\(\int (-1+e^{2 x} (5+2 x)+e^{2 x+\frac {2}{5} (5 e^{12} x^4+\log (2))} (2+8 e^{12} x^3)) \, dx\) [9306]

Optimal result

Integrand size = 45, antiderivative size = 29 \[ \int \left (-1+e^{2 x} (5+2 x)+e^{2 x+\frac {2}{5} \left (5 e^{12} x^4+\log (2)\right )} \left (2+8 e^{12} x^3\right )\right ) \, dx=-x+e^{2 x} \left (2+2^{2/5} e^{2 e^{12} x^4}+x\right ) \]

[Out]

(2+x+exp(1/5*ln(2)+x^4*exp(3)^4)^2)*exp(x)^2-x

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2207, 2225, 6838} \[ \int \left (-1+e^{2 x} (5+2 x)+e^{2 x+\frac {2}{5} \left (5 e^{12} x^4+\log (2)\right )} \left (2+8 e^{12} x^3\right )\right ) \, dx=e^{\frac {2}{5} \left (5 e^{12} x^4+\log (2)\right )+2 x}-x-\frac {e^{2 x}}{2}+\frac {1}{2} e^{2 x} (2 x+5) \]

[In]

Int[-1 + E^(2*x)*(5 + 2*x) + E^(2*x + (2*(5*E^12*x^4 + Log[2]))/5)*(2 + 8*E^12*x^3),x]

[Out]

-1/2*E^(2*x) + E^(2*x + (2*(5*E^12*x^4 + Log[2]))/5) - x + (E^(2*x)*(5 + 2*x))/2

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

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 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}& = -x+\int e^{2 x} (5+2 x) \, dx+\int e^{2 x+\frac {2}{5} \left (5 e^{12} x^4+\log (2)\right )} \left (2+8 e^{12} x^3\right ) \, dx \\ & = e^{2 x+\frac {2}{5} \left (5 e^{12} x^4+\log (2)\right )}-x+\frac {1}{2} e^{2 x} (5+2 x)-\int e^{2 x} \, dx \\ & = -\frac {e^{2 x}}{2}+e^{2 x+\frac {2}{5} \left (5 e^{12} x^4+\log (2)\right )}-x+\frac {1}{2} e^{2 x} (5+2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \left (-1+e^{2 x} (5+2 x)+e^{2 x+\frac {2}{5} \left (5 e^{12} x^4+\log (2)\right )} \left (2+8 e^{12} x^3\right )\right ) \, dx=2^{2/5} e^{2 \left (x+e^{12} x^4\right )}-x+e^{2 x} (2+x) \]

[In]

Integrate[-1 + E^(2*x)*(5 + 2*x) + E^(2*x + (2*(5*E^12*x^4 + Log[2]))/5)*(2 + 8*E^12*x^3),x]

[Out]

2^(2/5)*E^(2*(x + E^12*x^4)) - x + E^(2*x)*(2 + x)

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00

method	result	size
risch	\(2^{\frac {2}{5}} {\mathrm e}^{2 x \left (x^{3} {\mathrm e}^{12}+1\right )}+{\mathrm e}^{2 x} \left (2+x \right )-x\)	\(29\)
default	\(-x +x \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{2 x}+{\mathrm e}^{2 x +\frac {2 \ln \left (2\right )}{5}+2 x^{4} {\mathrm e}^{12}}\)	\(33\)
parallelrisch	\({\mathrm e}^{2 x} {\mathrm e}^{\frac {2 \ln \left (2\right )}{5}+2 x^{4} {\mathrm e}^{12}}+x \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{2 x}-x\)	\(38\)

[In]

int((8*x^3*exp(3)^4+2)*exp(x)^2*exp(1/5*ln(2)+x^4*exp(3)^4)^2+(5+2*x)*exp(x)^2-1,x,method=_RETURNVERBOSE)

[Out]

