\(\int \frac {\sin ^2(a+b \log (c x^n))}{x^2} \, dx\) [11]

Optimal result

Integrand size = 17, antiderivative size = 95 \[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {2 b^2 n^2}{\left (1+4 b^2 n^2\right ) x}-\frac {2 b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{\left (1+4 b^2 n^2\right ) x}-\frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{\left (1+4 b^2 n^2\right ) x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 95, 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, 30} \[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x \left (4 b^2 n^2+1\right )}-\frac {2 b n \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{x \left (4 b^2 n^2+1\right )}-\frac {2 b^2 n^2}{x \left (4 b^2 n^2+1\right )} \]

[In]

Int[Sin[a + b*Log[c*x^n]]^2/x^2,x]

[Out]

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

Rule 30

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

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.60 \[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\frac {-1-4 b^2 n^2+\cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-2 b n \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )}{2 \left (x+4 b^2 n^2 x\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.62

method	result	size
parallelrisch	\(\frac {-4 b^{2} n^{2}-2 b n \sin \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )+\cos \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )-1}{8 b^{2} n^{2} x +2 x}\)	\(59\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.75 \[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {2 \, b^{2} n^{2} + 2 \, b n \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 1}{{\left (4 \, b^{2} n^{2} + 1\right )} x} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.80 (sec) , antiderivative size = 303, normalized size of antiderivative = 3.19 \[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\begin {cases} \frac {\cos {\left (2 a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{4 x} - \frac {1}{2 x} - \frac {i \log {\left (c x^{n} \right )} \sin {\left (2 a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{4 n x} - \frac {\log {\left (c x^{n} \right )} \cos {\left (2 a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{4 n x} & \text {for}\: b = - \frac {i}{2 n} \\\frac {i \sin {\left (2 a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{4 x} - \frac {1}{2 x} + \frac {i \log {\left (c x^{n} \right )} \sin {\left (2 a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{4 n x} - \frac {\log {\left (c x^{n} \right )} \cos {\left (2 a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{4 n x} & \text {for}\: b = \frac {i}{2 n} \\- \frac {2 b^{2} n^{2} \sin ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} x + x} - \frac {2 b^{2} n^{2} \cos ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} x + x} - \frac {2 b n \sin {\left (a + b \log {\left (c x^{n} \right )} \right )} \cos {\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} x + x} - \frac {\sin ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} x + x} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((cos(2*a - I*log(c*x**n)/n)/(4*x) - 1/(2*x) - I*log(c*x**n)*sin(2*a - I*log(c*x**n)/n)/(4*n*x) - log
(c*x**n)*cos(2*a - I*log(c*x**n)/n)/(4*n*x), Eq(b, -I/(2*n))), (I*sin(2*a + I*log(c*x**n)/n)/(4*x) - 1/(2*x) +
I*log(c*x**n)*sin(2*a + I*log(c*x**n)/n)/(4*n*x) - log(c*x**n)*cos(2*a + I*log(c*x**n)/n)/(4*n*x), Eq(b, I/(2
*n))), (-2*b**2*n**2*sin(a + b*log(c*x**n))**2/(4*b**2*n**2*x + x) - 2*b**2*n**2*cos(a + b*log(c*x**n))**2/(4*
b**2*n**2*x + x) - 2*b*n*sin(a + b*log(c*x**n))*cos(a + b*log(c*x**n))/(4*b**2*n**2*x + x) - sin(a + b*log(c*x
**n))**2/(4*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. 283 vs. \(2 (95) = 190\).

Time = 0.22 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.98 \[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {8 \, {\left (b^{2} \cos \left (2 \, b \log \left (c\right )\right )^{2} + b^{2} \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )} n^{2} + 2 \, \cos \left (2 \, b \log \left (c\right )\right )^{2} + {\left (2 \, {\left (b \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (4 \, b \log \left (c\right )\right ) - b \cos \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + b \sin \left (2 \, b \log \left (c\right )\right )\right )} n - \cos \left (4 \, b \log \left (c\right )\right ) \cos \left (2 \, b \log \left (c\right )\right ) - \sin \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) - \cos \left (2 \, b \log \left (c\right )\right )\right )} \cos \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right ) + 2 \, \sin \left (2 \, b \log \left (c\right )\right )^{2} + {\left (2 \, {\left (b \cos \left (4 \, b \log \left (c\right )\right ) \cos \left (2 \, b \log \left (c\right )\right ) + b \sin \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + b \cos \left (2 \, b \log \left (c\right )\right )\right )} n + \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (4 \, b \log \left (c\right )\right ) - \cos \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + \sin \left (2 \, b \log \left (c\right )\right )\right )} \sin \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )}{4 \, {\left (4 \, {\left (b^{2} \cos \left (2 \, b \log \left (c\right )\right )^{2} + b^{2} \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )} n^{2} + \cos \left (2 \, b \log \left (c\right )\right )^{2} + \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )} x} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(sin(a + b*log(c*x^n))^2/x^2,x)

[Out]

int(sin(a + b*log(c*x^n))^2/x^2, x)