3.8 \(\int x \sin ^2(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=98 \[ \frac{x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (b^2 n^2+1\right )}-\frac{b n x^2 \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{2 \left (b^2 n^2+1\right )}+\frac{b^2 n^2 x^2}{4 \left (b^2 n^2+1\right )} \]

[Out]

(b^2*n^2*x^2)/(4*(1 + b^2*n^2)) - (b*n*x^2*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]])/(2*(1 + b^2*n^2)) + (x
^2*Sin[a + b*Log[c*x^n]]^2)/(2*(1 + b^2*n^2))

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Rubi [A]  time = 0.0222139, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4487, 30} \[ \frac{x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (b^2 n^2+1\right )}-\frac{b n x^2 \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{2 \left (b^2 n^2+1\right )}+\frac{b^2 n^2 x^2}{4 \left (b^2 n^2+1\right )} \]

Antiderivative was successfully verified.

[In]

Int[x*Sin[a + b*Log[c*x^n]]^2,x]

[Out]

(b^2*n^2*x^2)/(4*(1 + b^2*n^2)) - (b*n*x^2*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]])/(2*(1 + b^2*n^2)) + (x
^2*Sin[a + b*Log[c*x^n]]^2)/(2*(1 + b^2*n^2))

Rule 4487

Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_), x_Symbol] :> Simp[((m + 1)*(e*x)
^(m + 1)*Sin[d*(a + b*Log[c*x^n])]^p)/(b^2*d^2*e*n^2*p^2 + e*(m + 1)^2), x] + (Dist[(b^2*d^2*n^2*p*(p - 1))/(b
^2*d^2*n^2*p^2 + (m + 1)^2), Int[(e*x)^m*Sin[d*(a + b*Log[c*x^n])]^(p - 2), x], x] - Simp[(b*d*n*p*(e*x)^(m +
1)*Cos[d*(a + b*Log[c*x^n])]*Sin[d*(a + b*Log[c*x^n])]^(p - 1))/(b^2*d^2*e*n^2*p^2 + e*(m + 1)^2), x]) /; Free
Q[{a, b, c, d, e, m, n}, x] && IGtQ[p, 1] && NeQ[b^2*d^2*n^2*p^2 + (m + 1)^2, 0]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x \sin ^2\left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac{b n x^2 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{2 \left (1+b^2 n^2\right )}+\frac{x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+b^2 n^2\right )}+\frac{\left (b^2 n^2\right ) \int x \, dx}{2 \left (1+b^2 n^2\right )}\\ &=\frac{b^2 n^2 x^2}{4 \left (1+b^2 n^2\right )}-\frac{b n x^2 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{2 \left (1+b^2 n^2\right )}+\frac{x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+b^2 n^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.109502, size = 57, normalized size = 0.58 \[ \frac{x^2 \left (-b n \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-\cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+b^2 n^2+1\right )}{4 b^2 n^2+4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(1 + b^2*n^2 - Cos[2*(a + b*Log[c*x^n])] - b*n*Sin[2*(a + b*Log[c*x^n])]))/(4 + 4*b^2*n^2)

