\(\int \frac {1}{x^2 (a+b \log (c x^n))} \, dx\) [70]

Optimal result

Integrand size = 16, antiderivative size = 48 \[ \int \frac {1}{x^2 \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {e^{\frac {a}{b n}} \left (c x^n\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (-\frac {a+b \log \left (c x^n\right )}{b n}\right )}{b n x} \]

[Out]

exp(a/b/n)*(c*x^n)^(1/n)*Ei((-a-b*ln(c*x^n))/b/n)/b/n/x

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2347, 2209} \[ \int \frac {1}{x^2 \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {e^{\frac {a}{b n}} \left (c x^n\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (-\frac {a+b \log \left (c x^n\right )}{b n}\right )}{b n x} \]

[In]

Int[1/(x^2*(a + b*Log[c*x^n])),x]

[Out]

(E^(a/(b*n))*(c*x^n)^n^(-1)*ExpIntegralEi[-((a + b*Log[c*x^n])/(b*n))])/(b*n*x)

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 (c x^n\right )^{\frac {1}{n}} \text {Subst}\left (\int \frac {e^{-\frac {x}{n}}}{a+b x} \, dx,x,\log \left (c x^n\right )\right )}{n x} \\ & = \frac {e^{\frac {a}{b n}} \left (c x^n\right )^{\frac {1}{n}} \text {Ei}\left (-\frac {a+b \log \left (c x^n\right )}{b n}\right )}{b n x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {e^{\frac {a}{b n}} \left (c x^n\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (-\frac {a+b \log \left (c x^n\right )}{b n}\right )}{b n x} \]

[In]

Integrate[1/(x^2*(a + b*Log[c*x^n])),x]

[Out]

(E^(a/(b*n))*(c*x^n)^n^(-1)*ExpIntegralEi[-((a + b*Log[c*x^n])/(b*n))])/(b*n*x)

Maple [C] (warning: unable to verify)

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

Time = 0.58 (sec) , antiderivative size = 236, normalized size of antiderivative = 4.92

method	result	size
risch	\(-\frac {c^{\frac {1}{n}} \left (x^{n}\right )^{\frac {1}{n}} {\mathrm e}^{\frac {-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 a}{2 b n}} \operatorname {Ei}_{1}\left (\ln \left (x \right )+\frac {-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+2 b \left (\ln \left (x^{n}\right )-n \ln \left (x \right )\right )+2 a}{2 b n}\right )}{b n x}\)	\(236\)

[In]

int(1/x^2/(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)

[Out]

-1/b/n/x*c^(1/n)*(x^n)^(1/n)*exp(1/2*(-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,ln(x)+1/2*(-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 = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^2 \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )} \operatorname {log\_integral}\left (\frac {e^{\left (-\frac {b \log \left (c\right ) + a}{b n}\right )}}{x}\right )}{b n} \]

[In]

integrate(1/x^2/(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{x^2 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {1}{x^{2} \left (a + b \log {\left (c x^{n} \right )}\right )}\, dx \]

[In]

integrate(1/x**2/(a+b*ln(c*x**n)),x)

[Out]

Integral(1/(x**2*(a + b*log(c*x**n))), x)

Maxima [F]

\[ \int \frac {1}{x^2 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}} \,d x } \]

[In]

integrate(1/x^2/(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

integrate(1/((b*log(c*x^n) + a)*x^2), x)

Giac [F]

\[ \int \frac {1}{x^2 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}} \,d x } \]

[In]

integrate(1/x^2/(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

integrate(1/((b*log(c*x^n) + a)*x^2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {1}{x^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )} \,d x \]

[In]

int(1/(x^2*(a + b*log(c*x^n))),x)

[Out]

int(1/(x^2*(a + b*log(c*x^n))), x)