3.74.62 \(\int \frac {1}{5} (5+5 x+e^{2 x^2} (-2 x-4 x^3)-2 x \log (x)) \, dx\)

Optimal. Leaf size=25 \[ -2+x+\frac {1}{5} x^2 \left (3-e^{2 x^2}-\log (x)\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 32, normalized size of antiderivative = 1.28, number of steps used = 9, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {12, 1593, 2226, 2209, 2212, 2304} \begin {gather*} -\frac {1}{5} e^{2 x^2} x^2+\frac {3 x^2}{5}-\frac {1}{5} x^2 \log (x)+x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, 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
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (5+5 x+e^{2 x^2} \left (-2 x-4 x^3\right )-2 x \log (x)\right ) \, dx\\ &=x+\frac {x^2}{2}+\frac {1}{5} \int e^{2 x^2} \left (-2 x-4 x^3\right ) \, dx-\frac {2}{5} \int x \log (x) \, dx\\ &=x+\frac {3 x^2}{5}-\frac {1}{5} x^2 \log (x)+\frac {1}{5} \int e^{2 x^2} x \left (-2-4 x^2\right ) \, dx\\ &=x+\frac {3 x^2}{5}-\frac {1}{5} x^2 \log (x)+\frac {1}{5} \int \left (-2 e^{2 x^2} x-4 e^{2 x^2} x^3\right ) \, dx\\ &=x+\frac {3 x^2}{5}-\frac {1}{5} x^2 \log (x)-\frac {2}{5} \int e^{2 x^2} x \, dx-\frac {4}{5} \int e^{2 x^2} x^3 \, dx\\ &=-\frac {1}{10} e^{2 x^2}+x+\frac {3 x^2}{5}-\frac {1}{5} e^{2 x^2} x^2-\frac {1}{5} x^2 \log (x)+\frac {2}{5} \int e^{2 x^2} x \, dx\\ &=x+\frac {3 x^2}{5}-\frac {1}{5} e^{2 x^2} x^2-\frac {1}{5} x^2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 32, normalized size = 1.28 \begin {gather*} x+\frac {3 x^2}{5}-\frac {1}{5} e^{2 x^2} x^2-\frac {1}{5} x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.94, size = 25, normalized size = 1.00 \begin {gather*} -\frac {1}{5} \, x^{2} e^{\left (2 \, x^{2}\right )} - \frac {1}{5} \, x^{2} \log \relax (x) + \frac {3}{5} \, x^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/5*x*log(x)+1/5*(-4*x^3-2*x)*exp(2*x^2)+x+1,x, algorithm="fricas")

[Out]

-1/5*x^2*e^(2*x^2) - 1/5*x^2*log(x) + 3/5*x^2 + x

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giac [B]  time = 0.13, size = 37, normalized size = 1.48 \begin {gather*} -\frac {1}{5} \, x^{2} \log \relax (x) + \frac {3}{5} \, x^{2} - \frac {1}{10} \, {\left (2 \, x^{2} - 1\right )} e^{\left (2 \, x^{2}\right )} + x - \frac {1}{10} \, e^{\left (2 \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/5*x*log(x)+1/5*(-4*x^3-2*x)*exp(2*x^2)+x+1,x, algorithm="giac")

[Out]

-1/5*x^2*log(x) + 3/5*x^2 - 1/10*(2*x^2 - 1)*e^(2*x^2) + x - 1/10*e^(2*x^2)

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maple [A]  time = 0.02, size = 26, normalized size = 1.04

method	result	size
default	\(x +\frac {3 x^{2}}{5}-\frac {x^{2} \ln \relax (x )}{5}-\frac {{\mathrm e}^{2 x^{2}} x^{2}}{5}\)	\(26\)
norman	\(x +\frac {3 x^{2}}{5}-\frac {x^{2} \ln \relax (x )}{5}-\frac {{\mathrm e}^{2 x^{2}} x^{2}}{5}\)	\(26\)
risch	\(x +\frac {3 x^{2}}{5}-\frac {x^{2} \ln \relax (x )}{5}-\frac {{\mathrm e}^{2 x^{2}} x^{2}}{5}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2/5*x*ln(x)+1/5*(-4*x^3-2*x)*exp(2*x^2)+x+1,x,method=_RETURNVERBOSE)

[Out]

3/5*x^2+x-1/5*x^2*ln(x)-1/5*x^2*exp(x^2)^2

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maxima [A]  time = 0.37, size = 25, normalized size = 1.00 \begin {gather*} -\frac {1}{5} \, x^{2} e^{\left (2 \, x^{2}\right )} - \frac {1}{5} \, x^{2} \log \relax (x) + \frac {3}{5} \, x^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/5*x*log(x)+1/5*(-4*x^3-2*x)*exp(2*x^2)+x+1,x, algorithm="maxima")

[Out]

-1/5*x^2*e^(2*x^2) - 1/5*x^2*log(x) + 3/5*x^2 + x

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mupad [B]  time = 6.55, size = 22, normalized size = 0.88 \begin {gather*} \frac {x\,\left (3\,x-x\,{\mathrm {e}}^{2\,x^2}-x\,\ln \relax (x)+5\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x - (exp(2*x^2)*(2*x + 4*x^3))/5 - (2*x*log(x))/5 + 1,x)

[Out]

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

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sympy [A]  time = 0.29, size = 27, normalized size = 1.08 \begin {gather*} - \frac {x^{2} e^{2 x^{2}}}{5} - \frac {x^{2} \log {\relax (x )}}{5} + \frac {3 x^{2}}{5} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/5*x*ln(x)+1/5*(-4*x**3-2*x)*exp(2*x**2)+x+1,x)

[Out]

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

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