Integrand size = 32, antiderivative size = 258 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=\frac {2 n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )^2}{\sqrt {b^2-4 a c} e}-\frac {4 n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (\frac {2}{1-\frac {b}{\sqrt {b^2-4 a c}}-\frac {2 c x}{\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} e}-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{\sqrt {b^2-4 a c} e}-\frac {2 n \operatorname {PolyLog}\left (2,-\frac {1+\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}}{1-\frac {b}{\sqrt {b^2-4 a c}}-\frac {2 c x}{\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} e} \] Output:
2*n*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))^2/(-4*a*c+b^2)^(1/2)/e-4*n*arcta nh((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)))/(-4*a*c+b^2)^(1/2)/e-2*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2) )*ln(d*(c*x^2+b*x+a)^n)/(-4*a*c+b^2)^(1/2)/e-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)))/(-4*a*c+b^2)^(1/2)/e
Time = 0.41 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.31 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=\frac {-n \log ^2\left (b-\sqrt {b^2-4 a c}+2 c x\right )+2 n \log \left (\frac {-b+\sqrt {b^2-4 a c}-2 c x}{2 \sqrt {b^2-4 a c}}\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c x\right )+n \log ^2\left (b+\sqrt {b^2-4 a c}+2 c x\right )-2 n \log \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \log \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{2 \sqrt {b^2-4 a c}}\right )+2 \log \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \log \left (d (a+x (b+c x))^n\right )-2 \log \left (b+\sqrt {b^2-4 a c}+2 c x\right ) \log \left (d (a+x (b+c x))^n\right )-2 n \operatorname {PolyLog}\left (2,\frac {-b+\sqrt {b^2-4 a c}-2 c x}{2 \sqrt {b^2-4 a c}}\right )+2 n \operatorname {PolyLog}\left (2,\frac {b+\sqrt {b^2-4 a c}+2 c x}{2 \sqrt {b^2-4 a c}}\right )}{2 \sqrt {b^2-4 a c} e} \] Input:
Integrate[Log[d*(a + b*x + c*x^2)^n]/(a*e + b*e*x + c*e*x^2),x]
Output:
(-(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])]*Log[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)
Time = 0.96 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.96, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {3007, 27, 6671, 27, 25, 6546, 6470, 2849, 2752}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx\) |
| \(\Big \downarrow \) 3007 |
| \(\displaystyle -n \int -\frac {2 (b+2 c x) \text {arctanh}\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} e \left (c x^2+b x+a\right )}dx-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e \sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 n \int \frac {(b+2 c x) \text {arctanh}\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )}{c x^2+b x+a}dx}{e \sqrt {b^2-4 a c}}-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e \sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 6671 |
| \(\displaystyle \frac {n \int \frac {4 c \sqrt {b^2-4 a c} \left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right ) \left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )^2+\left (4 a-\frac {b^2}{c}\right ) c}d\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )}{c e}-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e \sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4 n \sqrt {b^2-4 a c} \int -\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )}{b^2-\left (b^2-4 a c\right ) \left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )^2-4 a c}d\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )}{e}-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e \sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {4 n \sqrt {b^2-4 a c} \int \frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )}{b^2-\left (b^2-4 a c\right ) \left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )^2-4 a c}d\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )}{e}-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e \sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 6546 |
