Integrand size = 16, antiderivative size = 51 \[ \int \frac {1}{x^4 \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {e^{\frac {3 a}{b n}} \left (c x^n\right )^{3/n} \operatorname {ExpIntegralEi}\left (-\frac {3 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b n x^3} \] Output:
exp(3*a/b/n)*(c*x^n)^(3/n)*Ei((-3*a-3*b*ln(c*x^n))/b/n)/b/n/x^3
Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {e^{\frac {3 a}{b n}} \left (c x^n\right )^{3/n} \operatorname {ExpIntegralEi}\left (-\frac {3 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b n x^3} \] Input:
Integrate[1/(x^4*(a + b*Log[c*x^n])),x]
Output:
(E^((3*a)/(b*n))*(c*x^n)^(3/n)*ExpIntegralEi[(-3*(a + b*Log[c*x^n]))/(b*n) ])/(b*n*x^3)
Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2747, 2609}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x^4 \left (a+b \log \left (c x^n\right )\right )} \, dx\) |
\(\Big \downarrow \) 2747 |
\(\displaystyle \frac {\left (c x^n\right )^{3/n} \int \frac {\left (c x^n\right )^{-3/n}}{a+b \log \left (c x^n\right )}d\log \left (c x^n\right )}{n x^3}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {e^{\frac {3 a}{b n}} \left (c x^n\right )^{3/n} \operatorname {ExpIntegralEi}\left (-\frac {3 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b n x^3}\) |
Input:
Int[1/(x^4*(a + b*Log[c*x^n])),x]
Output:
(E^((3*a)/(b*n))*(c*x^n)^(3/n)*ExpIntegralEi[(-3*(a + b*Log[c*x^n]))/(b*n) ])/(b*n*x^3)
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F reeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol ] :> Simp[(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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.80 (sec) , antiderivative size = 242, normalized size of antiderivative = 4.75
method | result | size |
risch | \(-\frac {\left (x^{n}\right )^{\frac {3}{n}} c^{\frac {3}{n}} {\mathrm e}^{\frac {\frac {3 i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {3 i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {3 i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {3 i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+3 a}{n b}} \operatorname {expIntegral}_{1}\left (3 \ln \left (x \right )+\frac {\frac {3 i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {3 i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {3 i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {3 i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+3 b \ln \left (c \right )+3 b \left (\ln \left (x^{n}\right )-n \ln \left (x \right )\right )+3 a}{n b}\right )}{n b \,x^{3}}\) | \(242\) |
Input:
int(1/x^4/(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)
Output:
-1/n/b/x^3*(x^n)^(3/n)*c^(3/n)*exp(3/2*(I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2 -I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I*Pi*b*csgn(I*c*x^n)^3+I*Pi*b* csgn(I*c*x^n)^2*csgn(I*c)+2*a)/n/b)*Ei(1,3*ln(x)+3/2*(I*Pi*b*csgn(I*x^n)*c sgn(I*c*x^n)^2-I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I*Pi*b*csgn(I*c* x^n)^3+I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+2*b*ln(c)+2*b*(ln(x^n)-n*ln(x))+2* a)/n/b)
Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^4 \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {e^{\left (\frac {3 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )} \operatorname {log\_integral}\left (\frac {e^{\left (-\frac {3 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}}{x^{3}}\right )}{b n} \] Input:
integrate(1/x^4/(a+b*log(c*x^n)),x, algorithm="fricas")
Output:
e^(3*(b*log(c) + a)/(b*n))*log_integral(e^(-3*(b*log(c) + a)/(b*n))/x^3)/( b*n)
\[ \int \frac {1}{x^4 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {1}{x^{4} \left (a + b \log {\left (c x^{n} \right )}\right )}\, dx \] Input:
integrate(1/x**4/(a+b*ln(c*x**n)),x)
Output:
Integral(1/(x**4*(a + b*log(c*x**n))), x)
\[ \int \frac {1}{x^4 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{{\left (b \log \left (c x^{n}\right ) + a\right )} x^{4}} \,d x } \] Input:
integrate(1/x^4/(a+b*log(c*x^n)),x, algorithm="maxima")
Output:
integrate(1/((b*log(c*x^n) + a)*x^4), x)
\[ \int \frac {1}{x^4 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{{\left (b \log \left (c x^{n}\right ) + a\right )} x^{4}} \,d x } \] Input:
integrate(1/x^4/(a+b*log(c*x^n)),x, algorithm="giac")
Output:
integrate(1/((b*log(c*x^n) + a)*x^4), x)
Timed out. \[ \int \frac {1}{x^4 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {1}{x^4\,\left (a+b\,\ln \left (c\,x^n\right )\right )} \,d x \] Input:
int(1/(x^4*(a + b*log(c*x^n))),x)
Output:
int(1/(x^4*(a + b*log(c*x^n))), x)
\[ \int \frac {1}{x^4 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {1}{\mathrm {log}\left (x^{n} c \right ) b \,x^{4}+a \,x^{4}}d x \] Input:
int(1/x^4/(a+b*log(c*x^n)),x)
Output:
int(1/(log(x**n*c)*b*x**4 + a*x**4),x)