3.3.23 \(\int \frac {1}{(a g+b g x)^3 (A+B \log (\frac {e (c+d x)^2}{(a+b x)^2}))} \, dx\) [223]

3.3.23.1 Optimal result

Integrand size = 34, antiderivative size = 151 \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )} \, dx=\frac {d e^{-\frac {A}{2 B}} (c+d x) \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{2 B}\right )}{2 B (b c-a d)^2 g^3 (a+b x) \sqrt {\frac {e (c+d x)^2}{(a+b x)^2}}}-\frac {b e^{-\frac {A}{B}} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{B}\right )}{2 B (b c-a d)^2 e g^3} \]

output

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

3.3.23.2 Mathematica [F]

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

input

Integrate[1/((a*g + b*g*x)^3*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])),x]
 

output

Integrate[1/((a*g + b*g*x)^3*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])), x]
 

3.3.23.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, 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.

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

\(\Big \downarrow \) 2952

\(\displaystyle \frac {\int \frac {d-\frac {b (c+d x)}{a+b x}}{A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}d\frac {c+d x}{a+b x}}{g^3 (b c-a d)^2}\)

\(\Big \downarrow \) 2767

\(\displaystyle \frac {\int \left (\frac {d}{A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}-\frac {b (c+d x)}{(a+b x) \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )}\right )d\frac {c+d x}{a+b x}}{g^3 (b c-a d)^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {d e^{-\frac {A}{2 B}} (c+d x) \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{2 B}\right )}{2 B (a+b x) \sqrt {\frac {e (c+d x)^2}{(a+b x)^2}}}-\frac {b e^{-\frac {A}{B}} \operatorname {ExpIntegralEi}\left (\frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{B}\right )}{2 B e}}{g^3 (b c-a d)^2}\)

input

Int[1/((a*g + b*g*x)^3*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])),x]
 

output

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

3.3.23.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2767

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]))
 

rule 2952

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])
 

3.3.23.4 Maple [F]

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

input

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

output

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

3.3.23.5 Fricas [F]

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

input

integrate(1/(b*g*x+a*g)^3/(A+B*log(e*(d*x+c)^2/(b*x+a)^2)),x, algorithm="f 
ricas")
 

output

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

3.3.23.6 Sympy [F]

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

input

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

3.3.23.7 Maxima [F]

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

input

integrate(1/(b*g*x+a*g)^3/(A+B*log(e*(d*x+c)^2/(b*x+a)^2)),x, algorithm="m 
axima")
 

output

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

3.3.23.8 Giac [F]

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

input

integrate(1/(b*g*x+a*g)^3/(A+B*log(e*(d*x+c)^2/(b*x+a)^2)),x, algorithm="g 
iac")
 

output

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

3.3.23.9 Mupad [F(-1)]

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

input

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

output

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