\(\int \frac {\sin ^3(a+b \log (c x^n))}{x^2} \, dx\) [17]

Optimal result

Integrand size = 17, antiderivative size = 158 \[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {6 b^3 n^3 \cos \left (a+b \log \left (c x^n\right )\right )}{\left (1+10 b^2 n^2+9 b^4 n^4\right ) x}-\frac {6 b^2 n^2 \sin \left (a+b \log \left (c x^n\right )\right )}{\left (1+10 b^2 n^2+9 b^4 n^4\right ) x}-\frac {3 b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^2\left (a+b \log \left (c x^n\right )\right )}{\left (1+9 b^2 n^2\right ) x}-\frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{\left (1+9 b^2 n^2\right ) x} \]

[Out]

-6*b^3*n^3*cos(a+b*ln(c*x^n))/(9*b^4*n^4+10*b^2*n^2+1)/x-6*b^2*n^2*sin(a+b*ln(c*x^n))/(9*b^4*n^4+10*b^2*n^2+1)
/x-3*b*n*cos(a+b*ln(c*x^n))*sin(a+b*ln(c*x^n))^2/(9*b^2*n^2+1)/x-sin(a+b*ln(c*x^n))^3/(9*b^2*n^2+1)/x

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4575, 4573} \[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x \left (9 b^2 n^2+1\right )}-\frac {3 b n \sin ^2\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{x \left (9 b^2 n^2+1\right )}-\frac {6 b^2 n^2 \sin \left (a+b \log \left (c x^n\right )\right )}{x \left (9 b^4 n^4+10 b^2 n^2+1\right )}-\frac {6 b^3 n^3 \cos \left (a+b \log \left (c x^n\right )\right )}{x \left (9 b^4 n^4+10 b^2 n^2+1\right )} \]

[In]

Int[Sin[a + b*Log[c*x^n]]^3/x^2,x]

[Out]

(-6*b^3*n^3*Cos[a + b*Log[c*x^n]])/((1 + 10*b^2*n^2 + 9*b^4*n^4)*x) - (6*b^2*n^2*Sin[a + b*Log[c*x^n]])/((1 +
10*b^2*n^2 + 9*b^4*n^4)*x) - (3*b*n*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]]^2)/((1 + 9*b^2*n^2)*x) - Sin[a
+ b*Log[c*x^n]]^3/((1 + 9*b^2*n^2)*x)

Rule 4573

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

Rule 4575

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^2\left (a+b \log \left (c x^n\right )\right )}{\left (1+9 b^2 n^2\right ) x}-\frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{\left (1+9 b^2 n^2\right ) x}+\frac {\left (6 b^2 n^2\right ) \int \frac {\sin \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx}{1+9 b^2 n^2} \\ & = -\frac {6 b^3 n^3 \cos \left (a+b \log \left (c x^n\right )\right )}{\left (1+10 b^2 n^2+9 b^4 n^4\right ) x}-\frac {6 b^2 n^2 \sin \left (a+b \log \left (c x^n\right )\right )}{\left (1+10 b^2 n^2+9 b^4 n^4\right ) x}-\frac {3 b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^2\left (a+b \log \left (c x^n\right )\right )}{\left (1+9 b^2 n^2\right ) x}-\frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{\left (1+9 b^2 n^2\right ) x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.79 \[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\frac {-3 b n \left (1+9 b^2 n^2\right ) \cos \left (a+b \log \left (c x^n\right )\right )+3 \left (b n+b^3 n^3\right ) \cos \left (3 \left (a+b \log \left (c x^n\right )\right )\right )+2 \left (-1-13 b^2 n^2+\left (1+b^2 n^2\right ) \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{4 \left (1+10 b^2 n^2+9 b^4 n^4\right ) x} \]

[In]

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

[Out]

(-3*b*n*(1 + 9*b^2*n^2)*Cos[a + b*Log[c*x^n]] + 3*(b*n + b^3*n^3)*Cos[3*(a + b*Log[c*x^n])] + 2*(-1 - 13*b^2*n
^2 + (1 + b^2*n^2)*Cos[2*(a + b*Log[c*x^n])])*Sin[a + b*Log[c*x^n]])/(4*(1 + 10*b^2*n^2 + 9*b^4*n^4)*x)

Maple [A] (verified)

