3.125 \(\int \frac{A+B \log (\frac{e (a+b x)^2}{(c+d x)^2})}{(a g+b g x)^3} \, dx\)

Optimal. Leaf size=138 \[ -\frac{B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A}{2 b g^3 (a+b x)^2}+\frac{B d^2 \log (a+b x)}{b g^3 (b c-a d)^2}-\frac{B d^2 \log (c+d x)}{b g^3 (b c-a d)^2}+\frac{B d}{b g^3 (a+b x) (b c-a d)}-\frac{B}{2 b g^3 (a+b x)^2} \]

[Out]

-B/(2*b*g^3*(a + b*x)^2) + (B*d)/(b*(b*c - a*d)*g^3*(a + b*x)) + (B*d^2*Log[a + b*x])/(b*(b*c - a*d)^2*g^3) -
(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])/(2*b*g^3*(a + b*x)^2) - (B*d^2*Log[c + d*x])/(b*(b*c - a*d)^2*g^3)

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Rubi [A]  time = 0.0934478, antiderivative size = 138, normalized size of antiderivative = 1., 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} \[ -\frac{B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A}{2 b g^3 (a+b x)^2}+\frac{B d^2 \log (a+b x)}{b g^3 (b c-a d)^2}-\frac{B d^2 \log (c+d x)}{b g^3 (b c-a d)^2}+\frac{B d}{b g^3 (a+b x) (b c-a d)}-\frac{B}{2 b g^3 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])/(a*g + b*g*x)^3,x]

[Out]

-B/(2*b*g^3*(a + b*x)^2) + (B*d)/(b*(b*c - a*d)*g^3*(a + b*x)) + (B*d^2*Log[a + b*x])/(b*(b*c - a*d)^2*g^3) -
(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])/(2*b*g^3*(a + b*x)^2) - (B*d^2*Log[c + d*x])/(b*(b*c - a*d)^2*g^3)

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)^3} \, dx &=-\frac{A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )}{2 b g^3 (a+b x)^2}+\frac{B \int \frac{2 (b c-a d)}{g^2 (a+b x)^3 (c+d x)} \, dx}{2 b g}\\ &=-\frac{A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )}{2 b g^3 (a+b x)^2}+\frac{(B (b c-a d)) \int \frac{1}{(a+b x)^3 (c+d x)} \, dx}{b g^3}\\ &=-\frac{A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )}{2 b g^3 (a+b x)^2}+\frac{(B (b c-a d)) \int \left (\frac{b}{(b c-a d) (a+b x)^3}-\frac{b d}{(b c-a d)^2 (a+b x)^2}+\frac{b d^2}{(b c-a d)^3 (a+b x)}-\frac{d^3}{(b c-a d)^3 (c+d x)}\right ) \, dx}{b g^3}\\ &=-\frac{B}{2 b g^3 (a+b x)^2}+\frac{B d}{b (b c-a d) g^3 (a+b x)}+\frac{B d^2 \log (a+b x)}{b (b c-a d)^2 g^3}-\frac{A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )}{2 b g^3 (a+b x)^2}-\frac{B d^2 \log (c+d x)}{b (b c-a d)^2 g^3}\\ \end{align*}

Mathematica [A]  time = 0.136493, size = 109, normalized size = 0.79 \[ -\frac{\frac{B \left (2 d^2 (a+b x)^2 \log (c+d x)+(b c-a d) (b (c-2 d x)-3 a d)-2 d^2 (a+b x)^2 \log (a+b x)\right )}{(b c-a d)^2}+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A}{2 b g^3 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])/(a*g + b*g*x)^3,x]

[Out]

-(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2] + (B*((b*c - a*d)*(-3*a*d + b*(c - 2*d*x)) - 2*d^2*(a + b*x)^2*Log[a
+ b*x] + 2*d^2*(a + b*x)^2*Log[c + d*x]))/(b*c - a*d)^2)/(2*b*g^3*(a + b*x)^2)

