\(\int \frac {e^6+e^3 (5 x+x^2)+(-e^6+e^3 (-10 x-3 x^2)) \log (x)}{e^6 x^2+25 x^4+10 x^5+x^6+e^3 (10 x^3+2 x^4)} \, dx\) [265]

Optimal result

Integrand size = 76, antiderivative size = 17 \[ \int \frac {e^6+e^3 \left (5 x+x^2\right )+\left (-e^6+e^3 \left (-10 x-3 x^2\right )\right ) \log (x)}{e^6 x^2+25 x^4+10 x^5+x^6+e^3 \left (10 x^3+2 x^4\right )} \, dx=\frac {\log (x)}{x+\frac {x^2 (5+x)}{e^3}} \] Output:

ln(x)/(x+(5+x)*x^2/exp(3))
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \frac {e^6+e^3 \left (5 x+x^2\right )+\left (-e^6+e^3 \left (-10 x-3 x^2\right )\right ) \log (x)}{e^6 x^2+25 x^4+10 x^5+x^6+e^3 \left (10 x^3+2 x^4\right )} \, dx=\frac {e^3 \log (x)}{x \left (e^3+5 x+x^2\right )} \] Input:

Integrate[(E^6 + E^3*(5*x + x^2) + (-E^6 + E^3*(-10*x - 3*x^2))*Log[x])/(E 
^6*x^2 + 25*x^4 + 10*x^5 + x^6 + E^3*(10*x^3 + 2*x^4)),x]
 

Output:

(E^3*Log[x])/(x*(E^3 + 5*x + x^2))
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 5.17 (sec) , antiderivative size = 1467, normalized size of antiderivative = 86.29, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {2026, 2463, 2009}

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.

\(\displaystyle \int \frac {e^3 \left (x^2+5 x\right )+\left (e^3 \left (-3 x^2-10 x\right )-e^6\right ) \log (x)+e^6}{x^6+10 x^5+25 x^4+e^6 x^2+e^3 \left (2 x^4+10 x^3\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {e^3 \left (x^2+5 x\right )+\left (e^3 \left (-3 x^2-10 x\right )-e^6\right ) \log (x)+e^6}{x^2 \left (x^4+10 x^3+\left (25+2 e^3\right ) x^2+10 e^3 x+e^6\right )}dx\)

