\(\int \frac {A+B \log (\frac {e (a+b x)^2}{(c+d x)^2})}{(f+g x)^2} \, dx\) [268]

Optimal result

Integrand size = 29, antiderivative size = 90 \[ \int \frac {A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{(f+g x)^2} \, dx=\frac {(a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(b f-a g) (f+g x)}+\frac {2 B (b c-a d) \log \left (\frac {f+g x}{c+d x}\right )}{(b f-a g) (d f-c g)} \]

[Out]

(b*x+a)*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))/(-a*g+b*f)/(g*x+f)+2*B*(-a*d+b*c)*ln((g*x+f)/(d*x+c))/(-a*g+b*f)/(-c*g
+d*f)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2554, 2351, 31} \[ \int \frac {A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{(f+g x)^2} \, dx=\frac {(a+b x) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{(f+g x) (b f-a g)}+\frac {2 B (b c-a d) \log \left (\frac {f+g x}{c+d x}\right )}{(b f-a g) (d f-c g)} \]

[In]

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

[Out]

((a + b*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]))/((b*f - a*g)*(f + g*x)) + (2*B*(b*c - a*d)*Log[(f + g*x)/
(c + d*x)])/((b*f - a*g)*(d*f - c*g))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[x*(d + e*x^r)^(q +
1)*((a + b*Log[c*x^n])/d), x] - Dist[b*(n/d), Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2554

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_)
)^(m_.), x_Symbol] :> Dist[b*c - a*d, Subst[Int[(b*f - a*g - (d*f - c*g)*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] && IGt
Q[n, 0] && NeQ[b*c - a*d, 0] && IntegerQ[m] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = (b c-a d) \text {Subst}\left (\int \frac {A+B \log \left (e x^2\right )}{(b f-a g+(-d f+c g) x)^2} \, dx,x,\frac {a+b x}{c+d x}\right ) \\ & = \frac {(a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(b f-a g) (f+g x)}-\frac {(2 B (b c-a d)) \text {Subst}\left (\int \frac {1}{b f-a g+(-d f+c g) x} \, dx,x,\frac {a+b x}{c+d x}\right )}{b f-a g} \\ & = \frac {(a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{(b f-a g) (f+g x)}+\frac {2 B (b c-a d) \log \left (\frac {f+g x}{c+d x}\right )}{(b f-a g) (d f-c g)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.20 \[ \int \frac {A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{(f+g x)^2} \, dx=\frac {-\frac {A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{f+g x}+\frac {2 B (b (d f-c g) \log (a+b x)+(-b d f+a d g) \log (c+d x)+(b c-a d) g \log (f+g x))}{(b f-a g) (d f-c g)}}{g} \]

[In]

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

[Out]

(-((A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])/(f + g*x)) + (2*B*(b*(d*f - c*g)*Log[a + b*x] + (-(b*d*f) + a*d*g)
*Log[c + d*x] + (b*c - a*d)*g*Log[f + g*x]))/((b*f - a*g)*(d*f - c*g)))/g

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(244\) vs. \(2(90)=180\).