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Maple [B]  time = 0.108, size = 355, normalized size = 2.6 \begin{align*} -{\frac{{d}^{2}Ab}{2\,{g}^{3} \left ( ad-bc \right ) ^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{-2}}+{\frac{{d}^{2}A}{{g}^{3} \left ( ad-bc \right ) ^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{-1}}-{\frac{{d}^{2}B}{{g}^{3} \left ( ad-bc \right ) \left ( dx+c \right ) } \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{-2}}-{\frac{3\,{d}^{2}B}{2\,{g}^{3}b \left ( dx+c \right ) ^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{-2}}+{\frac{b{d}^{2}B}{2\,{g}^{3} \left ({a}^{2}{d}^{2}-2\,abcd+{b}^{2}{c}^{2} \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 ) ^{-2}}+{\frac{{d}^{2}B}{{g}^{3} \left ( ad-bc \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{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{-2}} \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)^3,x)

[Out]

-1/2*d^2/g^3*A*b/(a*d-b*c)^2/(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2+d^2/g^3*A/(a*d-b*c)^2/(1/(d*x+c)*a*d-b*c/(d*x+c)+
b)-d^2/g^3/(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2*B/(a*d-b*c)/(d*x+c)-3/2*d^2/g^3/(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2*B/b
/(d*x+c)^2+1/2*d^2/g^3/(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2*b*B/(a^2*d^2-2*a*b*c*d+b^2*c^2)*ln(e*(1/(d*x+c)*a*d-b*c
/(d*x+c)+b)^2/d^2)+d^2/g^3/(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2*B/(a*d-b*c)/(d*x+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 = 1.29435, size = 414, normalized size = 3. \begin{align*} \frac{1}{2} \, B{\left (\frac{2 \, b d x - b c + 3 \, a d}{{\left (b^{4} c - a b^{3} d\right )} g^{3} x^{2} + 2 \,{\left (a b^{3} c - a^{2} b^{2} d\right )} g^{3} x +{\left (a^{2} b^{2} c - a^{3} b d\right )} g^{3}} - \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^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}} + \frac{2 \, d^{2} \log \left (b x + a\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}} - \frac{2 \, d^{2} \log \left (d x + c\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}}\right )} - \frac{A}{2 \,{\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} \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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.05097, size = 495, normalized size = 3.59 \begin{align*} -\frac{{\left (A + B\right )} b^{2} c^{2} - 2 \,{\left (A + 2 \, B\right )} a b c d +{\left (A + 3 \, B\right )} a^{2} d^{2} - 2 \,{\left (B b^{2} c d - B a b d^{2}\right )} x -{\left (B b^{2} d^{2} x^{2} + 2 \, B a b d^{2} x - B b^{2} c^{2} + 2 \, B a b c d\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 )}{2 \,{\left ({\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} g^{3} x^{2} + 2 \,{\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} g^{3} x +{\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} g^{3}\right )}} \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)^3,x, algorithm="fricas")

[Out]

-1/2*((A + B)*b^2*c^2 - 2*(A + 2*B)*a*b*c*d + (A + 3*B)*a^2*d^2 - 2*(B*b^2*c*d - B*a*b*d^2)*x - (B*b^2*d^2*x^2
+ 2*B*a*b*d^2*x - B*b^2*c^2 + 2*B*a*b*c*d)*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^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*g^3*x^2 + 2*(a*b^4*c^2 - 2*a^2*b^3*c*d + a^3*b^2*d^2)*g^3*x + (a^2*b^3*c^
2 - 2*a^3*b^2*c*d + a^4*b*d^2)*g^3)

