\(\int \frac {(e x)^q}{(a+b \log (c (d x^m)^n))^2} \, dx\) [247]

Optimal result

Integrand size = 22, antiderivative size = 127 \[ \int \frac {(e x)^q}{\left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2} \, dx=\frac {e^{-\frac {a (1+q)}{b m n}} (1+q) (e x)^{1+q} \left (c \left (d x^m\right )^n\right )^{-\frac {1+q}{m n}} \operatorname {ExpIntegralEi}\left (\frac {(1+q) \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{b m n}\right )}{b^2 e m^2 n^2}-\frac {(e x)^{1+q}}{b e m n \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )} \] Output:

(1+q)*(e*x)^(1+q)*Ei((1+q)*(a+b*ln(c*(d*x^m)^n))/b/m/n)/b^2/e/exp(a*(1+q)/ 
b/m/n)/m^2/n^2/((c*(d*x^m)^n)^((1+q)/m/n))-(e*x)^(1+q)/b/e/m/n/(a+b*ln(c*( 
d*x^m)^n))
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.88 \[ \int \frac {(e x)^q}{\left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2} \, dx=\frac {(e x)^q \left (e^{-\frac {(1+q) \left (a-b m n \log (x)+b \log \left (c \left (d x^m\right )^n\right )\right )}{b m n}} (1+q) x^{-q} \operatorname {ExpIntegralEi}\left (\frac {(1+q) \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{b m n}\right )-\frac {b m n x}{a+b \log \left (c \left (d x^m\right )^n\right )}\right )}{b^2 m^2 n^2} \] Input:

Integrate[(e*x)^q/(a + b*Log[c*(d*x^m)^n])^2,x]
 

Output:

((e*x)^q*(((1 + q)*ExpIntegralEi[((1 + q)*(a + b*Log[c*(d*x^m)^n]))/(b*m*n 
)])/(E^(((1 + q)*(a - b*m*n*Log[x] + b*Log[c*(d*x^m)^n]))/(b*m*n))*x^q) - 
(b*m*n*x)/(a + b*Log[c*(d*x^m)^n])))/(b^2*m^2*n^2)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2895, 2743, 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.

\(\displaystyle \int \frac {(e x)^q}{\left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2} \, dx\)

\(\Big \downarrow \) 2895

\(\displaystyle \int \frac {(e x)^q}{\left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2}dx\)

\(\Big \downarrow \) 2743

\(\displaystyle \frac {(q+1) \int \frac {(e x)^q}{a+b \log \left (c \left (d x^m\right )^n\right )}dx}{b m n}-\frac {(e x)^{q+1}}{b e m n \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}\)

\(\Big \downarrow \) 2747

\(\displaystyle \frac {(q+1) (e x)^{q+1} \left (c \left (d x^m\right )^n\right )^{-\frac {q+1}{m n}} \int \frac {\left (c \left (d x^m\right )^n\right )^{\frac {q+1}{m n}}}{a+b \log \left (c \left (d x^m\right )^n\right )}d\log \left (c \left (d x^m\right )^n\right )}{b e m^2 n^2}-\frac {(e x)^{q+1}}{b e m n \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}\)

\(\Big \downarrow \) 2609

\(\displaystyle \frac {(q+1) (e x)^{q+1} e^{-\frac {a (q+1)}{b m n}} \left (c \left (d x^m\right )^n\right )^{-\frac {q+1}{m n}} \operatorname {ExpIntegralEi}\left (\frac {(q+1) \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{b m n}\right )}{b^2 e m^2 n^2}-\frac {(e x)^{q+1}}{b e m n \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}\)

Input:

Int[(e*x)^q/(a + b*Log[c*(d*x^m)^n])^2,x]
 

Output:

((1 + q)*(e*x)^(1 + q)*ExpIntegralEi[((1 + q)*(a + b*Log[c*(d*x^m)^n]))/(b 
*m*n)])/(b^2*e*E^((a*(1 + q))/(b*m*n))*m^2*n^2*(c*(d*x^m)^n)^((1 + q)/(m*n 
))) - (e*x)^(1 + q)/(b*e*m*n*(a + b*Log[c*(d*x^m)^n]))
 

