3.60.86 \(\int \frac {1}{5} (10+4 e^{x^2} x+10 x \log ^2(3)) \, dx\)

Optimal. Leaf size=22 \[ -3+\frac {2 e^{x^2}}{5}+2 x+x^2 \log ^2(3) \]

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 2209} \begin {gather*} \frac {2 e^{x^2}}{5}+x^2 \log ^2(3)+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^x^2)/5 + 2*x + x^2*Log[3]^2

Rule 12

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.95 \begin {gather*} \frac {2 e^{x^2}}{5}+2 x+x^2 \log ^2(3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^x^2)/5 + 2*x + x^2*Log[3]^2

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fricas [A]  time = 0.79, size = 18, normalized size = 0.82 \begin {gather*} x^{2} \log \relax (3)^{2} + 2 \, x + \frac {2}{5} \, e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*log(3)^2 + 2*x + 2/5*e^(x^2)

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giac [A]  time = 0.14, size = 18, normalized size = 0.82 \begin {gather*} x^{2} \log \relax (3)^{2} + 2 \, x + \frac {2}{5} \, e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*log(3)^2 + 2*x + 2/5*e^(x^2)

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maple [A]  time = 0.04, size = 19, normalized size = 0.86

method	result	size
default	\(2 x +x^{2} \ln \relax (3)^{2}+\frac {2 \,{\mathrm e}^{x^{2}}}{5}\)	\(19\)
norman	\(2 x +x^{2} \ln \relax (3)^{2}+\frac {2 \,{\mathrm e}^{x^{2}}}{5}\)	\(19\)
risch	\(2 x +x^{2} \ln \relax (3)^{2}+\frac {2 \,{\mathrm e}^{x^{2}}}{5}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 18, normalized size = 0.82 \begin {gather*} x^{2} \log \relax (3)^{2} + 2 \, x + \frac {2}{5} \, e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*log(3)^2 + 2*x + 2/5*e^(x^2)

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mupad [B]  time = 4.31, size = 18, normalized size = 0.82 \begin {gather*} 2\,x+\frac {2\,{\mathrm {e}}^{x^2}}{5}+x^2\,{\ln \relax (3)}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x + (2*exp(x^2))/5 + x^2*log(3)^2

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sympy [A]  time = 0.09, size = 19, normalized size = 0.86 \begin {gather*} x^{2} \log {\relax (3 )}^{2} + 2 x + \frac {2 e^{x^{2}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*log(3)**2 + 2*x + 2*exp(x**2)/5

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