\(\Big \downarrow \) 2463

\(\displaystyle \int \left (\frac {4 i \left (e^3 \left (x^2+5 x\right )+\left (e^3 \left (-3 x^2-10 x\right )-e^6\right ) \log (x)+e^6\right )}{\left (4 e^3-25\right )^{3/2} x^2 \left (2 x+i \sqrt {4 e^3-25}+5\right )}+\frac {4 i \left (e^3 \left (x^2+5 x\right )+\left (e^3 \left (-3 x^2-10 x\right )-e^6\right ) \log (x)+e^6\right )}{\left (4 e^3-25\right )^{3/2} \left (-2 x+i \sqrt {4 e^3-25}-5\right ) x^2}-\frac {4 \left (e^3 \left (x^2+5 x\right )+\left (e^3 \left (-3 x^2-10 x\right )-e^6\right ) \log (x)+e^6\right )}{\left (4 e^3-25\right ) \left (-2 x+i \sqrt {4 e^3-25}-5\right )^2 x^2}-\frac {4 \left (e^3 \left (x^2+5 x\right )+\left (e^3 \left (-3 x^2-10 x\right )-e^6\right ) \log (x)+e^6\right )}{\left (4 e^3-25\right ) x^2 \left (2 x+i \sqrt {4 e^3-25}+5\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 e^3 \left (25 i-2 i e^3+5 \sqrt {-25+4 e^3}\right ) \log ^2(x)}{\left (25-4 e^3\right ) \left (5 i+\sqrt {-25+4 e^3}\right )^3}-\frac {4 e^3 \left (25 i-i e^3+5 \sqrt {-25+4 e^3}\right ) \log ^2(x)}{\left (-25+4 e^3\right )^{3/2} \left (5 i+\sqrt {-25+4 e^3}\right )^2}+\frac {4 e^3 \left (i \left (25-e^3\right )-5 \sqrt {-25+4 e^3}\right ) \log ^2(x)}{\left (-25+4 e^3\right )^{3/2} \left (5 i-\sqrt {-25+4 e^3}\right )^2}+\frac {4 e^3 \left (25 i-2 i e^3-5 \sqrt {-25+4 e^3}\right ) \log ^2(x)}{\left (25-4 e^3\right ) \left (5 i-\sqrt {-25+4 e^3}\right )^3}-\frac {8 e^3 \left (25-4 e^3+5 i \sqrt {-25+4 e^3}\right ) x \log (x)}{\left (25-4 e^3\right ) \left (5 i-\sqrt {-25+4 e^3}\right )^3 \left (2 i x-\sqrt {-25+4 e^3}+5 i\right )}-\frac {8 e^3 \left (25-4 e^3-5 i \sqrt {-25+4 e^3}\right ) x \log (x)}{\left (25-4 e^3\right ) \left (5 i+\sqrt {-25+4 e^3}\right )^3 \left (2 i x+\sqrt {-25+4 e^3}+5 i\right )}-\frac {4 e^3 \left (25 i-4 i e^3+5 \sqrt {-25+4 e^3}\right ) \log \left (\frac {2 x}{5-i \sqrt {-25+4 e^3}}+1\right ) \log (x)}{\left (-25+4 e^3\right )^{3/2} \left (5 i+\sqrt {-25+4 e^3}\right )^2}-\frac {8 e^3 \left (25 i-2 i e^3+5 \sqrt {-25+4 e^3}\right ) \log \left (\frac {2 x}{5-i \sqrt {-25+4 e^3}}+1\right ) \log (x)}{\left (25-4 e^3\right ) \left (5 i+\sqrt {-25+4 e^3}\right )^3}+\frac {4 e^3 \left (25 i-4 i e^3-5 \sqrt {-25+4 e^3}\right ) \log \left (\frac {2 x}{5+i \sqrt {-25+4 e^3}}+1\right ) \log (x)}{\left (-25+4 e^3\right )^{3/2} \left (5 i-\sqrt {-25+4 e^3}\right )^2}-\frac {8 e^3 \left (25 i-2 i e^3-5 \sqrt {-25+4 e^3}\right ) \log \left (\frac {2 x}{5+i \sqrt {-25+4 e^3}}+1\right ) \log (x)}{\left (25-4 e^3\right ) \left (5 i-\sqrt {-25+4 e^3}\right )^3}+\frac {4 e^6 \log (x)}{\left (-25+4 e^3\right )^{3/2} \left (5 i+\sqrt {-25+4 e^3}\right ) x}-\frac {4 e^6 \log (x)}{\left (25-4 e^3\right ) \left (5 i+\sqrt {-25+4 e^3}\right )^2 x}-\frac {4 e^6 \log (x)}{\left (-25+4 e^3\right )^{3/2} \left (5 i-\sqrt {-25+4 e^3}\right ) x}-\frac {4 e^6 \log (x)}{\left (25-4 e^3\right ) \left (5 i-\sqrt {-25+4 e^3}\right )^2 x}+\frac {4 e^3 \left (4 i e^3-5 \left (5 i+\sqrt {-25+4 e^3}\right )\right ) \log (x)}{\left (25-4 e^3\right ) \left (5 i+\sqrt {-25+4 e^3}\right )^3}-\frac {4 e^3 \left (25 i-4 i e^3-5 \sqrt {-25+4 e^3}\right ) \log (x)}{\left (25-4 e^3\right ) \left (5 i-\sqrt {-25+4 e^3}\right )^3}-\frac {4 e^3 \left (25 i-4 i e^3+5 \sqrt {-25+4 e^3}\right ) \log \left (2 i x+\sqrt {-25+4 e^3}+5 i\right )}{\left (25-4 e^3\right ) \left (5 i+\sqrt {-25+4 e^3}\right )^3}+\frac {4 i e^3 \log \left (2 i x+\sqrt {-25+4 e^3}+5 i\right )}{\sqrt {-25+4 e^3} \left (5 i+\sqrt {-25+4 e^3}\right )^2}-\frac {4 e^3 \left (25 i-4 i e^3+5 \sqrt {-25+4 e^3}\right ) \operatorname {PolyLog}\left (2,-\frac {2 x}{5-i \sqrt {-25+4 e^3}}\right )}{\left (-25+4 e^3\right )^{3/2} \left (5 i+\sqrt {-25+4 e^3}\right )^2}-\frac {8 e^3 \left (25 i-2 i e^3+5 \sqrt {-25+4 e^3}\right ) \operatorname {PolyLog}\left (2,-\frac {2 x}{5-i \sqrt {-25+4 e^3}}\right )}{\left (25-4 e^3\right ) \left (5 i+\sqrt {-25+4 e^3}\right )^3}+\frac {4 e^3 \left (25 i-4 i e^3-5 \sqrt {-25+4 e^3}\right ) \operatorname {PolyLog}\left (2,-\frac {2 x}{5+i \sqrt {-25+4 e^3}}\right )}{\left (-25+4 e^3\right )^{3/2} \left (5 i-\sqrt {-25+4 e^3}\right )^2}-\frac {8 e^3 \left (25 i-2 i e^3-5 \sqrt {-25+4 e^3}\right ) \operatorname {PolyLog}\left (2,-\frac {2 x}{5+i \sqrt {-25+4 e^3}}\right )}{\left (25-4 e^3\right ) \left (5 i-\sqrt {-25+4 e^3}\right )^3}\)

