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

Optimal. Leaf size=139 \[ -\frac {B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A}{2 b g^3 (a+b x)^2}-\frac {B d^2 \log (a+b x)}{b g^3 (b c-a d)^2}+\frac {B d^2 \log (c+d x)}{b g^3 (b c-a d)^2}-\frac {B d}{b g^3 (a+b x) (b c-a d)}+\frac {B}{2 b g^3 (a+b x)^2} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 139, normalized size of antiderivative = 1.00, 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} \[ -\frac {B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A}{2 b g^3 (a+b x)^2}-\frac {B d^2 \log (a+b x)}{b g^3 (b c-a d)^2}+\frac {B d^2 \log (c+d x)}{b g^3 (b c-a d)^2}-\frac {B d}{b g^3 (a+b x) (b c-a d)}+\frac {B}{2 b g^3 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

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Mathematica [A]  time = 0.09, size = 128, normalized size = 0.92 \[ -\frac {(b c-a d) \left (-a A d+B (b c-a d) \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+3 a B d+A b c-b B c+2 b B d x\right )-2 B d^2 (a+b x)^2 \log (c+d x)+2 B d^2 (a+b x)^2 \log (a+b x)}{2 b g^3 (a+b x)^2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.65, size = 240, normalized size = 1.73 \[ -\frac {{\left (A - B\right )} b^{2} c^{2} - 2 \, {\left (A - 2 \, B\right )} a b c d + {\left (A - 3 \, B\right )} a^{2} d^{2} + 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} x - {\left (B b^{2} d^{2} x^{2} + 2 \, B a b d^{2} x - B b^{2} c^{2} + 2 \, B a b c d\right )} \log \left (\frac {d^{2} e x^{2} + 2 \, c d e x + c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}{2 \, {\left ({\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} g^{3} x^{2} + 2 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} g^{3} x + {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} g^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(d*x+c)^2/(b*x+a)^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((d^2*e*x^2 + 2*c*d*e*x + c^2*e)/(b^2*x^2 + 2*a*b*x + a^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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giac [A]  time = 0.34, size = 259, normalized size = 1.86 \[ -\frac {B d^{2} \log \left (b x + a\right )}{b^{3} c^{2} g^{3} - 2 \, a b^{2} c d g^{3} + a^{2} b d^{2} g^{3}} + \frac {B d^{2} \log \left (d x + c\right )}{b^{3} c^{2} g^{3} - 2 \, a b^{2} c d g^{3} + a^{2} b d^{2} g^{3}} - \frac {B \log \left (\frac {d^{2} x^{2} + 2 \, c d x + c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}{2 \, {\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} - \frac {2 \, B b d x + A b c - A a d + 2 \, B a d}{2 \, {\left (b^{4} c g^{3} x^{2} - a b^{3} d g^{3} x^{2} + 2 \, a b^{3} c g^{3} x - 2 \, a^{2} b^{2} d g^{3} x + a^{2} b^{2} c g^{3} - a^{3} b d g^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.06, size = 300, normalized size = 2.16 \[ \frac {B a \,d^{3} \ln \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )}{\left (a d -b c \right )^{3} b \,g^{3}}-\frac {B c \,d^{2} \ln \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )}{\left (a d -b c \right )^{3} g^{3}}+\frac {B \,a^{2} d^{2}}{2 \left (a d -b c \right )^{2} \left (b x +a \right )^{2} b \,g^{3}}-\frac {B a c d}{\left (a d -b c \right )^{2} \left (b x +a \right )^{2} g^{3}}+\frac {B b \,c^{2}}{2 \left (a d -b c \right )^{2} \left (b x +a \right )^{2} g^{3}}+\frac {B a \,d^{2}}{\left (a d -b c \right )^{2} \left (b x +a \right ) b \,g^{3}}-\frac {B c d}{\left (a d -b c \right )^{2} \left (b x +a \right ) g^{3}}-\frac {B \ln \left (\frac {\left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )^{2} e}{b^{2}}\right )}{2 \left (b x +a \right )^{2} b \,g^{3}}-\frac {A}{2 \left (b x +a \right )^{2} b \,g^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*ln(e*(d*x+c)^2/(b*x+a)^2))/(b*g*x+a*g)^3,x)

