\(\int \frac {1}{(a g+b g x)^3 (A+B \log (\frac {e (c+d x)}{a+b x}))} \, dx\) [195]

Optimal result

Integrand size = 32, antiderivative size = 109 \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\frac {d e^{-\frac {A}{B}} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{B}\right )}{B (b c-a d)^2 e g^3}-\frac {b e^{-\frac {2 A}{B}} \operatorname {ExpIntegralEi}\left (\frac {2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{B}\right )}{B (b c-a d)^2 e^2 g^3} \]

[Out]

d*Ei((A+B*ln(e*(d*x+c)/(b*x+a)))/B)/B/(-a*d+b*c)^2/e/exp(A/B)/g^3-b*Ei(2*(A+B*ln(e*(d*x+c)/(b*x+a)))/B)/B/(-a*
d+b*c)^2/e^2/exp(2*A/B)/g^3

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2552, 2367, 2336, 2209, 2346} \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\frac {d e^{-\frac {A}{B}} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{B}\right )}{B e g^3 (b c-a d)^2}-\frac {b e^{-\frac {2 A}{B}} \operatorname {ExpIntegralEi}\left (\frac {2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{B}\right )}{B e^2 g^3 (b c-a d)^2} \]

[In]

Int[1/((a*g + b*g*x)^3*(A + B*Log[(e*(c + d*x))/(a + b*x)])),x]

[Out]

(d*ExpIntegralEi[(A + B*Log[(e*(c + d*x))/(a + b*x)])/B])/(B*(b*c - a*d)^2*e*E^(A/B)*g^3) - (b*ExpIntegralEi[(
2*(A + B*Log[(e*(c + d*x))/(a + b*x)]))/B])/(B*(b*c - a*d)^2*e^2*E^((2*A)/B)*g^3)

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 2336

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

Rule 2346

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

Rule 2367

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

Rule 2552

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(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] && EqQ[n + mn, 0] && IGtQ[n, 0] && 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 {d-b x}{A+B \log (e x)} \, dx,x,\frac {c+d x}{a+b x}\right )}{(b c-a d)^2 g^3} \\ & = \frac {\text {Subst}\left (\int \left (\frac {d}{A+B \log (e x)}-\frac {b x}{A+B \log (e x)}\right ) \, dx,x,\frac {c+d x}{a+b x}\right )}{(b c-a d)^2 g^3} \\ & = -\frac {b \text {Subst}\left (\int \frac {x}{A+B \log (e x)} \, dx,x,\frac {c+d x}{a+b x}\right )}{(b c-a d)^2 g^3}+\frac {d \text {Subst}\left (\int \frac {1}{A+B \log (e x)} \, dx,x,\frac {c+d x}{a+b x}\right )}{(b c-a d)^2 g^3} \\ & = -\frac {b \text {Subst}\left (\int \frac {e^{2 x}}{A+B x} \, dx,x,\log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{(b c-a d)^2 e^2 g^3}+\frac {d \text {Subst}\left (\int \frac {e^x}{A+B x} \, dx,x,\log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{(b c-a d)^2 e g^3} \\ & = \frac {d e^{-\frac {A}{B}} \text {Ei}\left (\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{B}\right )}{B (b c-a d)^2 e g^3}-\frac {b e^{-\frac {2 A}{B}} \text {Ei}\left (\frac {2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{B}\right )}{B (b c-a d)^2 e^2 g^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.82 \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\frac {e^{-\frac {2 A}{B}} \left (d e e^{A/B} \operatorname {ExpIntegralEi}\left (\frac {A}{B}+\log \left (\frac {e (c+d x)}{a+b x}\right )\right )-b \operatorname {ExpIntegralEi}\left (\frac {2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{B}\right )\right )}{B (b c-a d)^2 e^2 g^3} \]

[In]

Integrate[1/((a*g + b*g*x)^3*(A + B*Log[(e*(c + d*x))/(a + b*x)])),x]

[Out]

(d*e*E^(A/B)*ExpIntegralEi[A/B + Log[(e*(c + d*x))/(a + b*x)]] - b*ExpIntegralEi[(2*(A + B*Log[(e*(c + d*x))/(
a + b*x)]))/B])/(B*(b*c - a*d)^2*e^2*E^((2*A)/B)*g^3)

Maple [A] (verified)