| \(\displaystyle -\frac {4 n \sqrt {b^2-4 a c} \left (\frac {\int \frac {\text {arctanh}\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )}{-\frac {b}{\sqrt {b^2-4 a c}}-\frac {2 c x}{\sqrt {b^2-4 a c}}+1}d\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )}{b^2-4 a c}-\frac {\text {arctanh}\left (\frac {2 c x}{\sqrt {b^2-4 a c}}+\frac {b}{\sqrt {b^2-4 a c}}\right )^2}{2 \left (b^2-4 a c\right )}\right )}{e}-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e \sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 6470 |
| \(\displaystyle -\frac {4 n \sqrt {b^2-4 a c} \left (\frac {\text {arctanh}\left (\frac {2 c x}{\sqrt {b^2-4 a c}}+\frac {b}{\sqrt {b^2-4 a c}}\right ) \log \left (\frac {2}{-\frac {2 c x}{\sqrt {b^2-4 a c}}-\frac {b}{\sqrt {b^2-4 a c}}+1}\right )-\int \frac {\log \left (\frac {2}{-\frac {b}{\sqrt {b^2-4 a c}}-\frac {2 c x}{\sqrt {b^2-4 a c}}+1}\right )}{1-\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )^2}d\left (\frac {b}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}\right )}{b^2-4 a c}-\frac {\text {arctanh}\left (\frac {2 c x}{\sqrt {b^2-4 a c}}+\frac {b}{\sqrt {b^2-4 a c}}\right )^2}{2 \left (b^2-4 a c\right )}\right )}{e}-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e \sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 2849 |
| \(\displaystyle -\frac {4 n \sqrt {b^2-4 a c} \left (\frac {\int \frac {\log \left (\frac {2}{-\frac {b}{\sqrt {b^2-4 a c}}-\frac {2 c x}{\sqrt {b^2-4 a c}}+1}\right )}{1-\frac {2}{-\frac {b}{\sqrt {b^2-4 a c}}-\frac {2 c x}{\sqrt {b^2-4 a c}}+1}}d\frac {1}{-\frac {b}{\sqrt {b^2-4 a c}}-\frac {2 c x}{\sqrt {b^2-4 a c}}+1}+\text {arctanh}\left (\frac {2 c x}{\sqrt {b^2-4 a c}}+\frac {b}{\sqrt {b^2-4 a c}}\right ) \log \left (\frac {2}{-\frac {2 c x}{\sqrt {b^2-4 a c}}-\frac {b}{\sqrt {b^2-4 a c}}+1}\right )}{b^2-4 a c}-\frac {\text {arctanh}\left (\frac {2 c x}{\sqrt {b^2-4 a c}}+\frac {b}{\sqrt {b^2-4 a c}}\right )^2}{2 \left (b^2-4 a c\right )}\right )}{e}-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e \sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 2752 |
| \(\displaystyle -\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e \sqrt {b^2-4 a c}}-\frac {4 n \sqrt {b^2-4 a c} \left (\frac {\text {arctanh}\left (\frac {2 c x}{\sqrt {b^2-4 a c}}+\frac {b}{\sqrt {b^2-4 a c}}\right ) \log \left (\frac {2}{-\frac {2 c x}{\sqrt {b^2-4 a c}}-\frac {b}{\sqrt {b^2-4 a c}}+1}\right )+\frac {1}{2} \operatorname {PolyLog}\left (2,1-\frac {2}{-\frac {b}{\sqrt {b^2-4 a c}}-\frac {2 c x}{\sqrt {b^2-4 a c}}+1}\right )}{b^2-4 a c}-\frac {\text {arctanh}\left (\frac {2 c x}{\sqrt {b^2-4 a c}}+\frac {b}{\sqrt {b^2-4 a c}}\right )^2}{2 \left (b^2-4 a c\right )}\right )}{e}\) |
Input:
Int[Log[d*(a + b*x + c*x^2)^n]/(a*e + b*e*x + c*e*x^2),x]
Output:
(-2*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]]*Log[d*(a + b*x + c*x^2)^n])/(Sq rt[b^2 - 4*a*c]*e) - (4*Sqrt[b^2 - 4*a*c]*n*(-1/2*ArcTanh[b/Sqrt[b^2 - 4*a *c] + (2*c*x)/Sqrt[b^2 - 4*a*c]]^2/(b^2 - 4*a*c) + (ArcTanh[b/Sqrt[b^2 - 4 *a*c] + (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])] + PolyLog[2, 1 - 2/(1 - b/Sqrt[b^2 - 4*a*c] - (2*c*x )/Sqrt[b^2 - 4*a*c])]/2)/(b^2 - 4*a*c)))/e
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp [-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]
Int[Log[(c_.)*(Px_)^(n_.)]/(Qx_), x_Symbol] :> With[{u = IntHide[1/Qx, x]}, Simp[u*Log[c*Px^n], x] - Simp[n Int[SimplifyIntegrand[u*(D[Px, x]/Px), x ], x], x]] /; FreeQ[{c, n}, x] && QuadraticQ[{Qx, Px}, x] && EqQ[D[Px/Qx, x ], 0]
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] + Simp[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]
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] + Simp[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]
Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( m_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/d Sub st[Int[((d*e - c*f)/d + f*(x/d))^m*(-C/d^2 + (C/d^2)*x^2)^q*(a + b*ArcTanh[ 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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.00 (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 a c -b^{2}}}\right ) n \ln \left (c \,x^{2}+b x +a \right )}{e \sqrt {4 a c -b^{2}}}+\frac {2 \arctan \left (\frac {2 x c +b}{\sqrt {4 a c -b^{2}}}\right ) \ln \left (\left (c \,x^{2}+b x +a \right )^{n}\right )}{e \sqrt {4 a c -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 c \underline {\hspace {1.25 ex}}\alpha +b}-\frac {2 \left (2 c \underline {\hspace {1.25 ex}}\alpha +b \right ) \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {2 c \underline {\hspace {1.25 ex}}\alpha +c \left (x -\underline {\hspace {1.25 ex}}\alpha \right )+b}{2 c \underline {\hspace {1.25 ex}}\alpha +b}\right )}{4 a c -b^{2}}-\frac {2 \left (2 c \underline {\hspace {1.25 ex}}\alpha +b \right ) \operatorname {dilog}\left (\frac {2 c \underline {\hspace {1.25 ex}}\alpha +c \left (x -\underline {\hspace {1.25 ex}}\alpha \right )+b}{2 c \underline {\hspace {1.25 ex}}\alpha +b}\right )}{4 a c -b^{2}}}{2 c \underline {\hspace {1.25 ex}}\alpha +b}\right )}{2 e}+\frac {\left (-i \pi \,\operatorname {csgn}\left (i d \right ) \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 )+i \pi \,\operatorname {csgn}\left (i d \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 )}^{2}-i \pi {\operatorname {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )}^{3}+2 \ln \left (d \right )\right ) \arctan \left (\frac {2 x c +b}{\sqrt {4 a c -b^{2}}}\right )}{e \sqrt {4 a c -b^{2}}}\) | \(433\) |
Input:
int(ln(d*(c*x^2+b*x+a)^n)/(c*e*x^2+b*e*x+a*e),x,method=_RETURNVERBOSE)
Output:
-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* c*_alpha+c*(x-_alpha)+b)/(2*_alpha*c+b))-2*(2*_alpha*c+b)/(4*a*c-b^2)*dilo g((2*c*_alpha+c*(x-_alpha)+b)/(2*_alpha*c+b))),_alpha=RootOf(_Z^2*c+_Z*b+a ))+(-I*Pi*csgn(I*d)*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)+I*Pi *csgn(I*d)*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)^2-I*Pi*csgn(I*d*(c*x^2+b*x+a)^n)^3+2*ln(d))/e/(4*a*c-b ^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))
\[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=\int { \frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{c e x^{2} + b e x + a e} \,d x } \] Input:
integrate(log(d*(c*x^2+b*x+a)^n)/(c*e*x^2+b*e*x+a*e),x, algorithm="fricas" )
Output:
integral(log((c*x^2 + b*x + a)^n*d)/(c*e*x^2 + b*e*x + a*e), x)
Timed out. \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=\text {Timed out} \] Input:
integrate(ln(d*(c*x**2+b*x+a)**n)/(c*e*x**2+b*e*x+a*e),x)
Output:
Timed out
Exception generated. \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(log(d*(c*x^2+b*x+a)^n)/(c*e*x^2+b*e*x+a*e),x, algorithm="maxima" )
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
\[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=\int { \frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{c e x^{2} + b e x + a e} \,d x } \] Input:
integrate(log(d*(c*x^2+b*x+a)^n)/(c*e*x^2+b*e*x+a*e),x, algorithm="giac")
Output:
integrate(log((c*x^2 + b*x + a)^n*d)/(c*e*x^2 + b*e*x + a*e), x)
Timed out. \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=\int \frac {\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{c\,e\,x^2+b\,e\,x+a\,e} \,d x \] Input:
int(log(d*(a + b*x + c*x^2)^n)/(a*e + b*e*x + c*e*x^2),x)
Output:
int(log(d*(a + b*x + c*x^2)^n)/(a*e + b*e*x + c*e*x^2), x)
\[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{a e+b e x+c e x^2} \, dx=\frac {\int \frac {\mathrm {log}\left (\left (c \,x^{2}+b x +a \right )^{n} d \right )}{c \,x^{2}+b x +a}d x}{e} \] Input:
int(log(d*(c*x^2+b*x+a)^n)/(c*e*x^2+b*e*x+a*e),x)
Output:
int(log((a + b*x + c*x**2)**n*d)/(a + b*x + c*x**2),x)/e