\(\int \frac {\sin ^3(a+b \log (c x^n))}{x^3} \, dx\) [18]

Optimal result

Integrand size = 17, antiderivative size = 158 \[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {6 b^3 n^3 \cos \left (a+b \log \left (c x^n\right )\right )}{\left (16+40 b^2 n^2+9 b^4 n^4\right ) x^2}-\frac {12 b^2 n^2 \sin \left (a+b \log \left (c x^n\right )\right )}{\left (16+40 b^2 n^2+9 b^4 n^4\right ) x^2}-\frac {3 b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^2\left (a+b \log \left (c x^n\right )\right )}{\left (4+9 b^2 n^2\right ) x^2}-\frac {2 \sin ^3\left (a+b \log \left (c x^n\right )\right )}{\left (4+9 b^2 n^2\right ) x^2} \]

[Out]

-6*b^3*n^3*cos(a+b*ln(c*x^n))/(9*b^4*n^4+40*b^2*n^2+16)/x^2-12*b^2*n^2*sin(a+b*ln(c*x^n))/(9*b^4*n^4+40*b^2*n^
2+16)/x^2-3*b*n*cos(a+b*ln(c*x^n))*sin(a+b*ln(c*x^n))^2/(9*b^2*n^2+4)/x^2-2*sin(a+b*ln(c*x^n))^3/(9*b^2*n^2+4)
/x^2

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} \[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {2 \sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^2 \left (9 b^2 n^2+4\right )}-\frac {3 b n \sin ^2\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{x^2 \left (9 b^2 n^2+4\right )}-\frac {12 b^2 n^2 \sin \left (a+b \log \left (c x^n\right )\right )}{x^2 \left (9 b^4 n^4+40 b^2 n^2+16\right )}-\frac {6 b^3 n^3 \cos \left (a+b \log \left (c x^n\right )\right )}{x^2 \left (9 b^4 n^4+40 b^2 n^2+16\right )} \]

[In]

Int[Sin[a + b*Log[c*x^n]]^3/x^3,x]

[Out]

(-6*b^3*n^3*Cos[a + b*Log[c*x^n]])/((16 + 40*b^2*n^2 + 9*b^4*n^4)*x^2) - (12*b^2*n^2*Sin[a + b*Log[c*x^n]])/((
16 + 40*b^2*n^2 + 9*b^4*n^4)*x^2) - (3*b*n*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]]^2)/((4 + 9*b^2*n^2)*x^2
) - (2*Sin[a + b*Log[c*x^n]]^3)/((4 + 9*b^2*n^2)*x^2)

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 (4+9 b^2 n^2\right ) x^2}-\frac {2 \sin ^3\left (a+b \log \left (c x^n\right )\right )}{\left (4+9 b^2 n^2\right ) x^2}+\frac {\left (6 b^2 n^2\right ) \int \frac {\sin \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx}{4+9 b^2 n^2} \\ & = -\frac {6 b^3 n^3 \cos \left (a+b \log \left (c x^n\right )\right )}{\left (16+40 b^2 n^2+9 b^4 n^4\right ) x^2}-\frac {12 b^2 n^2 \sin \left (a+b \log \left (c x^n\right )\right )}{\left (16+40 b^2 n^2+9 b^4 n^4\right ) x^2}-\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 (4+9 b^2 n^2\right ) x^2}-\frac {2 \sin ^3\left (a+b \log \left (c x^n\right )\right )}{\left (4+9 b^2 n^2\right ) x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.79 \[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\frac {-3 b n \left (4+9 b^2 n^2\right ) \cos \left (a+b \log \left (c x^n\right )\right )+3 b n \left (4+b^2 n^2\right ) \cos \left (3 \left (a+b \log \left (c x^n\right )\right )\right )+4 \left (-4-13 b^2 n^2+\left (4+b^2 n^2\right ) \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{4 \left (16+40 b^2 n^2+9 b^4 n^4\right ) x^2} \]

[In]

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

[Out]

