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

Optimal. Leaf size=183 \[ -\frac{B \log \left (\frac{e (a+b x)}{c+d x}\right )+A}{2 g (f+g x)^2}+\frac{b^2 B \log (a+b x)}{2 g (b f-a g)^2}-\frac{B (b c-a d)}{2 (f+g x) (b f-a g) (d f-c g)}+\frac{B (b c-a d) \log (f+g x) (-a d g-b c g+2 b d f)}{2 (b f-a g)^2 (d f-c g)^2}-\frac{B d^2 \log (c+d x)}{2 g (d f-c g)^2} \]

[Out]

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

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Rubi [A]  time = 0.185454, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2525, 12, 72} \[ -\frac{B \log \left (\frac{e (a+b x)}{c+d x}\right )+A}{2 g (f+g x)^2}+\frac{b^2 B \log (a+b x)}{2 g (b f-a g)^2}-\frac{B (b c-a d)}{2 (f+g x) (b f-a g) (d f-c g)}+\frac{B (b c-a d) \log (f+g x) (-a d g-b c g+2 b d f)}{2 (b f-a g)^2 (d f-c g)^2}-\frac{B d^2 \log (c+d x)}{2 g (d f-c g)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.645492, size = 169, normalized size = 0.92 \[ \frac{B (b c-a d) \left (\frac{b^2 \log (a+b x)}{(b c-a d) (b f-a g)^2}+\frac{\frac{d^2 \log (c+d x)}{a d-b c}+\frac{g (c g-d f)}{(f+g x) (b f-a g)}-\frac{g \log (f+g x) (a d g+b c g-2 b d f)}{(b f-a g)^2}}{(d f-c g)^2}\right )-\frac{B \log \left (\frac{e (a+b x)}{c+d x}\right )+A}{(f+g x)^2}}{2 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.175, size = 5274, normalized size = 28.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*ln(e*(b*x+a)/(d*x+c)))/(g*x+f)^3,x)

[Out]

result too large to display

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Maxima [B]  time = 1.24118, size = 474, normalized size = 2.59 \begin{align*} \frac{1}{2} \,{\left (\frac{b^{2} \log \left (b x + a\right )}{b^{2} f^{2} g - 2 \, a b f g^{2} + a^{2} g^{3}} - \frac{d^{2} \log \left (d x + c\right )}{d^{2} f^{2} g - 2 \, c d f g^{2} + c^{2} g^{3}} + \frac{{\left (2 \,{\left (b^{2} c d - a b d^{2}\right )} f -{\left (b^{2} c^{2} - a^{2} d^{2}\right )} g\right )} \log \left (g x + f\right )}{b^{2} d^{2} f^{4} + a^{2} c^{2} g^{4} - 2 \,{\left (b^{2} c d + a b d^{2}\right )} f^{3} g +{\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} f^{2} g^{2} - 2 \,{\left (a b c^{2} + a^{2} c d\right )} f g^{3}} - \frac{b c - a d}{b d f^{3} + a c f g^{2} -{\left (b c + a d\right )} f^{2} g +{\left (b d f^{2} g + a c g^{3} -{\left (b c + a d\right )} f g^{2}\right )} x} - \frac{\log \left (\frac{b e x}{d x + c} + \frac{a e}{d x + c}\right )}{g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g}\right )} B - \frac{A}{2 \,{\left (g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(g*x+f)^3,x, algorithm="maxima")

[Out]

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

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Fricas [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*log(e*(b*x+a)/(d*x+c)))/(g*x+f)^3,x, algorithm="fricas")

[Out]

Timed out

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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)/(d*x+c)))/(g*x+f)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.74853, size = 1110, normalized size = 6.07 \begin{align*} \frac{{\left (2 \, B b^{2} c d f - 2 \, B a b d^{2} f - B b^{2} c^{2} g + B a^{2} d^{2} g\right )} \log \left (g x + f\right )}{2 \,{\left (b^{2} d^{2} f^{4} - 2 \, b^{2} c d f^{3} g - 2 \, a b d^{2} f^{3} g + b^{2} c^{2} f^{2} g^{2} + 4 \, a b c d f^{2} g^{2} + a^{2} d^{2} f^{2} g^{2} - 2 \, a b c^{2} f g^{3} - 2 \, a^{2} c d f g^{3} + a^{2} c^{2} g^{4}\right )}} - \frac{B \log \left (\frac{b x + a}{d x + c}\right )}{2 \,{\left (g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g\right )}} - \frac{{\left (2 \, B b^{2} c d f - 2 \, B a b d^{2} f - B b^{2} c^{2} g + B a^{2} d^{2} g\right )} \log \left ({\left | b d x^{2} + b c x + a d x + a c \right |}\right )}{4 \,{\left (b^{2} d^{2} f^{4} - 2 \, b^{2} c d f^{3} g - 2 \, a b d^{2} f^{3} g + b^{2} c^{2} f^{2} g^{2} + 4 \, a b c d f^{2} g^{2} + a^{2} d^{2} f^{2} g^{2} - 2 \, a b c^{2} f g^{3} - 2 \, a^{2} c d f g^{3} + a^{2} c^{2} g^{4}\right )}} - \frac{B b c g^{2} x - B a d g^{2} x + A b d f^{2} + B b d f^{2} - A b c f g - A a d f g - 2 \, B a d f g + A a c g^{2} + B a c g^{2}}{2 \,{\left (b d f^{2} g^{3} x^{2} - b c f g^{4} x^{2} - a d f g^{4} x^{2} + a c g^{5} x^{2} + 2 \, b d f^{3} g^{2} x - 2 \, b c f^{2} g^{3} x - 2 \, a d f^{2} g^{3} x + 2 \, a c f g^{4} x + b d f^{4} g - b c f^{3} g^{2} - a d f^{3} g^{2} + a c f^{2} g^{3}\right )}} + \frac{{\left (2 \, B b^{3} c d^{2} f^{2} - 2 \, B a b^{2} d^{3} f^{2} - 2 \, B b^{3} c^{2} d f g + 2 \, B a^{2} b d^{3} f g + B b^{3} c^{3} g^{2} - B a b^{2} c^{2} d g^{2} + B a^{2} b c d^{2} g^{2} - B a^{3} d^{3} g^{2}\right )} \log \left ({\left | \frac{2 \, b d x + b c + a d -{\left | -b c + a d \right |}}{2 \, b d x + b c + a d +{\left | -b c + a d \right |}} \right |}\right )}{4 \,{\left (b^{2} d^{2} f^{4} g - 2 \, b^{2} c d f^{3} g^{2} - 2 \, a b d^{2} f^{3} g^{2} + b^{2} c^{2} f^{2} g^{3} + 4 \, a b c d f^{2} g^{3} + a^{2} d^{2} f^{2} g^{3} - 2 \, a b c^{2} f g^{4} - 2 \, a^{2} c d f g^{4} + a^{2} c^{2} g^{5}\right )}{\left | -b c + a d \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(g*x+f)^3,x, algorithm="giac")

[Out]

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