3.238 \(\int \frac {A+B \log (\frac {e (a+b x)}{c+d x})}{(f+g x)^4} \, dx\)

Optimal. Leaf size=275 \[ \frac {B (b c-a d) \log (f+g x) \left (a^2 d^2 g^2-a b d g (3 d f-c g)+b^2 \left (c^2 g^2-3 c d f g+3 d^2 f^2\right )\right )}{3 (b f-a g)^3 (d f-c g)^3}-\frac {B \log \left (\frac {e (a+b x)}{c+d x}\right )+A}{3 g (f+g x)^3}+\frac {b^3 B \log (a+b x)}{3 g (b f-a g)^3}-\frac {B (b c-a d) (-a d g-b c g+2 b d f)}{3 (f+g x) (b f-a g)^2 (d f-c g)^2}-\frac {B (b c-a d)}{6 (f+g x)^2 (b f-a g) (d f-c g)}-\frac {B d^3 \log (c+d x)}{3 g (d f-c g)^3} \]

[Out]

-1/6*B*(-a*d+b*c)/(-a*g+b*f)/(-c*g+d*f)/(g*x+f)^2-1/3*B*(-a*d+b*c)*(-a*d*g-b*c*g+2*b*d*f)/(-a*g+b*f)^2/(-c*g+d
*f)^2/(g*x+f)+1/3*b^3*B*ln(b*x+a)/g/(-a*g+b*f)^3+1/3*(-A-B*ln(e*(b*x+a)/(d*x+c)))/g/(g*x+f)^3-1/3*B*d^3*ln(d*x
+c)/g/(-c*g+d*f)^3+1/3*B*(-a*d+b*c)*(a^2*d^2*g^2-a*b*d*g*(-c*g+3*d*f)+b^2*(c^2*g^2-3*c*d*f*g+3*d^2*f^2))*ln(g*
x+f)/(-a*g+b*f)^3/(-c*g+d*f)^3

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Rubi [A]  time = 0.40, antiderivative size = 275, normalized size of antiderivative = 1.00, 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} \[ \frac {B (b c-a d) \log (f+g x) \left (a^2 d^2 g^2-a b d g (3 d f-c g)+b^2 \left (c^2 g^2-3 c d f g+3 d^2 f^2\right )\right )}{3 (b f-a g)^3 (d f-c g)^3}-\frac {B \log \left (\frac {e (a+b x)}{c+d x}\right )+A}{3 g (f+g x)^3}+\frac {b^3 B \log (a+b x)}{3 g (b f-a g)^3}-\frac {B (b c-a d) (-a d g-b c g+2 b d f)}{3 (f+g x) (b f-a g)^2 (d f-c g)^2}-\frac {B (b c-a d)}{6 (f+g x)^2 (b f-a g) (d f-c g)}-\frac {B d^3 \log (c+d x)}{3 g (d f-c g)^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Log[(e*(a + b*x))/(c + d*x)])/(f + g*x)^4,x]

[Out]

-(B*(b*c - a*d))/(6*(b*f - a*g)*(d*f - c*g)*(f + g*x)^2) - (B*(b*c - a*d)*(2*b*d*f - b*c*g - a*d*g))/(3*(b*f -
a*g)^2*(d*f - c*g)^2*(f + g*x)) + (b^3*B*Log[a + b*x])/(3*g*(b*f - a*g)^3) - (A + B*Log[(e*(a + b*x))/(c + d*
x)])/(3*g*(f + g*x)^3) - (B*d^3*Log[c + d*x])/(3*g*(d*f - c*g)^3) + (B*(b*c - a*d)*(a^2*d^2*g^2 - a*b*d*g*(3*d
*f - c*g) + b^2*(3*d^2*f^2 - 3*c*d*f*g + c^2*g^2))*Log[f + g*x])/(3*(b*f - a*g)^3*(d*f - c*g)^3)

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]

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]

