\(\int \frac {\sin ^4(a+b \log (c x^n))}{x^2} \, dx\) [23]

Optimal result

Integrand size = 17, antiderivative size = 202 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {24 b^4 n^4}{\left (1+20 b^2 n^2+64 b^4 n^4\right ) x}-\frac {24 b^3 n^3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{\left (1+20 b^2 n^2+64 b^4 n^4\right ) x}-\frac {12 b^2 n^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{\left (1+20 b^2 n^2+64 b^4 n^4\right ) x}-\frac {4 b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{\left (1+16 b^2 n^2\right ) x}-\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{\left (1+16 b^2 n^2\right ) x} \]

[Out]

-24*b^4*n^4/(64*b^4*n^4+20*b^2*n^2+1)/x-24*b^3*n^3*cos(a+b*ln(c*x^n))*sin(a+b*ln(c*x^n))/(64*b^4*n^4+20*b^2*n^
2+1)/x-12*b^2*n^2*sin(a+b*ln(c*x^n))^2/(64*b^4*n^4+20*b^2*n^2+1)/x-4*b*n*cos(a+b*ln(c*x^n))*sin(a+b*ln(c*x^n))
^3/(16*b^2*n^2+1)/x-sin(a+b*ln(c*x^n))^4/(16*b^2*n^2+1)/x

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4575, 30} \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x \left (16 b^2 n^2+1\right )}-\frac {4 b n \sin ^3\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{x \left (16 b^2 n^2+1\right )}-\frac {12 b^2 n^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{x \left (64 b^4 n^4+20 b^2 n^2+1\right )}-\frac {24 b^3 n^3 \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{x \left (64 b^4 n^4+20 b^2 n^2+1\right )}-\frac {24 b^4 n^4}{x \left (64 b^4 n^4+20 b^2 n^2+1\right )} \]

[In]

Int[Sin[a + b*Log[c*x^n]]^4/x^2,x]

[Out]

(-24*b^4*n^4)/((1 + 20*b^2*n^2 + 64*b^4*n^4)*x) - (24*b^3*n^3*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]])/((1
+ 20*b^2*n^2 + 64*b^4*n^4)*x) - (12*b^2*n^2*Sin[a + b*Log[c*x^n]]^2)/((1 + 20*b^2*n^2 + 64*b^4*n^4)*x) - (4*b
*n*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]]^3)/((1 + 16*b^2*n^2)*x) - Sin[a + b*Log[c*x^n]]^4/((1 + 16*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 {4 b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{\left (1+16 b^2 n^2\right ) x}-\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{\left (1+16 b^2 n^2\right ) x}+\frac {\left (12 b^2 n^2\right ) \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx}{1+16 b^2 n^2} \\ & = -\frac {24 b^3 n^3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{\left (1+20 b^2 n^2+64 b^4 n^4\right ) x}-\frac {12 b^2 n^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{\left (1+20 b^2 n^2+64 b^4 n^4\right ) x}-\frac {4 b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{\left (1+16 b^2 n^2\right ) x}-\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{\left (1+16 b^2 n^2\right ) x}+\frac {\left (24 b^4 n^4\right ) \int \frac {1}{x^2} \, dx}{1+20 b^2 n^2+64 b^4 n^4} \\ & = -\frac {24 b^4 n^4}{\left (1+20 b^2 n^2+64 b^4 n^4\right ) x}-\frac {24 b^3 n^3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{\left (1+20 b^2 n^2+64 b^4 n^4\right ) x}-\frac {12 b^2 n^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{\left (1+20 b^2 n^2+64 b^4 n^4\right ) x}-\frac {4 b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{\left (1+16 b^2 n^2\right ) x}-\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{\left (1+16 b^2 n^2\right ) x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.84 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {3+60 b^2 n^2+192 b^4 n^4-4 \left (1+16 b^2 n^2\right ) \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+\left (1+4 b^2 n^2\right ) \cos \left (4 \left (a+b \log \left (c x^n\right )\right )\right )+8 b n \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+128 b^3 n^3 \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-4 b n \sin \left (4 \left (a+b \log \left (c x^n\right )\right )\right )-16 b^3 n^3 \sin \left (4 \left (a+b \log \left (c x^n\right )\right )\right )}{8 \left (1+20 b^2 n^2+64 b^4 n^4\right ) x} \]

