3.28.5 \(\int \frac {-2 x^3+e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2} (-4 e^{-2+2 x} x^2-16 x^3+e^{-1+x} (-8 x^2-8 x^3))+e^{e^{-2+2 x}+4 e^{-1+x} x+4 x^2} (-2 x^2-4 e^{-2+2 x} x^3-16 x^4+e^{-1+x} (-8 x^3-8 x^4))+2 \log (x)-\log ^2(x)}{x^2} \, dx\)

Optimal. Leaf size=28 \[ -\left (e^{\left (e^{-1+x}+2 x\right )^2}+x\right )^2+\frac {\log ^2(x)}{x} \]

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Rubi [B]  time = 1.08, antiderivative size = 129, normalized size of antiderivative = 4.61, number of steps used = 9, number of rules used = 5, integrand size = 145, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {14, 6706, 2288, 2304, 2305} \begin {gather*} -x^2-e^{8 x^2+8 e^{x-1} x+2 e^{2 x-2}}-\frac {2 e^{4 x^2+4 e^{x-1} x+e^{2 x-2}-2} \left (2 e^{x+1} x^2+4 e^2 x^2+e^{2 x} x+2 e^{x+1} x\right )}{2 e^{x-1} x+4 x+2 e^{x-1}+e^{2 x-2}}+\frac {\log ^2(x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x^3 + E^(2*E^(-2 + 2*x) + 8*E^(-1 + x)*x + 8*x^2)*(-4*E^(-2 + 2*x)*x^2 - 16*x^3 + E^(-1 + x)*(-8*x^2 -
8*x^3)) + E^(E^(-2 + 2*x) + 4*E^(-1 + x)*x + 4*x^2)*(-2*x^2 - 4*E^(-2 + 2*x)*x^3 - 16*x^4 + E^(-1 + x)*(-8*x^
3 - 8*x^4)) + 2*Log[x] - Log[x]^2)/x^2,x]

[Out]

-E^(2*E^(-2 + 2*x) + 8*E^(-1 + x)*x + 8*x^2) - x^2 - (2*E^(-2 + E^(-2 + 2*x) + 4*E^(-1 + x)*x + 4*x^2)*(E^(2*x
)*x + 2*E^(1 + x)*x + 4*E^2*x^2 + 2*E^(1 + x)*x^2))/(2*E^(-1 + x) + E^(-2 + 2*x) + 4*x + 2*E^(-1 + x)*x) + Log
[x]^2/x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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 2305

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-4 e^{-2+2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2} \left (2 e+e^x\right ) \left (e^x+2 e x\right )-2 e^{-2+e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (e^2+2 e^{2 x} x+4 e^{1+x} x+8 e^2 x^2+4 e^{1+x} x^2\right )+\frac {-2 x^3+2 \log (x)-\log ^2(x)}{x^2}\right ) \, dx\\ &=-\left (2 \int e^{-2+e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (e^2+2 e^{2 x} x+4 e^{1+x} x+8 e^2 x^2+4 e^{1+x} x^2\right ) \, dx\right )-4 \int e^{-2+2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2} \left (2 e+e^x\right ) \left (e^x+2 e x\right ) \, dx+\int \frac {-2 x^3+2 \log (x)-\log ^2(x)}{x^2} \, dx\\ &=-e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2}-\frac {2 e^{-2+e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (e^{2 x} x+2 e^{1+x} x+4 e^2 x^2+2 e^{1+x} x^2\right )}{2 e^{-1+x}+e^{-2+2 x}+4 x+2 e^{-1+x} x}+\int \left (-2 x+\frac {2 \log (x)}{x^2}-\frac {\log ^2(x)}{x^2}\right ) \, dx\\ &=-e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2}-x^2-\frac {2 e^{-2+e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (e^{2 x} x+2 e^{1+x} x+4 e^2 x^2+2 e^{1+x} x^2\right )}{2 e^{-1+x}+e^{-2+2 x}+4 x+2 e^{-1+x} x}+2 \int \frac {\log (x)}{x^2} \, dx-\int \frac {\log ^2(x)}{x^2} \, dx\\ &=-e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2}-\frac {2}{x}-x^2-\frac {2 e^{-2+e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (e^{2 x} x+2 e^{1+x} x+4 e^2 x^2+2 e^{1+x} x^2\right )}{2 e^{-1+x}+e^{-2+2 x}+4 x+2 e^{-1+x} x}-\frac {2 \log (x)}{x}+\frac {\log ^2(x)}{x}-2 \int \frac {\log (x)}{x^2} \, dx\\ &=-e^{2 e^{-2+2 x}+8 e^{-1+x} x+8 x^2}-x^2-\frac {2 e^{-2+e^{-2+2 x}+4 e^{-1+x} x+4 x^2} \left (e^{2 x} x+2 e^{1+x} x+4 e^2 x^2+2 e^{1+x} x^2\right )}{2 e^{-1+x}+e^{-2+2 x}+4 x+2 e^{-1+x} x}+\frac {\log ^2(x)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 32, normalized size = 1.14 \begin {gather*} \frac {-x \left (e^{\frac {\left (e^x+2 e x\right )^2}{e^2}}+x\right )^2+\log ^2(x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^3 + E^(2*E^(-2 + 2*x) + 8*E^(-1 + x)*x + 8*x^2)*(-4*E^(-2 + 2*x)*x^2 - 16*x^3 + E^(-1 + x)*(-8
*x^2 - 8*x^3)) + E^(E^(-2 + 2*x) + 4*E^(-1 + x)*x + 4*x^2)*(-2*x^2 - 4*E^(-2 + 2*x)*x^3 - 16*x^4 + E^(-1 + x)*
(-8*x^3 - 8*x^4)) + 2*Log[x] - Log[x]^2)/x^2,x]

