3.99.74 \(\int \frac {e^{-x} (-20 x^2-2 e^{2 x} x^2-2 x^4+e^x (4 x^2-4 x^3)+(-50-50 x+15 x^2+5 x^3-x^4+x^5+e^x (10-2 x^2)+e^{2 x} (-5+5 x+x^2-x^3)) \log (10-2 x^2))}{-5 x^2+x^4} \, dx\)

Optimal. Leaf size=32 \[ \frac {\left (2-e^{-x} \left (10+\left (e^x+x\right )^2\right )\right ) \log \left (2 \left (5-x^2\right )\right )}{x} \]

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Rubi [B]  time = 8.44, antiderivative size = 121, normalized size of antiderivative = 3.78, number of steps used = 36, number of rules used = 16, integrand size = 115, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {1593, 6688, 6742, 6725, 2269, 2178, 633, 31, 2194, 2197, 2199, 2177, 2176, 2554, 12, 207} \begin {gather*} -e^{-x} x \log \left (10-2 x^2\right )-\frac {10 e^{-x} \log \left (10-2 x^2\right )}{x}-\frac {e^x \log \left (10-2 x^2\right )}{x}+\frac {2 \log \left (10-2 x^2\right )}{x}-\frac {2}{5} \left (5-\sqrt {5}\right ) \log \left (\sqrt {5}-x\right )-\frac {2}{5} \left (5+\sqrt {5}\right ) \log \left (x+\sqrt {5}\right )+\frac {4 \tanh ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{\sqrt {5}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20*x^2 - 2*E^(2*x)*x^2 - 2*x^4 + E^x*(4*x^2 - 4*x^3) + (-50 - 50*x + 15*x^2 + 5*x^3 - x^4 + x^5 + E^x*(1
0 - 2*x^2) + E^(2*x)*(-5 + 5*x + x^2 - x^3))*Log[10 - 2*x^2])/(E^x*(-5*x^2 + x^4)),x]

[Out]

(4*ArcTanh[x/Sqrt[5]])/Sqrt[5] - (2*(5 - Sqrt[5])*Log[Sqrt[5] - x])/5 - (2*(5 + Sqrt[5])*Log[Sqrt[5] + x])/5 +
(2*Log[10 - 2*x^2])/x - (10*Log[10 - 2*x^2])/(E^x*x) - (E^x*Log[10 - 2*x^2])/x - (x*Log[10 - 2*x^2])/E^x

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 207

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

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2269

Int[(F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[F^(g*(d +
e*x)^n), 1/(a + c*x^2), x], x] /; FreeQ[{F, a, c, d, e, g, n}, x]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6688

