3.1.80 \(\int x^2 \sin ^p(a+b \log (c x^n)) \, dx\) [80]

Optimal. Leaf size=114 \[ \frac {x^3 \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{-p} \, _2F_1\left (-p,-\frac {3 i+b n p}{2 b n};\frac {1}{2} \left (2-\frac {3 i}{b n}-p\right );e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sin ^p\left (a+b \log \left (c x^n\right )\right )}{3-i b n p} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4581, 4579, 371} \begin {gather*} \frac {x^3 \left (1-e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{-p} \, _2F_1\left (-p,-\frac {b n p+3 i}{2 b n};\frac {1}{2} \left (-p-\frac {3 i}{b n}+2\right );e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sin ^p\left (a+b \log \left (c x^n\right )\right )}{3-i b n p} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Sin[a + b*Log[c*x^n]]^p,x]

[Out]

(x^3*Hypergeometric2F1[-p, -1/2*(3*I + b*n*p)/(b*n), (2 - (3*I)/(b*n) - p)/2, E^((2*I)*a)*(c*x^n)^((2*I)*b)]*S
in[a + b*Log[c*x^n]]^p)/((3 - I*b*n*p)*(1 - E^((2*I)*a)*(c*x^n)^((2*I)*b))^p)

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

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Mathematica [A]
time = 0.98, size = 148, normalized size = 1.30 \begin {gather*} \frac {i x^3 \left (2-2 e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{-p} \left (-i e^{-i a} \left (c x^n\right )^{-i b} \left (-1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )\right )^p \, _2F_1\left (-p,-\frac {3 i+b n p}{2 b n};1-\frac {3 i}{2 b n}-\frac {p}{2};e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{3 i+b n p} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sin[a + b*Log[c*x^n]]^p,x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [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^2*sin(a+b*log(c*x^n))^p,x, algorithm="fricas")

[Out]

integral(x^2*sin(b*log(c*x^n) + a)^p, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \sin ^{p}{\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**2*sin(a+b*ln(c*x**n))**p,x)

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sin(a + b*log(c*x^n))^p,x)

[Out]

int(x^2*sin(a + b*log(c*x^n))^p, x)

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