\(\int \frac {1}{3} e^{\frac {1}{3} (e^{31/5} x^2+e^{\frac {6}{5}+x} x^2)} (2 e^{31/5} x+e^{\frac {6}{5}+x} (2 x+x^2)) \, dx\) [9215]

Optimal result

Integrand size = 55, antiderivative size = 21 \[ \int \frac {1}{3} e^{\frac {1}{3} \left (e^{31/5} x^2+e^{\frac {6}{5}+x} x^2\right )} \left (2 e^{31/5} x+e^{\frac {6}{5}+x} \left (2 x+x^2\right )\right ) \, dx=e^{\frac {1}{3} e^{6/5} \left (e^5+e^x\right ) x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {12, 6838} \[ \int \frac {1}{3} e^{\frac {1}{3} \left (e^{31/5} x^2+e^{\frac {6}{5}+x} x^2\right )} \left (2 e^{31/5} x+e^{\frac {6}{5}+x} \left (2 x+x^2\right )\right ) \, dx=e^{\frac {1}{3} \left (e^{x+\frac {6}{5}} x^2+e^{31/5} x^2\right )} \]

[In]

Int[(E^((E^(31/5)*x^2 + E^(6/5 + x)*x^2)/3)*(2*E^(31/5)*x + E^(6/5 + x)*(2*x + x^2)))/3,x]

[Out]

E^((E^(31/5)*x^2 + E^(6/5 + x)*x^2)/3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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}& = \frac {1}{3} \int e^{\frac {1}{3} \left (e^{31/5} x^2+e^{\frac {6}{5}+x} x^2\right )} \left (2 e^{31/5} x+e^{\frac {6}{5}+x} \left (2 x+x^2\right )\right ) \, dx \\ & = e^{\frac {1}{3} \left (e^{31/5} x^2+e^{\frac {6}{5}+x} x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1}{3} e^{\frac {1}{3} \left (e^{31/5} x^2+e^{\frac {6}{5}+x} x^2\right )} \left (2 e^{31/5} x+e^{\frac {6}{5}+x} \left (2 x+x^2\right )\right ) \, dx=e^{\frac {1}{3} e^{6/5} \left (e^5+e^x\right ) x^2} \]

[In]

Integrate[(E^((E^(31/5)*x^2 + E^(6/5 + x)*x^2)/3)*(2*E^(31/5)*x + E^(6/5 + x)*(2*x + x^2)))/3,x]

[Out]

E^((E^(6/5)*(E^5 + E^x)*x^2)/3)

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
risch	\({\mathrm e}^{\frac {x^{2} \left ({\mathrm e}^{\frac {6}{5}+x}+{\mathrm e}^{\frac {31}{5}}\right )}{3}}\)	\(14\)
parallelrisch	\({\mathrm e}^{\frac {\left ({\mathrm e}^{5}+{\mathrm e}^{x}\right ) {\mathrm e}^{\frac {6}{5}} x^{2}}{3}}\)	\(18\)
norman	\({\mathrm e}^{\frac {x^{2} {\mathrm e}^{\frac {6}{5}} {\mathrm e}^{x}}{3}+\frac {x^{2} {\mathrm e}^{\frac {6}{5}} {\mathrm e}^{5}}{3}}\)	\(27\)

[In]

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

[Out]

exp(1/3*x^2*(exp(6/5+x)+exp(31/5)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {1}{3} e^{\frac {1}{3} \left (e^{31/5} x^2+e^{\frac {6}{5}+x} x^2\right )} \left (2 e^{31/5} x+e^{\frac {6}{5}+x} \left (2 x+x^2\right )\right ) \, dx=e^{\left (\frac {1}{3} \, x^{2} e^{\frac {31}{5}} + \frac {1}{3} \, x^{2} e^{\left (x + \frac {6}{5}\right )}\right )} \]

[In]

integrate(1/3*((x^2+2*x)*exp(3/5)^2*exp(x)+2*x*exp(3/5)^2*exp(5/2)^2)*exp(1/3*x^2*exp(3/5)^2*exp(x)+1/3*x^2*ex
p(3/5)^2*exp(5/2)^2),x, algorithm="fricas")