Rubi steps

\begin {align*} \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(f+g x)^4} \, dx &=-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{3 g (f+g x)^3}+\frac {B \int \frac {b c-a d}{(a+b x) (c+d x) (f+g x)^3} \, dx}{3 g}\\ &=-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{3 g (f+g x)^3}+\frac {(B (b c-a d)) \int \frac {1}{(a+b x) (c+d x) (f+g x)^3} \, dx}{3 g}\\ &=-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{3 g (f+g x)^3}+\frac {(B (b c-a d)) \int \left (\frac {b^4}{(b c-a d) (b f-a g)^3 (a+b x)}+\frac {d^4}{(b c-a d) (-d f+c g)^3 (c+d x)}+\frac {g^2}{(b f-a g) (d f-c g) (f+g x)^3}-\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)^2}+\frac {g^2 \left (a^2 d^2 g^2-a b d g (3 d f-c g)+b^2 \left (3 d^2 f^2-3 c d f g+c^2 g^2\right )\right )}{(b f-a g)^3 (d f-c g)^3 (f+g x)}\right ) \, dx}{3 g}\\ &=-\frac {B (b c-a d)}{6 (b f-a g) (d f-c g) (f+g x)^2}-\frac {B (b c-a d) (2 b d f-b c g-a d g)}{3 (b f-a g)^2 (d f-c g)^2 (f+g x)}+\frac {b^3 B \log (a+b x)}{3 g (b f-a g)^3}-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{3 g (f+g x)^3}-\frac {B d^3 \log (c+d x)}{3 g (d f-c g)^3}+\frac {B (b c-a d) \left (a^2 d^2 g^2-a b d g (3 d f-c g)+b^2 \left (3 d^2 f^2-3 c d f g+c^2 g^2\right )\right ) \log (f+g x)}{3 (b f-a g)^3 (d f-c g)^3}\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 260, normalized size = 0.95 \[ \frac {B (b c-a d) \left (\frac {g \log (f+g x) \left (a^2 d^2 g^2+a b d g (c g-3 d f)+b^2 \left (c^2 g^2-3 c d f g+3 d^2 f^2\right )\right )}{(b f-a g)^3 (d f-c g)^3}+\frac {b^3 \log (a+b x)}{(b c-a d) (b f-a g)^3}+\frac {d^3 \log (c+d x)}{(b c-a d) (c g-d f)^3}+\frac {g (a d g+b c g-2 b d f)}{(f+g x) (b f-a g)^2 (d f-c g)^2}-\frac {g}{2 (f+g x)^2 (b f-a g) (d f-c g)}\right )-\frac {B \log \left (\frac {e (a+b x)}{c+d x}\right )+A}{(f+g x)^3}}{3 g} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Log[(e*(a + b*x))/(c + d*x)])/(f + g*x)^4,x]

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)^4,x, algorithm="fricas")

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)^4,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.18, size = 18285, normalized size = 66.49 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*ln((b*x+a)/(d*x+c)*e)+A)/(g*x+f)^4,x)

[Out]

result too large to display

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maxima [B]  time = 1.29, size = 848, normalized size = 3.08 \[ \frac {1}{6} \, {\left (\frac {2 \, b^{3} \log \left (b x + a\right )}{b^{3} f^{3} g - 3 \, a b^{2} f^{2} g^{2} + 3 \, a^{2} b f g^{3} - a^{3} g^{4}} - \frac {2 \, d^{3} \log \left (d x + c\right )}{d^{3} f^{3} g - 3 \, c d^{2} f^{2} g^{2} + 3 \, c^{2} d f g^{3} - c^{3} g^{4}} + \frac {2 \, {\left (3 \, {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )} f^{2} - 3 \, {\left (b^{3} c^{2} d - a^{2} b d^{3}\right )} f g + {\left (b^{3} c^{3} - a^{3} d^{3}\right )} g^{2}\right )} \log \left (g x + f\right )}{b^{3} d^{3} f^{6} + a^{3} c^{3} g^{6} - 3 \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} f^{5} g + 3 \, {\left (b^{3} c^{2} d + 3 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} f^{4} g^{2} - {\left (b^{3} c^{3} + 9 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} f^{3} g^{3} + 3 \, {\left (a b^{2} c^{3} + 3 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} f^{2} g^{4} - 3 \, {\left (a^{2} b c^{3} + a^{3} c^{2} d\right )} f g^{5}} - \frac {5 \, {\left (b^{2} c d - a b d^{2}\right )} f^{2} - 3 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} f g + {\left (a b c^{2} - a^{2} c d\right )} g^{2} + 2 \, {\left (2 \, {\left (b^{2} c d - a b d^{2}\right )} f g - {\left (b^{2} c^{2} - a^{2} d^{2}\right )} g^{2}\right )} x}{b^{2} d^{2} f^{6} + a^{2} c^{2} f^{2} g^{4} - 2 \, {\left (b^{2} c d + a b d^{2}\right )} f^{5} g + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} f^{4} g^{2} - 2 \, {\left (a b c^{2} + a^{2} c d\right )} f^{3} g^{3} + {\left (b^{2} d^{2} f^{4} g^{2} + a^{2} c^{2} g^{6} - 2 \, {\left (b^{2} c d + a b d^{2}\right )} f^{3} g^{3} + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} f^{2} g^{4} - 2 \, {\left (a b c^{2} + a^{2} c d\right )} f g^{5}\right )} x^{2} + 2 \, {\left (b^{2} d^{2} f^{5} g + a^{2} c^{2} f g^{5} - 2 \, {\left (b^{2} c d + a b d^{2}\right )} f^{4} g^{2} + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} f^{3} g^{3} - 2 \, {\left (a b c^{2} + a^{2} c d\right )} f^{2} g^{4}\right )} x} - \frac {2 \, \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right )}{g^{4} x^{3} + 3 \, f g^{3} x^{2} + 3 \, f^{2} g^{2} x + f^{3} g}\right )} B - \frac {A}{3 \, {\left (g^{4} x^{3} + 3 \, f g^{3} x^{2} + 3 \, f^{2} g^{2} x + f^{3} g\right )}} \]

