3.2.45 \(\int \frac {1}{(a g+b g x)^2 (A+B \log (\frac {e (a+b x)^2}{(c+d x)^2}))^2} \, dx\) [145]

3.2.45.1 Optimal result

Integrand size = 34, antiderivative size = 150 \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2} \, dx=-\frac {e^{\frac {A}{2 B}} \sqrt {\frac {e (a+b x)^2}{(c+d x)^2}} (c+d x) \operatorname {ExpIntegralEi}\left (\frac {-A-B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{2 B}\right )}{4 B^2 (b c-a d) g^2 (a+b x)}-\frac {c+d x}{2 B (b c-a d) g^2 (a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )} \]

output

1/2*(-d*x-c)/B/(-a*d+b*c)/g^2/(b*x+a)/(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))-1/4* 
exp(1/2*A/B)*(d*x+c)*Ei(1/2*(-A-B*ln(e*(b*x+a)^2/(d*x+c)^2))/B)*(e*(b*x+a) 
^2/(d*x+c)^2)^(1/2)/B^2/(-a*d+b*c)/g^2/(b*x+a)
 

3.2.45.2 Mathematica [F]

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

input

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

output

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

3.2.45.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2950, 2743, 2747, 2609}

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.

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

\(\Big \downarrow \) 2950

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

\(\Big \downarrow \) 2743

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

\(\Big \downarrow \) 2747

\(\displaystyle \frac {-\frac {(c+d x) \sqrt {\frac {e (a+b x)^2}{(c+d x)^2}} \int \frac {1}{\sqrt {\frac {e (a+b x)^2}{(c+d x)^2}} \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}d\log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{4 B (a+b x)}-\frac {c+d x}{2 B (a+b x) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}}{g^2 (b c-a d)}\)

\(\Big \downarrow \) 2609

\(\displaystyle \frac {-\frac {e^{\frac {A}{2 B}} (c+d x) \sqrt {\frac {e (a+b x)^2}{(c+d x)^2}} \operatorname {ExpIntegralEi}\left (-\frac {A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{2 B}\right )}{4 B^2 (a+b x)}-\frac {c+d x}{2 B (a+b x) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}}{g^2 (b c-a d)}\)

input

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

output

(-1/4*(E^(A/(2*B))*Sqrt[(e*(a + b*x)^2)/(c + d*x)^2]*(c + d*x)*ExpIntegral 
Ei[-1/2*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])/B])/(B^2*(a + b*x)) - (c 
+ d*x)/(2*B*(a + b*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])))/((b*c - a 
*d)*g^2)
 

3.2.45.3.1 Defintions of rubi rules used

rule 2609

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si 
mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F 
reeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]
 

rule 2743

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol 
] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])^(p + 1)/(b*d*n*(p + 1))), x] - 
Simp[(m + 1)/(b*n*(p + 1))   Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x], x] 
 /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]
 

rule 2747

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

rule 2950

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])
 

3.2.45.4 Maple [F]

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

input

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

output

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

3.2.45.5 Fricas [F]

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

input

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

output

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

3.2.45.6 Sympy [F]

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

input

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

output

(c + d*x)/(2*A*B*a**2*d*g**2 - 2*A*B*a*b*c*g**2 + 2*A*B*a*b*d*g**2*x - 2*A 
*B*b**2*c*g**2*x + (2*B**2*a**2*d*g**2 - 2*B**2*a*b*c*g**2 + 2*B**2*a*b*d* 
g**2*x - 2*B**2*b**2*c*g**2*x)*log(e*(a + b*x)**2/(c + d*x)**2)) - Integra 
l(1/(A*a**2 + 2*A*a*b*x + A*b**2*x**2 + B*a**2*log(a**2*e/(c**2 + 2*c*d*x 
+ d**2*x**2) + 2*a*b*e*x/(c**2 + 2*c*d*x + d**2*x**2) + b**2*e*x**2/(c**2 
+ 2*c*d*x + d**2*x**2)) + 2*B*a*b*x*log(a**2*e/(c**2 + 2*c*d*x + d**2*x**2 
) + 2*a*b*e*x/(c**2 + 2*c*d*x + d**2*x**2) + b**2*e*x**2/(c**2 + 2*c*d*x + 
 d**2*x**2)) + B*b**2*x**2*log(a**2*e/(c**2 + 2*c*d*x + d**2*x**2) + 2*a*b 
*e*x/(c**2 + 2*c*d*x + d**2*x**2) + b**2*e*x**2/(c**2 + 2*c*d*x + d**2*x** 
2))), x)/(2*B*g**2)
 

3.2.45.7 Maxima [F]

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

input

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

output

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

3.2.45.8 Giac [F]

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

input

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

output

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

3.2.45.9 Mupad [F(-1)]

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

input

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

output

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