[Out]

e^(1/3*x^2*e^(31/5) + 1/3*x^2*e^(x + 6/5))

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {1}{3} e^{\frac {1}{3} \left (e^{31/5} x^2+e^{\frac {6}{5}+x} x^2\right )} \left (2 e^{31/5} x+e^{\frac {6}{5}+x} \left (2 x+x^2\right )\right ) \, dx=e^{\frac {x^{2} e^{\frac {6}{5}} e^{x}}{3} + \frac {x^{2} e^{\frac {31}{5}}}{3}} \]

[In]

integrate(1/3*((x**2+2*x)*exp(3/5)**2*exp(x)+2*x*exp(3/5)**2*exp(5/2)**2)*exp(1/3*x**2*exp(3/5)**2*exp(x)+1/3*
x**2*exp(3/5)**2*exp(5/2)**2),x)

[Out]

exp(x**2*exp(6/5)*exp(x)/3 + x**2*exp(31/5)/3)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {1}{3} e^{\frac {1}{3} \left (e^{31/5} x^2+e^{\frac {6}{5}+x} x^2\right )} \left (2 e^{31/5} x+e^{\frac {6}{5}+x} \left (2 x+x^2\right )\right ) \, dx=e^{\left (\frac {1}{3} \, x^{2} e^{\frac {31}{5}} + \frac {1}{3} \, x^{2} e^{\left (x + \frac {6}{5}\right )}\right )} \]

[In]

integrate(1/3*((x^2+2*x)*exp(3/5)^2*exp(x)+2*x*exp(3/5)^2*exp(5/2)^2)*exp(1/3*x^2*exp(3/5)^2*exp(x)+1/3*x^2*ex
p(3/5)^2*exp(5/2)^2),x, algorithm="maxima")

[Out]

e^(1/3*x^2*e^(31/5) + 1/3*x^2*e^(x + 6/5))

Giac [F]

\[ \int \frac {1}{3} e^{\frac {1}{3} \left (e^{31/5} x^2+e^{\frac {6}{5}+x} x^2\right )} \left (2 e^{31/5} x+e^{\frac {6}{5}+x} \left (2 x+x^2\right )\right ) \, dx=\int { \frac {1}{3} \, {\left (2 \, x e^{\frac {31}{5}} + {\left (x^{2} + 2 \, x\right )} e^{\left (x + \frac {6}{5}\right )}\right )} e^{\left (\frac {1}{3} \, x^{2} e^{\frac {31}{5}} + \frac {1}{3} \, x^{2} e^{\left (x + \frac {6}{5}\right )}\right )} \,d x } \]

[In]

integrate(1/3*((x^2+2*x)*exp(3/5)^2*exp(x)+2*x*exp(3/5)^2*exp(5/2)^2)*exp(1/3*x^2*exp(3/5)^2*exp(x)+1/3*x^2*ex
p(3/5)^2*exp(5/2)^2),x, algorithm="giac")

[Out]

integrate(1/3*(2*x*e^(31/5) + (x^2 + 2*x)*e^(x + 6/5))*e^(1/3*x^2*e^(31/5) + 1/3*x^2*e^(x + 6/5)), x)

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {1}{3} e^{\frac {1}{3} \left (e^{31/5} x^2+e^{\frac {6}{5}+x} x^2\right )} \left (2 e^{31/5} x+e^{\frac {6}{5}+x} \left (2 x+x^2\right )\right ) \, dx={\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{x+\frac {6}{5}}}{3}+\frac {x^2\,{\mathrm {e}}^{31/5}}{3}} \]

[In]

int((exp((x^2*exp(31/5))/3 + (x^2*exp(6/5)*exp(x))/3)*(2*x*exp(31/5) + exp(6/5)*exp(x)*(2*x + x^2)))/3,x)

[Out]

exp((x^2*exp(x + 6/5))/3 + (x^2*exp(31/5))/3)