2^(2/5)*exp(2*x*(x^3*exp(12)+1))+exp(2*x)*(2+x)-x

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \left (-1+e^{2 x} (5+2 x)+e^{2 x+\frac {2}{5} \left (5 e^{12} x^4+\log (2)\right )} \left (2+8 e^{12} x^3\right )\right ) \, dx={\left (x + 2\right )} e^{\left (2 \, x\right )} - x + e^{\left (2 \, x^{4} e^{12} + 2 \, x + \frac {2}{5} \, \log \left (2\right )\right )} \]

[In]

integrate((8*x^3*exp(3)^4+2)*exp(x)^2*exp(1/5*log(2)+x^4*exp(3)^4)^2+(5+2*x)*exp(x)^2-1,x, algorithm="fricas")

[Out]

(x + 2)*e^(2*x) - x + e^(2*x^4*e^12 + 2*x + 2/5*log(2))

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \left (-1+e^{2 x} (5+2 x)+e^{2 x+\frac {2}{5} \left (5 e^{12} x^4+\log (2)\right )} \left (2+8 e^{12} x^3\right )\right ) \, dx=- x + \left (x + 2\right ) e^{2 x} + 2^{\frac {2}{5}} e^{2 x} e^{2 x^{4} e^{12}} \]

[In]

integrate((8*x**3*exp(3)**4+2)*exp(x)**2*exp(1/5*ln(2)+x**4*exp(3)**4)**2+(5+2*x)*exp(x)**2-1,x)

[Out]

-x + (x + 2)*exp(2*x) + 2**(2/5)*exp(2*x)*exp(2*x**4*exp(12))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \left (-1+e^{2 x} (5+2 x)+e^{2 x+\frac {2}{5} \left (5 e^{12} x^4+\log (2)\right )} \left (2+8 e^{12} x^3\right )\right ) \, dx=\frac {1}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - x + e^{\left (2 \, x^{4} e^{12} + 2 \, x + \frac {2}{5} \, \log \left (2\right )\right )} + \frac {5}{2} \, e^{\left (2 \, x\right )} \]

[In]

integrate((8*x^3*exp(3)^4+2)*exp(x)^2*exp(1/5*log(2)+x^4*exp(3)^4)^2+(5+2*x)*exp(x)^2-1,x, algorithm="maxima")

[Out]

1/2*(2*x - 1)*e^(2*x) - x + e^(2*x^4*e^12 + 2*x + 2/5*log(2)) + 5/2*e^(2*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \left (-1+e^{2 x} (5+2 x)+e^{2 x+\frac {2}{5} \left (5 e^{12} x^4+\log (2)\right )} \left (2+8 e^{12} x^3\right )\right ) \, dx={\left (x + 2\right )} e^{\left (2 \, x\right )} + 2^{\frac {2}{5}} e^{\left (2 \, x^{4} e^{12} + 2 \, x\right )} - x \]

[In]

integrate((8*x^3*exp(3)^4+2)*exp(x)^2*exp(1/5*log(2)+x^4*exp(3)^4)^2+(5+2*x)*exp(x)^2-1,x, algorithm="giac")

[Out]

(x + 2)*e^(2*x) + 2^(2/5)*e^(2*x^4*e^12 + 2*x) - x

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \left (-1+e^{2 x} (5+2 x)+e^{2 x+\frac {2}{5} \left (5 e^{12} x^4+\log (2)\right )} \left (2+8 e^{12} x^3\right )\right ) \, dx=2\,{\mathrm {e}}^{2\,x}-x+x\,{\mathrm {e}}^{2\,x}+2^{2/5}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{12}\,x^4+2\,x} \]

[In]

int(exp(2*x)*(2*x + 5) + exp((2*log(2))/5 + 2*x^4*exp(12))*exp(2*x)*(8*x^3*exp(12) + 2) - 1,x)

[Out]

2*exp(2*x) - x + x*exp(2*x) + 2^(2/5)*exp(2*x + 2*x^4*exp(12))