[In]

Integrate[Sin[a + b*Log[c*x^n]]^4/x^2,x]

[Out]

-1/8*(3 + 60*b^2*n^2 + 192*b^4*n^4 - 4*(1 + 16*b^2*n^2)*Cos[2*(a + b*Log[c*x^n])] + (1 + 4*b^2*n^2)*Cos[4*(a +
b*Log[c*x^n])] + 8*b*n*Sin[2*(a + b*Log[c*x^n])] + 128*b^3*n^3*Sin[2*(a + b*Log[c*x^n])] - 4*b*n*Sin[4*(a + b
*Log[c*x^n])] - 16*b^3*n^3*Sin[4*(a + b*Log[c*x^n])])/((1 + 20*b^2*n^2 + 64*b^4*n^4)*x)

Maple [A] (verified)

Time = 11.25 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.99

method	result	size
parallelrisch	\(\frac {-192 b^{4} n^{4}+16 b^{3} n^{3} \sin \left (4 b \ln \left (c \,x^{n}\right )+4 a \right )-128 b^{3} n^{3} \sin \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )+64 b^{2} n^{2} \cos \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )-4 b^{2} n^{2} \cos \left (4 b \ln \left (c \,x^{n}\right )+4 a \right )-60 b^{2} n^{2}+4 b n \sin \left (4 b \ln \left (c \,x^{n}\right )+4 a \right )-8 b n \sin \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )+4 \cos \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )-\cos \left (4 b \ln \left (c \,x^{n}\right )+4 a \right )-3}{8 x \left (64 b^{4} n^{4}+20 b^{2} n^{2}+1\right )}\)	\(200\)

[In]

int(sin(a+b*ln(c*x^n))^4/x^2,x,method=_RETURNVERBOSE)

[Out]

1/8*(-192*b^4*n^4+16*b^3*n^3*sin(4*b*ln(c*x^n)+4*a)-128*b^3*n^3*sin(2*b*ln(c*x^n)+2*a)+64*b^2*n^2*cos(2*b*ln(c
*x^n)+2*a)-4*b^2*n^2*cos(4*b*ln(c*x^n)+4*a)-60*b^2*n^2+4*b*n*sin(4*b*ln(c*x^n)+4*a)-8*b*n*sin(2*b*ln(c*x^n)+2*
a)+4*cos(2*b*ln(c*x^n)+2*a)-cos(4*b*ln(c*x^n)+4*a)-3)/x/(64*b^4*n^4+20*b^2*n^2+1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.80 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {24 \, b^{4} n^{4} + {\left (4 \, b^{2} n^{2} + 1\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{4} + 16 \, b^{2} n^{2} - 2 \, {\left (10 \, b^{2} n^{2} + 1\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - 4 \, {\left ({\left (4 \, b^{3} n^{3} + b n\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - {\left (10 \, b^{3} n^{3} + b n\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + 1}{{\left (64 \, b^{4} n^{4} + 20 \, b^{2} n^{2} + 1\right )} x} \]

[In]

integrate(sin(a+b*log(c*x^n))^4/x^2,x, algorithm="fricas")

[Out]

-(24*b^4*n^4 + (4*b^2*n^2 + 1)*cos(b*n*log(x) + b*log(c) + a)^4 + 16*b^2*n^2 - 2*(10*b^2*n^2 + 1)*cos(b*n*log(
x) + b*log(c) + a)^2 - 4*((4*b^3*n^3 + b*n)*cos(b*n*log(x) + b*log(c) + a)^3 - (10*b^3*n^3 + b*n)*cos(b*n*log(
x) + b*log(c) + a))*sin(b*n*log(x) + b*log(c) + a) + 1)/((64*b^4*n^4 + 20*b^2*n^2 + 1)*x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 99.62 (sec) , antiderivative size = 959, normalized size of antiderivative = 4.75 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\text {Too large to display} \]

[In]

integrate(sin(a+b*ln(c*x**n))**4/x**2,x)

[Out]

