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

Optimal. Leaf size=144 \[ -\frac {B \log \left (\frac {e (a+b x)}{c+d x}\right )+A}{2 b g^3 (a+b x)^2}+\frac {B d^2 \log (a+b x)}{2 b g^3 (b c-a d)^2}-\frac {B d^2 \log (c+d x)}{2 b g^3 (b c-a d)^2}+\frac {B d}{2 b g^3 (a+b x) (b c-a d)}-\frac {B}{4 b g^3 (a+b x)^2} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2525, 12, 44} \[ -\frac {B \log \left (\frac {e (a+b x)}{c+d x}\right )+A}{2 b g^3 (a+b x)^2}+\frac {B d^2 \log (a+b x)}{2 b g^3 (b c-a d)^2}-\frac {B d^2 \log (c+d x)}{2 b g^3 (b c-a d)^2}+\frac {B d}{2 b g^3 (a+b x) (b c-a d)}-\frac {B}{4 b g^3 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-B/(4*b*g^3*(a + b*x)^2) + (B*d)/(2*b*(b*c - a*d)*g^3*(a + b*x)) + (B*d^2*Log[a + b*x])/(2*b*(b*c - a*d)^2*g^3
) - (A + B*Log[(e*(a + b*x))/(c + d*x)])/(2*b*g^3*(a + b*x)^2) - (B*d^2*Log[c + d*x])/(2*b*(b*c - a*d)^2*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 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])

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

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Mathematica [A]  time = 0.13, size = 110, normalized size = 0.76 \[ -\frac {2 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )+\frac {B \left (2 d^2 (a+b x)^2 \log (c+d x)+(b c-a d) (b (c-2 d x)-3 a d)-2 d^2 (a+b x)^2 \log (a+b x)\right )}{(b c-a d)^2}}{4 b g^3 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(2*(A + B*Log[(e*(a + b*x))/(c + d*x)]) + (B*((b*c - a*d)*(-3*a*d + b*(c - 2*d*x)) - 2*d^2*(a + b*x)^2*Lo
g[a + b*x] + 2*d^2*(a + b*x)^2*Log[c + d*x]))/(b*c - a*d)^2)/(b*g^3*(a + b*x)^2)

