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

Optimal. Leaf size=17 \[ \frac {\log (x)}{x+\frac {x^2 (5+x)}{e^3}} \]

________________________________________________________________________________________

Rubi [C]  time = 1.50, antiderivative size = 962, normalized size of antiderivative = 56.59, number of steps used = 42, number of rules used = 15, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.197, Rules used = {6688, 12, 6742, 709, 800, 634, 618, 204, 628, 2357, 2304, 2314, 31, 2317, 2391} \begin {gather*} \frac {\left (25-2 e^3\right ) \tan ^{-1}\left (\frac {2 x+5}{\sqrt {-25+4 e^3}}\right )}{e^3 \sqrt {-25+4 e^3}}-\frac {2 \left (25-2 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 )}+\frac {5 \log \left (2 i x+\sqrt {-25+4 e^3}+5 i\right )}{25-4 e^3}+\frac {\log (x)}{x}+\frac {4 \left (25-2 e^3\right ) x \log (x)}{\left (25-4 e^3\right ) \left (5-i \sqrt {-25+4 e^3}\right ) \left (2 x-i \sqrt {-25+4 e^3}+5\right )}-\frac {10 x \log (x)}{\left (25-4 e^3\right ) \left (2 x-i \sqrt {-25+4 e^3}+5\right )}+\frac {4 \left (25-2 e^3\right ) x \log (x)}{\left (25-4 e^3\right ) \left (5+i \sqrt {-25+4 e^3}\right ) \left (2 x+i \sqrt {-25+4 e^3}+5\right )}-\frac {10 x \log (x)}{\left (25-4 e^3\right ) \left (2 x+i \sqrt {-25+4 e^3}+5\right )}-\frac {5 \log (x)}{e^3}-\frac {2 \left (25-2 e^3\right ) \log \left (2 x+i \sqrt {-25+4 e^3}+5\right )}{\left (25-4 e^3\right ) \left (5+i \sqrt {-25+4 e^3}\right )}+\frac {5 \log \left (2 x+i \sqrt {-25+4 e^3}+5\right )}{25-4 e^3}-\frac {i \log (x) \log \left (\frac {2 x}{5-i \sqrt {-25+4 e^3}}+1\right )}{\sqrt {-25+4 e^3}}-\frac {2 i \left (25-2 e^3\right ) \log (x) \log \left (\frac {2 x}{5-i \sqrt {-25+4 e^3}}+1\right )}{\left (-25+4 e^3\right )^{3/2}}+\frac {25 i \log (x) \log \left (\frac {2 x}{5-i \sqrt {-25+4 e^3}}+1\right )}{\left (-25+4 e^3\right )^{3/2}}+\frac {i \log (x) \log \left (\frac {2 x}{5+i \sqrt {-25+4 e^3}}+1\right )}{\sqrt {-25+4 e^3}}+\frac {2 i \left (25-2 e^3\right ) \log (x) \log \left (\frac {2 x}{5+i \sqrt {-25+4 e^3}}+1\right )}{\left (-25+4 e^3\right )^{3/2}}-\frac {25 i \log (x) \log \left (\frac {2 x}{5+i \sqrt {-25+4 e^3}}+1\right )}{\left (-25+4 e^3\right )^{3/2}}+\frac {5 \log \left (x^2+5 x+e^3\right )}{2 e^3}-\frac {i \text {Li}_2\left (-\frac {2 x}{5-i \sqrt {-25+4 e^3}}\right )}{\sqrt {-25+4 e^3}}-\frac {2 i \left (25-2 e^3\right ) \text {Li}_2\left (-\frac {2 x}{5-i \sqrt {-25+4 e^3}}\right )}{\left (-25+4 e^3\right )^{3/2}}+\frac {25 i \text {Li}_2\left (-\frac {2 x}{5-i \sqrt {-25+4 e^3}}\right )}{\left (-25+4 e^3\right )^{3/2}}+\frac {i \text {Li}_2\left (-\frac {2 x}{5+i \sqrt {-25+4 e^3}}\right )}{\sqrt {-25+4 e^3}}+\frac {2 i \left (25-2 e^3\right ) \text {Li}_2\left (-\frac {2 x}{5+i \sqrt {-25+4 e^3}}\right )}{\left (-25+4 e^3\right )^{3/2}}-\frac {25 i \text {Li}_2\left (-\frac {2 x}{5+i \sqrt {-25+4 e^3}}\right )}{\left (-25+4 e^3\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