Time = 3.70 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.40

method	result	size
parallelrisch	\(\frac {6 {\tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{6} b^{3} n^{3}-12 {\tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{5} b^{2} n^{2}+\left (18 b^{3} n^{3}+12 b n \right ) {\tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{4}+\left (-32 b^{2} n^{2}-8\right ) {\tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{3}+\left (-18 b^{3} n^{3}-12 b n \right ) {\tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2}-12 \tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right ) b^{2} n^{2}-6 b^{3} n^{3}}{9 \left (b^{2} n^{2}+1\right ) x \left (b^{2} n^{2}+\frac {1}{9}\right ) {\left (1+{\tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2}\right )}^{3}}\)	\(221\)

[In]

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

[Out]

1/9*(6*tan(1/2*a+b*ln((c*x^n)^(1/2)))^6*b^3*n^3-12*tan(1/2*a+b*ln((c*x^n)^(1/2)))^5*b^2*n^2+(18*b^3*n^3+12*b*n
)*tan(1/2*a+b*ln((c*x^n)^(1/2)))^4+(-32*b^2*n^2-8)*tan(1/2*a+b*ln((c*x^n)^(1/2)))^3+(-18*b^3*n^3-12*b*n)*tan(1
/2*a+b*ln((c*x^n)^(1/2)))^2-12*tan(1/2*a+b*ln((c*x^n)^(1/2)))*b^2*n^2-6*b^3*n^3)/(b^2*n^2+1)/x/(b^2*n^2+1/9)/(
1+tan(1/2*a+b*ln((c*x^n)^(1/2)))^2)^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.80 \[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\frac {3 \, {\left (b^{3} n^{3} + b n\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - 3 \, {\left (3 \, b^{3} n^{3} + b n\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - {\left (7 \, b^{2} n^{2} - {\left (b^{2} n^{2} + 1\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 1\right )} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{{\left (9 \, b^{4} n^{4} + 10 \, b^{2} n^{2} + 1\right )} x} \]

[In]

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

[Out]

