Integrand size = 35, antiderivative size = 197 \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {b e^{\frac {2 A}{B n}} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{2/n} (c+d x)^2 \operatorname {ExpIntegralEi}\left (-\frac {2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )}{B (b c-a d)^2 g^3 n (a+b x)^2}-\frac {d e^{\frac {A}{B n}} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} (c+d x) \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B (b c-a d)^2 g^3 n (a+b x)} \]
b*exp(2*A/B/n)*(e*((b*x+a)/(d*x+c))^n)^(2/n)*(d*x+c)^2*Ei(-2*(A+B*ln(e*((b *x+a)/(d*x+c))^n))/B/n)/B/(-a*d+b*c)^2/g^3/n/(b*x+a)^2-d*exp(A/B/n)*(e*((b *x+a)/(d*x+c))^n)^(1/n)*(d*x+c)*Ei((-A-B*ln(e*((b*x+a)/(d*x+c))^n))/B/n)/B /(-a*d+b*c)^2/g^3/n/(b*x+a)
Time = 0.28 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\frac {e^{\frac {A}{B n}} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} (c+d x) \left (b e^{\frac {A}{B n}} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} (c+d x) \operatorname {ExpIntegralEi}\left (-\frac {2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )-d (a+b x) \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )\right )}{B (b c-a d)^2 g^3 n (a+b x)^2} \]
(E^(A/(B*n))*(e*((a + b*x)/(c + d*x))^n)^n^(-1)*(c + d*x)*(b*E^(A/(B*n))*( e*((a + b*x)/(c + d*x))^n)^n^(-1)*(c + d*x)*ExpIntegralEi[(-2*(A + B*Log[e *((a + b*x)/(c + d*x))^n]))/(B*n)] - d*(a + b*x)*ExpIntegralEi[-((A + B*Lo g[e*((a + b*x)/(c + d*x))^n])/(B*n))]))/(B*(b*c - a*d)^2*g^3*n*(a + b*x)^2 )
Time = 0.46 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {2949, 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.
| \(\displaystyle \int \frac {1}{(a g+b g x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )} \, dx\) |
| \(\Big \downarrow \) 2949 |
| \(\displaystyle \frac {\int \frac {(c+d x)^3 \left (b-\frac {d (a+b x)}{c+d x}\right )}{(a+b x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}d\frac {a+b x}{c+d x}}{g^3 (b c-a d)^2}\) |
| \(\Big \downarrow \) 2795 |
| \(\displaystyle \frac {\int \left (\frac {b (c+d x)^3}{(a+b x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}-\frac {d (c+d x)^2}{(a+b x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}\right )d\frac {a+b x}{c+d x}}{g^3 (b c-a d)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b e^{\frac {2 A}{B n}} (c+d x)^2 \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{2/n} \operatorname {ExpIntegralEi}\left (-\frac {2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )}{B n (a+b x)^2}-\frac {d e^{\frac {A}{B n}} (c+d x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B n (a+b x)}}{g^3 (b c-a d)^2}\) |
((b*E^((2*A)/(B*n))*(e*((a + b*x)/(c + d*x))^n)^(2/n)*(c + d*x)^2*ExpInteg ralEi[(-2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(B*n)])/(B*n*(a + b*x)^2 ) - (d*E^(A/(B*n))*(e*((a + b*x)/(c + d*x))^n)^n^(-1)*(c + d*x)*ExpIntegra lEi[-((A + B*Log[e*((a + b*x)/(c + d*x))^n])/(B*n))])/(B*n*(a + b*x)))/((b *c - a*d)^2*g^3)
3.1.23.3.1 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_))/((c_.) + (d_.)*(x_)))^(n_.)]*( 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] && Ne Q[b*c - a*d, 0] && IntegersQ[m, p] && EqQ[b*f - a*g, 0] && (GtQ[p, 0] || Lt Q[m, -1])
\[\int \frac {1}{\left (b g x +a g \right )^{3} \left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}d x\]
Time = 0.28 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=-\frac {d e^{\left (\frac {B \log \left (e\right ) + A}{B n}\right )} \operatorname {log\_integral}\left (\frac {{\left (d x + c\right )} e^{\left (-\frac {B \log \left (e\right ) + A}{B n}\right )}}{b x + a}\right ) - b e^{\left (\frac {2 \, {\left (B \log \left (e\right ) + A\right )}}{B n}\right )} \operatorname {log\_integral}\left (\frac {{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} e^{\left (-\frac {2 \, {\left (B \log \left (e\right ) + A\right )}}{B n}\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}{{\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} g^{3} n} \]
-(d*e^((B*log(e) + A)/(B*n))*log_integral((d*x + c)*e^(-(B*log(e) + A)/(B* n))/(b*x + a)) - b*e^(2*(B*log(e) + A)/(B*n))*log_integral((d^2*x^2 + 2*c* d*x + c^2)*e^(-2*(B*log(e) + A)/(B*n))/(b^2*x^2 + 2*a*b*x + a^2)))/((B*b^2 *c^2 - 2*B*a*b*c*d + B*a^2*d^2)*g^3*n)
Timed out. \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{3} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}} \,d x } \]
\[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{3} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )} \, dx=\int \frac {1}{{\left (a\,g+b\,g\,x\right )}^3\,\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )} \,d x \]