\(\int \frac {\sin ^4(a+b \log (c x^n))}{x^3} \, dx\) [24]

Optimal result

Integrand size = 17, antiderivative size = 210 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {3 b^4 n^4}{4 \left (1+5 b^2 n^2+4 b^4 n^4\right ) x^2}-\frac {3 b^3 n^3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right ) x^2}-\frac {3 b^2 n^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right ) x^2}-\frac {b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{\left (1+4 b^2 n^2\right ) x^2}-\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+4 b^2 n^2\right ) x^2} \]

[Out]

-3/4*b^4*n^4/(4*b^4*n^4+5*b^2*n^2+1)/x^2-3/2*b^3*n^3*cos(a+b*ln(c*x^n))*sin(a+b*ln(c*x^n))/(4*b^4*n^4+5*b^2*n^
2+1)/x^2-3/2*b^2*n^2*sin(a+b*ln(c*x^n))^2/(4*b^4*n^4+5*b^2*n^2+1)/x^2-b*n*cos(a+b*ln(c*x^n))*sin(a+b*ln(c*x^n)
)^3/(4*b^2*n^2+1)/x^2-1/2*sin(a+b*ln(c*x^n))^4/(4*b^2*n^2+1)/x^2

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 210, 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} \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 x^2 \left (4 b^2 n^2+1\right )}-\frac {b n \sin ^3\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{x^2 \left (4 b^2 n^2+1\right )}-\frac {3 b^2 n^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 x^2 \left (4 b^4 n^4+5 b^2 n^2+1\right )}-\frac {3 b^3 n^3 \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{2 x^2 \left (4 b^4 n^4+5 b^2 n^2+1\right )}-\frac {3 b^4 n^4}{4 x^2 \left (4 b^4 n^4+5 b^2 n^2+1\right )} \]

[In]

Int[Sin[a + b*Log[c*x^n]]^4/x^3,x]

[Out]

(-3*b^4*n^4)/(4*(1 + 5*b^2*n^2 + 4*b^4*n^4)*x^2) - (3*b^3*n^3*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]])/(2*
(1 + 5*b^2*n^2 + 4*b^4*n^4)*x^2) - (3*b^2*n^2*Sin[a + b*Log[c*x^n]]^2)/(2*(1 + 5*b^2*n^2 + 4*b^4*n^4)*x^2) - (
b*n*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]]^3)/((1 + 4*b^2*n^2)*x^2) - Sin[a + b*Log[c*x^n]]^4/(2*(1 + 4*b
^2*n^2)*x^2)

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 {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+4 b^2 n^2\right ) x^2}-\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+4 b^2 n^2\right ) x^2}+\frac {\left (3 b^2 n^2\right ) \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx}{1+4 b^2 n^2} \\ & = -\frac {3 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 )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right ) x^2}-\frac {3 b^2 n^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right ) x^2}-\frac {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+4 b^2 n^2\right ) x^2}-\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+4 b^2 n^2\right ) x^2}+\frac {\left (3 b^4 n^4\right ) \int \frac {1}{x^3} \, dx}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right )} \\ & = -\frac {3 b^4 n^4}{4 \left (1+5 b^2 n^2+4 b^4 n^4\right ) x^2}-\frac {3 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 )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right ) x^2}-\frac {3 b^2 n^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right ) x^2}-\frac {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+4 b^2 n^2\right ) x^2}-\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+4 b^2 n^2\right ) x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.80 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {3+15 b^2 n^2+12 b^4 n^4-4 \left (1+4 b^2 n^2\right ) \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+\left (1+b^2 n^2\right ) \cos \left (4 \left (a+b \log \left (c x^n\right )\right )\right )+4 b n \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+16 b^3 n^3 \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-2 b n \sin \left (4 \left (a+b \log \left (c x^n\right )\right )\right )-2 b^3 n^3 \sin \left (4 \left (a+b \log \left (c x^n\right )\right )\right )}{16 \left (1+5 b^2 n^2+4 b^4 n^4\right ) x^2} \]

[In]

Integrate[Sin[a + b*Log[c*x^n]]^4/x^3,x]

[Out]

-1/16*(3 + 15*b^2*n^2 + 12*b^4*n^4 - 4*(1 + 4*b^2*n^2)*Cos[2*(a + b*Log[c*x^n])] + (1 + b^2*n^2)*Cos[4*(a + b*
Log[c*x^n])] + 4*b*n*Sin[2*(a + b*Log[c*x^n])] + 16*b^3*n^3*Sin[2*(a + b*Log[c*x^n])] - 2*b*n*Sin[4*(a + b*Log
[c*x^n])] - 2*b^3*n^3*Sin[4*(a + b*Log[c*x^n])])/((1 + 5*b^2*n^2 + 4*b^4*n^4)*x^2)

