3.48.45 \(\int (e^{4+x^2} (-5-10 x^2)+e^{2 x^2} (5+20 x^2)+5 \log (3)) \, dx\)

Optimal. Leaf size=22 \[ 5 x \left (e^{2 x^2}-e^{4+x^2}+\log (3)\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 11, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {2226, 2204, 2212} \begin {gather*} 5 e^{2 x^2} x-5 e^{x^2+4} x+5 x \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(4 + x^2)*(-5 - 10*x^2) + E^(2*x^2)*(5 + 20*x^2) + 5*Log[3],x]

[Out]

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

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=5 x \log (3)+\int e^{4+x^2} \left (-5-10 x^2\right ) \, dx+\int e^{2 x^2} \left (5+20 x^2\right ) \, dx\\ &=5 x \log (3)+\int \left (5 e^{2 x^2}+20 e^{2 x^2} x^2\right ) \, dx+\int \left (-5 e^{4+x^2}-10 e^{4+x^2} x^2\right ) \, dx\\ &=5 x \log (3)+5 \int e^{2 x^2} \, dx-5 \int e^{4+x^2} \, dx-10 \int e^{4+x^2} x^2 \, dx+20 \int e^{2 x^2} x^2 \, dx\\ &=5 e^{2 x^2} x-5 e^{4+x^2} x-\frac {5}{2} e^4 \sqrt {\pi } \text {erfi}(x)+\frac {5}{2} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\sqrt {2} x\right )+5 x \log (3)-5 \int e^{2 x^2} \, dx+5 \int e^{4+x^2} \, dx\\ &=5 e^{2 x^2} x-5 e^{4+x^2} x+5 x \log (3)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 25, normalized size = 1.14 \begin {gather*} 5 e^{2 x^2} x-5 e^{4+x^2} x+x \log (243) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(4 + x^2)*(-5 - 10*x^2) + E^(2*x^2)*(5 + 20*x^2) + 5*Log[3],x]

[Out]

5*E^(2*x^2)*x - 5*E^(4 + x^2)*x + x*Log[243]

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fricas [A]  time = 0.80, size = 30, normalized size = 1.36 \begin {gather*} 5 \, {\left (x e^{8} \log \relax (3) + x e^{\left (2 \, x^{2} + 8\right )} - x e^{\left (x^{2} + 12\right )}\right )} e^{\left (-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x^2+5)*exp(2*x^2)+(-10*x^2-5)*exp(x^2+4)+5*log(3),x, algorithm="fricas")

[Out]

5*(x*e^8*log(3) + x*e^(2*x^2 + 8) - x*e^(x^2 + 12))*e^(-8)

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giac [A]  time = 0.19, size = 24, normalized size = 1.09 \begin {gather*} 5 \, x e^{\left (2 \, x^{2}\right )} - 5 \, x e^{\left (x^{2} + 4\right )} + 5 \, x \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x^2+5)*exp(2*x^2)+(-10*x^2-5)*exp(x^2+4)+5*log(3),x, algorithm="giac")

[Out]

5*x*e^(2*x^2) - 5*x*e^(x^2 + 4) + 5*x*log(3)

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maple [A]  time = 0.05, size = 25, normalized size = 1.14

method	result	size
norman	\(5 x \,{\mathrm e}^{2 x^{2}}+5 x \ln \relax (3)-5 x \,{\mathrm e}^{4} {\mathrm e}^{x^{2}}\)	\(25\)
risch	\(5 x \,{\mathrm e}^{2 x^{2}}-5 x \,{\mathrm e}^{x^{2}+4}+5 x \ln \relax (3)\)	\(25\)
default	\(-\frac {5 \,{\mathrm e}^{4} \sqrt {\pi }\, \erfi \relax (x )}{2}-10 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{x^{2}} x}{2}-\frac {\sqrt {\pi }\, \erfi \relax (x )}{4}\right )+5 x \,{\mathrm e}^{2 x^{2}}+5 x \ln \relax (3)\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x^2+5)*exp(2*x^2)+(-10*x^2-5)*exp(x^2+4)+5*ln(3),x,method=_RETURNVERBOSE)

[Out]

5*x*exp(x^2)^2+5*x*ln(3)-5*x*exp(4)*exp(x^2)

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maxima [A]  time = 0.46, size = 24, normalized size = 1.09 \begin {gather*} 5 \, x e^{\left (2 \, x^{2}\right )} - 5 \, x e^{\left (x^{2} + 4\right )} + 5 \, x \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x^2+5)*exp(2*x^2)+(-10*x^2-5)*exp(x^2+4)+5*log(3),x, algorithm="maxima")

[Out]

5*x*e^(2*x^2) - 5*x*e^(x^2 + 4) + 5*x*log(3)

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mupad [B]  time = 0.08, size = 20, normalized size = 0.91 \begin {gather*} 5\,x\,\left (\ln \relax (3)+{\mathrm {e}}^{2\,x^2}-{\mathrm {e}}^{x^2+4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(5*log(3) + exp(2*x^2)*(20*x^2 + 5) - exp(x^2 + 4)*(10*x^2 + 5),x)

[Out]

5*x*(log(3) + exp(2*x^2) - exp(x^2 + 4))

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sympy [A]  time = 0.27, size = 32, normalized size = 1.45 \begin {gather*} 5 x e^{2 x^{2}} - 5 x e^{4} \sqrt {e^{2 x^{2}}} + 5 x \log {\relax (3 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x**2+5)*exp(2*x**2)+(-10*x**2-5)*exp(x**2+4)+5*ln(3),x)

[Out]

5*x*exp(2*x**2) - 5*x*exp(4)*sqrt(exp(2*x**2)) + 5*x*log(3)

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