((25 - 2*E^3)*ArcTan[(5 + 2*x)/Sqrt[-25 + 4*E^3]])/(E^3*Sqrt[-25 + 4*E^3]) + (5*Log[5*I + Sqrt[-25 + 4*E^3] +
(2*I)*x])/(25 - 4*E^3) - (2*(25 - 2*E^3)*Log[5*I + Sqrt[-25 + 4*E^3] + (2*I)*x])/((25 - 4*E^3)*(5 - I*Sqrt[-25
+ 4*E^3])) - (5*Log[x])/E^3 + Log[x]/x - (10*x*Log[x])/((25 - 4*E^3)*(5 - I*Sqrt[-25 + 4*E^3] + 2*x)) + (4*(2
5 - 2*E^3)*x*Log[x])/((25 - 4*E^3)*(5 - I*Sqrt[-25 + 4*E^3])*(5 - I*Sqrt[-25 + 4*E^3] + 2*x)) - (10*x*Log[x])/
((25 - 4*E^3)*(5 + I*Sqrt[-25 + 4*E^3] + 2*x)) + (4*(25 - 2*E^3)*x*Log[x])/((25 - 4*E^3)*(5 + I*Sqrt[-25 + 4*E
^3])*(5 + I*Sqrt[-25 + 4*E^3] + 2*x)) + (5*Log[5 + I*Sqrt[-25 + 4*E^3] + 2*x])/(25 - 4*E^3) - (2*(25 - 2*E^3)*
Log[5 + I*Sqrt[-25 + 4*E^3] + 2*x])/((25 - 4*E^3)*(5 + I*Sqrt[-25 + 4*E^3])) + ((25*I)*Log[x]*Log[1 + (2*x)/(5
- I*Sqrt[-25 + 4*E^3])])/(-25 + 4*E^3)^(3/2) - ((2*I)*(25 - 2*E^3)*Log[x]*Log[1 + (2*x)/(5 - I*Sqrt[-25 + 4*E
^3])])/(-25 + 4*E^3)^(3/2) - (I*Log[x]*Log[1 + (2*x)/(5 - I*Sqrt[-25 + 4*E^3])])/Sqrt[-25 + 4*E^3] - ((25*I)*L
og[x]*Log[1 + (2*x)/(5 + I*Sqrt[-25 + 4*E^3])])/(-25 + 4*E^3)^(3/2) + ((2*I)*(25 - 2*E^3)*Log[x]*Log[1 + (2*x)
/(5 + I*Sqrt[-25 + 4*E^3])])/(-25 + 4*E^3)^(3/2) + (I*Log[x]*Log[1 + (2*x)/(5 + I*Sqrt[-25 + 4*E^3])])/Sqrt[-2
5 + 4*E^3] + (5*Log[E^3 + 5*x + x^2])/(2*E^3) + ((25*I)*PolyLog[2, (-2*x)/(5 - I*Sqrt[-25 + 4*E^3])])/(-25 + 4
*E^3)^(3/2) - ((2*I)*(25 - 2*E^3)*PolyLog[2, (-2*x)/(5 - I*Sqrt[-25 + 4*E^3])])/(-25 + 4*E^3)^(3/2) - (I*PolyL
og[2, (-2*x)/(5 - I*Sqrt[-25 + 4*E^3])])/Sqrt[-25 + 4*E^3] - ((25*I)*PolyLog[2, (-2*x)/(5 + I*Sqrt[-25 + 4*E^3
])])/(-25 + 4*E^3)^(3/2) + ((2*I)*(25 - 2*E^3)*PolyLog[2, (-2*x)/(5 + I*Sqrt[-25 + 4*E^3])])/(-25 + 4*E^3)^(3/
2) + (I*PolyLog[2, (-2*x)/(5 + I*Sqrt[-25 + 4*E^3])])/Sqrt[-25 + 4*E^3]

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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 628