Time = 0.67 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.72

method	result	size
derivativedivides	\(-\frac {-\frac {d^{2} A}{\left (\frac {c g -d f}{d x +c}-g \right ) \left (c g -d f \right )}+\frac {-\frac {b B d \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )}{a g -b f}-\frac {B d \left (a d -c b \right ) \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )}{\left (a g -b f \right ) \left (d x +c \right )}}{\frac {c g}{d x +c}-\frac {f d}{d x +c}-g}+\frac {2 B d \left (a d -c b \right ) \ln \left (\frac {c g}{d x +c}-\frac {f d}{d x +c}-g \right )}{a c \,g^{2}-a d f g -b c f g +b d \,f^{2}}}{d}\)	\(245\)
default	\(-\frac {-\frac {d^{2} A}{\left (\frac {c g -d f}{d x +c}-g \right ) \left (c g -d f \right )}+\frac {-\frac {b B d \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )}{a g -b f}-\frac {B d \left (a d -c b \right ) \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )}{\left (a g -b f \right ) \left (d x +c \right )}}{\frac {c g}{d x +c}-\frac {f d}{d x +c}-g}+\frac {2 B d \left (a d -c b \right ) \ln \left (\frac {c g}{d x +c}-\frac {f d}{d x +c}-g \right )}{a c \,g^{2}-a d f g -b c f g +b d \,f^{2}}}{d}\)	\(245\)
risch	\(-\frac {B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right )}{g \left (g x +f \right )}-\frac {-2 B \ln \left (-d x -c \right ) a d \,g^{2} x +2 B \ln \left (-d x -c \right ) b d f g x +2 B \ln \left (g x +f \right ) a d \,g^{2} x -2 B \ln \left (g x +f \right ) b c \,g^{2} x +2 B \ln \left (-b x -a \right ) b c \,g^{2} x -2 B \ln \left (-b x -a \right ) b d f g x -2 B \ln \left (-d x -c \right ) a d f g +2 B \ln \left (-d x -c \right ) b d \,f^{2}+2 B \ln \left (g x +f \right ) a d f g -2 B \ln \left (g x +f \right ) b c f g +2 B \ln \left (-b x -a \right ) b c f g -2 B \ln \left (-b x -a \right ) b d \,f^{2}+A a c \,g^{2}-A a d f g -A b c f g +A b d \,f^{2}}{\left (a g -b f \right ) \left (c g -d f \right ) \left (g x +f \right ) g}\)	\(285\)
parallelrisch	\(\frac {4 B \ln \left (g x +f \right ) x a b \,c^{2} f g +4 B \ln \left (b x +a \right ) x \,a^{2} c d f g -2 B x \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) a^{2} c d f g +2 B x \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) a b c d \,f^{2}-4 B \ln \left (b x +a \right ) x a b \,c^{2} f g -4 B \ln \left (g x +f \right ) x \,a^{2} c d f g +2 A x \,a^{2} c^{2} g^{2}-2 A x a b \,c^{2} f g -2 B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) a^{2} c^{2} f g +2 B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) a b \,c^{2} f^{2}+2 A x a b c d \,f^{2}-2 A x \,a^{2} c d f g +4 B \ln \left (b x +a \right ) a^{2} c d \,f^{2}-4 B \ln \left (b x +a \right ) a b \,c^{2} f^{2}-4 B \ln \left (g x +f \right ) a^{2} c d \,f^{2}+4 B \ln \left (g x +f \right ) a b \,c^{2} f^{2}}{2 \left (a c \,g^{2}-a d f g -b c f g +b d \,f^{2}\right ) \left (g x +f \right ) a c f}\)	\(341\)

[In]

int((A+B*ln(e*(b*x+a)^2/(d*x+c)^2))/(g*x+f)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (90) = 180\).

Time = 3.50 (sec) , antiderivative size = 279, normalized size of antiderivative = 3.10 \[ \int \frac {A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{(f+g x)^2} \, dx=-\frac {A b d f^{2} + A a c g^{2} - {\left (A b c + A a d\right )} f g - 2 \, {\left (B b d f^{2} - B b c f g + {\left (B b d f g - B b c g^{2}\right )} x\right )} \log \left (b x + a\right ) + 2 \, {\left (B b d f^{2} - B a d f g + {\left (B b d f g - B a d g^{2}\right )} x\right )} \log \left (d x + c\right ) - 2 \, {\left ({\left (B b c - B a d\right )} g^{2} x + {\left (B b c - B a d\right )} f g\right )} \log \left (g x + f\right ) + {\left (B b d f^{2} + B a c g^{2} - {\left (B b c + B a d\right )} f g\right )} \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{b d f^{3} g + a c f g^{3} - {\left (b c + a d\right )} f^{2} g^{2} + {\left (b d f^{2} g^{2} + a c g^{4} - {\left (b c + a d\right )} f g^{3}\right )} x} \]

[In]

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

[Out]

-(A*b*d*f^2 + A*a*c*g^2 - (A*b*c + A*a*d)*f*g - 2*(B*b*d*f^2 - B*b*c*f*g + (B*b*d*f*g - B*b*c*g^2)*x)*log(b*x
+ a) + 2*(B*b*d*f^2 - B*a*d*f*g + (B*b*d*f*g - B*a*d*g^2)*x)*log(d*x + c) - 2*((B*b*c - B*a*d)*g^2*x + (B*b*c
- B*a*d)*f*g)*log(g*x + f) + (B*b*d*f^2 + B*a*c*g^2 - (B*b*c + B*a*d)*f*g)*log((b^2*e*x^2 + 2*a*b*e*x + a^2*e)
/(d^2*x^2 + 2*c*d*x + c^2)))/(b*d*f^3*g + a*c*f*g^3 - (b*c + a*d)*f^2*g^2 + (b*d*f^2*g^2 + a*c*g^4 - (b*c + a*
d)*f*g^3)*x)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (90) = 180\).

Time = 0.21 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.13 \[ \int \frac {A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{(f+g x)^2} \, dx=B {\left (\frac {2 \, b \log \left (b x + a\right )}{b f g - a g^{2}} - \frac {2 \, d \log \left (d x + c\right )}{d f g - c g^{2}} + \frac {2 \, {\left (b c - a d\right )} \log \left (g x + f\right )}{b d f^{2} + a c g^{2} - {\left (b c + a d\right )} f g} - \frac {\log \left (\frac {b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{g^{2} x + f g}\right )} - \frac {A}{g^{2} x + f g} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 454 vs. \(2 (90) = 180\).

