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

Optimal. Leaf size=102 \[ -\frac {A (c+d x)}{g^2 (a+b x) (b c-a d)}-\frac {B (c+d x) \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{g^2 (a+b x) (b c-a d)}+\frac {2 B (c+d x)}{g^2 (a+b x) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 105, normalized size of antiderivative = 1.03, 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} \[ -\frac {B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A}{b g^2 (a+b x)}+\frac {2 B d \log (a+b x)}{b g^2 (b c-a d)}-\frac {2 B d \log (c+d x)}{b g^2 (b c-a d)}+\frac {2 B}{b g^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 89, normalized size = 0.87 \[ \frac {-(b c-a d) \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A-2 B\right )-2 B d (a+b x) \log (c+d x)+2 B d (a+b x) \log (a+b x)}{b g^2 (a+b x) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 110, normalized size = 1.08 \[ -\frac {{\left (A - 2 \, B\right )} b c - {\left (A - 2 \, B\right )} a d + {\left (B b d x + B b c\right )} \log \left (\frac {d^{2} e x^{2} + 2 \, c d e x + c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}{{\left (b^{3} c - a b^{2} d\right )} g^{2} x + {\left (a b^{2} c - a^{2} b d\right )} g^{2}} \]

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)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.40, size = 188, normalized size = 1.84 \[ -{\left (2 \, {\left (b^{2} c g^{2} - a b d g^{2}\right )} {\left (\frac {d \log \left ({\left | \frac {b c g}{b g x + a g} - \frac {a d g}{b g x + a g} + d \right |}\right )}{b^{4} c^{2} g^{4} - 2 \, a b^{3} c d g^{4} + a^{2} b^{2} d^{2} g^{4}} - \frac {1}{{\left (b^{2} c g^{2} - a b d g^{2}\right )} {\left (b g x + a g\right )} b g}\right )} + \frac {\log \left (\frac {{\left (d x + c\right )}^{2} e}{{\left (b x + a\right )}^{2}}\right )}{{\left (b g x + a g\right )} b g}\right )} B - \frac {A}{{\left (b g x + a g\right )} b g} \]

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)^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 212, normalized size = 2.08 \[ \frac {2 B a \,d^{2} \ln \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )}{\left (a d -b c \right )^{2} b \,g^{2}}-\frac {2 B c d \ln \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )}{\left (a d -b c \right )^{2} g^{2}}+\frac {2 B a d}{\left (a d -b c \right ) \left (b x +a \right ) b \,g^{2}}-\frac {2 B c}{\left (a d -b c \right ) \left (b x +a \right ) g^{2}}-\frac {B \ln \left (\frac {\left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )^{2} e}{b^{2}}\right )}{\left (b x +a \right ) b \,g^{2}}-\frac {A}{\left (b x +a \right ) b \,g^{2}} \]

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)^2,x)

[Out]

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

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maxima [A]  time = 1.08, size = 187, normalized size = 1.83 \[ -B {\left (\frac {\log \left (\frac {d^{2} e x^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac {2 \, c d e x}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac {c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}{b^{2} g^{2} x + a b g^{2}} - \frac {2}{b^{2} g^{2} x + a b g^{2}} - \frac {2 \, d \log \left (b x + a\right )}{{\left (b^{2} c - a b d\right )} g^{2}} + \frac {2 \, d \log \left (d x + c\right )}{{\left (b^{2} c - a b d\right )} g^{2}}\right )} - \frac {A}{b^{2} g^{2} x + a b g^{2}} \]

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)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.94, size = 108, normalized size = 1.06 \[ -\frac {A-2\,B}{x\,b^2\,g^2+a\,b\,g^2}-\frac {B\,\ln \left (\frac {e\,{\left (c+d\,x\right )}^2}{{\left (a+b\,x\right )}^2}\right )}{b^2\,g^2\,\left (x+\frac {a}{b}\right )}+\frac {B\,d\,\mathrm {atan}\left (\frac {b\,c\,2{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}+1{}\mathrm {i}\right )\,4{}\mathrm {i}}{b\,g^2\,\left (a\,d-b\,c\right )} \]

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)^2,x)

[Out]

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

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sympy [B]  time = 1.64, size = 253, normalized size = 2.48 \[ - \frac {B \log {\left (\frac {e \left (c + d x\right )^{2}}{\left (a + b x\right )^{2}} \right )}}{a b g^{2} + b^{2} g^{2} x} + \frac {2 B d \log {\left (x + \frac {- \frac {2 B a^{2} d^{3}}{a d - b c} + \frac {4 B a b c d^{2}}{a d - b c} + 2 B a d^{2} - \frac {2 B b^{2} c^{2} d}{a d - b c} + 2 B b c d}{4 B b d^{2}} \right )}}{b g^{2} \left (a d - b c\right )} - \frac {2 B d \log {\left (x + \frac {\frac {2 B a^{2} d^{3}}{a d - b c} - \frac {4 B a b c d^{2}}{a d - b c} + 2 B a d^{2} + \frac {2 B b^{2} c^{2} d}{a d - b c} + 2 B b c d}{4 B b d^{2}} \right )}}{b g^{2} \left (a d - b c\right )} + \frac {- A + 2 B}{a b g^{2} + b^{2} g^{2} x} \]

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)**2,x)

[Out]

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

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