Input:

Int[(E^6 + E^3*(5*x + x^2) + (-E^6 + E^3*(-10*x - 3*x^2))*Log[x])/(E^6*x^2 
 + 25*x^4 + 10*x^5 + x^6 + E^3*(10*x^3 + 2*x^4)),x]
 

Output:

((4*I)*E^3*Log[5*I + Sqrt[-25 + 4*E^3] + (2*I)*x])/(Sqrt[-25 + 4*E^3]*(5*I 
 + Sqrt[-25 + 4*E^3])^2) - (4*E^3*(25*I - (4*I)*E^3 + 5*Sqrt[-25 + 4*E^3]) 
*Log[5*I + Sqrt[-25 + 4*E^3] + (2*I)*x])/((25 - 4*E^3)*(5*I + Sqrt[-25 + 4 
*E^3])^3) - (4*E^3*(25*I - (4*I)*E^3 - 5*Sqrt[-25 + 4*E^3])*Log[x])/((25 - 
 4*E^3)*(5*I - Sqrt[-25 + 4*E^3])^3) + (4*E^3*((4*I)*E^3 - 5*(5*I + Sqrt[- 
25 + 4*E^3]))*Log[x])/((25 - 4*E^3)*(5*I + Sqrt[-25 + 4*E^3])^3) - (4*E^6* 
Log[x])/((25 - 4*E^3)*(5*I - Sqrt[-25 + 4*E^3])^2*x) - (4*E^6*Log[x])/((-2 
5 + 4*E^3)^(3/2)*(5*I - Sqrt[-25 + 4*E^3])*x) - (4*E^6*Log[x])/((25 - 4*E^ 
3)*(5*I + Sqrt[-25 + 4*E^3])^2*x) + (4*E^6*Log[x])/((-25 + 4*E^3)^(3/2)*(5 
*I + Sqrt[-25 + 4*E^3])*x) - (8*E^3*(25 - 4*E^3 + (5*I)*Sqrt[-25 + 4*E^3]) 
*x*Log[x])/((25 - 4*E^3)*(5*I - Sqrt[-25 + 4*E^3])^3*(5*I - Sqrt[-25 + 4*E 
^3] + (2*I)*x)) - (8*E^3*(25 - 4*E^3 - (5*I)*Sqrt[-25 + 4*E^3])*x*Log[x])/ 
((25 - 4*E^3)*(5*I + Sqrt[-25 + 4*E^3])^3*(5*I + Sqrt[-25 + 4*E^3] + (2*I) 
*x)) + (4*E^3*(25*I - (2*I)*E^3 - 5*Sqrt[-25 + 4*E^3])*Log[x]^2)/((25 - 4* 
E^3)*(5*I - Sqrt[-25 + 4*E^3])^3) + (4*E^3*(I*(25 - E^3) - 5*Sqrt[-25 + 4* 
E^3])*Log[x]^2)/((-25 + 4*E^3)^(3/2)*(5*I - Sqrt[-25 + 4*E^3])^2) - (4*E^3 
*(25*I - I*E^3 + 5*Sqrt[-25 + 4*E^3])*Log[x]^2)/((-25 + 4*E^3)^(3/2)*(5*I 
+ Sqrt[-25 + 4*E^3])^2) + (4*E^3*(25*I - (2*I)*E^3 + 5*Sqrt[-25 + 4*E^3])* 
Log[x]^2)/((25 - 4*E^3)*(5*I + Sqrt[-25 + 4*E^3])^3) - (8*E^3*(25*I - (2*I 
)*E^3 + 5*Sqrt[-25 + 4*E^3])*Log[x]*Log[1 + (2*x)/(5 - I*Sqrt[-25 + 4*E...
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2026

Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[u, x])
 

rule 2463

Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr 
and[u, Qx^p, x], x] /;  !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt 
Q[Expon[Px, x], 2] &&  !BinomialQ[Px, x] &&  !TrinomialQ[Px, x] && ILtQ[p, 
0]
 

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18

method	result	size
norman	\(\frac {\ln \left (x \right ) {\mathrm e}^{3}}{x \left (x^{2}+{\mathrm e}^{3}+5 x \right )}\)	\(20\)
risch	\(\frac {\ln \left (x \right ) {\mathrm e}^{3}}{x \left (x^{2}+{\mathrm e}^{3}+5 x \right )}\)	\(20\)
parallelrisch	\(\frac {\ln \left (x \right ) {\mathrm e}^{3}}{x \left (x^{2}+{\mathrm e}^{3}+5 x \right )}\)	\(20\)
default	\({\mathrm e}^{3} \left (-{\mathrm e}^{-6} \left (-\frac {5 \ln \left (x^{2}+{\mathrm e}^{3}+5 x \right )}{2}+\frac {2 \left (-\frac {25}{2}+{\mathrm e}^{3}\right ) \arctan \left (\frac {5+2 x}{\sqrt {4 \,{\mathrm e}^{3}-25}}\right )}{\sqrt {4 \,{\mathrm e}^{3}-25}}\right )-\frac {{\mathrm e}^{-6} {\mathrm e}^{3}}{x}-5 \,{\mathrm e}^{-6} \ln \left (x \right )\right )-{\mathrm e}^{3} \left (-\frac {{\mathrm e}^{-12} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+10 \textit {\_Z}^{3}+\left (2 \,{\mathrm e}^{3}+25\right ) \textit {\_Z}^{2}+10 \textit {\_Z} \,{\mathrm e}^{3}+{\mathrm e}^{6}\right )}{\sum }\frac {\left (10 \,{\mathrm e}^{9} \textit {\_R1} +{\mathrm e}^{9} \textit {\_R1}^{2}+25 \,{\mathrm e}^{9}-{\mathrm e}^{12}\right ) \left (\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )\right )}{2 \textit {\_R1}^{3}+2 \,{\mathrm e}^{3} \textit {\_R1} +15 \textit {\_R1}^{2}+5 \,{\mathrm e}^{3}+25 \textit {\_R1}}\right )}{2}+{\mathrm e}^{-6} {\mathrm e}^{3} \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )\right )\)	\(217\)
parts	\({\mathrm e}^{3} \left (-{\mathrm e}^{-6} \left (-\frac {5 \ln \left (x^{2}+{\mathrm e}^{3}+5 x \right )}{2}+\frac {2 \left (-\frac {25}{2}+{\mathrm e}^{3}\right ) \arctan \left (\frac {5+2 x}{\sqrt {4 \,{\mathrm e}^{3}-25}}\right )}{\sqrt {4 \,{\mathrm e}^{3}-25}}\right )-\frac {{\mathrm e}^{-6} {\mathrm e}^{3}}{x}-5 \,{\mathrm e}^{-6} \ln \left (x \right )\right )-{\mathrm e}^{3} \left (-\frac {{\mathrm e}^{-12} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+10 \textit {\_Z}^{3}+\left (2 \,{\mathrm e}^{3}+25\right ) \textit {\_Z}^{2}+10 \textit {\_Z} \,{\mathrm e}^{3}+{\mathrm e}^{6}\right )}{\sum }\frac {\left (10 \,{\mathrm e}^{9} \textit {\_R1} +{\mathrm e}^{9} \textit {\_R1}^{2}+25 \,{\mathrm e}^{9}-{\mathrm e}^{12}\right ) \left (\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )\right )}{2 \textit {\_R1}^{3}+2 \,{\mathrm e}^{3} \textit {\_R1} +15 \textit {\_R1}^{2}+5 \,{\mathrm e}^{3}+25 \textit {\_R1}}\right )}{2}+{\mathrm e}^{-6} {\mathrm e}^{3} \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )\right )\)	\(217\)