(-3*b*n*(4 + 9*b^2*n^2)*Cos[a + b*Log[c*x^n]] + 3*b*n*(4 + b^2*n^2)*Cos[3*(a + b*Log[c*x^n])] + 4*(-4 - 13*b^2
*n^2 + (4 + b^2*n^2)*Cos[2*(a + b*Log[c*x^n])])*Sin[a + b*Log[c*x^n]])/(4*(16 + 40*b^2*n^2 + 9*b^4*n^4)*x^2)

Maple [A] (verified)

Time = 6.38 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.40

method	result	size
parallelrisch	\(\frac {6 {\tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{6} b^{3} n^{3}-24 {\tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{5} b^{2} n^{2}+\left (18 b^{3} n^{3}+48 b n \right ) {\tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{4}+\left (-64 b^{2} n^{2}-64\right ) {\tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{3}+\left (-18 b^{3} n^{3}-48 b n \right ) {\tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2}-24 \tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right ) b^{2} n^{2}-6 b^{3} n^{3}}{9 x^{2} \left (b^{2} n^{2}+4\right ) \left (b^{2} n^{2}+\frac {4}{9}\right ) {\left (1+{\tan \left (\frac {a}{2}+b \ln \left (\sqrt {c \,x^{n}}\right )\right )}^{2}\right )}^{3}}\)	\(221\)

[In]

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

[Out]

1/9*(6*tan(1/2*a+b*ln((c*x^n)^(1/2)))^6*b^3*n^3-24*tan(1/2*a+b*ln((c*x^n)^(1/2)))^5*b^2*n^2+(18*b^3*n^3+48*b*n
)*tan(1/2*a+b*ln((c*x^n)^(1/2)))^4+(-64*b^2*n^2-64)*tan(1/2*a+b*ln((c*x^n)^(1/2)))^3+(-18*b^3*n^3-48*b*n)*tan(
1/2*a+b*ln((c*x^n)^(1/2)))^2-24*tan(1/2*a+b*ln((c*x^n)^(1/2)))*b^2*n^2-6*b^3*n^3)/x^2/(b^2*n^2+4)/(b^2*n^2+4/9
)/(1+tan(1/2*a+b*ln((c*x^n)^(1/2)))^2)^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.82 \[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\frac {3 \, {\left (b^{3} n^{3} + 4 \, b n\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - 3 \, {\left (3 \, b^{3} n^{3} + 4 \, b n\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - 2 \, {\left (7 \, b^{2} n^{2} - {\left (b^{2} n^{2} + 4\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 4\right )} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{{\left (9 \, b^{4} n^{4} + 40 \, b^{2} n^{2} + 16\right )} x^{2}} \]

[In]

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

[Out]

