3.67 \(\int \frac {1}{\sin ^{\frac {5}{2}}(a+b \log (c x^n))} \, dx\)

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

[Out]

2*x*(1-exp(2*I*a)*(c*x^n)^(2*I*b))^(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)/sin(a+b*ln(c*x^n))^(5/2)

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Rubi [A]  time = 0.07, 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 = {4483, 4491, 364} \[ \frac {2 x \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \, _2F_1\left (\frac {5}{2},\frac {1}{4} \left (5-\frac {2 i}{b n}\right );\frac {1}{4} \left (9-\frac {2 i}{b n}\right );e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2+5 i b n) \sin ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sin[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)*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)*Sin[a + b*Log[c*x^n]]^(5/2))

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])

Rule 4483

Int[Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[x
^(1/n - 1)*Sin[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 4491

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

Rubi steps

\begin {align*} \int \frac {1}{\sin ^{\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 ) \operatorname {Subst}\left (\int \frac {x^{-1+\frac {1}{n}}}{\sin ^{\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}\right ) \operatorname {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 \sin ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}\\ &=\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 )}{(2+5 i b n) \sin ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}\\ \end {align*}

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Mathematica [A]  time = 1.51, size = 125, normalized size = 1.15 \[ \frac {2 x \left (i (b n+2 i) \left (-1+e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \, _2F_1\left (1,\frac {3}{4}-\frac {i}{2 b n};\frac {5}{4}-\frac {i}{2 b n};e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )-b n \cot \left (a+b \log \left (c x^n\right )\right )-2\right )}{3 b^2 n^2 \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*x*(-2 - b*n*Cot[a + b*Log[c*x^n]] + I*(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*n^2*Sqrt[Sin[a + b*Log[c*x^n]]])

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sin \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(a+b*log(c*x^n))^(5/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sin \left (a +b \ln \left (c \,x^{n}\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sin \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(a+b*log(c*x^n))^(5/2),x, algorithm="maxima")

[Out]

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

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(a+b*ln(c*x**n))**(5/2),x)

[Out]

Timed out

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