Input:

int(((-exp(3)^2+(-3*x^2-10*x)*exp(3))*ln(x)+exp(3)^2+(x^2+5*x)*exp(3))/(x^ 
2*exp(3)^2+(2*x^4+10*x^3)*exp(3)+x^6+10*x^5+25*x^4),x,method=_RETURNVERBOS 
E)
 

Output:

ln(x)*exp(3)/x/(x^2+exp(3)+5*x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {e^6+e^3 \left (5 x+x^2\right )+\left (-e^6+e^3 \left (-10 x-3 x^2\right )\right ) \log (x)}{e^6 x^2+25 x^4+10 x^5+x^6+e^3 \left (10 x^3+2 x^4\right )} \, dx=\frac {e^{3} \log \left (x\right )}{x^{3} + 5 \, x^{2} + x e^{3}} \] Input:

integrate(((-exp(3)^2+(-3*x^2-10*x)*exp(3))*log(x)+exp(3)^2+(x^2+5*x)*exp( 
3))/(x^2*exp(3)^2+(2*x^4+10*x^3)*exp(3)+x^6+10*x^5+25*x^4),x, algorithm="f 
ricas")
 

Output:

e^3*log(x)/(x^3 + 5*x^2 + x*e^3)
 

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {e^6+e^3 \left (5 x+x^2\right )+\left (-e^6+e^3 \left (-10 x-3 x^2\right )\right ) \log (x)}{e^6 x^2+25 x^4+10 x^5+x^6+e^3 \left (10 x^3+2 x^4\right )} \, dx=\frac {e^{3} \log {\left (x \right )}}{x^{3} + 5 x^{2} + x e^{3}} \] Input:

integrate(((-exp(3)**2+(-3*x**2-10*x)*exp(3))*ln(x)+exp(3)**2+(x**2+5*x)*e 
xp(3))/(x**2*exp(3)**2+(2*x**4+10*x**3)*exp(3)+x**6+10*x**5+25*x**4),x)
 

Output:

exp(3)*log(x)/(x**3 + 5*x**2 + x*exp(3))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (16) = 32\).