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

Rule 634

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

Rule 709

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

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2317

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

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2391

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^3 \left (e^3+x (5+x)-\left (e^3+x (10+3 x)\right ) \log (x)\right )}{x^2 \left (e^3+5 x+x^2\right )^2} \, dx\\ &=e^3 \int \frac {e^3+x (5+x)-\left (e^3+x (10+3 x)\right ) \log (x)}{x^2 \left (e^3+5 x+x^2\right )^2} \, dx\\ &=e^3 \int \left (\frac {1}{x^2 \left (e^3+5 x+x^2\right )}-\frac {\left (e^3+10 x+3 x^2\right ) \log (x)}{x^2 \left (e^3+5 x+x^2\right )^2}\right ) \, dx\\ &=e^3 \int \frac {1}{x^2 \left (e^3+5 x+x^2\right )} \, dx-e^3 \int \frac {\left (e^3+10 x+3 x^2\right ) \log (x)}{x^2 \left (e^3+5 x+x^2\right )^2} \, dx\\ &=-\frac {1}{x}-e^3 \int \left (\frac {\log (x)}{e^3 x^2}+\frac {\left (-25+2 e^3-5 x\right ) \log (x)}{e^3 \left (e^3+5 x+x^2\right )^2}-\frac {\log (x)}{e^3 \left (e^3+5 x+x^2\right )}\right ) \, dx+\int \frac {-5-x}{x \left (e^3+5 x+x^2\right )} \, dx\\ &=-\frac {1}{x}+\int \left (-\frac {5}{e^3 x}+\frac {25-e^3+5 x}{e^3 \left (e^3+5 x+x^2\right )}\right ) \, dx-\int \frac {\log (x)}{x^2} \, dx-\int \frac {\left (-25+2 e^3-5 x\right ) \log (x)}{\left (e^3+5 x+x^2\right )^2} \, dx+\int \frac {\log (x)}{e^3+5 x+x^2} \, dx\\ &=-\frac {5 \log (x)}{e^3}+\frac {\log (x)}{x}+\frac {\int \frac {25-e^3+5 x}{e^3+5 x+x^2} \, dx}{e^3}+\int \left (\frac {2 i \log (x)}{\sqrt {-25+4 e^3} \left (-5+i \sqrt {-25+4 e^3}-2 x\right )}+\frac {2 i \log (x)}{\sqrt {-25+4 e^3} \left (5+i \sqrt {-25+4 e^3}+2 x\right )}\right ) \, dx-\int \left (-\frac {25 \left (1-\frac {2 e^3}{25}\right ) \log (x)}{\left (e^3+5 x+x^2\right )^2}-\frac {5 x \log (x)}{\left (e^3+5 x+x^2\right )^2}\right ) \, dx\\ &=-\frac {5 \log (x)}{e^3}+\frac {\log (x)}{x}+5 \int \frac {x \log (x)}{\left (e^3+5 x+x^2\right )^2} \, dx+\frac {5 \int \frac {5+2 x}{e^3+5 x+x^2} \, dx}{2 e^3}+\left (25-2 e^3\right ) \int \frac {\log (x)}{\left (e^3+5 x+x^2\right )^2} \, dx+\frac {\left (25-2 e^3\right ) \int \frac {1}{e^3+5 x+x^2} \, dx}{2 e^3}+\frac {(2 i) \int \frac {\log (x)}{-5+i \sqrt {-25+4 e^3}-2 x} \, dx}{\sqrt {-25+4 e^3}}+\frac {(2 i) \int \frac {\log (x)}{5+i \sqrt {-25+4 e^3}+2 x} \, dx}{\sqrt {-25+4 e^3}}\\ &=-\frac {5 \log (x)}{e^3}+\frac {\log (x)}{x}-\frac {i \log (x) \log \left (1+\frac {2 x}{5-i \sqrt {-25+4 e^3}}\right )}{\sqrt {-25+4 e^3}}+\frac {i \log (x) \log \left (1+\frac {2 x}{5+i \sqrt {-25+4 e^3}}\right )}{\sqrt {-25+4 e^3}}+\frac {5 \log \left (e^3+5 x+x^2\right )}{2 e^3}+5 \int \left (-\frac {2 \left (-5+i \sqrt {-25+4 e^3}\right ) \log (x)}{\left (-25+4 e^3\right ) \left (-5+i \sqrt {-25+4 e^3}-2 x\right )^2}-\frac {10 i \log (x)}{\left (-25+4 e^3\right )^{3/2} \left (-5+i \sqrt {-25+4 e^3}-2 x\right )}-\frac {2 \left (-5-i \sqrt {-25+4 e^3}\right ) \log (x)}{\left (-25+4 e^3\right ) \left (5+i \sqrt {-25+4 e^3}+2 x\right )^2}-\frac {10 i \log (x)}{\left (-25+4 e^3\right )^{3/2} \left (5+i \sqrt {-25+4 e^3}+2 x\right )}\right ) \, dx+\left (25-2 e^3\right ) \int \left (-\frac {4 \log (x)}{\left (-25+4 e^3\right ) \left (-5+i \sqrt {-25+4 e^3}-2 x\right )^2}+\frac {4 i \log (x)}{\left (-25+4 e^3\right )^{3/2} \left (-5+i \sqrt {-25+4 e^3}-2 x\right )}-\frac {4 \log (x)}{\left (-25+4 e^3\right ) \left (5+i \sqrt {-25+4 e^3}+2 x\right )^2}+\frac {4 i \log (x)}{\left (-25+4 e^3\right )^{3/2} \left (5+i \sqrt {-25+4 e^3}+2 x\right )}\right ) \, dx-\frac {\left (25-2 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{25-4 e^3-x^2} \, dx,x,5+2 x\right )}{e^3}+\frac {i \int \frac {\log \left (1-\frac {2 x}{-5+i \sqrt {-25+4 e^3}}\right )}{x} \, dx}{\sqrt {-25+4 e^3}}-\frac {i \int \frac {\log \left (1+\frac {2 x}{5+i \sqrt {-25+4 e^3}}\right )}{x} \, dx}{\sqrt {-25+4 e^3}}\\ &=\frac {\left (25-2 e^3\right ) \tan ^{-1}\left (\frac {5+2 x}{\sqrt {-25+4 e^3}}\right )}{e^3 \sqrt {-25+4 e^3}}-\frac {5 \log (x)}{e^3}+\frac {\log (x)}{x}-\frac {i \log (x) \log \left (1+\frac {2 x}{5-i \sqrt {-25+4 e^3}}\right )}{\sqrt {-25+4 e^3}}+\frac {i \log (x) \log \left (1+\frac {2 x}{5+i \sqrt {-25+4 e^3}}\right )}{\sqrt {-25+4 e^3}}+\frac {5 \log \left (e^3+5 x+x^2\right )}{2 e^3}-\frac {i \text {Li}_2\left (-\frac {2 x}{5-i \sqrt {-25+4 e^3}}\right )}{\sqrt {-25+4 e^3}}+\frac {i \text {Li}_2\left (-\frac {2 x}{5+i \sqrt {-25+4 e^3}}\right )}{\sqrt {-25+4 e^3}}+\frac {\left (4 \left (25-2 e^3\right )\right ) \int \frac {\log (x)}{\left (-5+i \sqrt {-25+4 e^3}-2 x\right )^2} \, dx}{25-4 e^3}+\frac {\left (4 \left (25-2 e^3\right )\right ) \int \frac {\log (x)}{\left (5+i \sqrt {-25+4 e^3}+2 x\right )^2} \, dx}{25-4 e^3}-\frac {(50 i) \int \frac {\log (x)}{-5+i \sqrt {-25+4 e^3}-2 x} \, dx}{\left (-25+4 e^3\right )^{3/2}}-\frac {(50 i) \int \frac {\log (x)}{5+i \sqrt {-25+4 e^3}+2 x} \, dx}{\left (-25+4 e^3\right )^{3/2}}+\frac {\left (4 i \left (25-2 e^3\right )\right ) \int \frac {\log (x)}{-5+i \sqrt {-25+4 e^3}-2 x} \, dx}{\left (-25+4 e^3\right )^{3/2}}+\frac {\left (4 i \left (25-2 e^3\right )\right ) \int \frac {\log (x)}{5+i \sqrt {-25+4 e^3}+2 x} \, dx}{\left (-25+4 e^3\right )^{3/2}}-\frac {\left (10 \left (5-i \sqrt {-25+4 e^3}\right )\right ) \int \frac {\log (x)}{\left (-5+i \sqrt {-25+4 e^3}-2 x\right )^2} \, dx}{25-4 e^3}-\frac {\left (10 \left (5+i \sqrt {-25+4 e^3}\right )\right ) \int \frac {\log (x)}{\left (5+i \sqrt {-25+4 e^3}+2 x\right )^2} \, dx}{25-4 e^3}\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.48, size = 21, normalized size = 1.24 \begin {gather*} \frac {e^3 \log (x)}{x \left (e^3+5 x+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

