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

Optimal. Leaf size=102 \[ -\frac{A (c+d x)}{g^2 (a+b x) (b c-a d)}-\frac{B (c+d x) \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )}{g^2 (a+b x) (b c-a d)}+\frac{2 B (c+d x)}{g^2 (a+b x) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.07582, antiderivative size = 105, normalized size of antiderivative = 1.03, 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} \[ -\frac{B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )+A}{b g^2 (a+b x)}+\frac{2 B d \log (a+b x)}{b g^2 (b c-a d)}-\frac{2 B d \log (c+d x)}{b g^2 (b c-a d)}+\frac{2 B}{b g^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

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

Mathematica [A]  time = 0.0515359, size = 89, normalized size = 0.87 \[ \frac{-(b c-a d) \left (B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )+A-2 B\right )-2 B d (a+b x) \log (c+d x)+2 B d (a+b x) \log (a+b x)}{b g^2 (a+b x) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.062, size = 212, normalized size = 2.1 \begin{align*} -{\frac{A}{b{g}^{2} \left ( bx+a \right ) }}-{\frac{B}{b{g}^{2} \left ( bx+a \right ) }\ln \left ({\frac{e}{{b}^{2}} \left ({\frac{ad}{bx+a}}-{\frac{bc}{bx+a}}-d \right ) ^{2}} \right ) }+2\,{\frac{Bad}{b{g}^{2} \left ( ad-bc \right ) \left ( bx+a \right ) }}-2\,{\frac{Bc}{{g}^{2} \left ( ad-bc \right ) \left ( bx+a \right ) }}+2\,{\frac{B{d}^{2}a}{b{g}^{2} \left ( ad-bc \right ) ^{2}}\ln \left ({\frac{ad}{bx+a}}-{\frac{bc}{bx+a}}-d \right ) }-2\,{\frac{Bdc}{{g}^{2} \left ( ad-bc \right ) ^{2}}\ln \left ({\frac{ad}{bx+a}}-{\frac{bc}{bx+a}}-d \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.23322, size = 252, normalized size = 2.47 \begin{align*} -B{\left (\frac{\log \left (\frac{d^{2} e x^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac{2 \, c d e x}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac{c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{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*(d*x+c)^2/(b*x+a)^2))/(b*g*x+a*g)^2,x, algorithm="maxima")

[Out]

-B*(log(d^2*e*x^2/(b^2*x^2 + 2*a*b*x + a^2) + 2*c*d*e*x/(b^2*x^2 + 2*a*b*x + a^2) + c^2*e/(b^2*x^2 + 2*a*b*x +
a^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 = 1.0378, size = 228, normalized size = 2.24 \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{d^{2} e x^{2} + 2 \, c d e x + c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{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*(d*x+c)^2/(b*x+a)^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((d^2*e*x^2 + 2*c*d*e*x + c^2*e)/(b^2*x^2 + 2*a*b*x + a
^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 = 1.93865, size = 253, normalized size = 2.48 \begin{align*} - \frac{B \log{\left (\frac{e \left (c + d x\right )^{2}}{\left (a + b 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*(d*x+c)**2/(b*x+a)**2))/(b*g*x+a*g)**2,x)

[Out]

-B*log(e*(c + d*x)**2/(a + b*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 [A]  time = 1.38116, size = 254, normalized size = 2.49 \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 (d x + c\right )}^{2} e}{{\left (b x + a\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*(d*x+c)^2/(b*x+a)^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((d*x + c)^2*e/(b*x + a)^2)/(
(b*g*x + a*g)*b*g))*B - A/((b*g*x + a*g)*b*g)