Piecewise((I*sin(4*a - 2*I*log(c*x**n)/n)/(12*x) + cos(2*a - I*log(c*x**n)/n)/(4*x) + cos(4*a - 2*I*log(c*x**n
)/n)/(24*x) - 3/(8*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)/(2*n))/(3*x) + I*sin(4*a - I*log(c*x**n)/n)/(16*x) +
2*cos(2*a - I*log(c*x**n)/(2*n))/(3*x) - 3/(8*x) + I*log(c*x**n)*sin(4*a - I*log(c*x**n)/n)/(16*n*x) + log(c*
x**n)*cos(4*a - I*log(c*x**n)/n)/(16*n*x), Eq(b, -I/(4*n))), (-I*sin(2*a + I*log(c*x**n)/(2*n))/(3*x) - I*sin(
4*a + I*log(c*x**n)/n)/(16*x) + 2*cos(2*a + I*log(c*x**n)/(2*n))/(3*x) - 3/(8*x) - I*log(c*x**n)*sin(4*a + I*l
og(c*x**n)/n)/(16*n*x) + log(c*x**n)*cos(4*a + I*log(c*x**n)/n)/(16*n*x), Eq(b, I/(4*n))), (I*sin(2*a + I*log(
c*x**n)/n)/(4*x) - I*sin(4*a + 2*I*log(c*x**n)/n)/(12*x) + cos(4*a + 2*I*log(c*x**n)/n)/(24*x) - 3/(8*x) + I*l
og(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))
), (-24*b**4*n**4*sin(a + b*log(c*x**n))**4/(64*b**4*n**4*x + 20*b**2*n**2*x + x) - 48*b**4*n**4*sin(a + b*log
(c*x**n))**2*cos(a + b*log(c*x**n))**2/(64*b**4*n**4*x + 20*b**2*n**2*x + x) - 24*b**4*n**4*cos(a + b*log(c*x*
*n))**4/(64*b**4*n**4*x + 20*b**2*n**2*x + x) - 40*b**3*n**3*sin(a + b*log(c*x**n))**3*cos(a + b*log(c*x**n))/
(64*b**4*n**4*x + 20*b**2*n**2*x + x) - 24*b**3*n**3*sin(a + b*log(c*x**n))*cos(a + b*log(c*x**n))**3/(64*b**4
*n**4*x + 20*b**2*n**2*x + x) - 16*b**2*n**2*sin(a + b*log(c*x**n))**4/(64*b**4*n**4*x + 20*b**2*n**2*x + x) -
12*b**2*n**2*sin(a + b*log(c*x**n))**2*cos(a + b*log(c*x**n))**2/(64*b**4*n**4*x + 20*b**2*n**2*x + x) - 4*b*
n*sin(a + b*log(c*x**n))**3*cos(a + b*log(c*x**n))/(64*b**4*n**4*x + 20*b**2*n**2*x + x) - sin(a + b*log(c*x**
n))**4/(64*b**4*n**4*x + 20*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. 1085 vs. \(2 (202) = 404\).

Time = 0.26 (sec) , antiderivative size = 1085, normalized size of antiderivative = 5.37 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\text {Too large to display} \]

[In]

integrate(sin(a+b*log(c*x^n))^4/x^2,x, algorithm="maxima")

[Out]

