\(\int \frac {1}{2} (-1+e^{e+\frac {1}{2} e^e (10 x+5 x^2)} (5+5 x)) \, dx\) [141]

Optimal result

Integrand size = 32, antiderivative size = 23 \[ \int \frac {1}{2} \left (-1+e^{e+\frac {1}{2} e^e \left (10 x+5 x^2\right )} (5+5 x)\right ) \, dx=\frac {1}{2} \left (e^{5 e^e \left (1+\frac {x}{2}\right ) x}-x\right ) \]

[Out]

1/2*exp(5*(1+1/2*x)*x*exp(exp(1)))-1/2*x

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {12, 2276, 2268} \[ \int \frac {1}{2} \left (-1+e^{e+\frac {1}{2} e^e \left (10 x+5 x^2\right )} (5+5 x)\right ) \, dx=\frac {1}{2} e^{\frac {5 e^e x^2}{2}+5 e^e x}-\frac {x}{2} \]

[In]

Int[(-1 + E^(E + (E^E*(10*x + 5*x^2))/2)*(5 + 5*x))/2,x]

[Out]

E^(5*E^E*x + (5*E^E*x^2)/2)/2 - x/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2268

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[e*(F^(a + b*x + c*x^2)/(2
*c*Log[F])), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2276

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \left (-1+e^{e+\frac {1}{2} e^e \left (10 x+5 x^2\right )} (5+5 x)\right ) \, dx \\ & = -\frac {x}{2}+\frac {1}{2} \int e^{e+\frac {1}{2} e^e \left (10 x+5 x^2\right )} (5+5 x) \, dx \\ & = -\frac {x}{2}+\frac {1}{2} \int e^{e+5 e^e x+\frac {5 e^e x^2}{2}} (5+5 x) \, dx \\ & = \frac {1}{2} e^{5 e^e x+\frac {5 e^e x^2}{2}}-\frac {x}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {1}{2} \left (-1+e^{e+\frac {1}{2} e^e \left (10 x+5 x^2\right )} (5+5 x)\right ) \, dx=\frac {1}{2} \left (e^{\frac {5}{2} e^e x (2+x)}-x\right ) \]

[In]

Integrate[(-1 + E^(E + (E^E*(10*x + 5*x^2))/2)*(5 + 5*x))/2,x]

[Out]

(E^((5*E^E*x*(2 + x))/2) - x)/2

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {x}{2}+\frac {{\mathrm e}^{\frac {5 x \left (2+x \right ) {\mathrm e}^{{\mathrm e}}}{2}}}{2}\)	\(17\)
default	\(-\frac {x}{2}+\frac {{\mathrm e}^{\frac {\left (5 x^{2}+10 x \right ) {\mathrm e}^{{\mathrm e}}}{2}}}{2}\)	\(22\)
norman	\(-\frac {x}{2}+\frac {{\mathrm e}^{\frac {\left (5 x^{2}+10 x \right ) {\mathrm e}^{{\mathrm e}}}{2}}}{2}\)	\(22\)
parallelrisch	\(-\frac {x}{2}+\frac {{\mathrm e}^{\frac {\left (5 x^{2}+10 x \right ) {\mathrm e}^{{\mathrm e}}}{2}}}{2}\)	\(22\)
parts	\(-\frac {x}{2}+\frac {{\mathrm e}^{\frac {\left (5 x^{2}+10 x \right ) {\mathrm e}^{{\mathrm e}}}{2}}}{2}\)	\(22\)

[In]

int(1/2*(5*x+5)*exp(exp(1))*exp(1/2*(5*x^2+10*x)*exp(exp(1)))-1/2,x,method=_RETURNVERBOSE)

[Out]

-1/2*x+1/2*exp(5/2*x*(2+x)*exp(exp(1)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {1}{2} \left (-1+e^{e+\frac {1}{2} e^e \left (10 x+5 x^2\right )} (5+5 x)\right ) \, dx=-\frac {1}{2} \, {\left (x e^{e} - e^{\left (\frac {5}{2} \, {\left (x^{2} + 2 \, x\right )} e^{e} + e\right )}\right )} e^{\left (-e\right )} \]