[Out]

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

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fricas [B]  time = 0.67, size = 64, normalized size = 2.29 \begin {gather*} -\frac {x^{3} + 2 \, x^{2} e^{\left (4 \, x^{2} + 4 \, x e^{\left (x - 1\right )} + e^{\left (2 \, x - 2\right )}\right )} + x e^{\left (8 \, x^{2} + 8 \, x e^{\left (x - 1\right )} + 2 \, e^{\left (2 \, x - 2\right )}\right )} - \log \relax (x)^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2*exp(x-1)^2+(-8*x^3-8*x^2)*exp(x-1)-16*x^3)*exp(exp(x-1)^2+4*x*exp(x-1)+4*x^2)^2+(-4*x^3*exp
(x-1)^2+(-8*x^4-8*x^3)*exp(x-1)-16*x^4-2*x^2)*exp(exp(x-1)^2+4*x*exp(x-1)+4*x^2)-log(x)^2+2*log(x)-2*x^3)/x^2,
x, algorithm="fricas")

[Out]

-(x^3 + 2*x^2*e^(4*x^2 + 4*x*e^(x - 1) + e^(2*x - 2)) + x*e^(8*x^2 + 8*x*e^(x - 1) + 2*e^(2*x - 2)) - log(x)^2
)/x

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giac [B]  time = 0.31, size = 64, normalized size = 2.29 \begin {gather*} -\frac {x^{3} + 2 \, x^{2} e^{\left (4 \, x^{2} + 4 \, x e^{\left (x - 1\right )} + e^{\left (2 \, x - 2\right )}\right )} + x e^{\left (8 \, x^{2} + 8 \, x e^{\left (x - 1\right )} + 2 \, e^{\left (2 \, x - 2\right )}\right )} - \log \relax (x)^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2*exp(x-1)^2+(-8*x^3-8*x^2)*exp(x-1)-16*x^3)*exp(exp(x-1)^2+4*x*exp(x-1)+4*x^2)^2+(-4*x^3*exp
(x-1)^2+(-8*x^4-8*x^3)*exp(x-1)-16*x^4-2*x^2)*exp(exp(x-1)^2+4*x*exp(x-1)+4*x^2)-log(x)^2+2*log(x)-2*x^3)/x^2,
x, algorithm="giac")

[Out]

-(x^3 + 2*x^2*e^(4*x^2 + 4*x*e^(x - 1) + e^(2*x - 2)) + x*e^(8*x^2 + 8*x*e^(x - 1) + 2*e^(2*x - 2)) - log(x)^2
)/x