Maple [A] (verified)

Time = 18.99 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(\frac {-12 b^{4} n^{4}-16 b^{3} n^{3} \sin \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )+2 b^{3} n^{3} \sin \left (4 b \ln \left (c \,x^{n}\right )+4 a \right )-b^{2} n^{2} \cos \left (4 b \ln \left (c \,x^{n}\right )+4 a \right )+16 b^{2} n^{2} \cos \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )-15 b^{2} n^{2}-4 b n \sin \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )+2 b n \sin \left (4 b \ln \left (c \,x^{n}\right )+4 a \right )-\cos \left (4 b \ln \left (c \,x^{n}\right )+4 a \right )+4 \cos \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )-3}{16 x^{2} \left (4 b^{4} n^{4}+5 b^{2} n^{2}+1\right )}\)	\(200\)

[In]

int(sin(a+b*ln(c*x^n))^4/x^3,x,method=_RETURNVERBOSE)

[Out]

1/16*(-12*b^4*n^4-16*b^3*n^3*sin(2*b*ln(c*x^n)+2*a)+2*b^3*n^3*sin(4*b*ln(c*x^n)+4*a)-b^2*n^2*cos(4*b*ln(c*x^n)
+4*a)+16*b^2*n^2*cos(2*b*ln(c*x^n)+2*a)-15*b^2*n^2-4*b*n*sin(2*b*ln(c*x^n)+2*a)+2*b*n*sin(4*b*ln(c*x^n)+4*a)-c
os(4*b*ln(c*x^n)+4*a)+4*cos(2*b*ln(c*x^n)+2*a)-3)/x^2/(4*b^4*n^4+5*b^2*n^2+1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {3 \, b^{4} n^{4} + 2 \, {\left (b^{2} n^{2} + 1\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{4} + 8 \, b^{2} n^{2} - 2 \, {\left (5 \, b^{2} n^{2} + 2\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - 2 \, {\left (2 \, {\left (b^{3} n^{3} + b n\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - {\left (5 \, b^{3} n^{3} + 2 \, b n\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + 2}{4 \, {\left (4 \, b^{4} n^{4} + 5 \, b^{2} n^{2} + 1\right )} x^{2}} \]

[In]

integrate(sin(a+b*log(c*x^n))^4/x^3,x, algorithm="fricas")

[Out]

-1/4*(3*b^4*n^4 + 2*(b^2*n^2 + 1)*cos(b*n*log(x) + b*log(c) + a)^4 + 8*b^2*n^2 - 2*(5*b^2*n^2 + 2)*cos(b*n*log
(x) + b*log(c) + a)^2 - 2*(2*(b^3*n^3 + b*n)*cos(b*n*log(x) + b*log(c) + a)^3 - (5*b^3*n^3 + 2*b*n)*cos(b*n*lo
g(x) + b*log(c) + a))*sin(b*n*log(x) + b*log(c) + a) + 2)/((4*b^4*n^4 + 5*b^2*n^2 + 1)*x^2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 98.44 (sec) , antiderivative size = 1066, normalized size of antiderivative = 5.08 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\text {Too large to display} \]

[In]

integrate(sin(a+b*ln(c*x**n))**4/x**3,x)

[Out]

