3.1.62 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{f+g x} \, dx\) [62]

Optimal. Leaf size=147 \[ -\frac {B n \log \left (-\frac {g (a+b x)}{b f-a g}\right ) \log (f+g x)}{g}+\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (f+g x)}{g}+\frac {B n \log \left (-\frac {g (c+d x)}{d f-c g}\right ) \log (f+g x)}{g}-\frac {B n \text {Li}_2\left (\frac {b (f+g x)}{b f-a g}\right )}{g}+\frac {B n \text {Li}_2\left (\frac {d (f+g x)}{d f-c g}\right )}{g} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2545, 2441, 2440, 2438} \begin {gather*} -\frac {B n \text {PolyLog}\left (2,\frac {b (f+g x)}{b f-a g}\right )}{g}+\frac {B n \text {PolyLog}\left (2,\frac {d (f+g x)}{d f-c g}\right )}{g}+\frac {\log (f+g x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{g}-\frac {B n \log (f+g x) \log \left (-\frac {g (a+b x)}{b f-a g}\right )}{g}+\frac {B n \log (f+g x) \log \left (-\frac {g (c+d x)}{d f-c g}\right )}{g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + c*e*(x/g)])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2545

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))/((f_.) + (g_.)*(x_)), x_Symbo
l] :> Simp[Log[f + g*x]*((A + B*Log[e*((a + b*x)/(c + d*x))^n])/g), x] + (-Dist[b*B*(n/g), Int[Log[f + g*x]/(a
+ b*x), x], x] + Dist[B*d*(n/g), Int[Log[f + g*x]/(c + d*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, A, B, n},
x] && NeQ[b*c - a*d, 0]

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} \, dx &=\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (f+g x)}{g}-\frac {(B n) \int \frac {(c+d x) \left (-\frac {d (a+b x)}{(c+d x)^2}+\frac {b}{c+d x}\right ) \log (f+g x)}{a+b x} \, dx}{g}\\ &=\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (f+g x)}{g}-\frac {(B n) \int \left (\frac {b \log (f+g x)}{a+b x}-\frac {d \log (f+g x)}{c+d x}\right ) \, dx}{g}\\ &=\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (f+g x)}{g}-\frac {(b B n) \int \frac {\log (f+g x)}{a+b x} \, dx}{g}+\frac {(B d n) \int \frac {\log (f+g x)}{c+d x} \, dx}{g}\\ &=-\frac {B n \log \left (-\frac {g (a+b x)}{b f-a g}\right ) \log (f+g x)}{g}+\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (f+g x)}{g}+\frac {B n \log \left (-\frac {g (c+d x)}{d f-c g}\right ) \log (f+g x)}{g}+(B n) \int \frac {\log \left (\frac {g (a+b x)}{-b f+a g}\right )}{f+g x} \, dx-(B n) \int \frac {\log \left (\frac {g (c+d x)}{-d f+c g}\right )}{f+g x} \, dx\\ &=-\frac {B n \log \left (-\frac {g (a+b x)}{b f-a g}\right ) \log (f+g x)}{g}+\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (f+g x)}{g}+\frac {B n \log \left (-\frac {g (c+d x)}{d f-c g}\right ) \log (f+g x)}{g}+\frac {(B n) \text {Subst}\left (\int \frac {\log \left (1+\frac {b x}{-b f+a g}\right )}{x} \, dx,x,f+g x\right )}{g}-\frac {(B n) \text {Subst}\left (\int \frac {\log \left (1+\frac {d x}{-d f+c g}\right )}{x} \, dx,x,f+g x\right )}{g}\\ &=-\frac {B n \log \left (-\frac {g (a+b x)}{b f-a g}\right ) \log (f+g x)}{g}+\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (f+g x)}{g}+\frac {B n \log \left (-\frac {g (c+d x)}{d f-c g}\right ) \log (f+g x)}{g}-\frac {B n \text {Li}_2\left (\frac {b (f+g x)}{b f-a g}\right )}{g}+\frac {B n \text {Li}_2\left (\frac {d (f+g x)}{d f-c g}\right )}{g}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 122, normalized size = 0.83 \begin {gather*} \frac {\left (A-B n \log \left (\frac {g (a+b x)}{-b f+a g}\right )+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+B n \log \left (\frac {g (c+d x)}{-d f+c g}\right )\right ) \log (f+g x)-B n \text {Li}_2\left (\frac {b (f+g x)}{b f-a g}\right )+B n \text {Li}_2\left (\frac {d (f+g x)}{d f-c g}\right )}{g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \frac {A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}{g x +f}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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),x, algorithm="maxima")

[Out]

-B*integrate(-(log((b*x + a)^n) - log((d*x + c)^n) + 1)/(g*x + f), x) + A*log(g*x + f)/g

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{f + g x}\, dx \end {gather*}

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),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{f+g\,x} \,d x \end {gather*}

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),x)

[Out]

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

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