(3*(b^3*n^3 + b*n)*cos(b*n*log(x) + b*log(c) + a)^3 - 3*(3*b^3*n^3 + b*n)*cos(b*n*log(x) + b*log(c) + a) - (7*
b^2*n^2 - (b^2*n^2 + 1)*cos(b*n*log(x) + b*log(c) + a)^2 + 1)*sin(b*n*log(x) + b*log(c) + a))/((9*b^4*n^4 + 10
*b^2*n^2 + 1)*x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 34.11 (sec) , antiderivative size = 775, normalized size of antiderivative = 4.91 \[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\begin {cases} - \frac {\sin {\left (3 a - \frac {3 i \log {\left (c x^{n} \right )}}{n} \right )}}{32 x} - \frac {3 i \cos {\left (a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{8 x} + \frac {3 i \cos {\left (3 a - \frac {3 i \log {\left (c x^{n} \right )}}{n} \right )}}{32 x} + \frac {3 \log {\left (c x^{n} \right )} \sin {\left (a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{8 n x} - \frac {3 i \log {\left (c x^{n} \right )} \cos {\left (a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{8 n x} & \text {for}\: b = - \frac {i}{n} \\- \frac {27 \sin {\left (a - \frac {i \log {\left (c x^{n} \right )}}{3 n} \right )}}{32 x} + \frac {\sin {\left (3 a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{8 x} + \frac {9 i \cos {\left (a - \frac {i \log {\left (c x^{n} \right )}}{3 n} \right )}}{32 x} - \frac {\log {\left (c x^{n} \right )} \sin {\left (3 a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{8 n x} + \frac {i \log {\left (c x^{n} \right )} \cos {\left (3 a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{8 n x} & \text {for}\: b = - \frac {i}{3 n} \\- \frac {27 \sin {\left (a + \frac {i \log {\left (c x^{n} \right )}}{3 n} \right )}}{32 x} - \frac {9 i \cos {\left (a + \frac {i \log {\left (c x^{n} \right )}}{3 n} \right )}}{32 x} - \frac {i \cos {\left (3 a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{8 x} - \frac {\log {\left (c x^{n} \right )} \sin {\left (3 a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{8 n x} - \frac {i \log {\left (c x^{n} \right )} \cos {\left (3 a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{8 n x} & \text {for}\: b = \frac {i}{3 n} \\- \frac {3 \sin {\left (a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{8 x} - \frac {\sin {\left (3 a + \frac {3 i \log {\left (c x^{n} \right )}}{n} \right )}}{32 x} - \frac {3 i \cos {\left (3 a + \frac {3 i \log {\left (c x^{n} \right )}}{n} \right )}}{32 x} + \frac {3 \log {\left (c x^{n} \right )} \sin {\left (a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{8 n x} + \frac {3 i \log {\left (c x^{n} \right )} \cos {\left (a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{8 n x} & \text {for}\: b = \frac {i}{n} \\- \frac {9 b^{3} n^{3} \sin ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )} \cos {\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} x + 10 b^{2} n^{2} x + x} - \frac {6 b^{3} n^{3} \cos ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} x + 10 b^{2} n^{2} x + x} - \frac {7 b^{2} n^{2} \sin ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} x + 10 b^{2} n^{2} x + x} - \frac {6 b^{2} n^{2} \sin {\left (a + b \log {\left (c x^{n} \right )} \right )} \cos ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} x + 10 b^{2} n^{2} x + x} - \frac {3 b n \sin ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )} \cos {\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} x + 10 b^{2} n^{2} x + x} - \frac {\sin ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} x + 10 b^{2} n^{2} x + x} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-sin(3*a - 3*I*log(c*x**n)/n)/(32*x) - 3*I*cos(a - I*log(c*x**n)/n)/(8*x) + 3*I*cos(3*a - 3*I*log(c
*x**n)/n)/(32*x) + 3*log(c*x**n)*sin(a - I*log(c*x**n)/n)/(8*n*x) - 3*I*log(c*x**n)*cos(a - I*log(c*x**n)/n)/(
8*n*x), Eq(b, -I/n)), (-27*sin(a - I*log(c*x**n)/(3*n))/(32*x) + sin(3*a - I*log(c*x**n)/n)/(8*x) + 9*I*cos(a
- I*log(c*x**n)/(3*n))/(32*x) - log(c*x**n)*sin(3*a - I*log(c*x**n)/n)/(8*n*x) + I*log(c*x**n)*cos(3*a - I*log
(c*x**n)/n)/(8*n*x), Eq(b, -I/(3*n))), (-27*sin(a + I*log(c*x**n)/(3*n))/(32*x) - 9*I*cos(a + I*log(c*x**n)/(3
*n))/(32*x) - I*cos(3*a + I*log(c*x**n)/n)/(8*x) - log(c*x**n)*sin(3*a + I*log(c*x**n)/n)/(8*n*x) - I*log(c*x*
*n)*cos(3*a + I*log(c*x**n)/n)/(8*n*x), Eq(b, I/(3*n))), (-3*sin(a + I*log(c*x**n)/n)/(8*x) - sin(3*a + 3*I*lo
g(c*x**n)/n)/(32*x) - 3*I*cos(3*a + 3*I*log(c*x**n)/n)/(32*x) + 3*log(c*x**n)*sin(a + I*log(c*x**n)/n)/(8*n*x)
+ 3*I*log(c*x**n)*cos(a + I*log(c*x**n)/n)/(8*n*x), Eq(b, I/n)), (-9*b**3*n**3*sin(a + b*log(c*x**n))**2*cos(
a + b*log(c*x**n))/(9*b**4*n**4*x + 10*b**2*n**2*x + x) - 6*b**3*n**3*cos(a + b*log(c*x**n))**3/(9*b**4*n**4*x
+ 10*b**2*n**2*x + x) - 7*b**2*n**2*sin(a + b*log(c*x**n))**3/(9*b**4*n**4*x + 10*b**2*n**2*x + x) - 6*b**2*n
**2*sin(a + b*log(c*x**n))*cos(a + b*log(c*x**n))**2/(9*b**4*n**4*x + 10*b**2*n**2*x + x) - 3*b*n*sin(a + b*lo
g(c*x**n))**2*cos(a + b*log(c*x**n))/(9*b**4*n**4*x + 10*b**2*n**2*x + x) - sin(a + b*log(c*x**n))**3/(9*b**4*
n**4*x + 10*b**2*n**2*x + x), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 995 vs. \(2 (158) = 316\).