(E^3*Log[x])/(x*(E^3 + 5*x + x^2))

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fricas [A]  time = 0.66, size = 20, normalized size = 1.18 \begin {gather*} \frac {e^{3} \log \relax (x)}{x^{3} + 5 \, x^{2} + x e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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="fricas")

[Out]

e^3*log(x)/(x^3 + 5*x^2 + x*e^3)

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giac [A]  time = 0.27, size = 20, normalized size = 1.18 \begin {gather*} \frac {e^{3} \log \relax (x)}{x^{3} + 5 \, x^{2} + x e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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="giac")

[Out]

e^3*log(x)/(x^3 + 5*x^2 + x*e^3)

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maple [A]  time = 0.15, size = 20, normalized size = 1.18

method	result	size
norman	\(\frac {\ln \relax (x ) {\mathrm e}^{3}}{x \left (x^{2}+{\mathrm e}^{3}+5 x \right )}\)	\(20\)
risch	\(\frac {\ln \relax (x ) {\mathrm e}^{3}}{x \left (x^{2}+{\mathrm e}^{3}+5 x \right )}\)	\(20\)
default	\(-5 \,{\mathrm e}^{3} {\mathrm e}^{-6} \ln \relax (x )+\frac {5 \,{\mathrm e}^{3} {\mathrm e}^{-6} \ln \left (x^{2}+{\mathrm e}^{3}+5 x \right )}{2}+\frac {25 \,{\mathrm e}^{-6} \arctan \left (\frac {5+2 x}{\sqrt {4 \,{\mathrm e}^{3}-25}}\right ) {\mathrm e}^{3}}{\sqrt {4 \,{\mathrm e}^{3}-25}}-\frac {2 \,{\mathrm e}^{-6} \arctan \left (\frac {5+2 x}{\sqrt {4 \,{\mathrm e}^{3}-25}}\right ) {\mathrm e}^{6}}{\sqrt {4 \,{\mathrm e}^{3}-25}}+5 \ln \relax (x )^{2} {\mathrm e}^{-12} {\mathrm e}^{9}-5 \,{\mathrm e}^{3} \ln \relax (x )^{2} {\mathrm e}^{-6}+\frac {{\mathrm e}^{6} {\mathrm e}^{-6} \ln \relax (x )}{x}+\frac {{\mathrm e}^{3} \left (\munderset {\textit {\_R1} =\RootOf \left (\textit {\_Z}^{4}+10 \textit {\_Z}^{3}+\left (2 \,{\mathrm e}^{3}+25\right ) \textit {\_Z}^{2}+10 \textit {\_Z} \,{\mathrm e}^{3}+{\mathrm e}^{6}\right )}{\sum }\frac {\left (\textit {\_R1}^{2} {\mathrm e}^{9}+10 \,{\mathrm e}^{9} \textit {\_R1} +25 \,{\mathrm e}^{9}-{\mathrm e}^{12}\right ) \left (\ln \relax (x ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )\right )}{2 \textit {\_R1}^{3}+2 \textit {\_R1} \,{\mathrm e}^{3}+15 \textit {\_R1}^{2}+5 \,{\mathrm e}^{3}+25 \textit {\_R1}}\right ) {\mathrm e}^{-12}}{2}\)	\(255\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 2.66, size = 423, normalized size = 24.88 \begin {gather*} \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 \relax (x) - \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 \relax (x) + \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 \relax (x) + e^{6}}{x^{3} e^{3} + 5 \, x^{2} e^{3} + x e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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="maxima")

[Out]

(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 + 625)*arctan((2*x + 5)/sqrt(4*e^3 - 25))/((4*e^12 - 25*e^9)*sqrt(4*e^3 - 2
5)) - (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*arctan((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)*log(x) + e^6)/(x^3*e^3 + 5*x^2*e^3 + x*e^6)

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mupad [B]  time = 5.76, size = 19, normalized size = 1.12 \begin {gather*} \frac {{\mathrm {e}}^3\,\ln \relax (x)}{x\,\left (x^2+5\,x+{\mathrm {e}}^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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

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sympy [A]  time = 0.20, size = 19, normalized size = 1.12 \begin {gather*} \frac {e^{3} \log {\relax (x )}}{x^{3} + 5 x^{2} + x e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-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)

[Out]

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

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