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

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

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^{-x} \left (-20 x^2-2 e^{2 x} x^2-2 x^4+e^x \left (4 x^2-4 x^3\right )+\left (-50-50 x+15 x^2+5 x^3-x^4+x^5+e^x \left (10-2 x^2\right )+e^{2 x} \left (-5+5 x+x^2-x^3\right )\right ) \log \left (10-2 x^2\right )\right )}{x^2 \left (-5+x^2\right )} \, dx\\ &=\int e^{-x} \left (-\frac {2 \left (10+e^{2 x}+2 e^x (-1+x)+x^2\right )}{-5+x^2}+\frac {\left (10-2 e^x-e^{2 x} (-1+x)+10 x-x^2+x^3\right ) \log \left (-2 \left (-5+x^2\right )\right )}{x^2}\right ) \, dx\\ &=\int \left (-\frac {2 e^{-x} \left (10-2 e^x+e^{2 x}+2 e^x x+x^2\right )}{-5+x^2}+\frac {e^{-x} \left (10-2 e^x+e^{2 x}+10 x-e^{2 x} x-x^2+x^3\right ) \log \left (10-2 x^2\right )}{x^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{-x} \left (10-2 e^x+e^{2 x}+2 e^x x+x^2\right )}{-5+x^2} \, dx\right )+\int \frac {e^{-x} \left (10-2 e^x+e^{2 x}+10 x-e^{2 x} x-x^2+x^3\right ) \log \left (10-2 x^2\right )}{x^2} \, dx\\ &=\frac {2 \log \left (10-2 x^2\right )}{x}-\frac {10 e^{-x} \log \left (10-2 x^2\right )}{x}-\frac {e^x \log \left (10-2 x^2\right )}{x}-e^{-x} x \log \left (10-2 x^2\right )-2 \int \left (\frac {e^x}{-5+x^2}+\frac {2 (-1+x)}{-5+x^2}+\frac {e^{-x} \left (10+x^2\right )}{-5+x^2}\right ) \, dx-\int \frac {2 e^{-x} \left (10-2 e^x+e^{2 x}+x^2\right )}{5-x^2} \, dx\\ &=\frac {2 \log \left (10-2 x^2\right )}{x}-\frac {10 e^{-x} \log \left (10-2 x^2\right )}{x}-\frac {e^x \log \left (10-2 x^2\right )}{x}-e^{-x} x \log \left (10-2 x^2\right )-2 \int \frac {e^x}{-5+x^2} \, dx-2 \int \frac {e^{-x} \left (10+x^2\right )}{-5+x^2} \, dx-2 \int \frac {e^{-x} \left (10-2 e^x+e^{2 x}+x^2\right )}{5-x^2} \, dx-4 \int \frac {-1+x}{-5+x^2} \, dx\\ &=\frac {2 \log \left (10-2 x^2\right )}{x}-\frac {10 e^{-x} \log \left (10-2 x^2\right )}{x}-\frac {e^x \log \left (10-2 x^2\right )}{x}-e^{-x} x \log \left (10-2 x^2\right )-2 \int \left (-\frac {e^x}{2 \sqrt {5} \left (\sqrt {5}-x\right )}-\frac {e^x}{2 \sqrt {5} \left (\sqrt {5}+x\right )}\right ) \, dx-2 \int \left (e^{-x}+\frac {15 e^{-x}}{-5+x^2}\right ) \, dx-2 \int \left (\frac {2}{-5+x^2}-\frac {e^x}{-5+x^2}+\frac {e^{-x} \left (-10-x^2\right )}{-5+x^2}\right ) \, dx-\frac {1}{5} \left (2 \left (5-\sqrt {5}\right )\right ) \int \frac {1}{-\sqrt {5}+x} \, dx-\frac {1}{5} \left (2 \left (5+\sqrt {5}\right )\right ) \int \frac {1}{\sqrt {5}+x} \, dx\\ &=-\frac {2}{5} \left (5-\sqrt {5}\right ) \log \left (\sqrt {5}-x\right )-\frac {2}{5} \left (5+\sqrt {5}\right ) \log \left (\sqrt {5}+x\right )+\frac {2 \log \left (10-2 x^2\right )}{x}-\frac {10 e^{-x} \log \left (10-2 x^2\right )}{x}-\frac {e^x \log \left (10-2 x^2\right )}{x}-e^{-x} x \log \left (10-2 x^2\right )-2 \int e^{-x} \, dx+2 \int \frac {e^x}{-5+x^2} \, dx-2 \int \frac {e^{-x} \left (-10-x^2\right )}{-5+x^2} \, dx-4 \int \frac {1}{-5+x^2} \, dx-30 \int \frac {e^{-x}}{-5+x^2} \, dx+\frac {\int \frac {e^x}{\sqrt {5}-x} \, dx}{\sqrt {5}}+\frac {\int \frac {e^x}{\sqrt {5}+x} \, dx}{\sqrt {5}}\\ &=2 e^{-x}+\frac {4 \tanh ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{\sqrt {5}}-\frac {e^{\sqrt {5}} \text {Ei}\left (-\sqrt {5}+x\right )}{\sqrt {5}}+\frac {e^{-\sqrt {5}} \text {Ei}\left (\sqrt {5}+x\right )}{\sqrt {5}}-\frac {2}{5} \left (5-\sqrt {5}\right ) \log \left (\sqrt {5}-x\right )-\frac {2}{5} \left (5+\sqrt {5}\right ) \log \left (\sqrt {5}+x\right )+\frac {2 \log \left (10-2 x^2\right )}{x}-\frac {10 e^{-x} \log \left (10-2 x^2\right )}{x}-\frac {e^x \log \left (10-2 x^2\right )}{x}-e^{-x} x \log \left (10-2 x^2\right )+2 \int \left (-\frac {e^x}{2 \sqrt {5} \left (\sqrt {5}-x\right )}-\frac {e^x}{2 \sqrt {5} \left (\sqrt {5}+x\right )}\right ) \, dx-2 \int \left (-e^{-x}-\frac {15 e^{-x}}{-5+x^2}\right ) \, dx-30 \int \left (-\frac {e^{-x}}{2 \sqrt {5} \left (\sqrt {5}-x\right )}-\frac {e^{-x}}{2 \sqrt {5} \left (\sqrt {5}+x\right )}\right ) \, dx\\ &=2 e^{-x}+\frac {4 \tanh ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{\sqrt {5}}-\frac {e^{\sqrt {5}} \text {Ei}\left (-\sqrt {5}+x\right )}{\sqrt {5}}+\frac {e^{-\sqrt {5}} \text {Ei}\left (\sqrt {5}+x\right )}{\sqrt {5}}-\frac {2}{5} \left (5-\sqrt {5}\right ) \log \left (\sqrt {5}-x\right )-\frac {2}{5} \left (5+\sqrt {5}\right ) \log \left (\sqrt {5}+x\right )+\frac {2 \log \left (10-2 x^2\right )}{x}-\frac {10 e^{-x} \log \left (10-2 x^2\right )}{x}-\frac {e^x \log \left (10-2 x^2\right )}{x}-e^{-x} x \log \left (10-2 x^2\right )+2 \int e^{-x} \, dx+30 \int \frac {e^{-x}}{-5+x^2} \, dx-\frac {\int \frac {e^x}{\sqrt {5}-x} \, dx}{\sqrt {5}}-\frac {\int \frac {e^x}{\sqrt {5}+x} \, dx}{\sqrt {5}}+\left (3 \sqrt {5}\right ) \int \frac {e^{-x}}{\sqrt {5}-x} \, dx+\left (3 \sqrt {5}\right ) \int \frac {e^{-x}}{\sqrt {5}+x} \, dx\\ &=\frac {4 \tanh ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{\sqrt {5}}+3 \sqrt {5} e^{\sqrt {5}} \text {Ei}\left (-\sqrt {5}-x\right )-3 \sqrt {5} e^{-\sqrt {5}} \text {Ei}\left (\sqrt {5}-x\right )-\frac {2}{5} \left (5-\sqrt {5}\right ) \log \left (\sqrt {5}-x\right )-\frac {2}{5} \left (5+\sqrt {5}\right ) \log \left (\sqrt {5}+x\right )+\frac {2 \log \left (10-2 x^2\right )}{x}-\frac {10 e^{-x} \log \left (10-2 x^2\right )}{x}-\frac {e^x \log \left (10-2 x^2\right )}{x}-e^{-x} x \log \left (10-2 x^2\right )+30 \int \left (-\frac {e^{-x}}{2 \sqrt {5} \left (\sqrt {5}-x\right )}-\frac {e^{-x}}{2 \sqrt {5} \left (\sqrt {5}+x\right )}\right ) \, dx\\ &=\frac {4 \tanh ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{\sqrt {5}}+3 \sqrt {5} e^{\sqrt {5}} \text {Ei}\left (-\sqrt {5}-x\right )-3 \sqrt {5} e^{-\sqrt {5}} \text {Ei}\left (\sqrt {5}-x\right )-\frac {2}{5} \left (5-\sqrt {5}\right ) \log \left (\sqrt {5}-x\right )-\frac {2}{5} \left (5+\sqrt {5}\right ) \log \left (\sqrt {5}+x\right )+\frac {2 \log \left (10-2 x^2\right )}{x}-\frac {10 e^{-x} \log \left (10-2 x^2\right )}{x}-\frac {e^x \log \left (10-2 x^2\right )}{x}-e^{-x} x \log \left (10-2 x^2\right )-\left (3 \sqrt {5}\right ) \int \frac {e^{-x}}{\sqrt {5}-x} \, dx-\left (3 \sqrt {5}\right ) \int \frac {e^{-x}}{\sqrt {5}+x} \, dx\\ &=\frac {4 \tanh ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{\sqrt {5}}-\frac {2}{5} \left (5-\sqrt {5}\right ) \log \left (\sqrt {5}-x\right )-\frac {2}{5} \left (5+\sqrt {5}\right ) \log \left (\sqrt {5}+x\right )+\frac {2 \log \left (10-2 x^2\right )}{x}-\frac {10 e^{-x} \log \left (10-2 x^2\right )}{x}-\frac {e^x \log \left (10-2 x^2\right )}{x}-e^{-x} x \log \left (10-2 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 1.44, size = 125, normalized size = 3.91 \begin {gather*} \frac {4 \tanh ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{\sqrt {5}}+\frac {2}{5} \left (-5+\sqrt {5}\right ) \log \left (\sqrt {5}-x\right )-2 \log \left (\sqrt {5}+x\right )-\frac {2 \log \left (\sqrt {5}+x\right )}{\sqrt {5}}+\frac {2 \log \left (-2 \left (-5+x^2\right )\right )}{x}-\frac {10 e^{-x} \log \left (-2 \left (-5+x^2\right )\right )}{x}-\frac {e^x \log \left (-2 \left (-5+x^2\right )\right )}{x}-e^{-x} x \log \left (-2 \left (-5+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*x^2 - 2*E^(2*x)*x^2 - 2*x^4 + E^x*(4*x^2 - 4*x^3) + (-50 - 50*x + 15*x^2 + 5*x^3 - x^4 + x^5 +
E^x*(10 - 2*x^2) + E^(2*x)*(-5 + 5*x + x^2 - x^3))*Log[10 - 2*x^2])/(E^x*(-5*x^2 + x^4)),x]