-1/16*(384*(b^4*cos(4*b*log(c))^2 + b^4*sin(4*b*log(c))^2)*n^4 + 120*(b^2*cos(4*b*log(c))^2 + b^2*sin(4*b*log(
c))^2)*n^2 + 6*cos(4*b*log(c))^2 - (16*(b^3*cos(4*b*log(c))*sin(8*b*log(c)) - b^3*cos(8*b*log(c))*sin(4*b*log(
c)) + b^3*sin(4*b*log(c)))*n^3 - 4*(b^2*cos(8*b*log(c))*cos(4*b*log(c)) + b^2*sin(8*b*log(c))*sin(4*b*log(c))
+ b^2*cos(4*b*log(c)))*n^2 + 4*(b*cos(4*b*log(c))*sin(8*b*log(c)) - b*cos(8*b*log(c))*sin(4*b*log(c)) + b*sin(
4*b*log(c)))*n - cos(8*b*log(c))*cos(4*b*log(c)) - sin(8*b*log(c))*sin(4*b*log(c)) - cos(4*b*log(c)))*cos(4*b*
log(x^n) + 4*a) + 4*(32*(b^3*cos(4*b*log(c))*sin(6*b*log(c)) - b^3*cos(6*b*log(c))*sin(4*b*log(c)) + b^3*cos(2
*b*log(c))*sin(4*b*log(c)) - b^3*cos(4*b*log(c))*sin(2*b*log(c)))*n^3 - 16*(b^2*cos(6*b*log(c))*cos(4*b*log(c)
) + b^2*cos(4*b*log(c))*cos(2*b*log(c)) + b^2*sin(6*b*log(c))*sin(4*b*log(c)) + b^2*sin(4*b*log(c))*sin(2*b*lo
g(c)))*n^2 + 2*(b*cos(4*b*log(c))*sin(6*b*log(c)) - b*cos(6*b*log(c))*sin(4*b*log(c)) + b*cos(2*b*log(c))*sin(
4*b*log(c)) - b*cos(4*b*log(c))*sin(2*b*log(c)))*n - cos(6*b*log(c))*cos(4*b*log(c)) - cos(4*b*log(c))*cos(2*b
*log(c)) - sin(6*b*log(c))*sin(4*b*log(c)) - sin(4*b*log(c))*sin(2*b*log(c)))*cos(2*b*log(x^n) + 2*a) + 6*sin(
4*b*log(c))^2 - (16*(b^3*cos(8*b*log(c))*cos(4*b*log(c)) + b^3*sin(8*b*log(c))*sin(4*b*log(c)) + b^3*cos(4*b*l
og(c)))*n^3 + 4*(b^2*cos(4*b*log(c))*sin(8*b*log(c)) - b^2*cos(8*b*log(c))*sin(4*b*log(c)) + b^2*sin(4*b*log(c
)))*n^2 + 4*(b*cos(8*b*log(c))*cos(4*b*log(c)) + b*sin(8*b*log(c))*sin(4*b*log(c)) + b*cos(4*b*log(c)))*n + co
s(4*b*log(c))*sin(8*b*log(c)) - cos(8*b*log(c))*sin(4*b*log(c)) + sin(4*b*log(c)))*sin(4*b*log(x^n) + 4*a) + 4
*(32*(b^3*cos(6*b*log(c))*cos(4*b*log(c)) + b^3*cos(4*b*log(c))*cos(2*b*log(c)) + b^3*sin(6*b*log(c))*sin(4*b*
log(c)) + b^3*sin(4*b*log(c))*sin(2*b*log(c)))*n^3 + 16*(b^2*cos(4*b*log(c))*sin(6*b*log(c)) - b^2*cos(6*b*log
(c))*sin(4*b*log(c)) + b^2*cos(2*b*log(c))*sin(4*b*log(c)) - b^2*cos(4*b*log(c))*sin(2*b*log(c)))*n^2 + 2*(b*c
os(6*b*log(c))*cos(4*b*log(c)) + b*cos(4*b*log(c))*cos(2*b*log(c)) + b*sin(6*b*log(c))*sin(4*b*log(c)) + b*sin
(4*b*log(c))*sin(2*b*log(c)))*n + cos(4*b*log(c))*sin(6*b*log(c)) - cos(6*b*log(c))*sin(4*b*log(c)) + cos(2*b*
log(c))*sin(4*b*log(c)) - cos(4*b*log(c))*sin(2*b*log(c)))*sin(2*b*log(x^n) + 2*a))/((64*(b^4*cos(4*b*log(c))^
2 + b^4*sin(4*b*log(c))^2)*n^4 + 20*(b^2*cos(4*b*log(c))^2 + b^2*sin(4*b*log(c))^2)*n^2 + cos(4*b*log(c))^2 +
sin(4*b*log(c))^2)*x)

Giac [F]

\[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int { \frac {\sin \left (b \log \left (c x^{n}\right ) + a\right )^{4}}{x^{2}} \,d x } \]

[In]

integrate(sin(a+b*log(c*x^n))^4/x^2,x, algorithm="giac")

[Out]

integrate(sin(b*log(c*x^n) + a)^4/x^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int \frac {{\sin \left (a+b\,\ln \left (c\,x^n\right )\right )}^4}{x^2} \,d x \]

[In]

int(sin(a + b*log(c*x^n))^4/x^2,x)

[Out]

int(sin(a + b*log(c*x^n))^4/x^2, x)