Time = 0.25 (sec) , antiderivative size = 995, normalized size of antiderivative = 6.30 \[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/8*((3*(b^3*cos(6*b*log(c))*cos(3*b*log(c)) + b^3*sin(6*b*log(c))*sin(3*b*log(c)) + b^3*cos(3*b*log(c)))*n^3
+ (b^2*cos(3*b*log(c))*sin(6*b*log(c)) - b^2*cos(6*b*log(c))*sin(3*b*log(c)) + b^2*sin(3*b*log(c)))*n^2 + 3*(b
*cos(6*b*log(c))*cos(3*b*log(c)) + b*sin(6*b*log(c))*sin(3*b*log(c)) + b*cos(3*b*log(c)))*n + cos(3*b*log(c))*
sin(6*b*log(c)) - cos(6*b*log(c))*sin(3*b*log(c)) + sin(3*b*log(c)))*cos(3*b*log(x^n) + 3*a) - 3*(9*(b^3*cos(4
*b*log(c))*cos(3*b*log(c)) + b^3*cos(3*b*log(c))*cos(2*b*log(c)) + b^3*sin(4*b*log(c))*sin(3*b*log(c)) + b^3*s
in(3*b*log(c))*sin(2*b*log(c)))*n^3 + 9*(b^2*cos(3*b*log(c))*sin(4*b*log(c)) - b^2*cos(4*b*log(c))*sin(3*b*log
(c)) + b^2*cos(2*b*log(c))*sin(3*b*log(c)) - b^2*cos(3*b*log(c))*sin(2*b*log(c)))*n^2 + (b*cos(4*b*log(c))*cos
(3*b*log(c)) + b*cos(3*b*log(c))*cos(2*b*log(c)) + b*sin(4*b*log(c))*sin(3*b*log(c)) + b*sin(3*b*log(c))*sin(2
*b*log(c)))*n + cos(3*b*log(c))*sin(4*b*log(c)) - cos(4*b*log(c))*sin(3*b*log(c)) + cos(2*b*log(c))*sin(3*b*lo
g(c)) - cos(3*b*log(c))*sin(2*b*log(c)))*cos(b*log(x^n) + a) - (3*(b^3*cos(3*b*log(c))*sin(6*b*log(c)) - b^3*c
os(6*b*log(c))*sin(3*b*log(c)) + b^3*sin(3*b*log(c)))*n^3 - (b^2*cos(6*b*log(c))*cos(3*b*log(c)) + b^2*sin(6*b
*log(c))*sin(3*b*log(c)) + b^2*cos(3*b*log(c)))*n^2 + 3*(b*cos(3*b*log(c))*sin(6*b*log(c)) - b*cos(6*b*log(c))
*sin(3*b*log(c)) + b*sin(3*b*log(c)))*n - cos(6*b*log(c))*cos(3*b*log(c)) - sin(6*b*log(c))*sin(3*b*log(c)) -
cos(3*b*log(c)))*sin(3*b*log(x^n) + 3*a) + 3*(9*(b^3*cos(3*b*log(c))*sin(4*b*log(c)) - b^3*cos(4*b*log(c))*sin
(3*b*log(c)) + b^3*cos(2*b*log(c))*sin(3*b*log(c)) - b^3*cos(3*b*log(c))*sin(2*b*log(c)))*n^3 - 9*(b^2*cos(4*b
*log(c))*cos(3*b*log(c)) + b^2*cos(3*b*log(c))*cos(2*b*log(c)) + b^2*sin(4*b*log(c))*sin(3*b*log(c)) + b^2*sin
(3*b*log(c))*sin(2*b*log(c)))*n^2 + (b*cos(3*b*log(c))*sin(4*b*log(c)) - b*cos(4*b*log(c))*sin(3*b*log(c)) + b
*cos(2*b*log(c))*sin(3*b*log(c)) - b*cos(3*b*log(c))*sin(2*b*log(c)))*n - cos(4*b*log(c))*cos(3*b*log(c)) - co
s(3*b*log(c))*cos(2*b*log(c)) - sin(4*b*log(c))*sin(3*b*log(c)) - sin(3*b*log(c))*sin(2*b*log(c)))*sin(b*log(x
^n) + a))/((9*(b^4*cos(3*b*log(c))^2 + b^4*sin(3*b*log(c))^2)*n^4 + 10*(b^2*cos(3*b*log(c))^2 + b^2*sin(3*b*lo
g(c))^2)*n^2 + cos(3*b*log(c))^2 + sin(3*b*log(c))^2)*x)

Giac [F]

\[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int { \frac {\sin \left (b \log \left (c x^{n}\right ) + a\right )^{3}}{x^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int \frac {{\sin \left (a+b\,\ln \left (c\,x^n\right )\right )}^3}{x^2} \,d x \]

[In]

int(sin(a + b*log(c*x^n))^3/x^2,x)

[Out]

int(sin(a + b*log(c*x^n))^3/x^2, x)