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]

((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))

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Rubi [A]  time = 0.092491, antiderivative size = 117, normalized size of antiderivative = 1.3, 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 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]

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]

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*}

Mathematica [A]  time = 0.164156, size = 108, normalized size = 1.2 \[ \frac{\frac{2 B (b \log (a+b x) (d f-c g)+\log (c+d x) (a d g-b d f)+g (b c-a d) \log (f+g x))}{(b f-a g) (d f-c g)}-\frac{B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A}{f+g x}}{g} \]

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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Maple [B]  time = 0.09, size = 388, normalized size = 4.3 \begin{align*}{\frac{dA}{cg-df} \left ({\frac{cg}{dx+c}}-{\frac{df}{dx+c}}-g \right ) ^{-1}}+{\frac{Bb}{ag-bf}\ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) \left ({\frac{cg}{dx+c}}-{\frac{df}{dx+c}}-g \right ) ^{-1}}+{\frac{Bda}{ \left ( ag-bf \right ) \left ( dx+c \right ) }\ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) \left ({\frac{cg}{dx+c}}-{\frac{df}{dx+c}}-g \right ) ^{-1}}-{\frac{Bbc}{ \left ( ag-bf \right ) \left ( dx+c \right ) }\ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) \left ({\frac{cg}{dx+c}}-{\frac{df}{dx+c}}-g \right ) ^{-1}}-2\,{\frac{Bda}{ac{g}^{2}-adfg-bcfg+bd{f}^{2}}\ln \left ({\frac{cg}{dx+c}}-{\frac{df}{dx+c}}-g \right ) }+2\,{\frac{Bbc}{ac{g}^{2}-adfg-bcfg+bd{f}^{2}}\ln \left ({\frac{cg}{dx+c}}-{\frac{df}{dx+c}}-g \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.17778, size = 259, normalized size = 2.88 \begin{align*} 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} \end{align*}

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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Fricas [B]  time = 22.2962, size = 613, normalized size = 6.81 \begin{align*} -\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} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

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