### 3.93 $$\int \frac{\log (d (a+b x+c x^2)^n)}{a e+b e x+c e x^2} \, dx$$

Optimal. Leaf size=258 $-\frac{2 n \text{PolyLog}\left (2,-\frac{\frac{2 c x}{\sqrt{b^2-4 a c}}+\frac{b}{\sqrt{b^2-4 a c}}+1}{-\frac{2 c x}{\sqrt{b^2-4 a c}}-\frac{b}{\sqrt{b^2-4 a c}}+1}\right )}{e \sqrt{b^2-4 a c}}-\frac{2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e \sqrt{b^2-4 a c}}+\frac{2 n \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )^2}{e \sqrt{b^2-4 a c}}-\frac{4 n \log \left (\frac{2}{-\frac{2 c x}{\sqrt{b^2-4 a c}}-\frac{b}{\sqrt{b^2-4 a c}}+1}\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{e \sqrt{b^2-4 a c}}$

[Out]

(2*n*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]]^2)/(Sqrt[b^2 - 4*a*c]*e) - (4*n*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a
*c]]*Log[2/(1 - b/Sqrt[b^2 - 4*a*c] - (2*c*x)/Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*e) - (2*ArcTanh[(b + 2*c
*x)/Sqrt[b^2 - 4*a*c]]*Log[d*(a + b*x + c*x^2)^n])/(Sqrt[b^2 - 4*a*c]*e) - (2*n*PolyLog[2, -((1 + b/Sqrt[b^2 -
4*a*c] + (2*c*x)/Sqrt[b^2 - 4*a*c])/(1 - b/Sqrt[b^2 - 4*a*c] - (2*c*x)/Sqrt[b^2 - 4*a*c]))])/(Sqrt[b^2 - 4*a*
c]*e)

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Rubi [A]  time = 0.345819, antiderivative size = 258, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 9, integrand size = 32, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.281, Rules used = {618, 206, 2527, 12, 6121, 5984, 5918, 2402, 2315} $-\frac{2 n \text{PolyLog}\left (2,-\frac{\frac{2 c x}{\sqrt{b^2-4 a c}}+\frac{b}{\sqrt{b^2-4 a c}}+1}{-\frac{2 c x}{\sqrt{b^2-4 a c}}-\frac{b}{\sqrt{b^2-4 a c}}+1}\right )}{e \sqrt{b^2-4 a c}}-\frac{2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e \sqrt{b^2-4 a c}}+\frac{2 n \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )^2}{e \sqrt{b^2-4 a c}}-\frac{4 n \log \left (\frac{2}{-\frac{2 c x}{\sqrt{b^2-4 a c}}-\frac{b}{\sqrt{b^2-4 a c}}+1}\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{e \sqrt{b^2-4 a c}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*n*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]]^2)/(Sqrt[b^2 - 4*a*c]*e) - (4*n*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a
*c]]*Log[2/(1 - b/Sqrt[b^2 - 4*a*c] - (2*c*x)/Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*e) - (2*ArcTanh[(b + 2*c
*x)/Sqrt[b^2 - 4*a*c]]*Log[d*(a + b*x + c*x^2)^n])/(Sqrt[b^2 - 4*a*c]*e) - (2*n*PolyLog[2, -((1 + b/Sqrt[b^2 -
4*a*c] + (2*c*x)/Sqrt[b^2 - 4*a*c])/(1 - b/Sqrt[b^2 - 4*a*c] - (2*c*x)/Sqrt[b^2 - 4*a*c]))])/(Sqrt[b^2 - 4*a*
c]*e)

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2527

Int[Log[(c_.)*(Px_)^(n_.)]/(Qx_), x_Symbol] :> With[{u = IntHide[1/Qx, x]}, Simp[u*Log[c*Px^n], x] - Dist[n, I
nt[SimplifyIntegrand[(u*D[Px, x])/Px, x], x], x]] /; FreeQ[{c, n}, x] && QuadraticQ[{Qx, Px}, x] && EqQ[D[Px/Q
x, x], 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6121

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.)*((A_.) + (B_.)*(x_) + (C_.)*(x
_)^2)^(q_.), x_Symbol] :> Dist[1/d, Subst[Int[((d*e - c*f)/d + (f*x)/d)^m*(-(C/d^2) + (C*x^2)/d^2)^q*(a + b*Ar
cTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, p, q}, x] && EqQ[B*(1 - c^2) + 2*A*c*
d, 0] && EqQ[2*c*C - B*d, 0]

Rule 5984

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 - c^2
*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

