\(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(f+g x)^2} \, dx\) [63]

Optimal result

Integrand size = 30, antiderivative size = 91 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^2} \, dx=\frac {(a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b f-a g) (f+g x)}+\frac {B (b c-a d) n \log \left (\frac {f+g x}{c+d x}\right )}{(b f-a g) (d f-c g)} \]

[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)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2553, 2351, 31} \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^2} \, dx=\frac {(a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{(f+g x) (b f-a g)}+\frac {B n (b c-a d) \log \left (\frac {f+g x}{c+d x}\right )}{(b f-a g) (d f-c g)} \]

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[x*(d + e*x^r)^(q +
1)*((a + b*Log[c*x^n])/d), x] - Dist[b*(n/d), Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2553

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.), x_Symbol] :> Dist[b*c - a*d, Subst[Int[(b*f - a*g - (d*f - c*g)*x)^m*((A + B*Log[e*x^n])^p/(b - d*x)^(m +
2)), x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && NeQ[b*c - a*d, 0] && Inte
gerQ[m] && IGtQ[p, 0]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.20 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^2} \, dx=\frac {-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{f+g x}+\frac {B n (b (d f-c g) \log (a+b x)+(-b d f+a d g) \log (c+d x)+(b c-a d) g \log (f+g x))}{(b f-a g) (d f-c g)}}{g} \]

[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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(363\) vs. \(2(91)=182\).

Time = 3.51 (sec) , antiderivative size = 364, normalized size of antiderivative = 4.00

method	result	size
parallelrisch	\(\frac {B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a b c d \,f^{2} n -B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{2} c d f g n +B \ln \left (b x +a \right ) x \,a^{2} c d f g \,n^{2}-B \ln \left (b x +a \right ) x a b \,c^{2} f g \,n^{2}-B \ln \left (g x +f \right ) x \,a^{2} c d f g \,n^{2}+B \ln \left (g x +f \right ) x a b \,c^{2} f g \,n^{2}+A x \,a^{2} c^{2} g^{2} n -A x \,a^{2} c d f g n -A x a b \,c^{2} f g n +A x a b c d \,f^{2} n -B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a^{2} c^{2} f g n +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a b \,c^{2} f^{2} n +B \ln \left (b x +a \right ) a^{2} c d \,f^{2} n^{2}-B \ln \left (b x +a \right ) a b \,c^{2} f^{2} n^{2}-B \ln \left (g x +f \right ) a^{2} c d \,f^{2} n^{2}+B \ln \left (g x +f \right ) a b \,c^{2} f^{2} n^{2}}{\left (a g -b f \right ) \left (g x +f \right ) n \left (c g -d f \right ) a c f}\)	\(364\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (91) = 182\).

Time = 3.14 (sec) , antiderivative size = 294, normalized size of antiderivative = 3.23 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^2} \, dx=-\frac {A b d f^{2} + A a c g^{2} - {\left (A b c + A a d\right )} f g + {\left (B b d f^{2} + B a c g^{2} - {\left (B b c + B a d\right )} f g\right )} n \log \left (\frac {b x + a}{d x + c}\right ) - {\left ({\left (B b d f g - B b c g^{2}\right )} n x + {\left (B b d f^{2} - B b c f g\right )} n\right )} \log \left (b x + a\right ) + {\left ({\left (B b d f g - B a d g^{2}\right )} n x + {\left (B b d f^{2} - B a d f g\right )} n\right )} \log \left (d x + c\right ) - {\left ({\left (B b c - B a d\right )} g^{2} n x + {\left (B b c - B a d\right )} f g n\right )} \log \left (g x + f\right ) + {\left (B b d f^{2} + B a c g^{2} - {\left (B b c + B a d\right )} f g\right )} \log \left (e\right )}{b d f^{3} g + a c f g^{3} - {\left (b c + a d\right )} f^{2} g^{2} + {\left (b d f^{2} g^{2} + a c g^{4} - {\left (b c + a d\right )} f g^{3}\right )} x} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.56 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^2} \, dx=B n {\left (\frac {b \log \left (b x + a\right )}{b f g - a g^{2}} - \frac {d \log \left (d x + c\right )}{d f g - c g^{2}} + \frac {{\left (b c - a d\right )} \log \left (g x + f\right )}{b d f^{2} + a c g^{2} - {\left (b c + a d\right )} f g}\right )} - \frac {B \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right )}{g^{2} x + f g} - \frac {A}{g^{2} x + f g} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 461 vs. \(2 (91) = 182\).

Time = 0.55 (sec) , antiderivative size = 461, normalized size of antiderivative = 5.07 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^2} \, dx={\left (\frac {{\left (B b^{2} c^{2} n - 2 \, B a b c d n + B a^{2} d^{2} n\right )} \log \left (-b f + \frac {{\left (b x + a\right )} d f}{d x + c} + a g - \frac {{\left (b x + a\right )} c g}{d x + c}\right )}{b d f^{2} - b c f g - a d f g + a c g^{2}} + \frac {{\left (B b^{2} c^{2} n - 2 \, B a b c d n + B a^{2} d^{2} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b d f^{2} - \frac {{\left (b x + a\right )} d^{2} f^{2}}{d x + c} - b c f g - a d f g + \frac {2 \, {\left (b x + a\right )} c d f g}{d x + c} + a c g^{2} - \frac {{\left (b x + a\right )} c^{2} g^{2}}{d x + c}} - \frac {{\left (B b^{2} c^{2} n - 2 \, B a b c d n + B a^{2} d^{2} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b d f^{2} - b c f g - a d f g + a c g^{2}} + \frac {B b^{2} c^{2} \log \left (e\right ) - 2 \, B a b c d \log \left (e\right ) + B a^{2} d^{2} \log \left (e\right ) + A b^{2} c^{2} - 2 \, A a b c d + A a^{2} d^{2}}{b d f^{2} - \frac {{\left (b x + a\right )} d^{2} f^{2}}{d x + c} - b c f g - a d f g + \frac {2 \, {\left (b x + a\right )} c d f g}{d x + c} + a c g^{2} - \frac {{\left (b x + a\right )} c^{2} g^{2}}{d x + c}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

[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) + (B*b^2*c^2*log(e) - 2*B*a*b*c*d*log(e) + B*a^2*d^2*log(e) + A*b^2*c^2 - 2*A*a*b*c
*d + A*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)

Mupad [B] (verification not implemented)

Time = 1.37 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.54 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^2} \, dx=\frac {B\,d\,n\,\ln \left (c+d\,x\right )}{c\,g^2-d\,f\,g}-\frac {\ln \left (f+g\,x\right )\,\left (B\,a\,d\,n-B\,b\,c\,n\right )}{a\,c\,g^2+b\,d\,f^2-a\,d\,f\,g-b\,c\,f\,g}-\frac {B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{g\,\left (f+g\,x\right )}-\frac {B\,b\,n\,\ln \left (a+b\,x\right )}{a\,g^2-b\,f\,g}-\frac {A}{x\,g^2+f\,g} \]

[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
)