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

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

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

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

Antiderivative was successfully verified.

[In]

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

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

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Mathematica [A]  time = 0.10, size = 128, normalized size = 0.89 \[ -\frac {(b c-a d) \left (-2 a A d+2 B (b c-a d) \log \left (\frac {e (c+d x)}{a+b x}\right )+3 a B d+2 A b c-b B c+2 b B d x\right )-2 B d^2 (a+b x)^2 \log (c+d x)+2 B d^2 (a+b x)^2 \log (a+b x)}{4 b g^3 (a+b x)^2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.70, size = 221, normalized size = 1.53 \[ -\frac {{\left (2 \, A - B\right )} b^{2} c^{2} - 4 \, {\left (A - B\right )} a b c d + {\left (2 \, A - 3 \, B\right )} a^{2} d^{2} + 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} x - 2 \, {\left (B b^{2} d^{2} x^{2} + 2 \, B a b d^{2} x - B b^{2} c^{2} + 2 \, B a b c d\right )} \log \left (\frac {d e x + c e}{b x + a}\right )}{4 \, {\left ({\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} g^{3} x^{2} + 2 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} g^{3} x + {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} g^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(d*x+c)/(b*x+a)))/(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((d*e*x + c*e)/(b*x + a)))/((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 = 0.86, size = 254, normalized size = 1.76 \[ \frac {{\left (\frac {4 \, {\left (d x e + c e\right )} B d e \log \left (\frac {d x e + c e}{b x + a}\right )}{b x + a} + \frac {4 \, {\left (d x e + c e\right )} A d e}{b x + a} - \frac {4 \, {\left (d x e + c e\right )} B d e}{b x + a} - \frac {2 \, {\left (d x e + c e\right )}^{2} B b \log \left (\frac {d x e + c e}{b x + a}\right )}{{\left (b x + a\right )}^{2}} - \frac {2 \, {\left (d x e + c e\right )}^{2} A b}{{\left (b x + a\right )}^{2}} + \frac {{\left (d x e + c e\right )}^{2} B b}{{\left (b x + a\right )}^{2}}\right )} {\left (\frac {b c}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}} - \frac {a d}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}}\right )}}{4 \, {\left (b c g^{3} e - a d g^{3} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 753, normalized size = 5.23 \[ -\frac {B \,a^{3} d^{3} \ln \left (\frac {d e}{b}-\frac {\left (a d -b c \right ) e}{\left (b x +a \right ) b}\right )}{2 \left (a d -b c \right )^{3} \left (b x +a \right )^{2} b \,g^{3}}+\frac {3 B \,a^{2} c \,d^{2} \ln \left (\frac {d e}{b}-\frac {\left (a d -b c \right ) e}{\left (b x +a \right ) b}\right )}{2 \left (a d -b c \right )^{3} \left (b x +a \right )^{2} g^{3}}-\frac {3 B a b \,c^{2} d \ln \left (\frac {d e}{b}-\frac {\left (a d -b c \right ) e}{\left (b x +a \right ) b}\right )}{2 \left (a d -b c \right )^{3} \left (b x +a \right )^{2} g^{3}}+\frac {B \,b^{2} c^{3} \ln \left (\frac {d e}{b}-\frac {\left (a d -b c \right ) e}{\left (b x +a \right ) b}\right )}{2 \left (a d -b c \right )^{3} \left (b x +a \right )^{2} g^{3}}-\frac {A \,a^{3} d^{3}}{2 \left (a d -b c \right )^{3} \left (b x +a \right )^{2} b \,g^{3}}+\frac {3 A \,a^{2} c \,d^{2}}{2 \left (a d -b c \right )^{3} \left (b x +a \right )^{2} g^{3}}-\frac {3 A a b \,c^{2} d}{2 \left (a d -b c \right )^{3} \left (b x +a \right )^{2} g^{3}}+\frac {A \,b^{2} c^{3}}{2 \left (a d -b c \right )^{3} \left (b x +a \right )^{2} g^{3}}+\frac {B \,a^{3} d^{3}}{4 \left (a d -b c \right )^{3} \left (b x +a \right )^{2} b \,g^{3}}-\frac {3 B \,a^{2} c \,d^{2}}{4 \left (a d -b c \right )^{3} \left (b x +a \right )^{2} g^{3}}+\frac {3 B a b \,c^{2} d}{4 \left (a d -b c \right )^{3} \left (b x +a \right )^{2} g^{3}}-\frac {B \,b^{2} c^{3}}{4 \left (a d -b c \right )^{3} \left (b x +a \right )^{2} g^{3}}+\frac {B \,a^{2} d^{3}}{2 \left (a d -b c \right )^{3} \left (b x +a \right ) b \,g^{3}}+\frac {B a \,d^{3} \ln \left (\frac {d e}{b}-\frac {\left (a d -b c \right ) e}{\left (b x +a \right ) b}\right )}{2 \left (a d -b c \right )^{3} b \,g^{3}}-\frac {B a c \,d^{2}}{\left (a d -b c \right )^{3} \left (b x +a \right ) g^{3}}+\frac {B b \,c^{2} d}{2 \left (a d -b c \right )^{3} \left (b x +a \right ) g^{3}}-\frac {B c \,d^{2} \ln \left (\frac {d e}{b}-\frac {\left (a d -b c \right ) e}{\left (b x +a \right ) b}\right )}{2 \left (a d -b c \right )^{3} g^{3}}+\frac {A a \,d^{3}}{2 \left (a d -b c \right )^{3} b \,g^{3}}-\frac {A c \,d^{2}}{2 \left (a d -b c \right )^{3} g^{3}}-\frac {3 B a \,d^{3}}{4 \left (a d -b c \right )^{3} b \,g^{3}}+\frac {3 B c \,d^{2}}{4 \left (a d -b c \right )^{3} g^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.28, size = 255, normalized size = 1.77 \[ -\frac {1}{4} \, B {\left (\frac {2 \, b d x - b c + 3 \, a d}{{\left (b^{4} c - a b^{3} d\right )} g^{3} x^{2} + 2 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} g^{3} x + {\left (a^{2} b^{2} c - a^{3} b d\right )} g^{3}} + \frac {2 \, \log \left (\frac {d e x}{b x + a} + \frac {c e}{b x + a}\right )}{b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}} + \frac {2 \, d^{2} \log \left (b x + a\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}} - \frac {2 \, d^{2} \log \left (d x + c\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}}\right )} - \frac {A}{2 \, {\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(d*x+c)/(b*x+a)))/(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(d*e*x/(b*x + a) + c*e/(b*x + a))/(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 = 5.19, size = 208, normalized size = 1.44 \[ \frac {B\,d^2\,\mathrm {atanh}\left (\frac {2\,b^3\,c^2\,g^3-2\,a^2\,b\,d^2\,g^3}{2\,b\,g^3\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{b\,g^3\,{\left (a\,d-b\,c\right )}^2}-\frac {B\,\ln \left (\frac {e\,\left (c+d\,x\right )}{a+b\,x}\right )}{2\,b^2\,g^3\,\left (2\,a\,x+b\,x^2+\frac {a^2}{b}\right )}-\frac {\frac {2\,A\,a\,d-2\,A\,b\,c-3\,B\,a\,d+B\,b\,c}{2\,\left (a\,d-b\,c\right )}-\frac {B\,b\,d\,x}{a\,d-b\,c}}{2\,a^2\,b\,g^3+4\,a\,b^2\,g^3\,x+2\,b^3\,g^3\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 2.68, size = 422, normalized size = 2.93 \[ - \frac {B \log {\left (\frac {e \left (c + d x\right )}{a + b x} \right )}}{2 a^{2} b g^{3} + 4 a b^{2} g^{3} x + 2 b^{3} g^{3} x^{2}} + \frac {B d^{2} \log {\left (x + \frac {- \frac {B a^{3} d^{5}}{\left (a d - b c\right )^{2}} + \frac {3 B a^{2} b c d^{4}}{\left (a d - b c\right )^{2}} - \frac {3 B a b^{2} c^{2} d^{3}}{\left (a d - b c\right )^{2}} + B a d^{3} + \frac {B b^{3} c^{3} d^{2}}{\left (a d - b c\right )^{2}} + B b c d^{2}}{2 B b d^{3}} \right )}}{2 b g^{3} \left (a d - b c\right )^{2}} - \frac {B d^{2} \log {\left (x + \frac {\frac {B a^{3} d^{5}}{\left (a d - b c\right )^{2}} - \frac {3 B a^{2} b c d^{4}}{\left (a d - b c\right )^{2}} + \frac {3 B a b^{2} c^{2} d^{3}}{\left (a d - b c\right )^{2}} + B a d^{3} - \frac {B b^{3} c^{3} d^{2}}{\left (a d - b c\right )^{2}} + B b c d^{2}}{2 B b d^{3}} \right )}}{2 b g^{3} \left (a d - b c\right )^{2}} + \frac {- 2 A a d + 2 A b c + 3 B a d - B b c + 2 B b d x}{4 a^{3} b d g^{3} - 4 a^{2} b^{2} c g^{3} + x^{2} \left (4 a b^{3} d g^{3} - 4 b^{4} c g^{3}\right ) + x \left (8 a^{2} b^{2} d g^{3} - 8 a b^{3} c g^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-B*log(e*(c + d*x)/(a + b*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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