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)^4,x, algorithm="maxima")

[Out]

1/6*(2*b^3*log(b*x + a)/(b^3*f^3*g - 3*a*b^2*f^2*g^2 + 3*a^2*b*f*g^3 - a^3*g^4) - 2*d^3*log(d*x + c)/(d^3*f^3*
g - 3*c*d^2*f^2*g^2 + 3*c^2*d*f*g^3 - c^3*g^4) + 2*(3*(b^3*c*d^2 - a*b^2*d^3)*f^2 - 3*(b^3*c^2*d - a^2*b*d^3)*
f*g + (b^3*c^3 - a^3*d^3)*g^2)*log(g*x + f)/(b^3*d^3*f^6 + a^3*c^3*g^6 - 3*(b^3*c*d^2 + a*b^2*d^3)*f^5*g + 3*(
b^3*c^2*d + 3*a*b^2*c*d^2 + a^2*b*d^3)*f^4*g^2 - (b^3*c^3 + 9*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*f^3*g^3 +
3*(a*b^2*c^3 + 3*a^2*b*c^2*d + a^3*c*d^2)*f^2*g^4 - 3*(a^2*b*c^3 + a^3*c^2*d)*f*g^5) - (5*(b^2*c*d - a*b*d^2)
*f^2 - 3*(b^2*c^2 - a^2*d^2)*f*g + (a*b*c^2 - a^2*c*d)*g^2 + 2*(2*(b^2*c*d - a*b*d^2)*f*g - (b^2*c^2 - a^2*d^2
)*g^2)*x)/(b^2*d^2*f^6 + a^2*c^2*f^2*g^4 - 2*(b^2*c*d + a*b*d^2)*f^5*g + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*f^4*g
^2 - 2*(a*b*c^2 + a^2*c*d)*f^3*g^3 + (b^2*d^2*f^4*g^2 + a^2*c^2*g^6 - 2*(b^2*c*d + a*b*d^2)*f^3*g^3 + (b^2*c^2
+ 4*a*b*c*d + a^2*d^2)*f^2*g^4 - 2*(a*b*c^2 + a^2*c*d)*f*g^5)*x^2 + 2*(b^2*d^2*f^5*g + a^2*c^2*f*g^5 - 2*(b^2
*c*d + a*b*d^2)*f^4*g^2 + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*f^3*g^3 - 2*(a*b*c^2 + a^2*c*d)*f^2*g^4)*x) - 2*log(
b*e*x/(d*x + c) + a*e/(d*x + c))/(g^4*x^3 + 3*f*g^3*x^2 + 3*f^2*g^2*x + f^3*g))*B - 1/3*A/(g^4*x^3 + 3*f*g^3*x
^2 + 3*f^2*g^2*x + f^3*g)

