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

Optimal result

Integrand size = 13, antiderivative size = 57 \[ \int x \sin \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {b n x^2 \cos \left (a+b \log \left (c x^n\right )\right )}{4+b^2 n^2}+\frac {2 x^2 \sin \left (a+b \log \left (c x^n\right )\right )}{4+b^2 n^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {4573} \[ \int x \sin \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 x^2 \sin \left (a+b \log \left (c x^n\right )\right )}{b^2 n^2+4}-\frac {b n x^2 \cos \left (a+b \log \left (c x^n\right )\right )}{b^2 n^2+4} \]

[In]

Int[x*Sin[a + b*Log[c*x^n]],x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = -\frac {b n x^2 \cos \left (a+b \log \left (c x^n\right )\right )}{4+b^2 n^2}+\frac {2 x^2 \sin \left (a+b \log \left (c x^n\right )\right )}{4+b^2 n^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77 \[ \int x \sin \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {x^2 \left (b n \cos \left (a+b \log \left (c x^n\right )\right )-2 \sin \left (a+b \log \left (c x^n\right )\right )\right )}{4+b^2 n^2} \]

[In]

Integrate[x*Sin[a + b*Log[c*x^n]],x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(469\) vs. \(2(57)=114\).

Time = 1.22 (sec) , antiderivative size = 470, normalized size of antiderivative = 8.25

method	result	size
parts	\(-\frac {x b \,{\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )}{n}-\frac {\ln \left (c \right )}{n}} \cos \left (a +b \ln \left (c \,x^{n}\right )\right )}{n \left (\frac {1}{n^{2}}+b^{2}\right )}+\frac {x \,{\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )}{n}-\frac {\ln \left (c \right )}{n}} \sin \left (a +b \ln \left (c \,x^{n}\right )\right )}{n^{2} \left (\frac {1}{n^{2}}+b^{2}\right )}-\frac {\frac {\frac {b n \,c^{-\frac {1}{n}} {\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )-n \ln \left (x \right )}{n}} x^{2} {\tan \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}{b^{2} n^{2}+4}+\frac {4 c^{-\frac {1}{n}} {\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )-n \ln \left (x \right )}{n}} x^{2} \tan \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}{b^{2} n^{2}+4}-\frac {b n \,c^{-\frac {1}{n}} {\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )-n \ln \left (x \right )}{n}} x^{2}}{b^{2} n^{2}+4}}{n \left (\frac {1}{n^{2}}+b^{2}\right ) \left (1+{\tan \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}\right )}-\frac {b \left (\frac {2 c^{-\frac {1}{n}} {\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )-n \ln \left (x \right )}{n}} x^{2}}{b^{2} n^{2}+4}-\frac {2 c^{-\frac {1}{n}} {\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )-n \ln \left (x \right )}{n}} x^{2} {\tan \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}{b^{2} n^{2}+4}+\frac {2 b n \,c^{-\frac {1}{n}} {\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )-n \ln \left (x \right )}{n}} x^{2} \tan \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}{b^{2} n^{2}+4}\right )}{\left (\frac {1}{n^{2}}+b^{2}\right ) \left (1+{\tan \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}\right )}}{n}\)	\(470\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int x \sin \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {b n x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - 2 \, x^{2} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{b^{2} n^{2} + 4} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int x \sin \left (a+b \log \left (c x^n\right )\right ) \, dx=\begin {cases} \int x \sin {\left (a - \frac {2 i \log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = - \frac {2 i}{n} \\\int x \sin {\left (a + \frac {2 i \log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = \frac {2 i}{n} \\- \frac {b n x^{2} \cos {\left (a + b \log {\left (c x^{n} \right )} \right )}}{b^{2} n^{2} + 4} + \frac {2 x^{2} \sin {\left (a + b \log {\left (c x^{n} \right )} \right )}}{b^{2} n^{2} + 4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((Integral(x*sin(a - 2*I*log(c*x**n)/n), x), Eq(b, -2*I/n)), (Integral(x*sin(a + 2*I*log(c*x**n)/n),
x), Eq(b, 2*I/n)), (-b*n*x**2*cos(a + b*log(c*x**n))/(b**2*n**2 + 4) + 2*x**2*sin(a + b*log(c*x**n))/(b**2*n**
2 + 4), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (57) = 114\).

