right )^{\frac {3}{2}}}\, dx\]

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 = 0.00, 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(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{m}}{\csc ^{\frac {3}{2}}{\left (a + b \log {\left (c x^{n} \right )} \right )}}\, dx \end {gather*}

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]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(1/sin(a + b*log(c*x^n)))^(3/2),x)

[Out]

int(x^m/(1/sin(a + b*log(c*x^n)))^(3/2), x)

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