Integrand size = 32, antiderivative size = 109 \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\frac {d e^{-\frac {A}{B}} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{B}\right )}{B (b c-a d)^2 e g^3}-\frac {b e^{-\frac {2 A}{B}} \operatorname {ExpIntegralEi}\left (\frac {2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{B}\right )}{B (b c-a d)^2 e^2 g^3} \] Output:
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
Time = 0.34 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.82 \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\frac {e^{-\frac {2 A}{B}} \left (d e e^{A/B} \operatorname {ExpIntegralEi}\left (\frac {A}{B}+\log \left (\frac {e (c+d x)}{a+b x}\right )\right )-b \operatorname {ExpIntegralEi}\left (\frac {2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{B}\right )\right )}{B (b c-a d)^2 e^2 g^3} \] Input:
Integrate[1/((a*g + b*g*x)^3*(A + B*Log[(e*(c + d*x))/(a + b*x)])),x]
Output:
(d*e*E^(A/B)*ExpIntegralEi[A/B + Log[(e*(c + d*x))/(a + b*x)]] - b*ExpInte gralEi[(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)
Time = 0.39 (sec) , antiderivative size = 97, 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 = {2952, 2767, 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.
| \(\displaystyle \int \frac {1}{(a g+b g x)^3 \left (B \log \left (\frac {e (c+d x)}{a+b x}\right )+A\right )} \, dx\) |
| \(\Big \downarrow \) 2952 |
| \(\displaystyle \frac {\int \frac {d-\frac {b (c+d x)}{a+b x}}{A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}d\frac {c+d x}{a+b x}}{g^3 (b c-a d)^2}\) |
| \(\Big \downarrow \) 2767 |
| \(\displaystyle \frac {\int \left (\frac {d}{A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}-\frac {b (c+d x)}{(a+b x) \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}\right )d\frac {c+d x}{a+b x}}{g^3 (b c-a d)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {d e^{-\frac {A}{B}} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{B}\right )}{B e}-\frac {b e^{-\frac {2 A}{B}} \operatorname {ExpIntegralEi}\left (\frac {2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{B}\right )}{B e^2}}{g^3 (b c-a d)^2}\) |
Input:
Int[1/((a*g + b*g*x)^3*(A + B*Log[(e*(c + d*x))/(a + b*x)])),x]
Output:
((d*ExpIntegralEi[(A + B*Log[(e*(c + d*x))/(a + b*x)])/B])/(B*e*E^(A/B)) - (b*ExpIntegralEi[(2*(A + B*Log[(e*(c + d*x))/(a + b*x)]))/B])/(B*e^2*E^(( 2*A)/B)))/((b*c - a*d)^2*g^3)
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^( q_.), x_Symbol] :> With[{u = ExpandIntegrand[(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]))
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/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])
Time = 4.30 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.16
| method | result | size |
| derivativedivides | \(-\frac {-\frac {b \,{\mathrm e}^{-\frac {2 A}{B}} \operatorname {expIntegral}_{1}\left (-2 \ln \left (\frac {d e}{b}-\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}\right )-\frac {2 A}{B}\right )}{B}+\frac {d e \,{\mathrm e}^{-\frac {A}{B}} \operatorname {expIntegral}_{1}\left (-\ln \left (\frac {d e}{b}-\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{B}}{e^{2} \left (d a -b c \right )^{2} g^{3}}\) | \(126\) |
| default | \(-\frac {-\frac {b \,{\mathrm e}^{-\frac {2 A}{B}} \operatorname {expIntegral}_{1}\left (-2 \ln \left (\frac {d e}{b}-\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}\right )-\frac {2 A}{B}\right )}{B}+\frac {d e \,{\mathrm e}^{-\frac {A}{B}} \operatorname {expIntegral}_{1}\left (-\ln \left (\frac {d e}{b}-\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{B}}{e^{2} \left (d a -b c \right )^{2} g^{3}}\) | \(126\) |
