\(\int (2 e^{x^2} x+2^{2 x-2 e^{50} x} (-2+2 e^{50}) \log (2)) \, dx\) [1398]

Optimal result

Integrand size = 31, antiderivative size = 20 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=-2^{2 \left (1-e^{50}\right ) x}+e^{x^2} \]

[Out]

exp(x^2)-exp(x*ln(2)*(1-exp(25)^2))^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2240, 2259, 2225} \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=e^{x^2}-2^{2 \left (1-e^{50}\right ) x} \]

[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 2225

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 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]

Rule 2259

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=-4^{x-e^{50} x}+e^{x^2} \]

[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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90

method	result	size
risch	\(-2^{-2 \left ({\mathrm e}^{50}-1\right ) x}+{\mathrm e}^{x^{2}}\)	\(18\)
norman	\(-{\mathrm e}^{2 \left (-x \,{\mathrm e}^{50}+x \right ) \ln \left (2\right )}+{\mathrm e}^{x^{2}}\)	\(23\)
parallelrisch	\(-{\mathrm e}^{2 \left (-x \,{\mathrm e}^{50}+x \right ) \ln \left (2\right )}+{\mathrm e}^{x^{2}}\)	\(23\)
default	\(\frac {\left ({\mathrm e}^{50}-1\right ) {\mathrm e}^{2 \left (-x \,{\mathrm e}^{50}+x \right ) \ln \left (2\right )}}{-{\mathrm e}^{50}+1}+{\mathrm e}^{x^{2}}\)	\(38\)
parts	\(\frac {\left ({\mathrm e}^{50}-1\right ) {\mathrm e}^{2 \left (-x \,{\mathrm e}^{50}+x \right ) \ln \left (2\right )}}{-{\mathrm e}^{50}+1}+{\mathrm e}^{x^{2}}\)	\(38\)
meijerg	\(-1+{\mathrm e}^{x^{2}}+\frac {{\mathrm e}^{50} \left (1-{\mathrm e}^{-2 x \left ({\mathrm e}^{50}-1\right ) \ln \left (2\right )}\right )}{{\mathrm e}^{50}-1}-\frac {1-{\mathrm e}^{-2 x \left ({\mathrm e}^{50}-1\right ) \ln \left (2\right )}}{{\mathrm e}^{50}-1}\)	\(52\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=-2^{-2 \, x e^{50} + 2 \, x} + e^{\left (x^{2}\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=e^{x^{2}} - e^{2 \left (- x e^{50} + x\right ) \log {\left (2 \right )}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=-\frac {1}{2} \cdot 2^{-2 \, x e^{50} + 2 \, x + 1} + e^{\left (x^{2}\right )} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (17) = 34\).

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=-\frac {2^{-2 \, x e^{50} + 2 \, x} {\left (e^{50} - 1\right )} \log \left (2\right )}{e^{50} \log \left (2\right ) - \log \left (2\right )} + e^{\left (x^{2}\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 7.60 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx={\mathrm {e}}^{x^2}-\frac {2^{2\,x}}{2^{2\,x\,{\mathrm {e}}^{50}}} \]

[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))