3.4.12 \(\int \csc ^{\frac {5}{2}}(a+b \log (c x^n)) \, dx\) [312]

Optimal. Leaf size=109 \[ \frac {2 x \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \csc ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right ) \, _2F_1\left (\frac {5}{2},\frac {1}{4} \left (5-\frac {2 i}{b n}\right );\frac {1}{4} \left (9-\frac {2 i}{b n}\right );e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{2+5 i b n} \]

[Out]

2*x*(1-exp(2*I*a)*(c*x^n)^(2*I*b))^(5/2)*csc(a+b*ln(c*x^n))^(5/2)*hypergeom([5/2, 5/4-1/2*I/b/n],[9/4-1/2*I/b/
n],exp(2*I*a)*(c*x^n)^(2*I*b))/(2+5*I*b*n)

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Rubi [A]
time = 0.05, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4600, 4604, 371} \begin {gather*} \frac {2 x \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \, _2F_1\left (\frac {5}{2},\frac {1}{4} \left (5-\frac {2 i}{b n}\right );\frac {1}{4} \left (9-\frac {2 i}{b n}\right );e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \csc ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{2+5 i b n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*Log[c*x^n]]^(5/2),x]

[Out]

(2*x*(1 - E^((2*I)*a)*(c*x^n)^((2*I)*b))^(5/2)*Csc[a + b*Log[c*x^n]]^(5/2)*Hypergeometric2F1[5/2, (5 - (2*I)/(
b*n))/4, (9 - (2*I)/(b*n))/4, E^((2*I)*a)*(c*x^n)^((2*I)*b)])/(2 + (5*I)*b*n)

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 4600

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

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]

Rubi steps

\begin {align*} \int \csc ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {\left (x \left (c x^n\right )^{-1/n}\right ) \text {Subst}\left (\int x^{-1+\frac {1}{n}} \csc ^{\frac {5}{2}}(a+b \log (x)) \, dx,x,c x^n\right )}{n}\\ &=\frac {\left (x \left (c x^n\right )^{-\frac {5 i b}{2}-\frac {1}{n}} \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \csc ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {5 i b}{2}+\frac {1}{n}}}{\left (1-e^{2 i a} x^{2 i b}\right )^{5/2}} \, dx,x,c x^n\right )}{n}\\ &=\frac {2 x \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \csc ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right ) \, _2F_1\left (\frac {5}{2},\frac {1}{4} \left (5-\frac {2 i}{b n}\right );\frac {1}{4} \left (9-\frac {2 i}{b n}\right );e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{2+5 i b n}\\ \end {align*}

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Mathematica [A]
time = 1.67, size = 174, normalized size = 1.60 \begin {gather*} \frac {2 e^{-2 i \left (a-b n \log (x)+b \log \left (c x^n\right )\right )} x^{1-2 i b n} \sqrt {\csc \left (a+b \log \left (c x^n\right )\right )} \left (-e^{2 i a} \left (c x^n\right )^{2 i b} \left (2+b n \cot \left (a+b \log \left (c x^n\right )\right )\right )+(2+i b n) \left (-1+e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \, _2F_1\left (1,\frac {3}{4}+\frac {i}{2 b n};\frac {5}{4}+\frac {i}{2 b n};e^{-2 i \left (a+b \log \left (c x^n\right )\right )}\right )\right )}{3 b^2 n^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*Log[c*x^n]]^(5/2),x]

[Out]

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

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Maple [F]
time = 0.13, size = 0, normalized size = 0.00 \[\int \csc ^{\frac {5}{2}}\left (a +b \ln \left (c \,x^{n}\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(a+b*ln(c*x^n))^(5/2),x)

[Out]

int(csc(a+b*ln(c*x^n))^(5/2),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(csc(a+b*log(c*x^n))^(5/2),x, algorithm="maxima")

[Out]

integrate(csc(b*log(c*x^n) + a)^(5/2), 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(csc(a+b*log(c*x^n))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(a+b*ln(c*x**n))**(5/2),x)

[Out]

Timed out

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(a+b*log(c*x^n))^(5/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {1}{\sin \left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/sin(a + b*log(c*x^n)))^(5/2),x)

[Out]

int((1/sin(a + b*log(c*x^n)))^(5/2), x)

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