\(\int (e^x+2 x+(628+2 x) \log (4)) \, dx\) [920]

Optimal result

Integrand size = 15, antiderivative size = 15 \[ \int \left (e^x+2 x+(628+2 x) \log (4)\right ) \, dx=e^x+x^2+(314+x)^2 \log (4) \]

[Out]

exp(x)+x^2+2*ln(2)*(314+x)^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2225} \[ \int \left (e^x+2 x+(628+2 x) \log (4)\right ) \, dx=x^2+e^x+(x+314)^2 \log (4) \]

[In]

Int[E^x + 2*x + (628 + 2*x)*Log[4],x]

[Out]

E^x + x^2 + (314 + x)^2*Log[4]

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]

Rubi steps \begin{align*} \text {integral}& = x^2+(314+x)^2 \log (4)+\int e^x \, dx \\ & = e^x+x^2+(314+x)^2 \log (4) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int \left (e^x+2 x+(628+2 x) \log (4)\right ) \, dx=e^x+x^2+628 x \log (4)+x^2 \log (4) \]

[In]

Integrate[E^x + 2*x + (628 + 2*x)*Log[4],x]

[Out]

E^x + x^2 + 628*x*Log[4] + x^2*Log[4]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27

method	result	size
norman	\(\left (1+2 \ln \left (2\right )\right ) x^{2}+1256 x \ln \left (2\right )+{\mathrm e}^{x}\)	\(19\)
risch	\({\mathrm e}^{x}+2 x^{2} \ln \left (2\right )+1256 x \ln \left (2\right )+x^{2}\)	\(19\)
parallelrisch	\({\mathrm e}^{x}+2 x^{2} \ln \left (2\right )+1256 x \ln \left (2\right )+x^{2}\)	\(19\)
parts	\({\mathrm e}^{x}+2 x^{2} \ln \left (2\right )+1256 x \ln \left (2\right )+x^{2}\)	\(19\)
default	\(x^{2}+4 \ln \left (2\right ) \left (\frac {1}{2} x^{2}+314 x \right )+{\mathrm e}^{x}\)	\(20\)

[In]

int(exp(x)+2*(2*x+628)*ln(2)+2*x,x,method=_RETURNVERBOSE)

[Out]

(1+2*ln(2))*x^2+1256*x*ln(2)+exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \left (e^x+2 x+(628+2 x) \log (4)\right ) \, dx=x^{2} + 2 \, {\left (x^{2} + 628 \, x\right )} \log \left (2\right ) + e^{x} \]

[In]

integrate(exp(x)+2*(2*x+628)*log(2)+2*x,x, algorithm="fricas")

[Out]

x^2 + 2*(x^2 + 628*x)*log(2) + e^x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \left (e^x+2 x+(628+2 x) \log (4)\right ) \, dx=x^{2} \cdot \left (1 + 2 \log {\left (2 \right )}\right ) + 1256 x \log {\left (2 \right )} + e^{x} \]

[In]

integrate(exp(x)+2*(2*x+628)*ln(2)+2*x,x)

[Out]

x**2*(1 + 2*log(2)) + 1256*x*log(2) + exp(x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \left (e^x+2 x+(628+2 x) \log (4)\right ) \, dx=x^{2} + 2 \, {\left (x^{2} + 628 \, x\right )} \log \left (2\right ) + e^{x} \]

[In]

integrate(exp(x)+2*(2*x+628)*log(2)+2*x,x, algorithm="maxima")

[Out]

x^2 + 2*(x^2 + 628*x)*log(2) + e^x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \left (e^x+2 x+(628+2 x) \log (4)\right ) \, dx=x^{2} + 2 \, {\left (x^{2} + 628 \, x\right )} \log \left (2\right ) + e^{x} \]

[In]

integrate(exp(x)+2*(2*x+628)*log(2)+2*x,x, algorithm="giac")

[Out]

x^2 + 2*(x^2 + 628*x)*log(2) + e^x

Mupad [B] (verification not implemented)

Time = 8.97 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \left (e^x+2 x+(628+2 x) \log (4)\right ) \, dx={\mathrm {e}}^x+x^2\,\left (\ln \left (4\right )+1\right )+1256\,x\,\ln \left (2\right ) \]

[In]

int(2*x + exp(x) + 2*log(2)*(2*x + 628),x)

[Out]

exp(x) + x^2*(log(4) + 1) + 1256*x*log(2)