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

Optimal. Leaf size=90 \[ \frac {(a+b x) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{(f+g x) (b f-a g)}+\frac {2 B (b c-a d) \log \left (\frac {f+g x}{c+d x}\right )}{(b f-a g) (d f-c g)} \]

[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)

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Rubi [A]  time = 0.09, antiderivative size = 117, normalized size of antiderivative = 1.30, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2525, 12, 72} \[ -\frac {B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A}{g (f+g x)}+\frac {2 B (b c-a d) \log (f+g x)}{(b f-a g) (d f-c g)}+\frac {2 b B \log (a+b x)}{g (b f-a g)}-\frac {2 B d \log (c+d x)}{g (d f-c g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 108, normalized size = 1.20 \[ \frac {\frac {2 B (b \log (a+b x) (d f-c g)+\log (c+d x) (a d g-b d f)+g (b c-a d) \log (f+g x))}{(b f-a g) (d f-c g)}-\frac {B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A}{f+g x}}{g} \]

Antiderivative was successfully verified.

[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

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fricas [B]  time = 10.61, size = 279, normalized size = 3.10 \[ -\frac {A b d f^{2} + A a c g^{2} - {\left (A b c + A a d\right )} f g - 2 \, {\left (B b d f^{2} - B b c f g + {\left (B b d f g - B b c g^{2}\right )} x\right )} \log \left (b x + a\right ) + 2 \, {\left (B b d f^{2} - B a d f g + {\left (B b d f g - B a d g^{2}\right )} x\right )} \log \left (d x + c\right ) - 2 \, {\left ({\left (B b c - B a d\right )} g^{2} x + {\left (B b c - B a d\right )} f g\right )} \log \left (g x + f\right ) + {\left (B b d f^{2} + B a c g^{2} - {\left (B b c + B a d\right )} f g\right )} \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{b d f^{3} g + a c f g^{3} - {\left (b c + a d\right )} f^{2} g^{2} + {\left (b d f^{2} g^{2} + a c g^{4} - {\left (b c + a d\right )} f g^{3}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Undef/Unsigned Inf encountered in limitU
ndef/Unsigned Inf encountered in limitB*(-(g*x+f)^-1/g*ln((b*(-f+1/g/(g*x+f)^-1*g)/g+a)^2*exp(1)/(d*(-f+1/g/(g
*x+f)^-1*g)/g+c)^2)-(-2*a*d*g^2+2*b*c*g^2)*(1/(2*a*c*g^4-2*a*g^3*d*f-2*c*g^3*f*b+2*g^2*d*f^2*b)*ln(abs((-(g*x+
f)^-1/g)^2*a*c*g^4-(-(g*x+f)^-1/g)^2*a*g^3*d*f-(-(g*x+f)^-1/g)^2*c*g^3*f*b+(-(g*x+f)^-1/g)^2*g^2*d*f^2*b+(g*x+
f)^-1/g*a*g^2*d+(g*x+f)^-1/g*c*g^2*b-2*(g*x+f)^-1/g*g*d*f*b+d*b))+(a*g*d+c*g*b-2*d*f*b)/(2*a*c*g^3-2*a*g^2*d*f
-2*c*g^2*f*b+2*g*d*f^2*b)/abs(a*g^2*d-g^2*c*b)*ln(abs(-2*(g*x+f)^-1/g*a*c*g^4+2*(g*x+f)^-1/g*a*g^3*d*f+2*(g*x+
f)^-1/g*c*g^3*f*b-2*(g*x+f)^-1/g*g^2*d*f^2*b-a*g^2*d-c*g^2*b+2*g*d*f*b-abs(a*g^2*d-g^2*c*b))/abs(-2*(g*x+f)^-1
/g*a*c*g^4+2*(g*x+f)^-1/g*a*g^3*d*f+2*(g*x+f)^-1/g*c*g^3*f*b-2*(g*x+f)^-1/g*g^2*d*f^2*b-a*g^2*d-c*g^2*b+2*g*d*
f*b+abs(a*g^2*d-g^2*c*b)))))-A*(g*x+f)^-1/g

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maple [B]  time = 0.09, size = 388, normalized size = 4.31 \[ \frac {B a d \ln \left (\frac {\left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2} e}{d^{2}}\right )}{\left (\frac {c g}{d x +c}-\frac {d f}{d x +c}-g \right ) \left (a g -b f \right ) \left (d x +c \right )}-\frac {2 B a d \ln \left (\frac {c g}{d x +c}-\frac {d f}{d x +c}-g \right )}{a c \,g^{2}-a d f g -b c f g +b d \,f^{2}}-\frac {B b c \ln \left (\frac {\left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2} e}{d^{2}}\right )}{\left (\frac {c g}{d x +c}-\frac {d f}{d x +c}-g \right ) \left (a g -b f \right ) \left (d x +c \right )}+\frac {2 B b c \ln \left (\frac {c g}{d x +c}-\frac {d f}{d x +c}-g \right )}{a c \,g^{2}-a d f g -b c f g +b d \,f^{2}}+\frac {B b \ln \left (\frac {\left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2} e}{d^{2}}\right )}{\left (\frac {c g}{d x +c}-\frac {d f}{d x +c}-g \right ) \left (a g -b f \right )}+\frac {A d}{\left (\frac {c g}{d x +c}-\frac {d f}{d x +c}-g \right ) \left (c g -d f \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.70, size = 192, normalized size = 2.13 \[ B {\left (\frac {2 \, b \log \left (b x + a\right )}{b f g - a g^{2}} - \frac {2 \, d \log \left (d x + c\right )}{d f g - c g^{2}} + \frac {2 \, {\left (b c - a d\right )} \log \left (g x + f\right )}{b d f^{2} + a c g^{2} - {\left (b c + a d\right )} f g} - \frac {\log \left (\frac {b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{g^{2} x + f g}\right )} - \frac {A}{g^{2} x + f g} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 5.34, size = 191, normalized size = 2.12 \[ \frac {2\,B\,d\,\ln \left (c+d\,x\right )}{c\,g^2-d\,f\,g}-\frac {B\,\ln \left (\frac {e\,a^2+2\,e\,a\,b\,x+e\,b^2\,x^2}{c^2+2\,c\,d\,x+d^2\,x^2}\right )}{x\,g^2+f\,g}-\frac {2\,B\,b\,\ln \left (a+b\,x\right )}{a\,g^2-b\,f\,g}-\frac {A}{x\,g^2+f\,g}-\frac {2\,B\,a\,d\,\ln \left (f+g\,x\right )}{a\,c\,g^2+b\,d\,f^2-a\,d\,f\,g-b\,c\,f\,g}+\frac {2\,B\,b\,c\,\ln \left (f+g\,x\right )}{a\,c\,g^2+b\,d\,f^2-a\,d\,f\,g-b\,c\,f\,g} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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