3.14.98 \(\int (2 e^{x^2} x+2^{2 x-2 e^{50} x} (-2+2 e^{50}) \log (2)) \, dx\)

Optimal. Leaf size=20 \[ -2^{2 \left (1-e^{50}\right ) x}+e^{x^2} \]

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Rubi [A]  time = 0.03, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2209, 2227, 2194} \begin {gather*} e^{x^2}-2^{2 \left (1-e^{50}\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2*E^x^2*x + 2^(2*x - 2*E^50*x)*(-2 + 2*E^50)*Log[2],x]

[Out]

-2^(2*(1 - E^50)*x) + E^x^2

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, 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]

Rule 2227

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=2 \int e^{x^2} x \, dx-\left (2 \left (1-e^{50}\right ) \log (2)\right ) \int 2^{2 x-2 e^{50} x} \, dx\\ &=e^{x^2}-\left (2 \left (1-e^{50}\right ) \log (2)\right ) \int 2^{2 \left (1-e^{50}\right ) x} \, dx\\ &=-2^{2 \left (1-e^{50}\right ) x}+e^{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.90 \begin {gather*} -4^{x-e^{50} x}+e^{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2*E^x^2*x + 2^(2*x - 2*E^50*x)*(-2 + 2*E^50)*Log[2],x]

[Out]

-4^(x - E^50*x) + E^x^2

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fricas [A]  time = 0.88, size = 18, normalized size = 0.90 \begin {gather*} -2^{-2 \, x e^{50} + 2 \, x} + e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(x^2)*x+(2*exp(25)^2-2)*log(2)*exp((-x*exp(25)^2+x)*log(2))^2,x, algorithm="fricas")

[Out]

-2^(-2*x*e^50 + 2*x) + e^(x^2)

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giac [B]  time = 0.24, size = 36, normalized size = 1.80 \begin {gather*} -\frac {2^{-2 \, x e^{50} + 2 \, x} {\left (e^{50} - 1\right )} \log \relax (2)}{e^{50} \log \relax (2) - \log \relax (2)} + e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(x^2)*x+(2*exp(25)^2-2)*log(2)*exp((-x*exp(25)^2+x)*log(2))^2,x, algorithm="giac")

[Out]

-2^(-2*x*e^50 + 2*x)*(e^50 - 1)*log(2)/(e^50*log(2) - log(2)) + e^(x^2)

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maple [A]  time = 0.07, size = 18, normalized size = 0.90

method	result	size
risch	\(-2^{-2 \left ({\mathrm e}^{50}-1\right ) x}+{\mathrm e}^{x^{2}}\)	\(18\)
norman	\(-{\mathrm e}^{2 \left (-x \,{\mathrm e}^{50}+x \right ) \ln \relax (2)}+{\mathrm e}^{x^{2}}\)	\(23\)
default	\(\frac {{\mathrm e}^{2 \left (-x \,{\mathrm e}^{50}+x \right ) \ln \relax (2)} {\mathrm e}^{50}}{-{\mathrm e}^{50}+1}-\frac {{\mathrm e}^{2 \left (-x \,{\mathrm e}^{50}+x \right ) \ln \relax (2)}}{-{\mathrm e}^{50}+1}+{\mathrm e}^{x^{2}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(x^2)*x+(2*exp(25)^2-2)*ln(2)*exp((-x*exp(25)^2+x)*ln(2))^2,x,method=_RETURNVERBOSE)

[Out]

-((1/2)^((exp(50)-1)*x))^2+exp(x^2)

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maxima [A]  time = 0.49, size = 19, normalized size = 0.95 \begin {gather*} -\frac {1}{2} \cdot 2^{-2 \, x e^{50} + 2 \, x + 1} + e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(x^2)*x+(2*exp(25)^2-2)*log(2)*exp((-x*exp(25)^2+x)*log(2))^2,x, algorithm="maxima")

[Out]

-1/2*2^(-2*x*e^50 + 2*x + 1) + e^(x^2)

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mupad [B]  time = 1.01, size = 21, normalized size = 1.05 \begin {gather*} {\mathrm {e}}^{x^2}-\frac {2^{2\,x}}{2^{2\,x\,{\mathrm {e}}^{50}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x*exp(x^2) + exp(2*log(2)*(x - x*exp(50)))*log(2)*(2*exp(50) - 2),x)

[Out]

exp(x^2) - 2^(2*x)/2^(2*x*exp(50))

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sympy [A]  time = 0.32, size = 17, normalized size = 0.85 \begin {gather*} e^{x^{2}} - e^{2 \left (- x e^{50} + x\right ) \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(x**2)*x+(2*exp(25)**2-2)*ln(2)*exp((-x*exp(25)**2+x)*ln(2))**2,x)

[Out]

exp(x**2) - exp(2*(-x*exp(50) + x)*log(2))

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