[Out]

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

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maxima [B]  time = 1.16, size = 306, normalized size = 2.20 \[ -\frac {1}{2} \, B {\left (\frac {2 \, b d x - b c + 3 \, a d}{{\left (b^{4} c - a b^{3} d\right )} g^{3} x^{2} + 2 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} g^{3} x + {\left (a^{2} b^{2} c - a^{3} b d\right )} g^{3}} + \frac {\log \left (\frac {d^{2} e x^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac {2 \, c d e x}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac {c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}{b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}} + \frac {2 \, d^{2} \log \left (b x + a\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}} - \frac {2 \, d^{2} \log \left (d x + c\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}}\right )} - \frac {A}{2 \, {\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(d*x+c)^2/(b*x+a)^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(d^2*e*x^2/(b^2*x^2 + 2*a*b*x + a^2) + 2*c*d*e*x/(b^2*x^2 + 2*a*b*x + a^2) + c^2*e/(b^2*x^2 + 2
*a*b*x + a^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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mupad [B]  time = 5.93, size = 206, normalized size = 1.48 \[ \frac {2\,B\,d^2\,\mathrm {atanh}\left (\frac {b^3\,c^2\,g^3-a^2\,b\,d^2\,g^3}{b\,g^3\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{b\,g^3\,{\left (a\,d-b\,c\right )}^2}-\frac {B\,\ln \left (\frac {e\,{\left (c+d\,x\right )}^2}{{\left (a+b\,x\right )}^2}\right )}{2\,b^2\,g^3\,\left (2\,a\,x+b\,x^2+\frac {a^2}{b}\right )}-\frac {\frac {A\,a\,d-A\,b\,c-3\,B\,a\,d+B\,b\,c}{2\,\left (a\,d-b\,c\right )}-\frac {B\,b\,d\,x}{a\,d-b\,c}}{a^2\,b\,g^3+2\,a\,b^2\,g^3\,x+b^3\,g^3\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 2.60, size = 418, normalized size = 3.01 \[ - \frac {B \log {\left (\frac {e \left (c + d x\right )^{2}}{\left (a + b x\right )^{2}} \right )}}{2 a^{2} b g^{3} + 4 a b^{2} g^{3} x + 2 b^{3} g^{3} x^{2}} + \frac {B d^{2} \log {\left (x + \frac {- \frac {B a^{3} d^{5}}{\left (a d - b c\right )^{2}} + \frac {3 B a^{2} b c d^{4}}{\left (a d - b c\right )^{2}} - \frac {3 B a b^{2} c^{2} d^{3}}{\left (a d - b c\right )^{2}} + B a d^{3} + \frac {B b^{3} c^{3} d^{2}}{\left (a d - b c\right )^{2}} + B b c d^{2}}{2 B b d^{3}} \right )}}{b g^{3} \left (a d - b c\right )^{2}} - \frac {B d^{2} \log {\left (x + \frac {\frac {B a^{3} d^{5}}{\left (a d - b c\right )^{2}} - \frac {3 B a^{2} b c d^{4}}{\left (a d - b c\right )^{2}} + \frac {3 B a b^{2} c^{2} d^{3}}{\left (a d - b c\right )^{2}} + B a d^{3} - \frac {B b^{3} c^{3} d^{2}}{\left (a d - b c\right )^{2}} + B b c d^{2}}{2 B b d^{3}} \right )}}{b g^{3} \left (a d - b c\right )^{2}} + \frac {- A a d + A b c + 3 B a d - B b c + 2 B b d x}{2 a^{3} b d g^{3} - 2 a^{2} b^{2} c g^{3} + x^{2} \left (2 a b^{3} d g^{3} - 2 b^{4} c g^{3}\right ) + x \left (4 a^{2} b^{2} d g^{3} - 4 a b^{3} c g^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*ln(e*(d*x+c)**2/(b*x+a)**2))/(b*g*x+a*g)**3,x)

[Out]

-B*log(e*(c + d*x)**2/(a + b*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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