Time = 0.52 (sec) , antiderivative size = 454, normalized size of antiderivative = 5.04 \[ \int \frac {A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{(f+g x)^2} \, dx={\left ({\left (b c g^{2} - a d g^{2}\right )} {\left (\frac {{\left (2 \, b d f - b c g - a d g\right )} \log \left (\frac {{\left | 2 \, b d f g - \frac {2 \, b d f^{2} g}{g x + f} - b c g^{2} - a d g^{2} + \frac {2 \, b c f g^{2}}{g x + f} + \frac {2 \, a d f g^{2}}{g x + f} - \frac {2 \, a c g^{3}}{g x + f} - {\left | -b c g^{2} + a d g^{2} \right |} \right |}}{{\left | 2 \, b d f g - \frac {2 \, b d f^{2} g}{g x + f} - b c g^{2} - a d g^{2} + \frac {2 \, b c f g^{2}}{g x + f} + \frac {2 \, a d f g^{2}}{g x + f} - \frac {2 \, a c g^{3}}{g x + f} + {\left | -b c g^{2} + a d g^{2} \right |} \right |}}\right )}{{\left (b d f^{2} g - b c f g^{2} - a d f g^{2} + a c g^{3}\right )} {\left | -b c g^{2} + a d g^{2} \right |}} - \frac {\log \left ({\left | b d - \frac {2 \, b d f}{g x + f} + \frac {b d f^{2}}{{\left (g x + f\right )}^{2}} + \frac {b c g}{g x + f} + \frac {a d g}{g x + f} - \frac {b c f g}{{\left (g x + f\right )}^{2}} - \frac {a d f g}{{\left (g x + f\right )}^{2}} + \frac {a c g^{2}}{{\left (g x + f\right )}^{2}} \right |}\right )}{b d f^{2} g^{2} - b c f g^{3} - a d f g^{3} + a c g^{4}}\right )} - \frac {\log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right )}{{\left (g x + f\right )} g}\right )} B - \frac {A}{{\left (g x + f\right )} g} \]

[In]

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

[Out]

((b*c*g^2 - a*d*g^2)*((2*b*d*f - b*c*g - a*d*g)*log(abs(2*b*d*f*g - 2*b*d*f^2*g/(g*x + f) - b*c*g^2 - a*d*g^2
+ 2*b*c*f*g^2/(g*x + f) + 2*a*d*f*g^2/(g*x + f) - 2*a*c*g^3/(g*x + f) - abs(-b*c*g^2 + a*d*g^2))/abs(2*b*d*f*g
- 2*b*d*f^2*g/(g*x + f) - b*c*g^2 - a*d*g^2 + 2*b*c*f*g^2/(g*x + f) + 2*a*d*f*g^2/(g*x + f) - 2*a*c*g^3/(g*x
+ f) + abs(-b*c*g^2 + a*d*g^2)))/((b*d*f^2*g - b*c*f*g^2 - a*d*f*g^2 + a*c*g^3)*abs(-b*c*g^2 + a*d*g^2)) - log
(abs(b*d - 2*b*d*f/(g*x + f) + b*d*f^2/(g*x + f)^2 + b*c*g/(g*x + f) + a*d*g/(g*x + f) - b*c*f*g/(g*x + f)^2 -
a*d*f*g/(g*x + f)^2 + a*c*g^2/(g*x + f)^2))/(b*d*f^2*g^2 - b*c*f*g^3 - a*d*f*g^3 + a*c*g^4)) - log((b*x + a)^
2*e/(d*x + c)^2)/((g*x + f)*g))*B - A/((g*x + f)*g)

Mupad [B] (verification not implemented)

Time = 1.90 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.12 \[ \int \frac {A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{(f+g x)^2} \, dx=\frac {2\,B\,d\,\ln \left (c+d\,x\right )}{c\,g^2-d\,f\,g}-\frac {B\,\ln \left (\frac {e\,a^2+2\,e\,a\,b\,x+e\,b^2\,x^2}{c^2+2\,c\,d\,x+d^2\,x^2}\right )}{x\,g^2+f\,g}-\frac {2\,B\,b\,\ln \left (a+b\,x\right )}{a\,g^2-b\,f\,g}-\frac {A}{x\,g^2+f\,g}-\frac {2\,B\,a\,d\,\ln \left (f+g\,x\right )}{a\,c\,g^2+b\,d\,f^2-a\,d\,f\,g-b\,c\,f\,g}+\frac {2\,B\,b\,c\,\ln \left (f+g\,x\right )}{a\,c\,g^2+b\,d\,f^2-a\,d\,f\,g-b\,c\,f\,g} \]

[In]

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

[Out]

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