\(\int x \cos (a+b \log (c x^n)) \, dx\) [87]

Optimal result

Integrand size = 13, antiderivative size = 56 \[ \int x \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 x^2 \cos \left (a+b \log \left (c x^n\right )\right )}{4+b^2 n^2}+\frac {b n x^2 \sin \left (a+b \log \left (c x^n\right )\right )}{4+b^2 n^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, 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 = {4574} \[ \int x \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b n x^2 \sin \left (a+b \log \left (c x^n\right )\right )}{b^2 n^2+4}+\frac {2 x^2 \cos \left (a+b \log \left (c x^n\right )\right )}{b^2 n^2+4} \]

[In]

Int[x*Cos[a + b*Log[c*x^n]],x]

[Out]

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

Rule 4574

Int[Cos[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[(m + 1)*(e*x)^(m +
1)*(Cos[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)*(Sin[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 {2 x^2 \cos \left (a+b \log \left (c x^n\right )\right )}{4+b^2 n^2}+\frac {b n x^2 \sin \left (a+b \log \left (c x^n\right )\right )}{4+b^2 n^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.77 \[ \int x \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^2 \left (2 \cos \left (a+b \log \left (c x^n\right )\right )+b n \sin \left (a+b \log \left (c x^n\right )\right )\right )}{4+b^2 n^2} \]

[In]

Integrate[x*Cos[a + b*Log[c*x^n]],x]

[Out]

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

Maple [A] (verified)

Time = 1.98 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.79

method	result	size
parallelrisch	\(\frac {x^{2} \left (\sin \left (a +b \ln \left (c \,x^{n}\right )\right ) b n +2 \cos \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}{b^{2} n^{2}+4}\)	\(44\)
parts	\(\frac {x \,{\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )}{n}-\frac {\ln \left (c \right )}{n}} \cos \left (a +b \ln \left (c \,x^{n}\right )\right )}{n^{2} \left (\frac {1}{n^{2}}+b^{2}\right )}+\frac {x b \,{\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )}{n}-\frac {\ln \left (c \right )}{n}} \sin \left (a +b \ln \left (c \,x^{n}\right )\right )}{n \left (\frac {1}{n^{2}}+b^{2}\right )}-\frac {\frac {b \left (\frac {b n \,c^{-\frac {1}{n}} {\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )-n \ln \left (x \right )}{n}} x^{2} {\tan \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}{b^{2} n^{2}+4}+\frac {4 c^{-\frac {1}{n}} {\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )-n \ln \left (x \right )}{n}} x^{2} \tan \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}{b^{2} n^{2}+4}-\frac {b n \,c^{-\frac {1}{n}} {\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )-n \ln \left (x \right )}{n}} x^{2}}{b^{2} n^{2}+4}\right )}{\left (\frac {1}{n^{2}}+b^{2}\right ) \left (1+{\tan \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}\right )}+\frac {\frac {2 c^{-\frac {1}{n}} {\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )-n \ln \left (x \right )}{n}} x^{2}}{b^{2} n^{2}+4}-\frac {2 c^{-\frac {1}{n}} {\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )-n \ln \left (x \right )}{n}} x^{2} {\tan \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}{b^{2} n^{2}+4}+\frac {2 b n \,c^{-\frac {1}{n}} {\mathrm e}^{\frac {\ln \left (c \,x^{n}\right )-n \ln \left (x \right )}{n}} x^{2} \tan \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}{b^{2} n^{2}+4}}{n \left (\frac {1}{n^{2}}+b^{2}\right ) \left (1+{\tan \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}\right )}}{n}\)	\(468\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.86 \[ \int x \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b n x^{2} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + 2 \, x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{b^{2} n^{2} + 4} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Piecewise((Integral(x*cos(a - 2*I*log(c*x**n)/n), x), Eq(b, -2*I/n)), (Integral(x*cos(a + 2*I*log(c*x**n)/n),
x), Eq(b, 2*I/n)), (b*n*x**2*sin(a + b*log(c*x**n))/(b**2*n**2 + 4) + 2*x**2*cos(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. 218 vs. \(2 (56) = 112\).

Time = 0.25 (sec) , antiderivative size = 218, normalized size of antiderivative = 3.89 \[ \int x \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {{\left ({\left (b \cos \left (b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) - b \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) + b \sin \left (b \log \left (c\right )\right )\right )} n + 2 \, \cos \left (2 \, b \log \left (c\right )\right ) \cos \left (b \log \left (c\right )\right ) + 2 \, \sin \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) + 2 \, \cos \left (b \log \left (c\right )\right )\right )} x^{2} \cos \left (b \log \left (x^{n}\right ) + a\right ) + {\left ({\left (b \cos \left (2 \, b \log \left (c\right )\right ) \cos \left (b \log \left (c\right )\right ) + b \sin \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) + b \cos \left (b \log \left (c\right )\right )\right )} n - 2 \, \cos \left (b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + 2 \, \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) - 2 \, \sin \left (b \log \left (c\right )\right )\right )} x^{2} \sin \left (b \log \left (x^{n}\right ) + a\right )}{2 \, {\left ({\left (b^{2} \cos \left (b \log \left (c\right )\right )^{2} + b^{2} \sin \left (b \log \left (c\right )\right )^{2}\right )} n^{2} + 4 \, \cos \left (b \log \left (c\right )\right )^{2} + 4 \, \sin \left (b \log \left (c\right )\right )^{2}\right )}} \]

[In]

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

[Out]

1/2*(((b*cos(b*log(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*cos(b*log(x^n) + a) + ((b*cos(2*b*l
og(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*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. 915 vs. \(2 (56) = 112\).

Time = 0.33 (sec) , antiderivative size = 915, normalized size of antiderivative = 16.34 \[ \int x \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

-(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) + 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*l
og(abs(x)) + 1/2*b*log(abs(c)))^2*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*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)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))*tan(1/2*a)^2 - 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
- 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))) - 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))) - 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) - 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) + 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 +
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*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*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*l
og(abs(x)) + 1/2*b*log(abs(c)))*tan(1/2*a) + 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 + 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 - 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) - 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^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*tan(1/2*a)^2 + 4)

Mupad [B] (verification not implemented)

Time = 26.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.77 \[ \int x \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^2\,\left (2\,\cos \left (a+b\,\ln \left (c\,x^n\right )\right )+b\,n\,\sin \left (a+b\,\ln \left (c\,x^n\right )\right )\right )}{b^2\,n^2+4} \]

[In]

int(x*cos(a + b*log(c*x^n)),x)

[Out]

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