\(\int \frac {1}{5} (10+4 e^{x^2} x+10 x \log ^2(3)) \, dx\) [5986]

Optimal result

Integrand size = 21, antiderivative size = 22 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=-3+\frac {2 e^{x^2}}{5}+2 x+x^2 \log ^2(3) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 2240} \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=\frac {2 e^{x^2}}{5}+x^2 \log ^2(3)+2 x \]

[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 2240

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=\frac {2 e^{x^2}}{5}+2 x+x^2 \log ^2(3) \]

[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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=x^{2} \log \left (3\right )^{2} + 2 \, x + \frac {2}{5} \, e^{\left (x^{2}\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=x^{2} \log {\left (3 \right )}^{2} + 2 x + \frac {2 e^{x^{2}}}{5} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=x^{2} \log \left (3\right )^{2} + 2 \, x + \frac {2}{5} \, e^{\left (x^{2}\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=x^{2} \log \left (3\right )^{2} + 2 \, x + \frac {2}{5} \, e^{\left (x^{2}\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{5} \left (10+4 e^{x^2} x+10 x \log ^2(3)\right ) \, dx=2\,x+\frac {2\,{\mathrm {e}}^{x^2}}{5}+x^2\,{\ln \left (3\right )}^2 \]

[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