(3*(b^3*n^3 + 4*b*n)*cos(b*n*log(x) + b*log(c) + a)^3 - 3*(3*b^3*n^3 + 4*b*n)*cos(b*n*log(x) + b*log(c) + a) -
2*(7*b^2*n^2 - (b^2*n^2 + 4)*cos(b*n*log(x) + b*log(c) + a)^2 + 4)*sin(b*n*log(x) + b*log(c) + a))/((9*b^4*n^
4 + 40*b^2*n^2 + 16)*x^2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 35.84 (sec) , antiderivative size = 886, normalized size of antiderivative = 5.61 \[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((-sin(3*a - 6*I*log(c*x**n)/n)/(64*x**2) - 3*I*cos(a - 2*I*log(c*x**n)/n)/(16*x**2) + 3*I*cos(3*a -
6*I*log(c*x**n)/n)/(64*x**2) + 3*log(c*x**n)*sin(a - 2*I*log(c*x**n)/n)/(8*n*x**2) - 3*I*log(c*x**n)*cos(a - 2
*I*log(c*x**n)/n)/(8*n*x**2), Eq(b, -2*I/n)), (-27*sin(a - 2*I*log(c*x**n)/(3*n))/(64*x**2) + sin(3*a - 2*I*lo
g(c*x**n)/n)/(16*x**2) + 9*I*cos(a - 2*I*log(c*x**n)/(3*n))/(64*x**2) - log(c*x**n)*sin(3*a - 2*I*log(c*x**n)/
n)/(8*n*x**2) + I*log(c*x**n)*cos(3*a - 2*I*log(c*x**n)/n)/(8*n*x**2), Eq(b, -2*I/(3*n))), (-27*sin(a + 2*I*lo
g(c*x**n)/(3*n))/(64*x**2) - 9*I*cos(a + 2*I*log(c*x**n)/(3*n))/(64*x**2) - I*cos(3*a + 2*I*log(c*x**n)/n)/(16
*x**2) - log(c*x**n)*sin(3*a + 2*I*log(c*x**n)/n)/(8*n*x**2) - I*log(c*x**n)*cos(3*a + 2*I*log(c*x**n)/n)/(8*n
*x**2), Eq(b, 2*I/(3*n))), (-3*sin(a + 2*I*log(c*x**n)/n)/(16*x**2) - sin(3*a + 6*I*log(c*x**n)/n)/(64*x**2) -
3*I*cos(3*a + 6*I*log(c*x**n)/n)/(64*x**2) + 3*log(c*x**n)*sin(a + 2*I*log(c*x**n)/n)/(8*n*x**2) + 3*I*log(c*
x**n)*cos(a + 2*I*log(c*x**n)/n)/(8*n*x**2), Eq(b, 2*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**2 + 40*b**2*n**2*x**2 + 16*x**2) - 6*b**3*n**3*cos(a + b*log(c*x**n))**3/(9*b**4*
n**4*x**2 + 40*b**2*n**2*x**2 + 16*x**2) - 14*b**2*n**2*sin(a + b*log(c*x**n))**3/(9*b**4*n**4*x**2 + 40*b**2*
n**2*x**2 + 16*x**2) - 12*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**2 + 40*b*
*2*n**2*x**2 + 16*x**2) - 12*b*n*sin(a + b*log(c*x**n))**2*cos(a + b*log(c*x**n))/(9*b**4*n**4*x**2 + 40*b**2*
n**2*x**2 + 16*x**2) - 8*sin(a + b*log(c*x**n))**3/(9*b**4*n**4*x**2 + 40*b**2*n**2*x**2 + 16*x**2), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1007 vs. \(2 (158) = 316\).

Time = 0.25 (sec) , antiderivative size = 1007, normalized size of antiderivative = 6.37 \[ \int \frac {\sin ^3\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\text {Too large to display} \]

[In]

integrate(sin(a+b*log(c*x^n))^3/x^3,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
+ 2*(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 + 12
*(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 + 8*cos(3*b*log
(c))*sin(6*b*log(c)) - 8*cos(6*b*log(c))*sin(3*b*log(c)) + 8*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*sin(3*b*log(c))*sin(2*b*log(c)))*n^3 + 18*(b^2*cos(3*b*log(c))*sin(4*b*log(c)) - b^2*cos(4*b*log(c))*s
in(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 + 4*(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*l
og(c))*sin(2*b*log(c)))*n + 8*cos(3*b*log(c))*sin(4*b*log(c)) - 8*cos(4*b*log(c))*sin(3*b*log(c)) + 8*cos(2*b*
log(c))*sin(3*b*log(c)) - 8*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*cos(6*b*log(c))*sin(3*b*log(c)) + b^3*sin(3*b*log(c)))*n^3 - 2*(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 + 12*(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 - 8*cos(6*b*log(c))*cos(3*b*log(c)) - 8*sin(6*b
*log(c))*sin(3*b*log(c)) - 8*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 - 18*(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 + 4*(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 - 8*co
s(4*b*log(c))*cos(3*b*log(c)) - 8*cos(3*b*log(c))*cos(2*b*log(c)) - 8*sin(4*b*log(c))*sin(3*b*log(c)) - 8*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 + 40
*(b^2*cos(3*b*log(c))^2 + b^2*sin(3*b*log(c))^2)*n^2 + 16*cos(3*b*log(c))^2 + 16*sin(3*b*log(c))^2)*x^2)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(sin(a + b*log(c*x^n))^3/x^3,x)

[Out]

int(sin(a + b*log(c*x^n))^3/x^3, x)