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mupad [B]  time = 10.67, size = 1154, normalized size = 4.20 \[ \frac {\ln \left (f+g\,x\right )\,\left (g\,\left (3\,B\,a^2\,b\,d^3\,f-3\,B\,b^3\,c^2\,d\,f\right )-g^2\,\left (B\,a^3\,d^3-B\,b^3\,c^3\right )-3\,B\,a\,b^2\,d^3\,f^2+3\,B\,b^3\,c\,d^2\,f^2\right )}{3\,a^3\,c^3\,g^6-9\,a^3\,c^2\,d\,f\,g^5+9\,a^3\,c\,d^2\,f^2\,g^4-3\,a^3\,d^3\,f^3\,g^3-9\,a^2\,b\,c^3\,f\,g^5+27\,a^2\,b\,c^2\,d\,f^2\,g^4-27\,a^2\,b\,c\,d^2\,f^3\,g^3+9\,a^2\,b\,d^3\,f^4\,g^2+9\,a\,b^2\,c^3\,f^2\,g^4-27\,a\,b^2\,c^2\,d\,f^3\,g^3+27\,a\,b^2\,c\,d^2\,f^4\,g^2-9\,a\,b^2\,d^3\,f^5\,g-3\,b^3\,c^3\,f^3\,g^3+9\,b^3\,c^2\,d\,f^4\,g^2-9\,b^3\,c\,d^2\,f^5\,g+3\,b^3\,d^3\,f^6}-\frac {\frac {2\,A\,a^2\,c^2\,g^4+2\,A\,b^2\,d^2\,f^4+2\,A\,a^2\,d^2\,f^2\,g^2+2\,A\,b^2\,c^2\,f^2\,g^2+3\,B\,a^2\,d^2\,f^2\,g^2-3\,B\,b^2\,c^2\,f^2\,g^2-4\,A\,a\,b\,c^2\,f\,g^3-4\,A\,a\,b\,d^2\,f^3\,g+B\,a\,b\,c^2\,f\,g^3-4\,A\,a^2\,c\,d\,f\,g^3-5\,B\,a\,b\,d^2\,f^3\,g-4\,A\,b^2\,c\,d\,f^3\,g-B\,a^2\,c\,d\,f\,g^3+5\,B\,b^2\,c\,d\,f^3\,g+8\,A\,a\,b\,c\,d\,f^2\,g^2}{2\,\left (a^2\,c^2\,g^4-2\,a^2\,c\,d\,f\,g^3+a^2\,d^2\,f^2\,g^2-2\,a\,b\,c^2\,f\,g^3+4\,a\,b\,c\,d\,f^2\,g^2-2\,a\,b\,d^2\,f^3\,g+b^2\,c^2\,f^2\,g^2-2\,b^2\,c\,d\,f^3\,g+b^2\,d^2\,f^4\right )}+\frac {x^2\,\left (B\,a^2\,d^2\,g^4-2\,B\,f\,a\,b\,d^2\,g^3-B\,b^2\,c^2\,g^4+2\,B\,f\,b^2\,c\,d\,g^3\right )}{a^2\,c^2\,g^4-2\,a^2\,c\,d\,f\,g^3+a^2\,d^2\,f^2\,g^2-2\,a\,b\,c^2\,f\,g^3+4\,a\,b\,c\,d\,f^2\,g^2-2\,a\,b\,d^2\,f^3\,g+b^2\,c^2\,f^2\,g^2-2\,b^2\,c\,d\,f^3\,g+b^2\,d^2\,f^4}+\frac {x\,\left (-B\,a^2\,c\,d\,g^4+5\,B\,a^2\,d^2\,f\,g^3+B\,a\,b\,c^2\,g^4-9\,B\,a\,b\,d^2\,f^2\,g^2-5\,B\,b^2\,c^2\,f\,g^3+9\,B\,b^2\,c\,d\,f^2\,g^2\right )}{2\,\left (a^2\,c^2\,g^4-2\,a^2\,c\,d\,f\,g^3+a^2\,d^2\,f^2\,g^2-2\,a\,b\,c^2\,f\,g^3+4\,a\,b\,c\,d\,f^2\,g^2-2\,a\,b\,d^2\,f^3\,g+b^2\,c^2\,f^2\,g^2-2\,b^2\,c\,d\,f^3\,g+b^2\,d^2\,f^4\right )}}{3\,f^3\,g+9\,f^2\,g^2\,x+9\,f\,g^3\,x^2+3\,g^4\,x^3}-\frac {B\,b^3\,\ln \left (a+b\,x\right )}{3\,a^3\,g^4-9\,a^2\,b\,f\,g^3+9\,a\,b^2\,f^2\,g^2-3\,b^3\,f^3\,g}+\frac {B\,d^3\,\ln \left (c+d\,x\right )}{3\,c^3\,g^4-9\,c^2\,d\,f\,g^3+9\,c\,d^2\,f^2\,g^2-3\,d^3\,f^3\,g}-\frac {B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )}{3\,g\,\left (f^3+3\,f^2\,g\,x+3\,f\,g^2\,x^2+g^3\,x^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*log((e*(a + b*x))/(c + d*x)))/(f + g*x)^4,x)