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fricas [A]  time = 0.61, size = 217, normalized size = 1.51 \[ -\frac {{\left (2 \, A + B\right )} b^{2} c^{2} - 4 \, {\left (A + B\right )} a b c d + {\left (2 \, A + 3 \, B\right )} a^{2} d^{2} - 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} x - 2 \, {\left (B b^{2} d^{2} x^{2} + 2 \, B a b d^{2} x - B b^{2} c^{2} + 2 \, B a b c d\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}{4 \, {\left ({\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} g^{3} x^{2} + 2 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} g^{3} x + {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} g^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((2*A + B)*b^2*c^2 - 4*(A + B)*a*b*c*d + (2*A + 3*B)*a^2*d^2 - 2*(B*b^2*c*d - B*a*b*d^2)*x - 2*(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*e*x + a*e)/(d*x + c)))/((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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giac [A]  time = 1.69, size = 237, normalized size = 1.65 \[ -\frac {{\left (2 \, B b e^{3} \log \left (\frac {b x e + a e}{d x + c}\right ) - \frac {4 \, {\left (b x e + a e\right )} B d e^{2} \log \left (\frac {b x e + a e}{d x + c}\right )}{d x + c} + 2 \, A b e^{3} + B b e^{3} - \frac {4 \, {\left (b x e + a e\right )} A d e^{2}}{d x + c} - \frac {4 \, {\left (b x e + a e\right )} B d e^{2}}{d x + c}\right )} {\left (\frac {b c}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}} - \frac {a d}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}}\right )}}{4 \, {\left (\frac {{\left (b x e + a e\right )}^{2} b c g^{3}}{{\left (d x + c\right )}^{2}} - \frac {{\left (b x e + a e\right )}^{2} a d g^{3}}{{\left (d x + c\right )}^{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 777, normalized size = 5.40 \[ -\frac {B a b d \,e^{2} \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{2 \left (a d -b c \right )^{3} \left (\frac {a e}{d x +c}-\frac {b c e}{\left (d x +c \right ) d}+\frac {b e}{d}\right )^{2} g^{3}}+\frac {B \,b^{2} c \,e^{2} \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{2 \left (a d -b c \right )^{3} \left (\frac {a e}{d x +c}-\frac {b c e}{\left (d x +c \right ) d}+\frac {b e}{d}\right )^{2} g^{3}}-\frac {A a b d \,e^{2}}{2 \left (a d -b c \right )^{3} \left (\frac {a e}{d x +c}-\frac {b c e}{\left (d x +c \right ) d}+\frac {b e}{d}\right )^{2} g^{3}}+\frac {A \,b^{2} c \,e^{2}}{2 \left (a d -b c \right )^{3} \left (\frac {a e}{d x +c}-\frac {b c e}{\left (d x +c \right ) d}+\frac {b e}{d}\right )^{2} g^{3}}-\frac {B a b d \,e^{2}}{4 \left (a d -b c \right )^{3} \left (\frac {a e}{d x +c}-\frac {b c e}{\left (d x +c \right ) d}+\frac {b e}{d}\right )^{2} g^{3}}+\frac {B a \,d^{2} e \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right )^{3} \left (\frac {a e}{d x +c}-\frac {b c e}{\left (d x +c \right ) d}+\frac {b e}{d}\right ) g^{3}}+\frac {B \,b^{2} c \,e^{2}}{4 \left (a d -b c \right )^{3} \left (\frac {a e}{d x +c}-\frac {b c e}{\left (d x +c \right ) d}+\frac {b e}{d}\right )^{2} g^{3}}-\frac {B b c d e \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right )^{3} \left (\frac {a e}{d x +c}-\frac {b c e}{\left (d x +c \right ) d}+\frac {b e}{d}\right ) g^{3}}+\frac {A a \,d^{2} e}{\left (a d -b c \right )^{3} \left (\frac {a e}{d x +c}-\frac {b c e}{\left (d x +c \right ) d}+\frac {b e}{d}\right ) g^{3}}-\frac {A b c d e}{\left (a d -b c \right )^{3} \left (\frac {a e}{d x +c}-\frac {b c e}{\left (d x +c \right ) d}+\frac {b e}{d}\right ) g^{3}}+\frac {B a \,d^{2} e}{\left (a d -b c \right )^{3} \left (\frac {a e}{d x +c}-\frac {b c e}{\left (d x +c \right ) d}+\frac {b e}{d}\right ) g^{3}}-\frac {B b c d e}{\left (a d -b c \right )^{3} \left (\frac {a e}{d x +c}-\frac {b c e}{\left (d x +c \right ) d}+\frac {b e}{d}\right ) g^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.26, size = 255, normalized size = 1.77 \[ \frac {1}{4} \, B {\left (\frac {2 \, b d x - b c + 3 \, a d}{{\left (b^{4} c - a b^{3} d\right )} g^{3} x^{2} + 2 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} g^{3} x + {\left (a^{2} b^{2} c - a^{3} b d\right )} g^{3}} - \frac {2 \, \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right )}{b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}} + \frac {2 \, d^{2} \log \left (b x + a\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}} - \frac {2 \, d^{2} \log \left (d x + c\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}}\right )} - \frac {A}{2 \, {\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*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) - 2*log(b*e*x/(d*x + c) + a*e/(d*x + c))/(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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mupad [B]  time = 5.04, size = 209, normalized size = 1.45 \[ -\frac {\frac {2\,A\,a\,d-2\,A\,b\,c+3\,B\,a\,d-B\,b\,c}{2\,\left (a\,d-b\,c\right )}+\frac {B\,b\,d\,x}{a\,d-b\,c}}{2\,a^2\,b\,g^3+4\,a\,b^2\,g^3\,x+2\,b^3\,g^3\,x^2}-\frac {B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )}{2\,b^2\,g^3\,\left (2\,a\,x+b\,x^2+\frac {a^2}{b}\right )}-\frac {B\,d^2\,\mathrm {atanh}\left (\frac {2\,b^3\,c^2\,g^3-2\,a^2\,b\,d^2\,g^3}{2\,b\,g^3\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{b\,g^3\,{\left (a\,d-b\,c\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 2.71, size = 422, normalized size = 2.93 \[ - \frac {B \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )}}{2 a^{2} b g^{3} + 4 a b^{2} g^{3} x + 2 b^{3} g^{3} x^{2}} - \frac {B d^{2} \log {\left (x + \frac {- \frac {B a^{3} d^{5}}{\left (a d - b c\right )^{2}} + \frac {3 B a^{2} b c d^{4}}{\left (a d - b c\right )^{2}} - \frac {3 B a b^{2} c^{2} d^{3}}{\left (a d - b c\right )^{2}} + B a d^{3} + \frac {B b^{3} c^{3} d^{2}}{\left (a d - b c\right )^{2}} + B b c d^{2}}{2 B b d^{3}} \right )}}{2 b g^{3} \left (a d - b c\right )^{2}} + \frac {B d^{2} \log {\left (x + \frac {\frac {B a^{3} d^{5}}{\left (a d - b c\right )^{2}} - \frac {3 B a^{2} b c d^{4}}{\left (a d - b c\right )^{2}} + \frac {3 B a b^{2} c^{2} d^{3}}{\left (a d - b c\right )^{2}} + B a d^{3} - \frac {B b^{3} c^{3} d^{2}}{\left (a d - b c\right )^{2}} + B b c d^{2}}{2 B b d^{3}} \right )}}{2 b g^{3} \left (a d - b c\right )^{2}} + \frac {- 2 A a d + 2 A b c - 3 B a d + B b c - 2 B b d x}{4 a^{3} b d g^{3} - 4 a^{2} b^{2} c g^{3} + x^{2} \left (4 a b^{3} d g^{3} - 4 b^{4} c g^{3}\right ) + x \left (8 a^{2} b^{2} d g^{3} - 8 a b^{3} c g^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-B*log(e*(a + b*x)/(c + d*x))/(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))/(2*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))/(2*b*g**3*(a*d - b*c)**2) + (-2*A*a*d + 2*A*b*c - 3*B*a*d
+ B*b*c - 2*B*b*d*x)/(4*a**3*b*d*g**3 - 4*a**2*b**2*c*g**3 + x**2*(4*a*b**3*d*g**3 - 4*b**4*c*g**3) + x*(8*a**
2*b**2*d*g**3 - 8*a*b**3*c*g**3))

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