\(\int (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)) \, dx\) [145]

Optimal result

Integrand size = 39, antiderivative size = 20 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=1+\left (4+2^{e^x x}\right )^2+2 x-4 \log (5) \]

[Out]

2*x+1+(exp(x*ln(2)*exp(x))+4)^2-4*ln(5)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6838} \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=2 x+2^{2 e^x x}+2^{e^x x+3} \]

[In]

Int[2 + 2^(2*E^x*x)*E^x*(2 + 2*x)*Log[2] + 2^(E^x*x)*E^x*(8 + 8*x)*Log[2],x]

[Out]

2^(2*E^x*x) + 2^(3 + E^x*x) + 2*x

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = 2 x+\log (2) \int 2^{2 e^x x} e^x (2+2 x) \, dx+\log (2) \int 2^{e^x x} e^x (8+8 x) \, dx \\ & = 2^{2 e^x x}+2^{3+e^x x}+2 x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=\left (4+2^{e^x x}\right )^2+2 x \]

[In]

Integrate[2 + 2^(2*E^x*x)*E^x*(2 + 2*x)*Log[2] + 2^(E^x*x)*E^x*(8 + 8*x)*Log[2],x]

[Out]

(4 + 2^(E^x*x))^2 + 2*x

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05

method	result	size
risch	\(2 x +2^{2 \,{\mathrm e}^{x} x}+8 \,2^{{\mathrm e}^{x} x}\)	\(21\)
default	\(2 x +{\mathrm e}^{2 x \ln \left (2\right ) {\mathrm e}^{x}}+8 \,{\mathrm e}^{x \ln \left (2\right ) {\mathrm e}^{x}}\)	\(23\)
norman	\(2 x +{\mathrm e}^{2 x \ln \left (2\right ) {\mathrm e}^{x}}+8 \,{\mathrm e}^{x \ln \left (2\right ) {\mathrm e}^{x}}\)	\(23\)
parallelrisch	\(2 x +{\mathrm e}^{2 x \ln \left (2\right ) {\mathrm e}^{x}}+8 \,{\mathrm e}^{x \ln \left (2\right ) {\mathrm e}^{x}}\)	\(23\)
parts	\(2 x +{\mathrm e}^{2 x \ln \left (2\right ) {\mathrm e}^{x}}+8 \,{\mathrm e}^{x \ln \left (2\right ) {\mathrm e}^{x}}\)	\(23\)

[In]

int((2+2*x)*ln(2)*exp(x)*exp(x*ln(2)*exp(x))^2+(8*x+8)*ln(2)*exp(x)*exp(x*ln(2)*exp(x))+2,x,method=_RETURNVERB
OSE)

[Out]

2*x+(2^(exp(x)*x))^2+8*2^(exp(x)*x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=2^{2 \, x e^{x}} + 8 \cdot 2^{x e^{x}} + 2 \, x \]

[In]

integrate((2+2*x)*log(2)*exp(x)*exp(x*log(2)*exp(x))^2+(8*x+8)*log(2)*exp(x)*exp(x*log(2)*exp(x))+2,x, algorit
hm="fricas")

[Out]

2^(2*x*e^x) + 8*2^(x*e^x) + 2*x

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=2 x + e^{2 x e^{x} \log {\left (2 \right )}} + 8 e^{x e^{x} \log {\left (2 \right )}} \]

[In]

integrate((2+2*x)*ln(2)*exp(x)*exp(x*ln(2)*exp(x))**2+(8*x+8)*ln(2)*exp(x)*exp(x*ln(2)*exp(x))+2,x)

[Out]

2*x + exp(2*x*exp(x)*log(2)) + 8*exp(x*exp(x)*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=2^{2 \, x e^{x}} + 8 \cdot 2^{x e^{x}} + 2 \, x \]

[In]

integrate((2+2*x)*log(2)*exp(x)*exp(x*log(2)*exp(x))^2+(8*x+8)*log(2)*exp(x)*exp(x*log(2)*exp(x))+2,x, algorit
hm="maxima")

[Out]

2^(2*x*e^x) + 8*2^(x*e^x) + 2*x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=2^{2 \, x e^{x}} + 8 \cdot 2^{x e^{x}} + 2 \, x \]

[In]

integrate((2+2*x)*log(2)*exp(x)*exp(x*log(2)*exp(x))^2+(8*x+8)*log(2)*exp(x)*exp(x*log(2)*exp(x))+2,x, algorit
hm="giac")

[Out]

2^(2*x*e^x) + 8*2^(x*e^x) + 2*x

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (2+2^{2 e^x x} e^x (2+2 x) \log (2)+2^{e^x x} e^x (8+8 x) \log (2)\right ) \, dx=2\,x+8\,2^{x\,{\mathrm {e}}^x}+2^{2\,x\,{\mathrm {e}}^x} \]

[In]

int(exp(2*x*exp(x)*log(2))*exp(x)*log(2)*(2*x + 2) + exp(x*exp(x)*log(2))*exp(x)*log(2)*(8*x + 8) + 2,x)

[Out]

2*x + 8*2^(x*exp(x)) + 2^(2*x*exp(x))