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

Optimal. Leaf size=65 \[ -\frac{(c+d x) \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{g^2 (a+b x) (b c-a d)}-\frac{2 B}{b g^2 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0774241, antiderivative size = 105, normalized size of antiderivative = 1.62, number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {2525, 12, 44} \[ -\frac{B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A}{b g^2 (a+b x)}-\frac{2 B d \log (a+b x)}{b g^2 (b c-a d)}+\frac{2 B d \log (c+d x)}{b g^2 (b c-a d)}-\frac{2 B}{b g^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.058319, size = 111, normalized size = 1.71 \[ \frac{2 B (b c-a d) \left (-\frac{1}{(a+b x) (b c-a d)}-\frac{d \log (a+b x)}{(b c-a d)^2}+\frac{d \log (c+d x)}{(b c-a d)^2}\right )}{b g^2}-\frac{B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A}{b g^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.079, size = 157, normalized size = 2.4 \begin{align*}{\frac{dA}{{g}^{2} \left ( ad-bc \right ) } \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{-1}}-2\,{\frac{dB}{{g}^{2}b \left ( dx+c \right ) } \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{-1}}+{\frac{dB}{{g}^{2} \left ( ad-bc \right ) }\ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{-1}} \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))/(b*g*x+a*g)^2,x)

[Out]

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

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Maxima [B]  time = 1.23987, size = 252, normalized size = 3.88 \begin{align*} -B{\left (\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 )}{b^{2} g^{2} x + a b g^{2}} + \frac{2}{b^{2} g^{2} x + a b g^{2}} + \frac{2 \, d \log \left (b x + a\right )}{{\left (b^{2} c - a b d\right )} g^{2}} - \frac{2 \, d \log \left (d x + c\right )}{{\left (b^{2} c - a b d\right )} g^{2}}\right )} - \frac{A}{b^{2} g^{2} x + a b g^{2}} \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))/(b*g*x+a*g)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.01482, size = 228, normalized size = 3.51 \begin{align*} -\frac{{\left (A + 2 \, B\right )} b c -{\left (A + 2 \, B\right )} a d +{\left (B b d x + B b c\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 )}{{\left (b^{3} c - a b^{2} d\right )} g^{2} x +{\left (a b^{2} c - a^{2} b d\right )} g^{2}} \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))/(b*g*x+a*g)^2,x, algorithm="fricas")

[Out]

-((A + 2*B)*b*c - (A + 2*B)*a*d + (B*b*d*x + B*b*c)*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^3*c - a*b^2*d)*g^2*x + (a*b^2*c - a^2*b*d)*g^2)

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Sympy [B]  time = 1.91317, size = 253, normalized size = 3.89 \begin{align*} - \frac{B \log{\left (\frac{e \left (a + b x\right )^{2}}{\left (c + d x\right )^{2}} \right )}}{a b g^{2} + b^{2} g^{2} x} - \frac{2 B d \log{\left (x + \frac{- \frac{2 B a^{2} d^{3}}{a d - b c} + \frac{4 B a b c d^{2}}{a d - b c} + 2 B a d^{2} - \frac{2 B b^{2} c^{2} d}{a d - b c} + 2 B b c d}{4 B b d^{2}} \right )}}{b g^{2} \left (a d - b c\right )} + \frac{2 B d \log{\left (x + \frac{\frac{2 B a^{2} d^{3}}{a d - b c} - \frac{4 B a b c d^{2}}{a d - b c} + 2 B a d^{2} + \frac{2 B b^{2} c^{2} d}{a d - b c} + 2 B b c d}{4 B b d^{2}} \right )}}{b g^{2} \left (a d - b c\right )} - \frac{A + 2 B}{a b g^{2} + b^{2} g^{2} x} \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))/(b*g*x+a*g)**2,x)

[Out]

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

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Giac [B]  time = 1.39159, size = 254, normalized size = 3.91 \begin{align*}{\left (2 \,{\left (b^{2} c g^{2} - a b d g^{2}\right )}{\left (\frac{d \log \left ({\left | \frac{b c g}{b g x + a g} - \frac{a d g}{b g x + a g} + d \right |}\right )}{b^{4} c^{2} g^{4} - 2 \, a b^{3} c d g^{4} + a^{2} b^{2} d^{2} g^{4}} - \frac{1}{{\left (b^{2} c g^{2} - a b d g^{2}\right )}{\left (b g x + a g\right )} b g}\right )} - \frac{\log \left (\frac{{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right )}{{\left (b g x + a g\right )} b g}\right )} B - \frac{A}{{\left (b g x + a g\right )} b 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))/(b*g*x+a*g)^2,x, algorithm="giac")

[Out]

(2*(b^2*c*g^2 - a*b*d*g^2)*(d*log(abs(b*c*g/(b*g*x + a*g) - a*d*g/(b*g*x + a*g) + d))/(b^4*c^2*g^4 - 2*a*b^3*c
*d*g^4 + a^2*b^2*d^2*g^4) - 1/((b^2*c*g^2 - a*b*d*g^2)*(b*g*x + a*g)*b*g)) - log((b*x + a)^2*e/(d*x + c)^2)/((
b*g*x + a*g)*b*g))*B - A/((b*g*x + a*g)*b*g)