Integrand size = 22, antiderivative size = 86 \[ \int \frac {(e x)^q}{a+b \log \left (c \left (d x^m\right )^n\right )} \, dx=\frac {e^{-\frac {a (1+q)}{b m n}} (e x)^{1+q} \left (c \left (d x^m\right )^n\right )^{-\frac {1+q}{m n}} \operatorname {ExpIntegralEi}\left (\frac {(1+q) \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{b m n}\right )}{b e m n} \] Output:
(e*x)^(1+q)*Ei((1+q)*(a+b*ln(c*(d*x^m)^n))/b/m/n)/b/e/exp(a*(1+q)/b/m/n)/m /n/((c*(d*x^m)^n)^((1+q)/m/n))
Time = 0.14 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99 \[ \int \frac {(e x)^q}{a+b \log \left (c \left (d x^m\right )^n\right )} \, dx=\frac {e^{-\frac {(1+q) \left (a-b m n \log (x)+b \log \left (c \left (d x^m\right )^n\right )\right )}{b m n}} x^{-q} (e x)^q \operatorname {ExpIntegralEi}\left (\frac {(1+q) \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{b m n}\right )}{b m n} \] Input:
Integrate[(e*x)^q/(a + b*Log[c*(d*x^m)^n]),x]
Output:
((e*x)^q*ExpIntegralEi[((1 + q)*(a + b*Log[c*(d*x^m)^n]))/(b*m*n)])/(b*E^( ((1 + q)*(a - b*m*n*Log[x] + b*Log[c*(d*x^m)^n]))/(b*m*n))*m*n*x^q)
Time = 0.50 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2895, 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.
\(\displaystyle \int \frac {(e x)^q}{a+b \log \left (c \left (d x^m\right )^n\right )} \, dx\) |
\(\Big \downarrow \) 2895 |
\(\displaystyle \int \frac {(e x)^q}{a+b \log \left (c \left (d x^m\right )^n\right )}dx\) |
\(\Big \downarrow \) 2747 |
\(\displaystyle \frac {(e x)^{q+1} \left (c \left (d x^m\right )^n\right )^{-\frac {q+1}{m n}} \int \frac {\left (c \left (d x^m\right )^n\right )^{\frac {q+1}{m n}}}{a+b \log \left (c \left (d x^m\right )^n\right )}d\log \left (c \left (d x^m\right )^n\right )}{e m n}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {(e x)^{q+1} e^{-\frac {a (q+1)}{b m n}} \left (c \left (d x^m\right )^n\right )^{-\frac {q+1}{m n}} \operatorname {ExpIntegralEi}\left (\frac {(q+1) \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{b m n}\right )}{b e m n}\) |
Input:
Int[(e*x)^q/(a + b*Log[c*(d*x^m)^n]),x]
Output:
((e*x)^(1 + q)*ExpIntegralEi[((1 + q)*(a + b*Log[c*(d*x^m)^n]))/(b*m*n)])/ (b*e*E^((a*(1 + q))/(b*m*n))*m*n*(c*(d*x^m)^n)^((1 + q)/(m*n)))
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]
Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_. )*(u_.), x_Symbol] :> Subst[Int[u*(a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[n] && !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[ IntHide[u*(a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x]]
\[\int \frac {\left (e x \right )^{q}}{a +b \ln \left (c \left (d \,x^{m}\right )^{n}\right )}d x\]
Input:
int((e*x)^q/(a+b*ln(c*(d*x^m)^n)),x)
Output:
int((e*x)^q/(a+b*ln(c*(d*x^m)^n)),x)
Time = 0.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.22 \[ \int \frac {(e x)^q}{a+b \log \left (c \left (d x^m\right )^n\right )} \, dx=\frac {{\rm Ei}\left (\frac {a q + {\left (b q + b\right )} \log \left (c\right ) + {\left (b n q + b n\right )} \log \left (d\right ) + {\left (b m n q + b m n\right )} \log \left (x\right ) + a}{b m n}\right ) e^{\left (\frac {b m n q \log \left (e\right ) - a q - {\left (b q + b\right )} \log \left (c\right ) - {\left (b n q + b n\right )} \log \left (d\right ) - a}{b m n}\right )}}{b m n} \] Input:
integrate((e*x)^q/(a+b*log(c*(d*x^m)^n)),x, algorithm="fricas")
Output:
Ei((a*q + (b*q + b)*log(c) + (b*n*q + b*n)*log(d) + (b*m*n*q + b*m*n)*log( x) + a)/(b*m*n))*e^((b*m*n*q*log(e) - a*q - (b*q + b)*log(c) - (b*n*q + b* n)*log(d) - a)/(b*m*n))/(b*m*n)
\[ \int \frac {(e x)^q}{a+b \log \left (c \left (d x^m\right )^n\right )} \, dx=\int \frac {\left (e x\right )^{q}}{a + b \log {\left (c \left (d x^{m}\right )^{n} \right )}}\, dx \] Input:
integrate((e*x)**q/(a+b*ln(c*(d*x**m)**n)),x)
Output:
Integral((e*x)**q/(a + b*log(c*(d*x**m)**n)), x)
\[ \int \frac {(e x)^q}{a+b \log \left (c \left (d x^m\right )^n\right )} \, dx=\int { \frac {\left (e x\right )^{q}}{b \log \left (\left (d x^{m}\right )^{n} c\right ) + a} \,d x } \] Input:
integrate((e*x)^q/(a+b*log(c*(d*x^m)^n)),x, algorithm="maxima")
Output:
integrate((e*x)^q/(b*log((d*x^m)^n*c) + a), x)
\[ \int \frac {(e x)^q}{a+b \log \left (c \left (d x^m\right )^n\right )} \, dx=\int { \frac {\left (e x\right )^{q}}{b \log \left (\left (d x^{m}\right )^{n} c\right ) + a} \,d x } \] Input:
integrate((e*x)^q/(a+b*log(c*(d*x^m)^n)),x, algorithm="giac")
Output:
integrate((e*x)^q/(b*log((d*x^m)^n*c) + a), x)
Timed out. \[ \int \frac {(e x)^q}{a+b \log \left (c \left (d x^m\right )^n\right )} \, dx=\int \frac {{\left (e\,x\right )}^q}{a+b\,\ln \left (c\,{\left (d\,x^m\right )}^n\right )} \,d x \] Input:
int((e*x)^q/(a + b*log(c*(d*x^m)^n)),x)
Output:
int((e*x)^q/(a + b*log(c*(d*x^m)^n)), x)
\[ \int \frac {(e x)^q}{a+b \log \left (c \left (d x^m\right )^n\right )} \, dx=e^{q} \left (\int \frac {x^{q}}{\mathrm {log}\left (x^{m n} d^{n} c \right ) b +a}d x \right ) \] Input:
int((e*x)^q/(a+b*log(c*(d*x^m)^n)),x)
Output:
e**q*int(x**q/(log(x**(m*n)*d**n*c)*b + a),x)