| risch | \(\frac {b \,{\mathrm e}^{-\frac {2 A}{B}} \operatorname {expIntegral}_{1}\left (-2 \ln \left (\frac {d e}{b}-\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}\right )-\frac {2 A}{B}\right )}{g^{3} \left (d a -b c \right )^{2} e^{2} B}-\frac {d \,{\mathrm e}^{-\frac {A}{B}} \operatorname {expIntegral}_{1}\left (-\ln \left (\frac {d e}{b}-\frac {e \left (d a -b c \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{g^{3} \left (d a -b c \right )^{2} e B}\) | \(139\) |
Input:
int(1/(b*g*x+a*g)^3/(A+B*ln(e*(d*x+c)/(b*x+a))),x,method=_RETURNVERBOSE)
Output:
-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))
Time = 0.07 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.18 \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\frac {{\left (d e e^{\frac {A}{B}} \operatorname {log\_integral}\left (\frac {{\left (d e x + c e\right )} e^{\frac {A}{B}}}{b x + a}\right ) - b \operatorname {log\_integral}\left (\frac {{\left (d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}\right )} e^{\left (\frac {2 \, A}{B}\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )\right )} e^{\left (-\frac {2 \, A}{B}\right )}}{{\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} e^{2} g^{3}} \] Input:
integrate(1/(b*g*x+a*g)^3/(A+B*log(e*(d*x+c)/(b*x+a))),x, algorithm="frica s")
Output:
(d*e*e^(A/B)*log_integral((d*e*x + c*e)*e^(A/B)/(b*x + a)) - b*log_integra l((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)
\[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\frac {\int \frac {1}{A a^{3} + 3 A a^{2} b x + 3 A a b^{2} x^{2} + A b^{3} x^{3} + B a^{3} \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )} + 3 B a^{2} b x \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )} + 3 B a b^{2} x^{2} \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )} + B b^{3} x^{3} \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )}}\, dx}{g^{3}} \] Input:
integrate(1/(b*g*x+a*g)**3/(A+B*ln(e*(d*x+c)/(b*x+a))),x)
Output:
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
\[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{3} {\left (B \log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) + A\right )}} \,d x } \] Input:
integrate(1/(b*g*x+a*g)^3/(A+B*log(e*(d*x+c)/(b*x+a))),x, algorithm="maxim a")
Output:
integrate(1/((b*g*x + a*g)^3*(B*log((d*x + c)*e/(b*x + a)) + A)), x)
\[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\int { \frac {1}{{\left (b g x + a g\right )}^{3} {\left (B \log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) + A\right )}} \,d x } \] Input:
integrate(1/(b*g*x+a*g)^3/(A+B*log(e*(d*x+c)/(b*x+a))),x, algorithm="giac" )
Output:
integrate(1/((b*g*x + a*g)^3*(B*log((d*x + c)*e/(b*x + a)) + A)), x)
Timed out. \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\int \frac {1}{{\left (a\,g+b\,g\,x\right )}^3\,\left (A+B\,\ln \left (\frac {e\,\left (c+d\,x\right )}{a+b\,x}\right )\right )} \,d x \] Input:
int(1/((a*g + b*g*x)^3*(A + B*log((e*(c + d*x))/(a + b*x)))),x)
Output:
int(1/((a*g + b*g*x)^3*(A + B*log((e*(c + d*x))/(a + b*x)))), x)
\[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \, dx=\frac {\int \frac {1}{\mathrm {log}\left (\frac {d e x +c e}{b x +a}\right ) a^{3} b +3 \,\mathrm {log}\left (\frac {d e x +c e}{b x +a}\right ) a^{2} b^{2} x +3 \,\mathrm {log}\left (\frac {d e x +c e}{b x +a}\right ) a \,b^{3} x^{2}+\mathrm {log}\left (\frac {d e x +c e}{b x +a}\right ) b^{4} x^{3}+a^{4}+3 a^{3} b x +3 a^{2} b^{2} x^{2}+a \,b^{3} x^{3}}d x}{g^{3}} \] Input:
int(1/(b*g*x+a*g)^3/(A+B*log(e*(d*x+c)/(b*x+a))),x)
Output:
int(1/(log((c*e + d*e*x)/(a + b*x))*a**3*b + 3*log((c*e + d*e*x)/(a + b*x) )*a**2*b**2*x + 3*log((c*e + d*e*x)/(a + b*x))*a*b**3*x**2 + log((c*e + d* e*x)/(a + b*x))*b**4*x**3 + a**4 + 3*a**3*b*x + 3*a**2*b**2*x**2 + a*b**3* x**3),x)/g**3