[Out]

(4*ArcTanh[x/Sqrt[5]])/Sqrt[5] + (2*(-5 + Sqrt[5])*Log[Sqrt[5] - x])/5 - 2*Log[Sqrt[5] + x] - (2*Log[Sqrt[5] +
x])/Sqrt[5] + (2*Log[-2*(-5 + x^2)])/x - (10*Log[-2*(-5 + x^2)])/(E^x*x) - (E^x*Log[-2*(-5 + x^2)])/x - (x*Lo
g[-2*(-5 + x^2)])/E^x

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fricas [A]  time = 0.59, size = 33, normalized size = 1.03 \begin {gather*} -\frac {{\left (x^{2} + 2 \, {\left (x - 1\right )} e^{x} + e^{\left (2 \, x\right )} + 10\right )} e^{\left (-x\right )} \log \left (-2 \, x^{2} + 10\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3+x^2+5*x-5)*exp(x)^2+(-2*x^2+10)*exp(x)+x^5-x^4+5*x^3+15*x^2-50*x-50)*log(-2*x^2+10)-2*exp(x)
^2*x^2+(-4*x^3+4*x^2)*exp(x)-2*x^4-20*x^2)/(x^4-5*x^2)/exp(x),x, algorithm="fricas")

[Out]

-(x^2 + 2*(x - 1)*e^x + e^(2*x) + 10)*e^(-x)*log(-2*x^2 + 10)/x

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giac [B]  time = 0.21, size = 66, normalized size = 2.06 \begin {gather*} -\frac {x^{2} e^{\left (-x\right )} \log \left (-2 \, x^{2} + 10\right ) + 2 \, x \log \left (x^{2} - 5\right ) + 10 \, e^{\left (-x\right )} \log \left (-2 \, x^{2} + 10\right ) + e^{x} \log \left (-2 \, x^{2} + 10\right ) - 2 \, \log \left (-2 \, x^{2} + 10\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3+x^2+5*x-5)*exp(x)^2+(-2*x^2+10)*exp(x)+x^5-x^4+5*x^3+15*x^2-50*x-50)*log(-2*x^2+10)-2*exp(x)
^2*x^2+(-4*x^3+4*x^2)*exp(x)-2*x^4-20*x^2)/(x^4-5*x^2)/exp(x),x, algorithm="giac")

[Out]

-(x^2*e^(-x)*log(-2*x^2 + 10) + 2*x*log(x^2 - 5) + 10*e^(-x)*log(-2*x^2 + 10) + e^x*log(-2*x^2 + 10) - 2*log(-
2*x^2 + 10))/x

