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

Optimal. Leaf size=95 \[ -\frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x \left (4 b^2 n^2+1\right )}-\frac {2 b n \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{x \left (4 b^2 n^2+1\right )}-\frac {2 b^2 n^2}{x \left (4 b^2 n^2+1\right )} \]

[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

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Rubi [A]  time = 0.03, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4487, 30} \[ -\frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x \left (4 b^2 n^2+1\right )}-\frac {2 b n \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{x \left (4 b^2 n^2+1\right )}-\frac {2 b^2 n^2}{x \left (4 b^2 n^2+1\right )} \]

Antiderivative was successfully verified.

[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 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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 57, normalized size = 0.60 \[ \frac {-2 b n \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+\cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-4 b^2 n^2-1}{2 \left (4 b^2 n^2 x+x\right )} \]

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.67, size = 71, normalized size = 0.75 \[ -\frac {2 \, b^{2} n^{2} + 2 \, b n \cos \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sin \left (b n \log \relax (x) + b \log \relax (c) + a\right ) - \cos \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + 1}{{\left (4 \, b^{2} n^{2} + 1\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (b \log \left (c x^{n}\right ) + a\right )^{2}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{2}\left (a +b \ln \left (c \,x^{n}\right )\right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.35, size = 283, normalized size = 2.98 \[ -\frac {8 \, {\left (b^{2} \cos \left (2 \, b \log \relax (c)\right )^{2} + b^{2} \sin \left (2 \, b \log \relax (c)\right )^{2}\right )} n^{2} + 2 \, \cos \left (2 \, b \log \relax (c)\right )^{2} + {\left (2 \, {\left (b \cos \left (2 \, b \log \relax (c)\right ) \sin \left (4 \, b \log \relax (c)\right ) - b \cos \left (4 \, b \log \relax (c)\right ) \sin \left (2 \, b \log \relax (c)\right ) + b \sin \left (2 \, b \log \relax (c)\right )\right )} n - \cos \left (4 \, b \log \relax (c)\right ) \cos \left (2 \, b \log \relax (c)\right ) - \sin \left (4 \, b \log \relax (c)\right ) \sin \left (2 \, b \log \relax (c)\right ) - \cos \left (2 \, b \log \relax (c)\right )\right )} \cos \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right ) + 2 \, \sin \left (2 \, b \log \relax (c)\right )^{2} + {\left (2 \, {\left (b \cos \left (4 \, b \log \relax (c)\right ) \cos \left (2 \, b \log \relax (c)\right ) + b \sin \left (4 \, b \log \relax (c)\right ) \sin \left (2 \, b \log \relax (c)\right ) + b \cos \left (2 \, b \log \relax (c)\right )\right )} n + \cos \left (2 \, b \log \relax (c)\right ) \sin \left (4 \, b \log \relax (c)\right ) - \cos \left (4 \, b \log \relax (c)\right ) \sin \left (2 \, b \log \relax (c)\right ) + \sin \left (2 \, b \log \relax (c)\right )\right )} \sin \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )}{4 \, {\left (4 \, {\left (b^{2} \cos \left (2 \, b \log \relax (c)\right )^{2} + b^{2} \sin \left (2 \, b \log \relax (c)\right )^{2}\right )} n^{2} + \cos \left (2 \, b \log \relax (c)\right )^{2} + \sin \left (2 \, b \log \relax (c)\right )^{2}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 24.23, size = 415, normalized size = 4.37 \[ \begin {cases} \frac {i \log {\relax (x )} \sin {\left (- 2 a + i \log {\relax (x )} + \frac {i \log {\relax (c )}}{n} \right )}}{4 x} - \frac {\log {\relax (x )} \cos {\left (- 2 a + i \log {\relax (x )} + \frac {i \log {\relax (c )}}{n} \right )}}{4 x} + \frac {i \sin {\left (- 2 a + i \log {\relax (x )} + \frac {i \log {\relax (c )}}{n} \right )}}{4 x} - \frac {1}{2 x} + \frac {i \log {\relax (c )} \sin {\left (- 2 a + i \log {\relax (x )} + \frac {i \log {\relax (c )}}{n} \right )}}{4 n x} - \frac {\log {\relax (c )} \cos {\left (- 2 a + i \log {\relax (x )} + \frac {i \log {\relax (c )}}{n} \right )}}{4 n x} & \text {for}\: b = - \frac {i}{2 n} \\\frac {i \log {\relax (x )} \sin {\left (2 a + i \log {\relax (x )} + \frac {i \log {\relax (c )}}{n} \right )}}{4 x} - \frac {\log {\relax (x )} \cos {\left (2 a + i \log {\relax (x )} + \frac {i \log {\relax (c )}}{n} \right )}}{4 x} + \frac {\cos {\left (2 a + i \log {\relax (x )} + \frac {i \log {\relax (c )}}{n} \right )}}{4 x} - \frac {1}{2 x} + \frac {i \log {\relax (c )} \sin {\left (2 a + i \log {\relax (x )} + \frac {i \log {\relax (c )}}{n} \right )}}{4 n x} - \frac {\log {\relax (c )} \cos {\left (2 a + i \log {\relax (x )} + \frac {i \log {\relax (c )}}{n} \right )}}{4 n x} & \text {for}\: b = \frac {i}{2 n} \\- \frac {2 b^{2} n^{2} \sin ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} n^{2} x + x} - \frac {2 b^{2} n^{2} \cos ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} n^{2} x + x} - \frac {2 b n \sin {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cos {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} n^{2} x + x} - \frac {\sin ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} n^{2} x + x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*log(x)*sin(-2*a + I*log(x) + I*log(c)/n)/(4*x) - log(x)*cos(-2*a + I*log(x) + I*log(c)/n)/(4*x) +
I*sin(-2*a + I*log(x) + I*log(c)/n)/(4*x) - 1/(2*x) + I*log(c)*sin(-2*a + I*log(x) + I*log(c)/n)/(4*n*x) - lo
g(c)*cos(-2*a + I*log(x) + I*log(c)/n)/(4*n*x), Eq(b, -I/(2*n))), (I*log(x)*sin(2*a + I*log(x) + I*log(c)/n)/(
4*x) - log(x)*cos(2*a + I*log(x) + I*log(c)/n)/(4*x) + cos(2*a + I*log(x) + I*log(c)/n)/(4*x) - 1/(2*x) + I*lo
g(c)*sin(2*a + I*log(x) + I*log(c)/n)/(4*n*x) - log(c)*cos(2*a + I*log(x) + I*log(c)/n)/(4*n*x), Eq(b, I/(2*n)
)), (-2*b**2*n**2*sin(a + b*n*log(x) + b*log(c))**2/(4*b**2*n**2*x + x) - 2*b**2*n**2*cos(a + b*n*log(x) + b*l
og(c))**2/(4*b**2*n**2*x + x) - 2*b*n*sin(a + b*n*log(x) + b*log(c))*cos(a + b*n*log(x) + b*log(c))/(4*b**2*n*
*2*x + x) - sin(a + b*n*log(x) + b*log(c))**2/(4*b**2*n**2*x + x), True))

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