Integrand size = 32, antiderivative size = 107 \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx=\frac {b e^2 e^{\frac {2 A}{B}} \operatorname {ExpIntegralEi}\left (-\frac {2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{B}\right )}{B (b c-a d)^2 g^3}-\frac {d e e^{A/B} \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{B}\right )}{B (b c-a d)^2 g^3} \]
b*e^2*exp(2*A/B)*Ei(-2*(A+B*ln(e*(b*x+a)/(d*x+c)))/B)/B/(-a*d+b*c)^2/g^3-d *e*exp(A/B)*Ei((-A-B*ln(e*(b*x+a)/(d*x+c)))/B)/B/(-a*d+b*c)^2/g^3
Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx=\frac {e e^{A/B} \left (b e e^{A/B} \operatorname {ExpIntegralEi}\left (-\frac {2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{B}\right )-d \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{B}\right )\right )}{B (b c-a d)^2 g^3} \]
(e*E^(A/B)*(b*e*E^(A/B)*ExpIntegralEi[(-2*(A + B*Log[(e*(a + b*x))/(c + d* x)]))/B] - d*ExpIntegralEi[-((A + B*Log[(e*(a + b*x))/(c + d*x)])/B)]))/(B *(b*c - a*d)^2*g^3)
Time = 0.42 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {2950, 2795, 2009}
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.
| \(\displaystyle \int \frac {1}{(a g+b g x)^3 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )} \, dx\) |
| \(\Big \downarrow \) 2950 |
| \(\displaystyle \frac {\int \frac {(c+d x)^3 \left (b-\frac {d (a+b x)}{c+d x}\right )}{(a+b x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}d\frac {a+b x}{c+d x}}{g^3 (b c-a d)^2}\) |
| \(\Big \downarrow \) 2795 |
| \(\displaystyle \frac {\int \left (\frac {b (c+d x)^3}{(a+b x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}-\frac {d (c+d x)^2}{(a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}\right )d\frac {a+b x}{c+d x}}{g^3 (b c-a d)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b e^2 e^{\frac {2 A}{B}} \operatorname {ExpIntegralEi}\left (-\frac {2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{B}\right )}{B}-\frac {d e e^{A/B} \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{B}\right )}{B}}{g^3 (b c-a d)^2}\) |
((b*e^2*E^((2*A)/B)*ExpIntegralEi[(-2*(A + B*Log[(e*(a + b*x))/(c + d*x)]) )/B])/B - (d*e*E^(A/B)*ExpIntegralEi[-((A + B*Log[(e*(a + b*x))/(c + d*x)] )/B)])/B)/((b*c - a*d)^2*g^3)
3.2.13.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[ c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b , c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0 ] && IntegerQ[m] && IntegerQ[r]))
Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ )]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(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] & & EqQ[n + mn, 0] && IGtQ[n, 0] && NeQ[b*c - a*d, 0] && IntegersQ[m, p] && E qQ[b*f - a*g, 0] && (GtQ[p, 0] || LtQ[m, -1])
Time = 2.81 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.09
| method | result | size |
| derivativedivides | \(\frac {e \left (\frac {d \,{\mathrm e}^{\frac {A}{B}} \operatorname {Ei}_{1}\left (\ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )+\frac {A}{B}\right )}{B}-\frac {b e \,{\mathrm e}^{\frac {2 A}{B}} \operatorname {Ei}_{1}\left (2 \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )+\frac {2 A}{B}\right )}{B}\right )}{\left (a d -c b \right )^{2} g^{3}}\) | \(117\) |
| default | \(\frac {e \left (\frac {d \,{\mathrm e}^{\frac {A}{B}} \operatorname {Ei}_{1}\left (\ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )+\frac {A}{B}\right )}{B}-\frac {b e \,{\mathrm e}^{\frac {2 A}{B}} \operatorname {Ei}_{1}\left (2 \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )+\frac {2 A}{B}\right )}{B}\right )}{\left (a d -c b \right )^{2} g^{3}}\) | \(117\) |
| risch | \(-\frac {e^{2} b \,{\mathrm e}^{\frac {2 A}{B}} \operatorname {Ei}_{1}\left (2 \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )+\frac {2 A}{B}\right )}{\left (a d -c b \right )^{2} g^{3} B}+\frac {e d \,{\mathrm e}^{\frac {A}{B}} \operatorname {Ei}_{1}\left (\ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )+\frac {A}{B}\right )}{\left (a d -c b \right )^{2} g^{3} B}\) | \(131\) |
e/(a*d-b*c)^2/g^3*(d/B*exp(A/B)*Ei(1,ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))+A/B)- b*e/B*exp(2*A/B)*Ei(1,2*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))+2*A/B))
Time = 0.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx=\frac {b e^{2} e^{\left (\frac {2 \, A}{B}\right )} \operatorname {log\_integral}\left (\frac {{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} e^{\left (-\frac {2 \, A}{B}\right )}}{b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}}\right ) - d e e^{\frac {A}{B}} \operatorname {log\_integral}\left (\frac {{\left (d x + c\right )} e^{\left (-\frac {A}{B}\right )}}{b e x + a e}\right )}{{\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} g^{3}} \]
(b*e^2*e^(2*A/B)*log_integral((d^2*x^2 + 2*c*d*x + c^2)*e^(-2*A/B)/(b^2*e^ 2*x^2 + 2*a*b*e^2*x + a^2*e^2)) - d*e*e^(A/B)*log_integral((d*x + c)*e^(-A /B)/(b*e*x + a*e)))/((B*b^2*c^2 - 2*B*a*b*c*d + B*a^2*d^2)*g^3)
\[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx=\frac {\int \frac {1}{A a^{3} + 3 A a^{2} b x + 3 A a b^{2} x^{2} + A b^{3} x^{3} + B a^{3} \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )} + 3 B a^{2} b x \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )} + 3 B a b^{2} x^{2} \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )} + B b^{3} x^{3} \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx}{g^{3}} \]
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(a*e/(c + d*x) + b*e*x/(c + d*x)) + 3*B*a**2*b*x*log(a*e/(c + d*x) + b *e*x/(c + d*x)) + 3*B*a*b**2*x**2*log(a*e/(c + d*x) + b*e*x/(c + d*x)) + B *b**3*x**3*log(a*e/(c + d*x) + b*e*x/(c + d*x))), x)/g**3
\[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{3} {\left (B \log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right ) + A\right )}} \,d x } \]
\[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{3} {\left (B \log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right ) + A\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx=\int \frac {1}{{\left (a\,g+b\,g\,x\right )}^3\,\left (A+B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\right )} \,d x \]