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

Optimal. Leaf size=91 \[ \frac {(a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{(f+g x) (b f-a g)}+\frac {B n (b c-a d) \log \left (\frac {f+g x}{c+d x}\right )}{(b f-a g) (d f-c g)} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 119, normalized size of antiderivative = 1.31, 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, 72} \[ -\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{g (f+g x)}+\frac {B n (b c-a d) \log (f+g x)}{(b f-a g) (d f-c g)}+\frac {b B n \log (a+b x)}{g (b f-a g)}-\frac {B d n \log (c+d x)}{g (d f-c g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.15, size = 109, normalized size = 1.20 \[ \frac {\frac {B n (b \log (a+b x) (d f-c g)+\log (c+d x) (a d g-b d f)+g (b c-a d) \log (f+g x))}{(b f-a g) (d f-c g)}-\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{f+g x}}{g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((A + B*Log[e*((a + b*x)/(c + d*x))^n])/(f + g*x)) + (B*n*(b*(d*f - c*g)*Log[a + b*x] + (-(b*d*f) + a*d*g)*L
og[c + d*x] + (b*c - a*d)*g*Log[f + g*x]))/((b*f - a*g)*(d*f - c*g)))/g

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fricas [B]  time = 11.25, size = 294, normalized size = 3.23 \[ -\frac {A b d f^{2} + A a c g^{2} - {\left (A b c + A a d\right )} f g + {\left (B b d f^{2} + B a c g^{2} - {\left (B b c + B a d\right )} f g\right )} n \log \left (\frac {b x + a}{d x + c}\right ) - {\left ({\left (B b d f g - B b c g^{2}\right )} n x + {\left (B b d f^{2} - B b c f g\right )} n\right )} \log \left (b x + a\right ) + {\left ({\left (B b d f g - B a d g^{2}\right )} n x + {\left (B b d f^{2} - B a d f g\right )} n\right )} \log \left (d x + c\right ) - {\left ({\left (B b c - B a d\right )} g^{2} n x + {\left (B b c - B a d\right )} f g n\right )} \log \left (g x + f\right ) + {\left (B b d f^{2} + B a c g^{2} - {\left (B b c + B a d\right )} f g\right )} \log \relax (e)}{b d f^{3} g + a c f g^{3} - {\left (b c + a d\right )} f^{2} g^{2} + {\left (b d f^{2} g^{2} + a c g^{4} - {\left (b c + a d\right )} f g^{3}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(g*x+f)^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 4.10, size = 455, normalized size = 5.00 \[ {\left (\frac {{\left (B b^{2} c^{2} n - 2 \, B a b c d n + B a^{2} d^{2} n\right )} \log \left (-b f + \frac {{\left (b x + a\right )} d f}{d x + c} + a g - \frac {{\left (b x + a\right )} c g}{d x + c}\right )}{b d f^{2} - b c f g - a d f g + a c g^{2}} + \frac {{\left (B b^{2} c^{2} n - 2 \, B a b c d n + B a^{2} d^{2} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b d f^{2} - \frac {{\left (b x + a\right )} d^{2} f^{2}}{d x + c} - b c f g - a d f g + \frac {2 \, {\left (b x + a\right )} c d f g}{d x + c} + a c g^{2} - \frac {{\left (b x + a\right )} c^{2} g^{2}}{d x + c}} - \frac {{\left (B b^{2} c^{2} n - 2 \, B a b c d n + B a^{2} d^{2} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b d f^{2} - b c f g - a d f g + a c g^{2}} + \frac {A b^{2} c^{2} + B b^{2} c^{2} - 2 \, A a b c d - 2 \, B a b c d + A a^{2} d^{2} + B a^{2} d^{2}}{b d f^{2} - \frac {{\left (b x + a\right )} d^{2} f^{2}}{d x + c} - b c f g - a d f g + \frac {2 \, {\left (b x + a\right )} c d f g}{d x + c} + a c g^{2} - \frac {{\left (b x + a\right )} c^{2} g^{2}}{d x + c}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(g*x+f)^2,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )+A}{\left (g x +f \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.81, size = 142, normalized size = 1.56 \[ B n {\left (\frac {b \log \left (b x + a\right )}{b f g - a g^{2}} - \frac {d \log \left (d x + c\right )}{d f g - c g^{2}} + \frac {{\left (b c - a d\right )} \log \left (g x + f\right )}{b d f^{2} + a c g^{2} - {\left (b c + a d\right )} f g}\right )} - \frac {B \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right )}{g^{2} x + f g} - \frac {A}{g^{2} x + f g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(g*x+f)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.64, size = 140, normalized size = 1.54 \[ \frac {B\,d\,n\,\ln \left (c+d\,x\right )}{c\,g^2-d\,f\,g}-\frac {\ln \left (f+g\,x\right )\,\left (B\,a\,d\,n-B\,b\,c\,n\right )}{a\,c\,g^2+b\,d\,f^2-a\,d\,f\,g-b\,c\,f\,g}-\frac {B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{g\,\left (f+g\,x\right )}-\frac {B\,b\,n\,\ln \left (a+b\,x\right )}{a\,g^2-b\,f\,g}-\frac {A}{x\,g^2+f\,g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError

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