[In]

integrate(1/2*(5*x+5)*exp(exp(1))*exp(1/2*(5*x^2+10*x)*exp(exp(1)))-1/2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {1}{2} \left (-1+e^{e+\frac {1}{2} e^e \left (10 x+5 x^2\right )} (5+5 x)\right ) \, dx=- \frac {x}{2} + \frac {e^{\left (\frac {5 x^{2}}{2} + 5 x\right ) e^{e}}}{2} \]

[In]

integrate(1/2*(5*x+5)*exp(exp(1))*exp(1/2*(5*x**2+10*x)*exp(exp(1)))-1/2,x)

[Out]

-x/2 + exp((5*x**2/2 + 5*x)*exp(E))/2

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.31 (sec) , antiderivative size = 174, normalized size of antiderivative = 7.57 \[ \int \frac {1}{2} \left (-1+e^{e+\frac {1}{2} e^e \left (10 x+5 x^2\right )} (5+5 x)\right ) \, dx=\frac {\sqrt {\frac {5}{2}} \sqrt {\pi } \operatorname {erf}\left (\sqrt {\frac {5}{2}} x \sqrt {-e^{e}} - \frac {\sqrt {\frac {5}{2}} e^{e}}{\sqrt {-e^{e}}}\right ) e^{\left (e - \frac {5}{2} \, e^{e}\right )}}{2 \, \sqrt {-e^{e}}} - \frac {1}{10} \, \sqrt {\frac {5}{2}} {\left (\frac {\sqrt {10} \sqrt {\frac {5}{2}} \sqrt {\pi } {\left (x e^{e} + e^{e}\right )} {\left (\operatorname {erf}\left (\sqrt {\frac {5}{2}} \sqrt {-{\left (x e^{e} + e^{e}\right )}^{2} e^{\left (-e\right )}}\right ) - 1\right )} e^{\left (-\frac {1}{2} \, e\right )}}{\sqrt {-{\left (x e^{e} + e^{e}\right )}^{2} e^{\left (-e\right )}}} - 2 \, \sqrt {\frac {5}{2}} e^{\left (\frac {5}{2} \, {\left (x e^{e} + e^{e}\right )}^{2} e^{\left (-e\right )} - \frac {1}{2} \, e\right )}\right )} e^{\left (\frac {1}{2} \, e - \frac {5}{2} \, e^{e}\right )} - \frac {1}{2} \, x \]

[In]

integrate(1/2*(5*x+5)*exp(exp(1))*exp(1/2*(5*x^2+10*x)*exp(exp(1)))-1/2,x, algorithm="maxima")

[Out]

1/2*sqrt(5/2)*sqrt(pi)*erf(sqrt(5/2)*x*sqrt(-e^e) - sqrt(5/2)*e^e/sqrt(-e^e))*e^(e - 5/2*e^e)/sqrt(-e^e) - 1/1
0*sqrt(5/2)*(sqrt(10)*sqrt(5/2)*sqrt(pi)*(x*e^e + e^e)*(erf(sqrt(5/2)*sqrt(-(x*e^e + e^e)^2*e^(-e))) - 1)*e^(-
1/2*e)/sqrt(-(x*e^e + e^e)^2*e^(-e)) - 2*sqrt(5/2)*e^(5/2*(x*e^e + e^e)^2*e^(-e) - 1/2*e))*e^(1/2*e - 5/2*e^e)
- 1/2*x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1}{2} \left (-1+e^{e+\frac {1}{2} e^e \left (10 x+5 x^2\right )} (5+5 x)\right ) \, dx=-\frac {1}{2} \, x + \frac {1}{2} \, e^{\left (\frac {5}{2} \, x^{2} e^{e} + 5 \, x e^{e}\right )} \]

[In]

integrate(1/2*(5*x+5)*exp(exp(1))*exp(1/2*(5*x^2+10*x)*exp(exp(1)))-1/2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1}{2} \left (-1+e^{e+\frac {1}{2} e^e \left (10 x+5 x^2\right )} (5+5 x)\right ) \, dx=\frac {{\mathrm {e}}^{\frac {5\,{\mathrm {e}}^{\mathrm {e}}\,x^2}{2}+5\,{\mathrm {e}}^{\mathrm {e}}\,x}}{2}-\frac {x}{2} \]

[In]

int((exp(exp(1))*exp((exp(exp(1))*(10*x + 5*x^2))/2)*(5*x + 5))/2 - 1/2,x)

[Out]

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