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maple [B]  time = 0.11, size = 62, normalized size = 2.21

method	result	size
risch	\(\frac {\ln \relax (x )^{2}}{x}-x^{2}-{\mathrm e}^{2 \,{\mathrm e}^{2 x -2}+8 x \,{\mathrm e}^{x -1}+8 x^{2}}-2 x \,{\mathrm e}^{{\mathrm e}^{2 x -2}+4 x \,{\mathrm e}^{x -1}+4 x^{2}}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^2*exp(x-1)^2+(-8*x^3-8*x^2)*exp(x-1)-16*x^3)*exp(exp(x-1)^2+4*x*exp(x-1)+4*x^2)^2+(-4*x^3*exp(x-1)^
2+(-8*x^4-8*x^3)*exp(x-1)-16*x^4-2*x^2)*exp(exp(x-1)^2+4*x*exp(x-1)+4*x^2)-ln(x)^2+2*ln(x)-2*x^3)/x^2,x,method
=_RETURNVERBOSE)

[Out]

ln(x)^2/x-x^2-exp(2*exp(2*x-2)+8*x*exp(x-1)+8*x^2)-2*x*exp(exp(2*x-2)+4*x*exp(x-1)+4*x^2)

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maxima [B]  time = 0.84, size = 67, normalized size = 2.39 \begin {gather*} -x^{2} - \frac {2 \, x^{2} e^{\left (4 \, x^{2} + 4 \, x e^{\left (x - 1\right )} + e^{\left (2 \, x - 2\right )}\right )} + x e^{\left (8 \, x^{2} + 8 \, x e^{\left (x - 1\right )} + 2 \, e^{\left (2 \, x - 2\right )}\right )} - \log \relax (x)^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2*exp(x-1)^2+(-8*x^3-8*x^2)*exp(x-1)-16*x^3)*exp(exp(x-1)^2+4*x*exp(x-1)+4*x^2)^2+(-4*x^3*exp
(x-1)^2+(-8*x^4-8*x^3)*exp(x-1)-16*x^4-2*x^2)*exp(exp(x-1)^2+4*x*exp(x-1)+4*x^2)-log(x)^2+2*log(x)-2*x^3)/x^2,
x, algorithm="maxima")

[Out]

-x^2 - (2*x^2*e^(4*x^2 + 4*x*e^(x - 1) + e^(2*x - 2)) + x*e^(8*x^2 + 8*x*e^(x - 1) + 2*e^(2*x - 2)) - log(x)^2
)/x

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mupad [B]  time = 1.88, size = 77, normalized size = 2.75 \begin {gather*} \frac {{\ln \relax (x)}^2+2\,\ln \relax (x)+2}{x}-{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-2}+8\,x^2+8\,x\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^x}-\frac {2\,\left (\ln \relax (x)+1\right )}{x}-x^2-2\,x\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-2}+4\,x^2+4\,x\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(2*x - 2) + 4*x*exp(x - 1) + 4*x^2)*(exp(x - 1)*(8*x^3 + 8*x^4) + 4*x^3*exp(2*x - 2) + 2*x^2 + 16
*x^4) - 2*log(x) + exp(2*exp(2*x - 2) + 8*x*exp(x - 1) + 8*x^2)*(exp(x - 1)*(8*x^2 + 8*x^3) + 4*x^2*exp(2*x -
2) + 16*x^3) + log(x)^2 + 2*x^3)/x^2,x)

[Out]

(2*log(x) + log(x)^2 + 2)/x - exp(2*exp(2*x)*exp(-2) + 8*x^2 + 8*x*exp(-1)*exp(x)) - (2*(log(x) + 1))/x - x^2
- 2*x*exp(exp(2*x)*exp(-2) + 4*x^2 + 4*x*exp(-1)*exp(x))

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sympy [B]  time = 0.57, size = 58, normalized size = 2.07 \begin {gather*} - x^{2} - 2 x e^{4 x^{2} + 4 x e^{x - 1} + e^{2 x - 2}} - e^{8 x^{2} + 8 x e^{x - 1} + 2 e^{2 x - 2}} + \frac {\log {\relax (x )}^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**2*exp(x-1)**2+(-8*x**3-8*x**2)*exp(x-1)-16*x**3)*exp(exp(x-1)**2+4*x*exp(x-1)+4*x**2)**2+(-4
*x**3*exp(x-1)**2+(-8*x**4-8*x**3)*exp(x-1)-16*x**4-2*x**2)*exp(exp(x-1)**2+4*x*exp(x-1)+4*x**2)-ln(x)**2+2*ln
(x)-2*x**3)/x**2,x)

[Out]

-x**2 - 2*x*exp(4*x**2 + 4*x*exp(x - 1) + exp(2*x - 2)) - exp(8*x**2 + 8*x*exp(x - 1) + 2*exp(2*x - 2)) + log(
x)**2/x

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