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int x \left ( \sin \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.16108, size = 381, normalized size = 3.89 \begin{align*} -\frac{{\left ({\left (b \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (4 \, b \log \left (c\right )\right ) - b \cos \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + b \sin \left (2 \, b \log \left (c\right )\right )\right )} n + \cos \left (4 \, b \log \left (c\right )\right ) \cos \left (2 \, b \log \left (c\right )\right ) + \sin \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + \cos \left (2 \, b \log \left (c\right )\right )\right )} x^{2} \cos \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right ) +{\left ({\left (b \cos \left (4 \, b \log \left (c\right )\right ) \cos \left (2 \, b \log \left (c\right )\right ) + b \sin \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + b \cos \left (2 \, b \log \left (c\right )\right )\right )} n - \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (4 \, b \log \left (c\right )\right ) + \cos \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) - \sin \left (2 \, b \log \left (c\right )\right )\right )} x^{2} \sin \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right ) - 2 \,{\left ({\left (b^{2} \cos \left (2 \, b \log \left (c\right )\right )^{2} + b^{2} \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )} n^{2} + \cos \left (2 \, b \log \left (c\right )\right )^{2} + \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )} x^{2}}{8 \,{\left ({\left (b^{2} \cos \left (2 \, b \log \left (c\right )\right )^{2} + b^{2} \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )} n^{2} + \cos \left (2 \, b \log \left (c\right )\right )^{2} + \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(((b*cos(2*b*log(c))*sin(4*b*log(c)) - b*cos(4*b*log(c))*sin(2*b*log(c)) + b*sin(2*b*log(c)))*n + cos(4*b
*log(c))*cos(2*b*log(c)) + sin(4*b*log(c))*sin(2*b*log(c)) + cos(2*b*log(c)))*x^2*cos(2*b*log(x^n) + 2*a) + ((
b*cos(4*b*log(c))*cos(2*b*log(c)) + b*sin(4*b*log(c))*sin(2*b*log(c)) + b*cos(2*b*log(c)))*n - cos(2*b*log(c))
*sin(4*b*log(c)) + cos(4*b*log(c))*sin(2*b*log(c)) - sin(2*b*log(c)))*x^2*sin(2*b*log(x^n) + 2*a) - 2*((b^2*co
s(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2)*n^2 + cos(2*b*log(c))^2 + sin(2*b*log(c))^2)*x^2)/((b^2*cos(2*b*log(c
))^2 + b^2*sin(2*b*log(c))^2)*n^2 + cos(2*b*log(c))^2 + sin(2*b*log(c))^2)

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Fricas [A]  time = 0.489977, size = 209, normalized size = 2.13 \begin{align*} -\frac{2 \, b n x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + 2 \, x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} -{\left (b^{2} n^{2} + 2\right )} x^{2}}{4 \,{\left (b^{2} n^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*b*n*x^2*cos(b*n*log(x) + b*log(c) + a)*sin(b*n*log(x) + b*log(c) + a) + 2*x^2*cos(b*n*log(x) + b*log(c
) + a)^2 - (b^2*n^2 + 2)*x^2)/(b^2*n^2 + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.44718, size = 1107, normalized size = 11.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^2 + 1/8*(2*b*n*x^2*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))^
2*tan(a) + 2*b*n*x^2*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*t
an(a) + 2*b*n*x^2*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))*tan(a)^
2 + 2*b*n*x^2*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))*tan(a)^2 -
x^2*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(a)^2 - x^2*e^(
-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(a)^2 - 2*b*n*x^2*e^(p
i*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c))) - 2*b*n*x^2*e^(-pi*b*n*sgn(x)
+ pi*b*n - pi*b*sgn(c) + pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c))) - 2*b*n*x^2*e^(pi*b*n*sgn(x) - pi*b*n + p
i*b*sgn(c) - pi*b)*tan(a) - 2*b*n*x^2*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(a) + x^2*e^(pi*b*n*
sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))^2 + x^2*e^(-pi*b*n*sgn(x) + pi*b*n
- pi*b*sgn(c) + pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))^2 + 4*x^2*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) -
pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))*tan(a) + 4*x^2*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*ta
n(b*n*log(abs(x)) + b*log(abs(c)))*tan(a) + x^2*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(a)^2 + x^2
*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(a)^2 - x^2*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*
b) - x^2*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b))/(b^2*n^2*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan
(a)^2 + b^2*n^2*tan(b*n*log(abs(x)) + b*log(abs(c)))^2 + b^2*n^2*tan(a)^2 + b^2*n^2 + tan(b*n*log(abs(x)) + b*
log(abs(c)))^2*tan(a)^2 + tan(b*n*log(abs(x)) + b*log(abs(c)))^2 + tan(a)^2 + 1)