Piecewise((I*sin(4*a - 4*I*log(c*x**n)/n)/(24*x**2) + cos(2*a - 2*I*log(c*x**n)/n)/(8*x**2) + cos(4*a - 4*I*lo
g(c*x**n)/n)/(48*x**2) - 3/(16*x**2) - I*log(c*x**n)*sin(2*a - 2*I*log(c*x**n)/n)/(4*n*x**2) - log(c*x**n)*cos
(2*a - 2*I*log(c*x**n)/n)/(4*n*x**2), Eq(b, -I/n)), (I*sin(2*a - I*log(c*x**n)/n)/(6*x**2) + I*sin(4*a - 2*I*l
og(c*x**n)/n)/(32*x**2) + cos(2*a - I*log(c*x**n)/n)/(3*x**2) - 3/(16*x**2) + I*log(c*x**n)*sin(4*a - 2*I*log(
c*x**n)/n)/(16*n*x**2) + log(c*x**n)*cos(4*a - 2*I*log(c*x**n)/n)/(16*n*x**2), Eq(b, -I/(2*n))), (-I*sin(2*a +
I*log(c*x**n)/n)/(6*x**2) + cos(2*a + I*log(c*x**n)/n)/(3*x**2) - cos(4*a + 2*I*log(c*x**n)/n)/(32*x**2) - 3/
(16*x**2) - I*log(c*x**n)*sin(4*a + 2*I*log(c*x**n)/n)/(16*n*x**2) + log(c*x**n)*cos(4*a + 2*I*log(c*x**n)/n)/
(16*n*x**2), Eq(b, I/(2*n))), (-I*sin(4*a + 4*I*log(c*x**n)/n)/(24*x**2) + cos(2*a + 2*I*log(c*x**n)/n)/(8*x**
2) + cos(4*a + 4*I*log(c*x**n)/n)/(48*x**2) - 3/(16*x**2) + I*log(c*x**n)*sin(2*a + 2*I*log(c*x**n)/n)/(4*n*x*
*2) - log(c*x**n)*cos(2*a + 2*I*log(c*x**n)/n)/(4*n*x**2), Eq(b, I/n)), (-3*b**4*n**4*sin(a + b*log(c*x**n))**
4/(16*b**4*n**4*x**2 + 20*b**2*n**2*x**2 + 4*x**2) - 6*b**4*n**4*sin(a + b*log(c*x**n))**2*cos(a + b*log(c*x**
n))**2/(16*b**4*n**4*x**2 + 20*b**2*n**2*x**2 + 4*x**2) - 3*b**4*n**4*cos(a + b*log(c*x**n))**4/(16*b**4*n**4*
x**2 + 20*b**2*n**2*x**2 + 4*x**2) - 10*b**3*n**3*sin(a + b*log(c*x**n))**3*cos(a + b*log(c*x**n))/(16*b**4*n*
*4*x**2 + 20*b**2*n**2*x**2 + 4*x**2) - 6*b**3*n**3*sin(a + b*log(c*x**n))*cos(a + b*log(c*x**n))**3/(16*b**4*
n**4*x**2 + 20*b**2*n**2*x**2 + 4*x**2) - 8*b**2*n**2*sin(a + b*log(c*x**n))**4/(16*b**4*n**4*x**2 + 20*b**2*n
**2*x**2 + 4*x**2) - 6*b**2*n**2*sin(a + b*log(c*x**n))**2*cos(a + b*log(c*x**n))**2/(16*b**4*n**4*x**2 + 20*b
**2*n**2*x**2 + 4*x**2) - 4*b*n*sin(a + b*log(c*x**n))**3*cos(a + b*log(c*x**n))/(16*b**4*n**4*x**2 + 20*b**2*
n**2*x**2 + 4*x**2) - 2*sin(a + b*log(c*x**n))**4/(16*b**4*n**4*x**2 + 20*b**2*n**2*x**2 + 4*x**2), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1082 vs. \(2 (202) = 404\).

Time = 0.26 (sec) , antiderivative size = 1082, normalized size of antiderivative = 5.15 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\text {Too large to display} \]

[In]

integrate(sin(a+b*log(c*x^n))^4/x^3,x, algorithm="maxima")

[Out]

-1/32*(24*(b^4*cos(4*b*log(c))^2 + b^4*sin(4*b*log(c))^2)*n^4 + 30*(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 - (2*(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 - (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 + 2*(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*l
og(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*(4*(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 - 4*(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*log(c)))*
n^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 - (2*(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*log(c)))*n^
3 + (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 + 2*
(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 + cos(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*(4*(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 + 4*(b^2*cos(4*b*log(c))*sin(6*b*log(c)) - b^2*cos(6*b*log(c))*sin(4*b*l
og(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 + (b*cos(6*b*log(c))*c
os(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))/((4*(b^4*cos(4*b*log(c))^2 + b^4*sin(4*b*l
og(c))^2)*n^4 + 5*(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^2)

Giac [F]

\[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {\sin \left (b \log \left (c x^{n}\right ) + a\right )^{4}}{x^{3}} \,d x } \]

[In]

integrate(sin(a+b*log(c*x^n))^4/x^3,x, algorithm="giac")

[Out]

integrate(sin(b*log(c*x^n) + a)^4/x^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int \frac {{\sin \left (a+b\,\ln \left (c\,x^n\right )\right )}^4}{x^3} \,d x \]

[In]

int(sin(a + b*log(c*x^n))^4/x^3,x)

[Out]

int(sin(a + b*log(c*x^n))^4/x^3, x)