[Out]

(log(f + g*x)*(g*(3*B*a^2*b*d^3*f - 3*B*b^3*c^2*d*f) - g^2*(B*a^3*d^3 - B*b^3*c^3) - 3*B*a*b^2*d^3*f^2 + 3*B*b
^3*c*d^2*f^2))/(3*a^3*c^3*g^6 + 3*b^3*d^3*f^6 - 3*a^3*d^3*f^3*g^3 - 3*b^3*c^3*f^3*g^3 - 9*a^2*b*c^3*f*g^5 - 9*
a*b^2*d^3*f^5*g - 9*a^3*c^2*d*f*g^5 - 9*b^3*c*d^2*f^5*g + 9*a*b^2*c^3*f^2*g^4 + 9*a^2*b*d^3*f^4*g^2 + 9*a^3*c*
d^2*f^2*g^4 + 9*b^3*c^2*d*f^4*g^2 + 27*a*b^2*c*d^2*f^4*g^2 - 27*a*b^2*c^2*d*f^3*g^3 - 27*a^2*b*c*d^2*f^3*g^3 +
27*a^2*b*c^2*d*f^2*g^4) - ((2*A*a^2*c^2*g^4 + 2*A*b^2*d^2*f^4 + 2*A*a^2*d^2*f^2*g^2 + 2*A*b^2*c^2*f^2*g^2 + 3
*B*a^2*d^2*f^2*g^2 - 3*B*b^2*c^2*f^2*g^2 - 4*A*a*b*c^2*f*g^3 - 4*A*a*b*d^2*f^3*g + B*a*b*c^2*f*g^3 - 4*A*a^2*c
*d*f*g^3 - 5*B*a*b*d^2*f^3*g - 4*A*b^2*c*d*f^3*g - B*a^2*c*d*f*g^3 + 5*B*b^2*c*d*f^3*g + 8*A*a*b*c*d*f^2*g^2)/
(2*(a^2*c^2*g^4 + b^2*d^2*f^4 + a^2*d^2*f^2*g^2 + b^2*c^2*f^2*g^2 - 2*a*b*c^2*f*g^3 - 2*a*b*d^2*f^3*g - 2*a^2*
c*d*f*g^3 - 2*b^2*c*d*f^3*g + 4*a*b*c*d*f^2*g^2)) + (x^2*(B*a^2*d^2*g^4 - B*b^2*c^2*g^4 - 2*B*a*b*d^2*f*g^3 +
2*B*b^2*c*d*f*g^3))/(a^2*c^2*g^4 + b^2*d^2*f^4 + a^2*d^2*f^2*g^2 + b^2*c^2*f^2*g^2 - 2*a*b*c^2*f*g^3 - 2*a*b*d
^2*f^3*g - 2*a^2*c*d*f*g^3 - 2*b^2*c*d*f^3*g + 4*a*b*c*d*f^2*g^2) + (x*(5*B*a^2*d^2*f*g^3 - 5*B*b^2*c^2*f*g^3
+ B*a*b*c^2*g^4 - B*a^2*c*d*g^4 - 9*B*a*b*d^2*f^2*g^2 + 9*B*b^2*c*d*f^2*g^2))/(2*(a^2*c^2*g^4 + b^2*d^2*f^4 +
a^2*d^2*f^2*g^2 + b^2*c^2*f^2*g^2 - 2*a*b*c^2*f*g^3 - 2*a*b*d^2*f^3*g - 2*a^2*c*d*f*g^3 - 2*b^2*c*d*f^3*g + 4*
a*b*c*d*f^2*g^2)))/(3*f^3*g + 3*g^4*x^3 + 9*f^2*g^2*x + 9*f*g^3*x^2) - (B*b^3*log(a + b*x))/(3*a^3*g^4 - 3*b^3
*f^3*g + 9*a*b^2*f^2*g^2 - 9*a^2*b*f*g^3) + (B*d^3*log(c + d*x))/(3*c^3*g^4 - 3*d^3*f^3*g + 9*c*d^2*f^2*g^2 -
9*c^2*d*f*g^3) - (B*log((e*(a + b*x))/(c + d*x)))/(3*g*(f^3 + g^3*x^3 + 3*f^2*g*x + 3*f*g^2*x^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**4,x)

[Out]

Timed out

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