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Sympy [B]  time = 3.17597, size = 418, normalized size = 3.03 \begin{align*} - \frac{B \log{\left (\frac{e \left (a + b x\right )^{2}}{\left (c + d x\right )^{2}} \right )}}{2 a^{2} b g^{3} + 4 a b^{2} g^{3} x + 2 b^{3} g^{3} x^{2}} - \frac{B d^{2} \log{\left (x + \frac{- \frac{B a^{3} d^{5}}{\left (a d - b c\right )^{2}} + \frac{3 B a^{2} b c d^{4}}{\left (a d - b c\right )^{2}} - \frac{3 B a b^{2} c^{2} d^{3}}{\left (a d - b c\right )^{2}} + B a d^{3} + \frac{B b^{3} c^{3} d^{2}}{\left (a d - b c\right )^{2}} + B b c d^{2}}{2 B b d^{3}} \right )}}{b g^{3} \left (a d - b c\right )^{2}} + \frac{B d^{2} \log{\left (x + \frac{\frac{B a^{3} d^{5}}{\left (a d - b c\right )^{2}} - \frac{3 B a^{2} b c d^{4}}{\left (a d - b c\right )^{2}} + \frac{3 B a b^{2} c^{2} d^{3}}{\left (a d - b c\right )^{2}} + B a d^{3} - \frac{B b^{3} c^{3} d^{2}}{\left (a d - b c\right )^{2}} + B b c d^{2}}{2 B b d^{3}} \right )}}{b g^{3} \left (a d - b c\right )^{2}} - \frac{A a d - A b c + 3 B a d - B b c + 2 B b d x}{2 a^{3} b d g^{3} - 2 a^{2} b^{2} c g^{3} + x^{2} \left (2 a b^{3} d g^{3} - 2 b^{4} c g^{3}\right ) + x \left (4 a^{2} b^{2} d g^{3} - 4 a b^{3} c g^{3}\right )} \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)**3,x)

[Out]

-B*log(e*(a + b*x)**2/(c + d*x)**2)/(2*a**2*b*g**3 + 4*a*b**2*g**3*x + 2*b**3*g**3*x**2) - B*d**2*log(x + (-B*
a**3*d**5/(a*d - b*c)**2 + 3*B*a**2*b*c*d**4/(a*d - b*c)**2 - 3*B*a*b**2*c**2*d**3/(a*d - b*c)**2 + B*a*d**3 +
B*b**3*c**3*d**2/(a*d - b*c)**2 + B*b*c*d**2)/(2*B*b*d**3))/(b*g**3*(a*d - b*c)**2) + B*d**2*log(x + (B*a**3*
d**5/(a*d - b*c)**2 - 3*B*a**2*b*c*d**4/(a*d - b*c)**2 + 3*B*a*b**2*c**2*d**3/(a*d - b*c)**2 + B*a*d**3 - B*b*
*3*c**3*d**2/(a*d - b*c)**2 + B*b*c*d**2)/(2*B*b*d**3))/(b*g**3*(a*d - b*c)**2) - (A*a*d - A*b*c + 3*B*a*d - B
*b*c + 2*B*b*d*x)/(2*a**3*b*d*g**3 - 2*a**2*b**2*c*g**3 + x**2*(2*a*b**3*d*g**3 - 2*b**4*c*g**3) + x*(4*a**2*b
**2*d*g**3 - 4*a*b**3*c*g**3))

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Giac [A]  time = 1.31868, size = 356, normalized size = 2.58 \begin{align*} \frac{B d^{2} \log \left (b x + a\right )}{b^{3} c^{2} g^{3} - 2 \, a b^{2} c d g^{3} + a^{2} b d^{2} g^{3}} - \frac{B d^{2} \log \left (d x + c\right )}{b^{3} c^{2} g^{3} - 2 \, a b^{2} c d g^{3} + a^{2} b d^{2} g^{3}} - \frac{B \log \left (\frac{b^{2} x^{2} + 2 \, a b x + a^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \,{\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} + \frac{2 \, B b d x - A b c - 2 \, B b c + A a d + 4 \, B a d}{2 \,{\left (b^{4} c g^{3} x^{2} - a b^{3} d g^{3} x^{2} + 2 \, a b^{3} c g^{3} x - 2 \, a^{2} b^{2} d g^{3} x + a^{2} b^{2} c g^{3} - a^{3} b d g^{3}\right )}} \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)^3,x, algorithm="giac")

[Out]

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