Time = 0.24 (sec) , antiderivative size = 219, normalized size of antiderivative = 3.84 \[ \int x \sin \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {{\left ({\left (b \cos \left (2 \, b \log \left (c\right )\right ) \cos \left (b \log \left (c\right )\right ) + b \sin \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) + b \cos \left (b \log \left (c\right )\right )\right )} n - 2 \, \cos \left (b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + 2 \, \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) - 2 \, \sin \left (b \log \left (c\right )\right )\right )} x^{2} \cos \left (b \log \left (x^{n}\right ) + a\right ) - {\left ({\left (b \cos \left (b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) - b \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) + b \sin \left (b \log \left (c\right )\right )\right )} n + 2 \, \cos \left (2 \, b \log \left (c\right )\right ) \cos \left (b \log \left (c\right )\right ) + 2 \, \sin \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) + 2 \, \cos \left (b \log \left (c\right )\right )\right )} x^{2} \sin \left (b \log \left (x^{n}\right ) + a\right )}{2 \, {\left ({\left (b^{2} \cos \left (b \log \left (c\right )\right )^{2} + b^{2} \sin \left (b \log \left (c\right )\right )^{2}\right )} n^{2} + 4 \, \cos \left (b \log \left (c\right )\right )^{2} + 4 \, \sin \left (b \log \left (c\right )\right )^{2}\right )}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 923 vs. \(2 (57) = 114\).

Time = 0.33 (sec) , antiderivative size = 923, normalized size of antiderivative = 16.19 \[ \int x \sin \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/2*(b*n*x^2*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*
log(abs(c)))^2*tan(1/2*a)^2 + b*n*x^2*e^(-1/2*pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(1/2
*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(1/2*a)^2 - b*n*x^2*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sg
n(c) - 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2 - b*n*x^2*e^(-1/2*pi*b*n*sgn(x) + 1/2*pi*b*n -
1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2 - 4*b*n*x^2*e^(1/2*pi*b*n*sgn(x) -
1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))*tan(1/2*a) - 4*b*n*x^2*
e^(-1/2*pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))*
tan(1/2*a) - b*n*x^2*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(1/2*a)^2 - b*n*x^2*e^
(-1/2*pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(1/2*a)^2 + 4*x^2*e^(1/2*pi*b*n*sgn(x) - 1/2
*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(1/2*a) + 4*x^2*e^(-1/
2*pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(
1/2*a) + 4*x^2*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b
*log(abs(c)))*tan(1/2*a)^2 + 4*x^2*e^(-1/2*pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(1/2*b*
n*log(abs(x)) + 1/2*b*log(abs(c)))*tan(1/2*a)^2 + b*n*x^2*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c)
- 1/2*pi*b) + b*n*x^2*e^(-1/2*pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b) - 4*x^2*e^(1/2*pi*b*n*s
gn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c))) - 4*x^2*e^(-1/2*
pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c))) - 4*x^2*
e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(1/2*a) - 4*x^2*e^(-1/2*pi*b*n*sgn(x) + 1/2
*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(1/2*a))/(b^2*n^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(
1/2*a)^2 + b^2*n^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2 + b^2*n^2*tan(1/2*a)^2 + b^2*n^2 + 4*tan(1/2
*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(1/2*a)^2 + 4*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2 + 4*ta
n(1/2*a)^2 + 4)

Mupad [B] (verification not implemented)

Time = 27.56 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77 \[ \int x \sin \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^2\,\left (2\,\sin \left (a+b\,\ln \left (c\,x^n\right )\right )-b\,n\,\cos \left (a+b\,\ln \left (c\,x^n\right )\right )\right )}{b^2\,n^2+4} \]

[In]

int(x*sin(a + b*log(c*x^n)),x)

[Out]

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