3.92.29 \(\int \frac {2-2 x+2 e^{x^2} x \log (4)}{\log (4)} \, dx\)

Optimal. Leaf size=16 \[ e^{x^2}-\frac {(-2+x) x}{\log (4)} \]

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.38, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 2209} \begin {gather*} e^{x^2}-\frac {x^2}{\log (4)}+\frac {2 x}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^x^2 + (2*x)/Log[4] - x^2/Log[4]

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

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.56 \begin {gather*} \frac {2 x-x^2+\frac {1}{2} e^{x^2} \log (16)}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x - x^2 + (E^x^2*Log[16])/2)/Log[4]

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fricas [A]  time = 0.65, size = 21, normalized size = 1.31 \begin {gather*} -\frac {x^{2} - 2 \, e^{\left (x^{2}\right )} \log \relax (2) - 2 \, x}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 21, normalized size = 1.31 \begin {gather*} -\frac {x^{2} - 2 \, e^{\left (x^{2}\right )} \log \relax (2) - 2 \, x}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 20, normalized size = 1.25

method	result	size
default	\(\frac {x -\frac {x^{2}}{2}+\ln \relax (2) {\mathrm e}^{x^{2}}}{\ln \relax (2)}\)	\(20\)
norman	\(\frac {x}{\ln \relax (2)}-\frac {x^{2}}{2 \ln \relax (2)}+{\mathrm e}^{x^{2}}\)	\(21\)
risch	\(\frac {x}{\ln \relax (2)}-\frac {x^{2}}{2 \ln \relax (2)}+{\mathrm e}^{x^{2}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 21, normalized size = 1.31 \begin {gather*} -\frac {x^{2} - 2 \, e^{\left (x^{2}\right )} \log \relax (2) - 2 \, x}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.06, size = 23, normalized size = 1.44 \begin {gather*} \frac {2\,x+2\,{\mathrm {e}}^{x^2}\,\ln \relax (2)-x^2}{2\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.13, size = 17, normalized size = 1.06 \begin {gather*} - \frac {x^{2}}{2 \log {\relax (2 )}} + \frac {x}{\log {\relax (2 )}} + e^{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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