3.34.31 \(\int \frac {1}{5} e^{\frac {2 e^4 x}{5}} (10 x+2 e^4 x^2) \, dx\)

Optimal. Leaf size=14 \[ e^{\frac {2 e^4 x}{5}} x^2 \]

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Rubi [A]  time = 0.09, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} e^{\frac {2 e^4 x}{5}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((2*E^4*x)/5)*(10*x + 2*E^4*x^2))/5,x]

[Out]

E^((2*E^4*x)/5)*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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2176

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] &&  !$UseGamma === True

Rule 2194

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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int e^{\frac {2 e^4 x}{5}} \left (10 x+2 e^4 x^2\right ) \, dx\\ &=\frac {1}{5} \int e^{\frac {2 e^4 x}{5}} x \left (10+2 e^4 x\right ) \, dx\\ &=\frac {1}{5} \int \left (10 e^{\frac {2 e^4 x}{5}} x+2 e^{4+\frac {2 e^4 x}{5}} x^2\right ) \, dx\\ &=\frac {2}{5} \int e^{4+\frac {2 e^4 x}{5}} x^2 \, dx+2 \int e^{\frac {2 e^4 x}{5}} x \, dx\\ &=5 e^{-4+\frac {2 e^4 x}{5}} x+e^{\frac {2 e^4 x}{5}} x^2-\frac {2 \int e^{4+\frac {2 e^4 x}{5}} x \, dx}{e^4}-\frac {5 \int e^{\frac {2 e^4 x}{5}} \, dx}{e^4}\\ &=-\frac {25}{2} e^{-8+\frac {2 e^4 x}{5}}+e^{\frac {2 e^4 x}{5}} x^2+\frac {5 \int e^{4+\frac {2 e^4 x}{5}} \, dx}{e^8}\\ &=e^{\frac {2 e^4 x}{5}} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 1.00 \begin {gather*} e^{\frac {2 e^4 x}{5}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*E^4*x)/5)*(10*x + 2*E^4*x^2))/5,x]

[Out]

E^((2*E^4*x)/5)*x^2

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fricas [A]  time = 0.53, size = 10, normalized size = 0.71 \begin {gather*} x^{2} e^{\left (\frac {2}{5} \, x e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*x^2*exp(1)^4+10*x)*exp(1/5*x*exp(1)^4)^2,x, algorithm="fricas")

[Out]

x^2*e^(2/5*x*e^4)

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giac [B]  time = 0.17, size = 42, normalized size = 3.00 \begin {gather*} \frac {1}{2} \, {\left (2 \, x^{2} e^{8} - 10 \, x e^{4} + 25\right )} e^{\left (\frac {2}{5} \, x e^{4} - 8\right )} + \frac {5}{2} \, {\left (2 \, x e^{4} - 5\right )} e^{\left (\frac {2}{5} \, x e^{4} - 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*x^2*exp(1)^4+10*x)*exp(1/5*x*exp(1)^4)^2,x, algorithm="giac")

[Out]

1/2*(2*x^2*e^8 - 10*x*e^4 + 25)*e^(2/5*x*e^4 - 8) + 5/2*(2*x*e^4 - 5)*e^(2/5*x*e^4 - 8)

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maple [A]  time = 0.04, size = 11, normalized size = 0.79

method	result	size
risch	\(x^{2} {\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}}\)	\(11\)
gosper	\(x^{2} {\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}}\)	\(15\)
norman	\(x^{2} {\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}}\)	\(15\)
meijerg	\(-\frac {25 \,{\mathrm e}^{-8} \left (2-\frac {\left (\frac {12 x^{2} {\mathrm e}^{8}}{25}-\frac {12 x \,{\mathrm e}^{4}}{5}+6\right ) {\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}}}{3}\right )}{4}+\frac {25 \,{\mathrm e}^{-8} \left (1-\frac {\left (2-\frac {4 x \,{\mathrm e}^{4}}{5}\right ) {\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}}}{2}\right )}{2}\)	\(51\)
derivativedivides	\(50 \,{\mathrm e}^{-4} \left ({\mathrm e}^{-4} \left (\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}} x \,{\mathrm e}^{4}}{10}-\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}}}{4}\right )+{\mathrm e}^{-4} \left (\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}} x^{2} {\mathrm e}^{8}}{50}-\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}} x \,{\mathrm e}^{4}}{10}+\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}}}{4}\right )\right )\)	\(97\)
default	\(50 \,{\mathrm e}^{-4} \left ({\mathrm e}^{-4} \left (\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}} x \,{\mathrm e}^{4}}{10}-\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}}}{4}\right )+{\mathrm e}^{-4} \left (\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}} x^{2} {\mathrm e}^{8}}{50}-\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}} x \,{\mathrm e}^{4}}{10}+\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}}}{4}\right )\right )\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(2*x^2*exp(1)^4+10*x)*exp(1/5*x*exp(1)^4)^2,x,method=_RETURNVERBOSE)

[Out]

x^2*exp(2/5*x*exp(4))

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maxima [B]  time = 0.40, size = 42, normalized size = 3.00 \begin {gather*} \frac {1}{2} \, {\left (2 \, x^{2} e^{8} - 10 \, x e^{4} + 25\right )} e^{\left (\frac {2}{5} \, x e^{4} - 8\right )} + \frac {5}{2} \, {\left (2 \, x e^{4} - 5\right )} e^{\left (\frac {2}{5} \, x e^{4} - 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*x^2*exp(1)^4+10*x)*exp(1/5*x*exp(1)^4)^2,x, algorithm="maxima")

[Out]

1/2*(2*x^2*e^8 - 10*x*e^4 + 25)*e^(2/5*x*e^4 - 8) + 5/2*(2*x*e^4 - 5)*e^(2/5*x*e^4 - 8)

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mupad [B]  time = 1.91, size = 10, normalized size = 0.71 \begin {gather*} x^2\,{\mathrm {e}}^{\frac {2\,x\,{\mathrm {e}}^4}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((2*x*exp(4))/5)*(10*x + 2*x^2*exp(4)))/5,x)

[Out]

x^2*exp((2*x*exp(4))/5)

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sympy [A]  time = 0.12, size = 12, normalized size = 0.86 \begin {gather*} x^{2} e^{\frac {2 x e^{4}}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*x**2*exp(1)**4+10*x)*exp(1/5*x*exp(1)**4)**2,x)

[Out]

x**2*exp(2*x*exp(4)/5)

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