3.298 \(\int \frac {A+B \log (e (a+b x)^n (c+d x)^{-n})}{g+h x} \, dx\)

Optimal. Leaf size=148 \[ \frac {\log (g+h x) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}{h}-\frac {B n \text {Li}_2\left (\frac {b (g+h x)}{b g-a h}\right )}{h}-\frac {B n \log (g+h x) \log \left (-\frac {h (a+b x)}{b g-a h}\right )}{h}+\frac {B n \text {Li}_2\left (\frac {d (g+h x)}{d g-c h}\right )}{h}+\frac {B n \log (g+h x) \log \left (-\frac {h (c+d x)}{d g-c h}\right )}{h} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 156, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {6742, 2494, 2394, 2393, 2391} \[ -\frac {B n \text {PolyLog}\left (2,\frac {b (g+h x)}{b g-a h}\right )}{h}+\frac {B n \text {PolyLog}\left (2,\frac {d (g+h x)}{d g-c h}\right )}{h}+\frac {B \log (g+h x) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{h}-\frac {B n \log (g+h x) \log \left (-\frac {h (a+b x)}{b g-a h}\right )}{h}+\frac {A \log (g+h x)}{h}+\frac {B n \log (g+h x) \log \left (-\frac {h (c+d x)}{d g-c h}\right )}{h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2391

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 2393

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 2394

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 2494

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]/((g_.) + (h_.)*(x_)), x_Sym
bol] :> Simp[(Log[g + h*x]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/h, x] + (-Dist[(b*p*r)/h, Int[Log[g + h*x]/(a
+ b*x), x], x] - Dist[(d*q*r)/h, Int[Log[g + h*x]/(c + d*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, p, q,
r}, x] && NeQ[b*c - a*d, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 150, normalized size = 1.01 \[ \frac {\log (g+h x) \left (B \left (\log \left (e (a+b x)^n (c+d x)^{-n}\right )-n \log (a+b x)+n \log (c+d x)\right )+A\right )+B n \left (\text {Li}_2\left (\frac {h (a+b x)}{a h-b g}\right )+\log (a+b x) \log \left (\frac {b (g+h x)}{b g-a h}\right )\right )-B n \left (\text {Li}_2\left (\frac {h (c+d x)}{c h-d g}\right )+\log (c+d x) \log \left (\frac {d (g+h x)}{d g-c h}\right )\right )}{h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A}{h x + g}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A}{h x + g}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.45, size = 597, normalized size = 4.03 \[ -\frac {i \pi B \,\mathrm {csgn}\left (i e \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \mathrm {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \ln \left (h x +g \right )}{2 h}+\frac {i \pi B \,\mathrm {csgn}\left (i e \right ) \mathrm {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2} \ln \left (h x +g \right )}{2 h}-\frac {i \pi B \,\mathrm {csgn}\left (i \left (b x +a \right )^{n}\right ) \mathrm {csgn}\left (i \left (d x +c \right )^{-n}\right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \ln \left (h x +g \right )}{2 h}+\frac {i \pi B \,\mathrm {csgn}\left (i \left (b x +a \right )^{n}\right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2} \ln \left (h x +g \right )}{2 h}+\frac {i \pi B \,\mathrm {csgn}\left (i \left (d x +c \right )^{-n}\right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2} \ln \left (h x +g \right )}{2 h}-\frac {i \pi B \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{3} \ln \left (h x +g \right )}{2 h}+\frac {i \pi B \,\mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \mathrm {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2} \ln \left (h x +g \right )}{2 h}-\frac {i \pi B \mathrm {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{3} \ln \left (h x +g \right )}{2 h}-\frac {B n \ln \left (\frac {a h -b g +\left (h x +g \right ) b}{a h -b g}\right ) \ln \left (h x +g \right )}{h}+\frac {B n \ln \left (\frac {c h -d g +\left (h x +g \right ) d}{c h -d g}\right ) \ln \left (h x +g \right )}{h}-\frac {B n \dilog \left (\frac {a h -b g +\left (h x +g \right ) b}{a h -b g}\right )}{h}+\frac {B n \dilog \left (\frac {c h -d g +\left (h x +g \right ) d}{c h -d g}\right )}{h}+\frac {B \ln \relax (e ) \ln \left (h x +g \right )}{h}+\frac {B \ln \left (\left (b x +a \right )^{n}\right ) \ln \left (h x +g \right )}{h}-\frac {B \ln \left (\left (d x +c \right )^{n}\right ) \ln \left (h x +g \right )}{h}+\frac {A \ln \left (h x +g \right )}{h} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*ln(h*x+g)/h*B*Pi*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3-1/2*I*ln(h*x+g)/h*B*Pi*csgn(I*e)*csgn(I*(b*x+a)^n/((
d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)+1/2*I*ln(h*x+g)/h*B*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x
+c)^n)*(b*x+a)^n)^2+1/2*I*ln(h*x+g)/h*B*Pi*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2-1/2*I*ln(h*x+g)
/h*B*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))^3+1/2*I*ln(h*x+g)/h*B*Pi*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+1/2
*I*ln(h*x+g)/h*B*Pi*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2-1/2*I*ln(h*x+g)/h*B*Pi*csgn(I*(b*x+a)^n)
*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))+ln(h*x+g)/h*B*ln(e)+A*ln(h*x+g)/h+B*ln(h*x+g)/h*ln((b*x+a)^
n)-B/h*n*dilog((b*(h*x+g)+a*h-b*g)/(a*h-b*g))-B/h*n*ln(h*x+g)*ln((b*(h*x+g)+a*h-b*g)/(a*h-b*g))-B*ln(h*x+g)/h*
ln((d*x+c)^n)+B/h*n*dilog(((h*x+g)*d+c*h-d*g)/(c*h-d*g))+B/h*n*ln(h*x+g)*ln(((h*x+g)*d+c*h-d*g)/(c*h-d*g))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -B \int -\frac {\log \left ({\left (b x + a\right )}^{n}\right ) - \log \left ({\left (d x + c\right )}^{n}\right ) + \log \relax (e)}{h x + g}\,{d x} + \frac {A \log \left (h x + g\right )}{h} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed

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