Defintions of rubi rules used

rule 2609

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]
 

rule 2743

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

rule 2747

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]
 

rule 2895

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]]
 

Maple [F]

\[\int \frac {\left (e x \right )^{q}}{{\left (a +b \ln \left (c \left (d \,x^{m}\right )^{n}\right )\right )}^{2}}d x\]

Input:

int((e*x)^q/(a+b*ln(c*(d*x^m)^n))^2,x)
 

Output:

int((e*x)^q/(a+b*ln(c*(d*x^m)^n))^2,x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.59 \[ \int \frac {(e x)^q}{\left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2} \, dx=-\frac {b m n x e^{\left (q \log \left (e\right ) + q \log \left (x\right )\right )} - {\left (a q + {\left (b q + b\right )} \log \left (c\right ) + {\left (b n q + b n\right )} \log \left (d\right ) + {\left (b m n q + b m n\right )} \log \left (x\right ) + a\right )} {\rm Ei}\left (\frac {a q + {\left (b q + b\right )} \log \left (c\right ) + {\left (b n q + b n\right )} \log \left (d\right ) + {\left (b m n q + b m n\right )} \log \left (x\right ) + a}{b m n}\right ) e^{\left (\frac {b m n q \log \left (e\right ) - a q - {\left (b q + b\right )} \log \left (c\right ) - {\left (b n q + b n\right )} \log \left (d\right ) - a}{b m n}\right )}}{b^{3} m^{3} n^{3} \log \left (x\right ) + b^{3} m^{2} n^{3} \log \left (d\right ) + b^{3} m^{2} n^{2} \log \left (c\right ) + a b^{2} m^{2} n^{2}} \] Input:

integrate((e*x)^q/(a+b*log(c*(d*x^m)^n))^2,x, algorithm="fricas")
 

Output:

-(b*m*n*x*e^(q*log(e) + q*log(x)) - (a*q + (b*q + b)*log(c) + (b*n*q + b*n 
)*log(d) + (b*m*n*q + b*m*n)*log(x) + a)*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^3*m^3 
*n^3*log(x) + b^3*m^2*n^3*log(d) + b^3*m^2*n^2*log(c) + a*b^2*m^2*n^2)
 

Sympy [F]

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

integrate((e*x)**q/(a+b*ln(c*(d*x**m)**n))**2,x)
 

Output:

Integral((e*x)**q/(a + b*log(c*(d*x**m)**n))**2, x)
 

Maxima [F]

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

integrate((e*x)^q/(a+b*log(c*(d*x^m)^n))^2,x, algorithm="maxima")
 

Output:

e^q*(q + 1)*integrate(x^q/(b^2*m*n*log((x^m)^n) + a*b*m*n + (m*n^2*log(d) 
+ m*n*log(c))*b^2), x) - e^q*x*x^q/(b^2*m*n*log((x^m)^n) + a*b*m*n + (m*n^ 
2*log(d) + m*n*log(c))*b^2)
 

Giac [F]

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

integrate((e*x)^q/(a+b*log(c*(d*x^m)^n))^2,x, algorithm="giac")
 

Output:

integrate((e*x)^q/(b*log((d*x^m)^n*c) + a)^2, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

int((e*x)^q/(a + b*log(c*(d*x^m)^n))^2,x)
 

Output:

int((e*x)^q/(a + b*log(c*(d*x^m)^n))^2, x)
 

Reduce [F]

\[ \int \frac {(e x)^q}{\left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2} \, dx=e^{q} \left (\int \frac {x^{q}}{\mathrm {log}\left (x^{m n} d^{n} c \right )^{2} b^{2}+2 \,\mathrm {log}\left (x^{m n} d^{n} c \right ) a b +a^{2}}d x \right ) \] Input:

int((e*x)^q/(a+b*log(c*(d*x^m)^n))^2,x)
 

Output:

e**q*int(x**q/(log(x**(m*n)*d**n*c)**2*b**2 + 2*log(x**(m*n)*d**n*c)*a*b + 
 a**2),x)