\(\int \frac {1}{(c g+d g x)^2 (A+B \log (e (\frac {a+b x}{c+d x})^n))} \, dx\) [50]

Optimal result

Integrand size = 35, antiderivative size = 96 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {e^{-\frac {A}{B n}} (a+b x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B (b c-a d) g^2 n (c+d x)} \]

[Out]

(b*x+a)*Ei((A+B*ln(e*((b*x+a)/(d*x+c))^n))/B/n)/B/(-a*d+b*c)/exp(A/B/n)/g^2/n/((e*((b*x+a)/(d*x+c))^n)^(1/n))/
(d*x+c)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {2551, 2337, 2209} \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {(a+b x) e^{-\frac {A}{B n}} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B g^2 n (c+d x) (b c-a d)} \]

[In]

Int[1/((c*g + d*g*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])),x]

[Out]

((a + b*x)*ExpIntegralEi[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(B*n)])/(B*(b*c - a*d)*E^(A/(B*n))*g^2*n*(e*((
a + b*x)/(c + d*x))^n)^n^(-1)*(c + d*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 2337

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

Rule 2551

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.), x_Symbol] :> Dist[(b*c - a*d)^(m + 1)*(g/d)^m, Subst[Int[(A + B*Log[e*x^n])^p/(b - d*x)^(m + 2), x], x, (
a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && NeQ[b*c - a*d, 0] && IntegersQ[m, p] &&
EqQ[d*f - c*g, 0] && (GtQ[p, 0] || LtQ[m, -1])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{A+B \log \left (e x^n\right )} \, dx,x,\frac {a+b x}{c+d x}\right )}{(b c-a d) g^2} \\ & = \frac {\left ((a+b x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n}\right ) \text {Subst}\left (\int \frac {e^{\frac {x}{n}}}{A+B x} \, dx,x,\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) g^2 n (c+d x)} \\ & = \frac {e^{-\frac {A}{B n}} (a+b x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \text {Ei}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B (b c-a d) g^2 n (c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {e^{-\frac {A}{B n}} (a+b x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B (b c-a d) g^2 n (c+d x)} \]

[In]

Integrate[1/((c*g + d*g*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])),x]

[Out]

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

Maple [F]

\[\int \frac {1}{\left (d g x +c g \right )^{2} \left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}d x\]

[In]

int(1/(d*g*x+c*g)^2/(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)

[Out]

int(1/(d*g*x+c*g)^2/(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.65 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {e^{\left (-\frac {B \log \left (e\right ) + A}{B n}\right )} \operatorname {log\_integral}\left (\frac {{\left (b x + a\right )} e^{\left (\frac {B \log \left (e\right ) + A}{B n}\right )}}{d x + c}\right )}{{\left (B b c - B a d\right )} g^{2} n} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(1/(d*g*x+c*g)**2/(A+B*ln(e*((b*x+a)/(d*x+c))**n)),x)

[Out]

Integral(1/(A*c**2 + 2*A*c*d*x + A*d**2*x**2 + B*c**2*log(e*(a/(c + d*x) + b*x/(c + d*x))**n) + 2*B*c*d*x*log(
e*(a/(c + d*x) + b*x/(c + d*x))**n) + B*d**2*x**2*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)), x)/g**2

Maxima [F]

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

[In]

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

[Out]

integrate(1/((d*g*x + c*g)^2*(B*log(e*((b*x + a)/(d*x + c))^n) + A)), x)

Giac [F]

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

[In]

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

[Out]

integrate(1/((d*g*x + c*g)^2*(B*log(e*((b*x + a)/(d*x + c))^n) + A)), x)

Mupad [F(-1)]

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

[In]

int(1/((c*g + d*g*x)^2*(A + B*log(e*((a + b*x)/(c + d*x))^n))),x)

[Out]

int(1/((c*g + d*g*x)^2*(A + B*log(e*((a + b*x)/(c + d*x))^n))), x)