\(\int \frac {1}{(c g+d g x)^2 (A+B \log (e (\frac {a+b x}{c+d x})^n))^2} \, dx\) [55]

Optimal result

Integrand size = 35, antiderivative size = 154 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=\frac {e^{-\frac {A}{B n}} (a+b x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B^2 (b c-a d) g^2 n^2 (c+d x)}-\frac {a+b x}{B (b c-a d) g^2 n (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {2551, 2334, 2337, 2209} \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=\frac {(a+b x) e^{-\frac {A}{B n}} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B^2 g^2 n^2 (c+d x) (b c-a d)}-\frac {a+b x}{B g^2 n (c+d x) (b c-a d) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )} \]

[In]

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

[Out]

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

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 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b*Log[c*x^n])^(p + 1)/(b*n*(p + 1)))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

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}{\left (A+B \log \left (e x^n\right )\right )^2} \, dx,x,\frac {a+b x}{c+d x}\right )}{(b c-a d) g^2} \\ & = -\frac {a+b x}{B (b c-a d) g^2 n (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}+\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 (b c-a d) g^2 n} \\ & = -\frac {a+b x}{B (b c-a d) g^2 n (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}+\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 (b c-a d) g^2 n^2 (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^2 (b c-a d) g^2 n^2 (c+d x)}-\frac {a+b x}{B (b c-a d) g^2 n (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.17 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=-\frac {e^{-\frac {A}{B n}} (a+b x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-1/n} \left (B e^{\frac {A}{B n}} n \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}}-\operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right ) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )\right )}{B^2 (b c-a d) g^2 n^2 (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.89 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=-\frac {{\left ({\left (B b n x + B a n\right )} e^{\left (\frac {B \log \left (e\right ) + A}{B n}\right )} - {\left (A d x + A c + {\left (B d x + B c\right )} \log \left (e\right ) + {\left (B d n x + B c n\right )} \log \left (\frac {b x + a}{d x + c}\right )\right )} \operatorname {log\_integral}\left (\frac {{\left (b x + a\right )} e^{\left (\frac {B \log \left (e\right ) + A}{B n}\right )}}{d x + c}\right )\right )} e^{\left (-\frac {B \log \left (e\right ) + A}{B n}\right )}}{{\left (A B^{2} b c d - A B^{2} a d^{2}\right )} g^{2} n^{2} x + {\left (A B^{2} b c^{2} - A B^{2} a c d\right )} g^{2} n^{2} + {\left ({\left (B^{3} b c d - B^{3} a d^{2}\right )} g^{2} n^{2} x + {\left (B^{3} b c^{2} - B^{3} a c d\right )} g^{2} n^{2}\right )} \log \left (e\right ) + {\left ({\left (B^{3} b c d - B^{3} a d^{2}\right )} g^{2} n^{3} x + {\left (B^{3} b c^{2} - B^{3} a c d\right )} g^{2} n^{3}\right )} \log \left (\frac {b x + a}{d x + c}\right )} \]

[In]

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

[Out]

-((B*b*n*x + B*a*n)*e^((B*log(e) + A)/(B*n)) - (A*d*x + A*c + (B*d*x + B*c)*log(e) + (B*d*n*x + B*c*n)*log((b*
x + a)/(d*x + c)))*log_integral((b*x + a)*e^((B*log(e) + A)/(B*n))/(d*x + c)))*e^(-(B*log(e) + A)/(B*n))/((A*B
^2*b*c*d - A*B^2*a*d^2)*g^2*n^2*x + (A*B^2*b*c^2 - A*B^2*a*c*d)*g^2*n^2 + ((B^3*b*c*d - B^3*a*d^2)*g^2*n^2*x +
(B^3*b*c^2 - B^3*a*c*d)*g^2*n^2)*log(e) + ((B^3*b*c*d - B^3*a*d^2)*g^2*n^3*x + (B^3*b*c^2 - B^3*a*c*d)*g^2*n^
3)*log((b*x + a)/(d*x + c)))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2} \, dx=-{\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} {\left (\frac {b x + a}{{\left (B^{2} g^{2} n^{2} \log \left (\frac {b x + a}{d x + c}\right ) + B^{2} g^{2} n \log \left (e\right ) + A B g^{2} n\right )} {\left (d x + c\right )}} - \frac {{\rm Ei}\left (\frac {\log \left (e\right )}{n} + \frac {A}{B n} + \log \left (\frac {b x + a}{d x + c}\right )\right ) e^{\left (-\frac {A}{B n}\right )}}{B^{2} e^{\left (\frac {1}{n}\right )} g^{2} n^{2}}\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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