Time = 1.10 (sec) , antiderivative size = 423, normalized size of antiderivative = 24.88 \[ \int \frac {e^6+e^3 \left (5 x+x^2\right )+\left (-e^6+e^3 \left (-10 x-3 x^2\right )\right ) \log (x)}{e^6 x^2+25 x^4+10 x^5+x^6+e^3 \left (10 x^3+2 x^4\right )} \, dx=\frac {{\left (2 \, e^{3} - 25\right )} \arctan \left (\frac {2 \, x + 5}{\sqrt {4 \, e^{3} - 25}}\right ) e^{\left (-3\right )}}{\sqrt {4 \, e^{3} - 25}} + {\left (5 \, e^{\left (-9\right )} \log \left (x^{2} + 5 \, x + e^{3}\right ) - 10 \, e^{\left (-9\right )} \log \left (x\right ) - \frac {2 \, {\left (6 \, e^{6} - 150 \, e^{3} + 625\right )} \arctan \left (\frac {2 \, x + 5}{\sqrt {4 \, e^{3} - 25}}\right )}{{\left (4 \, e^{12} - 25 \, e^{9}\right )} \sqrt {4 \, e^{3} - 25}} - \frac {2 \, x^{2} {\left (3 \, e^{3} - 25\right )} + 5 \, x {\left (7 \, e^{3} - 50\right )} + 4 \, e^{6} - 25 \, e^{3}}{x^{3} {\left (4 \, e^{9} - 25 \, e^{6}\right )} + 5 \, x^{2} {\left (4 \, e^{9} - 25 \, e^{6}\right )} + x {\left (4 \, e^{12} - 25 \, e^{9}\right )}}\right )} e^{6} - \frac {5}{2} \, {\left (e^{\left (-6\right )} \log \left (x^{2} + 5 \, x + e^{3}\right ) - 2 \, e^{\left (-6\right )} \log \left (x\right ) + \frac {10 \, {\left (6 \, e^{3} - 25\right )} \arctan \left (\frac {2 \, x + 5}{\sqrt {4 \, e^{3} - 25}}\right )}{{\left (4 \, e^{9} - 25 \, e^{6}\right )} \sqrt {4 \, e^{3} - 25}} + \frac {2 \, {\left (5 \, x - 2 \, e^{3} + 25\right )}}{x^{2} {\left (4 \, e^{6} - 25 \, e^{3}\right )} + 5 \, x {\left (4 \, e^{6} - 25 \, e^{3}\right )} + 4 \, e^{9} - 25 \, e^{6}}\right )} e^{3} + {\left (\frac {2 \, x + 5}{x^{2} {\left (4 \, e^{3} - 25\right )} + 5 \, x {\left (4 \, e^{3} - 25\right )} + 4 \, e^{6} - 25 \, e^{3}} + \frac {4 \, \arctan \left (\frac {2 \, x + 5}{\sqrt {4 \, e^{3} - 25}}\right )}{{\left (4 \, e^{3} - 25\right )}^{\frac {3}{2}}}\right )} e^{3} - \frac {5}{2} \, e^{\left (-3\right )} \log \left (x^{2} + 5 \, x + e^{3}\right ) + \frac {x^{2} e^{3} + 5 \, x e^{3} + {\left (5 \, x^{3} + 25 \, x^{2} + 5 \, x e^{3} + e^{6}\right )} \log \left (x\right ) + e^{6}}{x^{3} e^{3} + 5 \, x^{2} e^{3} + x e^{6}} \] Input:

integrate(((-exp(3)^2+(-3*x^2-10*x)*exp(3))*log(x)+exp(3)^2+(x^2+5*x)*exp( 
3))/(x^2*exp(3)^2+(2*x^4+10*x^3)*exp(3)+x^6+10*x^5+25*x^4),x, algorithm="m 
axima")
 

Output:

(2*e^3 - 25)*arctan((2*x + 5)/sqrt(4*e^3 - 25))*e^(-3)/sqrt(4*e^3 - 25) + 
(5*e^(-9)*log(x^2 + 5*x + e^3) - 10*e^(-9)*log(x) - 2*(6*e^6 - 150*e^3 + 6 
25)*arctan((2*x + 5)/sqrt(4*e^3 - 25))/((4*e^12 - 25*e^9)*sqrt(4*e^3 - 25) 
) - (2*x^2*(3*e^3 - 25) + 5*x*(7*e^3 - 50) + 4*e^6 - 25*e^3)/(x^3*(4*e^9 - 
 25*e^6) + 5*x^2*(4*e^9 - 25*e^6) + x*(4*e^12 - 25*e^9)))*e^6 - 5/2*(e^(-6 
)*log(x^2 + 5*x + e^3) - 2*e^(-6)*log(x) + 10*(6*e^3 - 25)*arctan((2*x + 5 
)/sqrt(4*e^3 - 25))/((4*e^9 - 25*e^6)*sqrt(4*e^3 - 25)) + 2*(5*x - 2*e^3 + 
 25)/(x^2*(4*e^6 - 25*e^3) + 5*x*(4*e^6 - 25*e^3) + 4*e^9 - 25*e^6))*e^3 + 
 ((2*x + 5)/(x^2*(4*e^3 - 25) + 5*x*(4*e^3 - 25) + 4*e^6 - 25*e^3) + 4*arc 
tan((2*x + 5)/sqrt(4*e^3 - 25))/(4*e^3 - 25)^(3/2))*e^3 - 5/2*e^(-3)*log(x 
^2 + 5*x + e^3) + (x^2*e^3 + 5*x*e^3 + (5*x^3 + 25*x^2 + 5*x*e^3 + e^6)*lo 
g(x) + e^6)/(x^3*e^3 + 5*x^2*e^3 + x*e^6)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {e^6+e^3 \left (5 x+x^2\right )+\left (-e^6+e^3 \left (-10 x-3 x^2\right )\right ) \log (x)}{e^6 x^2+25 x^4+10 x^5+x^6+e^3 \left (10 x^3+2 x^4\right )} \, dx=\frac {e^{3} \log \left (x\right )}{x^{3} + 5 \, x^{2} + x e^{3}} \] Input:

integrate(((-exp(3)^2+(-3*x^2-10*x)*exp(3))*log(x)+exp(3)^2+(x^2+5*x)*exp( 
3))/(x^2*exp(3)^2+(2*x^4+10*x^3)*exp(3)+x^6+10*x^5+25*x^4),x, algorithm="g 
iac")
 

Output:

e^3*log(x)/(x^3 + 5*x^2 + x*e^3)
 

Mupad [B] (verification not implemented)

Time = 3.88 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {e^6+e^3 \left (5 x+x^2\right )+\left (-e^6+e^3 \left (-10 x-3 x^2\right )\right ) \log (x)}{e^6 x^2+25 x^4+10 x^5+x^6+e^3 \left (10 x^3+2 x^4\right )} \, dx=\frac {{\mathrm {e}}^3\,\ln \left (x\right )}{x\,\left (x^2+5\,x+{\mathrm {e}}^3\right )} \] Input:

int((exp(6) - log(x)*(exp(6) + exp(3)*(10*x + 3*x^2)) + exp(3)*(5*x + x^2) 
)/(exp(3)*(10*x^3 + 2*x^4) + x^2*exp(6) + 25*x^4 + 10*x^5 + x^6),x)
 

Output:

(exp(3)*log(x))/(x*(5*x + exp(3) + x^2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.41 \[ \int \frac {e^6+e^3 \left (5 x+x^2\right )+\left (-e^6+e^3 \left (-10 x-3 x^2\right )\right ) \log (x)}{e^6 x^2+25 x^4+10 x^5+x^6+e^3 \left (10 x^3+2 x^4\right )} \, dx=\frac {5 \,\mathrm {log}\left (x \right ) e^{3}-e^{3} x -x^{3}-5 x^{2}}{5 x \left (e^{3}+x^{2}+5 x \right )} \] Input:

int(((-exp(3)^2+(-3*x^2-10*x)*exp(3))*log(x)+exp(3)^2+(x^2+5*x)*exp(3))/(x 
^2*exp(3)^2+(2*x^4+10*x^3)*exp(3)+x^6+10*x^5+25*x^4),x)
 

Output:

(5*log(x)*e**3 - e**3*x - x**3 - 5*x**2)/(5*x*(e**3 + x**2 + 5*x))