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 \[ -\frac {2 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e \sqrt {b^2-4 a c}}-\frac {2 n \text {Li}_2\left (-\frac {\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}+1}{-\frac {b}{\sqrt {b^2-4 a c}}-\frac {2 c x}{\sqrt {b^2-4 a c}}+1}\right )}{e \sqrt {b^2-4 a c}}+\frac {2 n \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )^2}{e \sqrt {b^2-4 a c}}-\frac {4 n \log \left (\frac {2}{-\frac {2 c x}{\sqrt {b^2-4 a c}}-\frac {b}{\sqrt {b^2-4 a c}}+1}\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{e \sqrt {b^2-4 a c}} \]

[Out]

2*n*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))^2/e/(-4*a*c+b^2)^(1/2)-2*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))*ln(d*
(c*x^2+b*x+a)^n)/e/(-4*a*c+b^2)^(1/2)-4*n*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))*ln(2/(1-b/(-4*a*c+b^2)^(1/2)-2
*c*x/(-4*a*c+b^2)^(1/2)))/e/(-4*a*c+b^2)^(1/2)-2*n*polylog(2,(-1-b/(-4*a*c+b^2)^(1/2)-2*c*x/(-4*a*c+b^2)^(1/2)
)/(1-b/(-4*a*c+b^2)^(1/2)-2*c*x/(-4*a*c+b^2)^(1/2)))/e/(-4*a*c+b^2)^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 258, normalized size of antiderivative = 1.00, 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 12

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

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 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 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]

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.19, size = 339, normalized size = 1.31 \[ \frac {2 \log \left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \log \left (d (a+x (b+c x))^n\right )-2 \log \left (\sqrt {b^2-4 a c}+b+2 c x\right ) \log \left (d (a+x (b+c x))^n\right )-2 n \text {Li}_2\left (\frac {-b-2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )+2 n \text {Li}_2\left (\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )-n \log ^2\left (-\sqrt {b^2-4 a c}+b+2 c x\right )+n \log ^2\left (\sqrt {b^2-4 a c}+b+2 c x\right )-2 n \log \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{2 \sqrt {b^2-4 a c}}\right ) \log \left (-\sqrt {b^2-4 a c}+b+2 c x\right )+2 n \log \left (\frac {\sqrt {b^2-4 a c}-b-2 c x}{2 \sqrt {b^2-4 a c}}\right ) \log \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{2 e \sqrt {b^2-4 a c}} \]

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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fricas [F]  time = 1.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{c e x^{2} + b e x + a e}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{c e x^{2} + b e x + a e}\,{d x} \]

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)

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maple [F]  time = 1.71, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right )}{c e \,x^{2}+b e x +a e}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{c\,e\,x^2+b\,e\,x+a\,e} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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