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maple [A]  time = 0.18, size = 40, normalized size = 1.25

method	result	size
risch	\(-\frac {\left (x^{2}+{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+10\right ) {\mathrm e}^{-x} \ln \left (-2 x^{2}+10\right )}{x}-2 \ln \left (x^{2}-5\right )\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^3+x^2+5*x-5)*exp(x)^2+(-2*x^2+10)*exp(x)+x^5-x^4+5*x^3+15*x^2-50*x-50)*ln(-2*x^2+10)-2*exp(x)^2*x^2+
(-4*x^3+4*x^2)*exp(x)-2*x^4-20*x^2)/(x^4-5*x^2)/exp(x),x,method=_RETURNVERBOSE)

[Out]

-(x^2+exp(2*x)-2*exp(x)+10)/x*exp(-x)*ln(-2*x^2+10)-2*ln(x^2-5)

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maxima [C]  time = 0.49, size = 59, normalized size = 1.84 \begin {gather*} -\frac {-2 i \, \pi + {\left (10 i \, \pi + {\left (i \, \pi + \log \relax (2)\right )} x^{2} + 10 \, \log \relax (2)\right )} e^{\left (-x\right )} + {\left (i \, \pi + \log \relax (2)\right )} e^{x} + {\left (2 \, x + e^{x} - 2\right )} \log \left (x^{2} - 5\right ) - 2 \, \log \relax (2)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3+x^2+5*x-5)*exp(x)^2+(-2*x^2+10)*exp(x)+x^5-x^4+5*x^3+15*x^2-50*x-50)*log(-2*x^2+10)-2*exp(x)
^2*x^2+(-4*x^3+4*x^2)*exp(x)-2*x^4-20*x^2)/(x^4-5*x^2)/exp(x),x, algorithm="maxima")

[Out]

-(-2*I*pi + (10*I*pi + (I*pi + log(2))*x^2 + 10*log(2))*e^(-x) + (I*pi + log(2))*e^x + (2*x + e^x - 2)*log(x^2
- 5) - 2*log(2))/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{-x}\,\left (2\,x^2\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (4\,x^2-4\,x^3\right )+\ln \left (10-2\,x^2\right )\,\left (50\,x+{\mathrm {e}}^x\,\left (2\,x^2-10\right )-15\,x^2-5\,x^3+x^4-x^5-{\mathrm {e}}^{2\,x}\,\left (-x^3+x^2+5\,x-5\right )+50\right )+20\,x^2+2\,x^4\right )}{5\,x^2-x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(2*x^2*exp(2*x) - exp(x)*(4*x^2 - 4*x^3) + log(10 - 2*x^2)*(50*x + exp(x)*(2*x^2 - 10) - 15*x^2 -
5*x^3 + x^4 - x^5 - exp(2*x)*(5*x + x^2 - x^3 - 5) + 50) + 20*x^2 + 2*x^4))/(5*x^2 - x^4),x)

[Out]

int((exp(-x)*(2*x^2*exp(2*x) - exp(x)*(4*x^2 - 4*x^3) + log(10 - 2*x^2)*(50*x + exp(x)*(2*x^2 - 10) - 15*x^2 -
5*x^3 + x^4 - x^5 - exp(2*x)*(5*x + x^2 - x^3 - 5) + 50) + 20*x^2 + 2*x^4))/(5*x^2 - x^4), x)

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sympy [B]  time = 0.51, size = 65, normalized size = 2.03 \begin {gather*} - 2 \log {\left (x^{2} - 5 \right )} + \frac {2 \log {\left (10 - 2 x^{2} \right )}}{x} + \frac {- x e^{x} \log {\left (10 - 2 x^{2} \right )} + \left (- x^{3} \log {\left (10 - 2 x^{2} \right )} - 10 x \log {\left (10 - 2 x^{2} \right )}\right ) e^{- x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**3+x**2+5*x-5)*exp(x)**2+(-2*x**2+10)*exp(x)+x**5-x**4+5*x**3+15*x**2-50*x-50)*ln(-2*x**2+10)-
2*exp(x)**2*x**2+(-4*x**3+4*x**2)*exp(x)-2*x**4-20*x**2)/(x**4-5*x**2)/exp(x),x)

[Out]

-2*log(x**2 - 5) + 2*log(10 - 2*x**2)/x + (-x*exp(x)*log(10 - 2*x**2) + (-x**3*log(10 - 2*x**2) - 10*x*log(10
- 2*x**2))*exp(-x))/x**2

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