\begin{align*} \int \frac{\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx &=-\frac{2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{\sqrt{b^2-4 a c} e}-n \int \frac{2 (-b-2 c x) \tanh ^{-1}\left (\frac{b}{\sqrt{b^2-4 a c}}+\frac{2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} e \left (a+b x+c x^2\right )} \, dx\\ &=-\frac{2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{\sqrt{b^2-4 a c} e}-\frac{(2 n) \int \frac{(-b-2 c x) \tanh ^{-1}\left (\frac{b}{\sqrt{b^2-4 a c}}+\frac{2 c x}{\sqrt{b^2-4 a c}}\right )}{a+b x+c x^2} \, dx}{\sqrt{b^2-4 a c} e}\\ &=-\frac{2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{\sqrt{b^2-4 a c} e}+\frac{n \operatorname{Subst}\left (\int \frac{\sqrt{b^2-4 a c} x \tanh ^{-1}(x)}{-\frac{b^2-4 a c}{4 c}+\frac{\left (b^2-4 a c\right ) x^2}{4 c}} \, dx,x,\frac{b}{\sqrt{b^2-4 a c}}+\frac{2 c x}{\sqrt{b^2-4 a c}}\right )}{c e}\\ &=-\frac{2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{\sqrt{b^2-4 a c} e}+\frac{\left (\sqrt{b^2-4 a c} n\right ) \operatorname{Subst}\left (\int \frac{x \tanh ^{-1}(x)}{-\frac{b^2-4 a c}{4 c}+\frac{\left (b^2-4 a c\right ) x^2}{4 c}} \, dx,x,\frac{b}{\sqrt{b^2-4 a c}}+\frac{2 c x}{\sqrt{b^2-4 a c}}\right )}{c e}\\ &=\frac{2 n \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )^2}{\sqrt{b^2-4 a c} e}-\frac{2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{\sqrt{b^2-4 a c} e}-\frac{(4 n) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)}{1-x} \, dx,x,\frac{b}{\sqrt{b^2-4 a c}}+\frac{2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} e}\\ &=\frac{2 n \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )^2}{\sqrt{b^2-4 a c} e}-\frac{4 n \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \log \left (\frac{2}{1-\frac{b}{\sqrt{b^2-4 a c}}-\frac{2 c x}{\sqrt{b^2-4 a c}}}\right )}{\sqrt{b^2-4 a c} e}-\frac{2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{\sqrt{b^2-4 a c} e}+\frac{(4 n) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1-x}\right )}{1-x^2} \, dx,x,\frac{b}{\sqrt{b^2-4 a c}}+\frac{2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} e}\\ &=\frac{2 n \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )^2}{\sqrt{b^2-4 a c} e}-\frac{4 n \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \log \left (\frac{2}{1-\frac{b}{\sqrt{b^2-4 a c}}-\frac{2 c x}{\sqrt{b^2-4 a c}}}\right )}{\sqrt{b^2-4 a c} e}-\frac{2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{\sqrt{b^2-4 a c} e}-\frac{(4 n) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-\frac{b}{\sqrt{b^2-4 a c}}-\frac{2 c x}{\sqrt{b^2-4 a c}}}\right )}{\sqrt{b^2-4 a c} e}\\ &=\frac{2 n \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )^2}{\sqrt{b^2-4 a c} e}-\frac{4 n \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \log \left (\frac{2}{1-\frac{b}{\sqrt{b^2-4 a c}}-\frac{2 c x}{\sqrt{b^2-4 a c}}}\right )}{\sqrt{b^2-4 a c} e}-\frac{2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{\sqrt{b^2-4 a c} e}-\frac{2 n \text{Li}_2\left (1-\frac{2}{1-\frac{b}{\sqrt{b^2-4 a c}}-\frac{2 c x}{\sqrt{b^2-4 a c}}}\right )}{\sqrt{b^2-4 a c} e}\\ \end{align*}

Mathematica [A]  time = 0.162295, size = 339, normalized size = 1.31 $\frac{-2 n \text{PolyLog}\left (2,\frac{\sqrt{b^2-4 a c}-b-2 c x}{2 \sqrt{b^2-4 a c}}\right )+2 n \text{PolyLog}\left (2,\frac{\sqrt{b^2-4 a c}+b+2 c x}{2 \sqrt{b^2-4 a c}}\right )+2 \log \left (-\sqrt{b^2-4 a c}+b+2 c x\right ) \log \left (d (a+x (b+c x))^n\right )-2 \log \left (\sqrt{b^2-4 a c}+b+2 c x\right ) \log \left (d (a+x (b+c x))^n\right )-n \log ^2\left (-\sqrt{b^2-4 a c}+b+2 c x\right )+n \log ^2\left (\sqrt{b^2-4 a c}+b+2 c x\right )-2 n \log \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{2 \sqrt{b^2-4 a c}}\right ) \log \left (-\sqrt{b^2-4 a c}+b+2 c x\right )+2 n \log \left (\frac{\sqrt{b^2-4 a c}-b-2 c x}{2 \sqrt{b^2-4 a c}}\right ) \log \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{2 e \sqrt{b^2-4 a c}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-(n*Log[b - Sqrt[b^2 - 4*a*c] + 2*c*x]^2) + 2*n*Log[(-b + Sqrt[b^2 - 4*a*c] - 2*c*x)/(2*Sqrt[b^2 - 4*a*c])]*L
og[b + Sqrt[b^2 - 4*a*c] + 2*c*x] + n*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x]^2 - 2*n*Log[b - Sqrt[b^2 - 4*a*c] + 2
*c*x]*Log[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(2*Sqrt[b^2 - 4*a*c])] + 2*Log[b - Sqrt[b^2 - 4*a*c] + 2*c*x]*Log[d*
(a + x*(b + c*x))^n] - 2*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x]*Log[d*(a + x*(b + c*x))^n] - 2*n*PolyLog[2, (-b +
Sqrt[b^2 - 4*a*c] - 2*c*x)/(2*Sqrt[b^2 - 4*a*c])] + 2*n*PolyLog[2, (b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(2*Sqrt[b^2
- 4*a*c])])/(2*Sqrt[b^2 - 4*a*c]*e)

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Maple [F]  time = 0.229, size = 0, normalized size = 0. \begin{align*} \int{\frac{\ln \left ( d \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) }{ce{x}^{2}+bex+ea}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{c e x^{2} + b e x + a e}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log((c*x^2 + b*x + a)^n*d)/(c*e*x^2 + b*e*x + a*e), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{c e x^{2} + b e x + a e}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log((c*x^2 + b*x + a)^n*d)/(c*e*x^2 + b*e*x + a*e), x)