\(\int \frac {\log (d (a+b x+c x^2)^n)}{a e+b e x+c e x^2} \, dx\) [93]

Optimal result

Integrand size = 32, antiderivative size = 258 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=\frac {2 n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )^2}{\sqrt {b^2-4 a c} e}-\frac {4 n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (\frac {2}{1-\frac {b}{\sqrt {b^2-4 a c}}-\frac {2 c x}{\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} e}-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{\sqrt {b^2-4 a c} e}-\frac {2 n \operatorname {PolyLog}\left (2,-\frac {1+\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}}{1-\frac {b}{\sqrt {b^2-4 a c}}-\frac {2 c x}{\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} e} \]

[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)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {632, 212, 2607, 12, 6256, 6131, 6055, 2449, 2352} \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e \sqrt {b^2-4 a c}}+\frac {2 n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )^2}{e \sqrt {b^2-4 a c}}-\frac {4 n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (\frac {2}{-\frac {2 c x}{\sqrt {b^2-4 a c}}-\frac {b}{\sqrt {b^2-4 a c}}+1}\right )}{e \sqrt {b^2-4 a c}}-\frac {2 n \operatorname {PolyLog}\left (2,-\frac {\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}+1}{-\frac {b}{\sqrt {b^2-4 a c}}-\frac {2 c x}{\sqrt {b^2-4 a c}}+1}\right )}{e \sqrt {b^2-4 a c}} \]

[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 212

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

Rule 632

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 2352

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

Rule 2449

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 2607

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 6055

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 6131

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 6256

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/d^2)*x^2)^q*(a + b*ArcT
anh[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*} \text {integral}& = -\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 \text {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 ) \text {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) \text {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) \text {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) \text {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] (verified)

Time = 0.22 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.31 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=\frac {-n \log ^2\left (b-\sqrt {b^2-4 a c}+2 c x\right )+2 n \log \left (\frac {-b+\sqrt {b^2-4 a c}-2 c x}{2 \sqrt {b^2-4 a c}}\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c x\right )+n \log ^2\left (b+\sqrt {b^2-4 a c}+2 c x\right )-2 n \log \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \log \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{2 \sqrt {b^2-4 a c}}\right )+2 \log \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \log \left (d (a+x (b+c x))^n\right )-2 \log \left (b+\sqrt {b^2-4 a c}+2 c x\right ) \log \left (d (a+x (b+c x))^n\right )-2 n \operatorname {PolyLog}\left (2,\frac {-b+\sqrt {b^2-4 a c}-2 c x}{2 \sqrt {b^2-4 a c}}\right )+2 n \operatorname {PolyLog}\left (2,\frac {b+\sqrt {b^2-4 a c}+2 c x}{2 \sqrt {b^2-4 a c}}\right )}{2 \sqrt {b^2-4 a c} e} \]

[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)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 2.46 (sec) , antiderivative size = 433, normalized size of antiderivative = 1.68

method	result	size
risch	\(-\frac {2 \arctan \left (\frac {2 x c +b}{\sqrt {4 c a -b^{2}}}\right ) n \ln \left (c \,x^{2}+b x +a \right )}{e \sqrt {4 c a -b^{2}}}+\frac {2 \arctan \left (\frac {2 x c +b}{\sqrt {4 c a -b^{2}}}\right ) \ln \left (\left (c \,x^{2}+b x +a \right )^{n}\right )}{e \sqrt {4 c a -b^{2}}}+\frac {n \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (c \,\textit {\_Z}^{2}+\textit {\_Z} b +a \right )}{\sum }\frac {2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (c \,x^{2}+b x +a \right )-\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{2 \underline {\hspace {1.25 ex}}\alpha c +b}-\frac {2 \left (2 \underline {\hspace {1.25 ex}}\alpha c +b \right ) \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {2 \underline {\hspace {1.25 ex}}\alpha c +\left (x -\underline {\hspace {1.25 ex}}\alpha \right ) c +b}{2 \underline {\hspace {1.25 ex}}\alpha c +b}\right )}{4 c a -b^{2}}-\frac {2 \left (2 \underline {\hspace {1.25 ex}}\alpha c +b \right ) \operatorname {dilog}\left (\frac {2 \underline {\hspace {1.25 ex}}\alpha c +\left (x -\underline {\hspace {1.25 ex}}\alpha \right ) c +b}{2 \underline {\hspace {1.25 ex}}\alpha c +b}\right )}{4 c a -b^{2}}}{2 \underline {\hspace {1.25 ex}}\alpha c +b}\right )}{2 e}+\frac {\left (i \pi \,\operatorname {csgn}\left (i \left (c \,x^{2}+b x +a \right )^{n}\right ) {\operatorname {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (c \,x^{2}+b x +a \right )^{n}\right ) \operatorname {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right ) \operatorname {csgn}\left (i d \right )-i \pi {\operatorname {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )}^{3}+i \pi {\operatorname {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )}^{2} \operatorname {csgn}\left (i d \right )+2 \ln \left (d \right )\right ) \arctan \left (\frac {2 x c +b}{\sqrt {4 c a -b^{2}}}\right )}{e \sqrt {4 c a -b^{2}}}\)	\(433\)

[In]

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

[Out]

-2/e/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*n*ln(c*x^2+b*x+a)+2/e/(4*a*c-b^2)^(1/2)*arctan((2*c
*x+b)/(4*a*c-b^2)^(1/2))*ln((c*x^2+b*x+a)^n)+1/2/e*n*sum(1/(2*_alpha*c+b)*(2*ln(x-_alpha)*ln(c*x^2+b*x+a)-1/(2
*_alpha*c+b)*ln(x-_alpha)^2-2*(2*_alpha*c+b)/(4*a*c-b^2)*ln(x-_alpha)*ln((2*_alpha*c+(x-_alpha)*c+b)/(2*_alpha
*c+b))-2*(2*_alpha*c+b)/(4*a*c-b^2)*dilog((2*_alpha*c+(x-_alpha)*c+b)/(2*_alpha*c+b))),_alpha=RootOf(_Z^2*c+_Z
*b+a))+(I*Pi*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)^2-I*Pi*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+
b*x+a)^n)*csgn(I*d)-I*Pi*csgn(I*d*(c*x^2+b*x+a)^n)^3+I*Pi*csgn(I*d*(c*x^2+b*x+a)^n)^2*csgn(I*d)+2*ln(d))/e/(4*
a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))

Fricas [F]

\[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=\int { \frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{c e x^{2} + b e x + a e} \,d x } \]

[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)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=\text {Exception raised: ValueError} \]

[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 deta

Giac [F]

\[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=\int { \frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{c e x^{2} + b e x + a e} \,d x } \]

[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)

Mupad [F(-1)]

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

[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)