Time = 3.36 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.16

method	result	size
derivativedivides	\(-\frac {-\frac {b \,{\mathrm e}^{-\frac {2 A}{B}} \operatorname {Ei}_{1}\left (-2 \ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {2 A}{B}\right )}{B}+\frac {d e \,{\mathrm e}^{-\frac {A}{B}} \operatorname {Ei}_{1}\left (-\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{B}}{e^{2} \left (a d -c b \right )^{2} g^{3}}\)	\(126\)
default	\(-\frac {-\frac {b \,{\mathrm e}^{-\frac {2 A}{B}} \operatorname {Ei}_{1}\left (-2 \ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {2 A}{B}\right )}{B}+\frac {d e \,{\mathrm e}^{-\frac {A}{B}} \operatorname {Ei}_{1}\left (-\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{B}}{e^{2} \left (a d -c b \right )^{2} g^{3}}\)	\(126\)
risch	\(\frac {b \,{\mathrm e}^{-\frac {2 A}{B}} \operatorname {Ei}_{1}\left (-2 \ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {2 A}{B}\right )}{g^{3} \left (a d -c b \right )^{2} e^{2} B}-\frac {d \,{\mathrm e}^{-\frac {A}{B}} \operatorname {Ei}_{1}\left (-\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{g^{3} \left (a d -c b \right )^{2} e B}\)	\(139\)

[In]

int(1/(b*g*x+a*g)^3/(A+B*ln(e*(d*x+c)/(b*x+a))),x,method=_RETURNVERBOSE)

[Out]

-1/e^2/(a*d-b*c)^2/g^3*(-b/B*exp(-2*A/B)*Ei(1,-2*ln(d*e/b-e*(a*d-b*c)/b/(b*x+a))-2*A/B)+d*e/B*exp(-A/B)*Ei(1,-
ln(d*e/b-e*(a*d-b*c)/b/(b*x+a))-A/B))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.18 \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\frac {{\left (d e e^{\frac {A}{B}} \operatorname {log\_integral}\left (\frac {{\left (d e x + c e\right )} e^{\frac {A}{B}}}{b x + a}\right ) - b \operatorname {log\_integral}\left (\frac {{\left (d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}\right )} e^{\left (\frac {2 \, A}{B}\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )\right )} e^{\left (-\frac {2 \, A}{B}\right )}}{{\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} e^{2} g^{3}} \]

[In]

integrate(1/(b*g*x+a*g)^3/(A+B*log(e*(d*x+c)/(b*x+a))),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\frac {\int \frac {1}{A a^{3} + 3 A a^{2} b x + 3 A a b^{2} x^{2} + A b^{3} x^{3} + B a^{3} \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )} + 3 B a^{2} b x \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )} + 3 B a b^{2} x^{2} \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )} + B b^{3} x^{3} \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )}}\, dx}{g^{3}} \]

[In]

integrate(1/(b*g*x+a*g)**3/(A+B*ln(e*(d*x+c)/(b*x+a))),x)

[Out]

Integral(1/(A*a**3 + 3*A*a**2*b*x + 3*A*a*b**2*x**2 + A*b**3*x**3 + B*a**3*log(c*e/(a + b*x) + d*e*x/(a + b*x)
) + 3*B*a**2*b*x*log(c*e/(a + b*x) + d*e*x/(a + b*x)) + 3*B*a*b**2*x**2*log(c*e/(a + b*x) + d*e*x/(a + b*x)) +
B*b**3*x**3*log(c*e/(a + b*x) + d*e*x/(a + b*x))), x)/g**3

Maxima [F]

\[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{3} {\left (B \log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) + A\right )}} \,d x } \]

[In]

integrate(1/(b*g*x+a*g)^3/(A+B*log(e*(d*x+c)/(b*x+a))),x, algorithm="maxima")

[Out]

integrate(1/((b*g*x + a*g)^3*(B*log((d*x + c)*e/(b*x + a)) + A)), x)

Giac [F]

\[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{3} {\left (B \log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) + A\right )}} \,d x } \]

[In]

integrate(1/(b*g*x+a*g)^3/(A+B*log(e*(d*x+c)/(b*x+a))),x, algorithm="giac")

[Out]

integrate(1/((b*g*x + a*g)^3*(B*log((d*x + c)*e/(b*x + a)) + A)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\int \frac {1}{{\left (a\,g+b\,g\,x\right )}^3\,\left (A+B\,\ln \left (\frac {e\,\left (c+d\,x\right )}{a+b\,x}\right )\right )} \,d x \]

[In]

int(1/((a*g + b*g*x)^3*(A + B*log((e*(c + d*x))/(a + b*x)))),x)

[Out]

int(1/((a*g + b*g*x)^3*(A + B*log((e*(c + d*x))/(a + b*x)))), x)