\(\int \frac {x}{a+b \log (c x^n)} \, dx\) [67]

Optimal result

Integrand size = 14, antiderivative size = 51 \[ \int \frac {x}{a+b \log \left (c x^n\right )} \, dx=\frac {e^{-\frac {2 a}{b n}} x^2 \left (c x^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b n} \]

[Out]

x^2*Ei(2*(a+b*ln(c*x^n))/b/n)/b/exp(2*a/b/n)/n/((c*x^n)^(2/n))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2347, 2209} \[ \int \frac {x}{a+b \log \left (c x^n\right )} \, dx=\frac {x^2 e^{-\frac {2 a}{b n}} \left (c x^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b n} \]

[In]

Int[x/(a + b*Log[c*x^n]),x]

[Out]

(x^2*ExpIntegralEi[(2*(a + b*Log[c*x^n]))/(b*n)])/(b*E^((2*a)/(b*n))*n*(c*x^n)^(2/n))

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2347

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^2 \left (c x^n\right )^{-2/n}\right ) \text {Subst}\left (\int \frac {e^{\frac {2 x}{n}}}{a+b x} \, dx,x,\log \left (c x^n\right )\right )}{n} \\ & = \frac {e^{-\frac {2 a}{b n}} x^2 \left (c x^n\right )^{-2/n} \text {Ei}\left (\frac {2 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {x}{a+b \log \left (c x^n\right )} \, dx=\frac {e^{-\frac {2 a}{b n}} x^2 \left (c x^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b n} \]

[In]

Integrate[x/(a + b*Log[c*x^n]),x]

[Out]

(x^2*ExpIntegralEi[(2*(a + b*Log[c*x^n]))/(b*n)])/(b*E^((2*a)/(b*n))*n*(c*x^n)^(2/n))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.28 (sec) , antiderivative size = 242, normalized size of antiderivative = 4.75

method	result	size
risch	\(-\frac {x^{2} \left (x^{n}\right )^{-\frac {2}{n}} c^{-\frac {2}{n}} {\mathrm e}^{-\frac {-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 a}{b n}} \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )-\frac {-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+2 b \left (\ln \left (x^{n}\right )-n \ln \left (x \right )\right )+2 a}{b n}\right )}{b n}\)	\(242\)

[In]

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

[Out]

-1/b/n*x^2*(x^n)^(-2/n)*c^(-2/n)*exp(-(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*b*Pi*csgn(I*c)*csgn(I*c*x
^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*b*Pi*csgn(I*c*x^n)^3+2*a)/b/n)*Ei(1,-2*ln(x)-(-I*b*Pi*csgn(I*c)*csg
n(I*x^n)*csgn(I*c*x^n)+I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*b*Pi*csgn(I*c*x^n
)^3+2*b*ln(c)+2*b*(ln(x^n)-n*ln(x))+2*a)/b/n)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.82 \[ \int \frac {x}{a+b \log \left (c x^n\right )} \, dx=\frac {e^{\left (-\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )} \operatorname {log\_integral}\left (x^{2} e^{\left (\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}\right )}{b n} \]

[In]

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

[Out]

e^(-2*(b*log(c) + a)/(b*n))*log_integral(x^2*e^(2*(b*log(c) + a)/(b*n)))/(b*n)

Sympy [F]

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

[In]

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

[Out]

Integral(x/(a + b*log(c*x**n)), x)

Maxima [F]

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

[In]

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

[Out]

integrate(x/(b*log(c*x^n) + a), x)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94 \[ \int \frac {x}{a+b \log \left (c x^n\right )} \, dx=\frac {{\rm Ei}\left (\frac {2 \, \log \left (c\right )}{n} + \frac {2 \, a}{b n} + 2 \, \log \left (x\right )\right ) e^{\left (-\frac {2 \, a}{b n}\right )}}{b c^{\frac {2}{n}} n} \]

[In]

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

[Out]

Ei(2*log(c)/n + 2*a/(b*n) + 2*log(x))*e^(-2*a/(b*n))/(b*c^(2/n)*n)

Mupad [F(-1)]

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

[In]

int(x/(a + b*log(c*x^n)),x)

[Out]

int(x/(a + b*log(c*x^n)), x)