\(\int \frac {1}{(a g+b g x)^2 (A+B \log (e (a+b x)^n (c+d x)^{-n}))} \, dx\) [172]

Optimal result

Integrand size = 36, antiderivative size = 96 \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {e^{\frac {A}{B n}} (c+d x) \left (e (a+b x)^n (c+d x)^{-n}\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{B n}\right )}{B (b c-a d) g^2 n (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2573, 2549, 2347, 2209} \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {e^{\frac {A}{B n}} (c+d x) \left (e (a+b x)^n (c+d x)^{-n}\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{B n}\right )}{B g^2 n (a+b x) (b c-a d)} \]

[In]

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

[Out]

(E^(A/(B*n))*(c + d*x)*((e*(a + b*x)^n)/(c + d*x)^n)^n^(-1)*ExpIntegralEi[-((A + B*Log[(e*(a + b*x)^n)/(c + d*
x)^n])/(B*n))])/(B*(b*c - a*d)*g^2*n*(a + b*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 2347

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(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 2549

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/b)^m, Subst[Int[x^m*((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[b*f - a*g, 0] && (GtQ[p, 0] || LtQ[m, -1])

Rule 2573

Int[((A_.) + Log[(e_.)*(u_)^(n_.)*(v_)^(mn_)]*(B_.))^(p_.)*(w_.), x_Symbol] :> Subst[Int[w*(A + B*Log[e*(u/v)^
n])^p, x], e*(u/v)^n, e*(u^n/v^n)] /; FreeQ[{e, A, B, n, p}, x] && EqQ[n + mn, 0] && LinearQ[{u, v}, x] &&  !I
ntegerQ[n]

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

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.65 \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \, dx=\frac {e^{\left (\frac {B \log \left (e\right ) + A}{B n}\right )} \operatorname {log\_integral}\left (\frac {{\left (d x + c\right )} e^{\left (-\frac {B \log \left (e\right ) + A}{B n}\right )}}{b x + a}\right )}{{\left (B b c - B a d\right )} g^{2} n} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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