3.1.62 \(\int \frac {\sin ^{\frac {3}{2}}(a+b \log (c x^n))}{x^3} \, dx\) [62]

Optimal. Leaf size=111 \[ -\frac {2 \, _2F_1\left (-\frac {3}{2},\frac {1}{4} \left (-3+\frac {4 i}{b n}\right );\frac {1}{4} \left (1+\frac {4 i}{b n}\right );e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{(4+3 i b n) x^2 \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 111, normalized size of antiderivative = 1.00, 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 = {4581, 4579, 371} \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {3}{2},\frac {1}{4} \left (\frac {4 i}{b n}-3\right );\frac {1}{4} \left (1+\frac {4 i}{b n}\right );e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{x^2 (4+3 i b n) \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Hypergeometric2F1[-3/2, (-3 + (4*I)/(b*n))/4, (1 + (4*I)/(b*n))/4, E^((2*I)*a)*(c*x^n)^((2*I)*b)]*Sin[a +
b*Log[c*x^n]]^(3/2))/((4 + (3*I)*b*n)*x^2*(1 - E^((2*I)*a)*(c*x^n)^((2*I)*b))^(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 4579

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]

Rule 4581

Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(e*x)^(m + 1)
/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Sin[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 {\sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx &=\frac {\left (c x^n\right )^{2/n} \text {Subst}\left (\int x^{-1-\frac {2}{n}} \sin ^{\frac {3}{2}}(a+b \log (x)) \, dx,x,c x^n\right )}{n x^2}\\ &=\frac {\left (\left (c x^n\right )^{\frac {3 i b}{2}+\frac {2}{n}} \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )\right ) \text {Subst}\left (\int x^{-1-\frac {3 i b}{2}-\frac {2}{n}} \left (1-e^{2 i a} x^{2 i b}\right )^{3/2} \, dx,x,c x^n\right )}{n x^2 \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2}}\\ &=-\frac {2 \, _2F_1\left (-\frac {3}{2},\frac {1}{4} \left (-3+\frac {4 i}{b n}\right );\frac {1}{4} \left (1+\frac {4 i}{b n}\right );e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{(4+3 i b n) x^2 \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 1.17, size = 216, normalized size = 1.95 \begin {gather*} \frac {6 b^2 \sqrt {2-2 e^{2 i \left (a+b \log \left (c x^n\right )\right )}} n^2 \, _2F_1\left (\frac {1}{2},\frac {1}{4}+\frac {i}{b n};\frac {5}{4}+\frac {i}{b n};e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt {-i e^{-i \left (a+b \log \left (c x^n\right )\right )} \left (-1+e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )} (4+3 i b n) (4 i+b n) (4 i+3 b n) x^2}-\frac {2 \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )} \left (3 b n \cos \left (a+b \log \left (c x^n\right )\right )+4 \sin \left (a+b \log \left (c x^n\right )\right )\right )}{\left (16+9 b^2 n^2\right ) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*b^2*Sqrt[2 - 2*E^((2*I)*(a + b*Log[c*x^n]))]*n^2*Hypergeometric2F1[1/2, 1/4 + I/(b*n), 5/4 + I/(b*n), E^((2
*I)*(a + b*Log[c*x^n]))])/(Sqrt[((-I)*(-1 + E^((2*I)*(a + b*Log[c*x^n]))))/E^(I*(a + b*Log[c*x^n]))]*(4 + (3*I
)*b*n)*(4*I + b*n)*(4*I + 3*b*n)*x^2) - (2*Sqrt[Sin[a + b*Log[c*x^n]]]*(3*b*n*Cos[a + b*Log[c*x^n]] + 4*Sin[a
+ b*Log[c*x^n]]))/((16 + 9*b^2*n^2)*x^2)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sin ^{\frac {3}{2}}\left (a +b \ln \left (c \,x^{n}\right )\right )}{x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sin(a+b*ln(c*x^n))^(3/2)/x^3,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(sin(a+b*log(c*x^n))^(3/2)/x^3,x, algorithm="maxima")

[Out]

integrate(sin(b*log(c*x^n) + a)^(3/2)/x^3, 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(sin(a+b*log(c*x^n))^(3/2)/x^3,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 {\sin ^{\frac {3}{2}}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b*ln(c*x**n))**(3/2)/x**3,x)

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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