3.63.11 \(\int \frac {1}{25} (-27 e^2 x^2+\frac {1}{25} e^{4+x} (30 x+15 x^2)) \, dx\)

Optimal. Leaf size=25 \[ \frac {3}{5} x^2 \left (\frac {e^{4+x}}{25}-\frac {3 e^2 x}{5}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} \frac {3}{125} e^{x+4} x^2-\frac {9 e^2 x^3}{25} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-27*E^2*x^2 + (E^(4 + x)*(30*x + 15*x^2))/25)/25,x]

[Out]

(3*E^(4 + x)*x^2)/125 - (9*E^2*x^3)/25

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}{25} \int \left (-27 e^2 x^2+\frac {1}{25} e^{4+x} \left (30 x+15 x^2\right )\right ) \, dx\\ &=-\frac {9}{25} e^2 x^3+\frac {1}{625} \int e^{4+x} \left (30 x+15 x^2\right ) \, dx\\ &=-\frac {9}{25} e^2 x^3+\frac {1}{625} \int e^{4+x} x (30+15 x) \, dx\\ &=-\frac {9}{25} e^2 x^3+\frac {1}{625} \int \left (30 e^{4+x} x+15 e^{4+x} x^2\right ) \, dx\\ &=-\frac {9}{25} e^2 x^3+\frac {3}{125} \int e^{4+x} x^2 \, dx+\frac {6}{125} \int e^{4+x} x \, dx\\ &=\frac {6}{125} e^{4+x} x+\frac {3}{125} e^{4+x} x^2-\frac {9 e^2 x^3}{25}-\frac {6}{125} \int e^{4+x} \, dx-\frac {6}{125} \int e^{4+x} x \, dx\\ &=-\frac {6 e^{4+x}}{125}+\frac {3}{125} e^{4+x} x^2-\frac {9 e^2 x^3}{25}+\frac {6}{125} \int e^{4+x} \, dx\\ &=\frac {3}{125} e^{4+x} x^2-\frac {9 e^2 x^3}{25}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.88 \begin {gather*} \frac {3}{125} e^2 \left (e^{2+x} x^2-15 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-27*E^2*x^2 + (E^(4 + x)*(30*x + 15*x^2))/25)/25,x]

[Out]

(3*E^2*(E^(2 + x)*x^2 - 15*x^3))/125

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fricas [A]  time = 0.94, size = 21, normalized size = 0.84 \begin {gather*} -\frac {9}{25} \, x^{3} e^{2} + \frac {3}{5} \, x^{2} e^{\left (x - 2 \, \log \relax (5) + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(15*x^2+30*x)*exp(-2*log(5)+4+x)-27/25*x^2*exp(1)^2,x, algorithm="fricas")

[Out]

-9/25*x^3*e^2 + 3/5*x^2*e^(x - 2*log(5) + 4)

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giac [A]  time = 0.16, size = 21, normalized size = 0.84 \begin {gather*} -\frac {9}{25} \, x^{3} e^{2} + \frac {3}{5} \, x^{2} e^{\left (x - 2 \, \log \relax (5) + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(15*x^2+30*x)*exp(-2*log(5)+4+x)-27/25*x^2*exp(1)^2,x, algorithm="giac")

[Out]

-9/25*x^3*e^2 + 3/5*x^2*e^(x - 2*log(5) + 4)

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maple [A]  time = 0.11, size = 18, normalized size = 0.72

method	result	size
risch	\(-\frac {9 x^{3} {\mathrm e}^{2}}{25}+\frac {3 x^{2} {\mathrm e}^{4+x}}{125}\)	\(18\)
norman	\(-\frac {9 x^{3} {\mathrm e}^{2}}{25}+\frac {3 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} x^{2}}{5}\)	\(24\)
default	\(-\frac {9 x^{3} {\mathrm e}^{2}}{25}+\frac {3 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \left (-2 \ln \relax (5)+4+x \right )^{2}}{5}-\frac {24 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \left (-2 \ln \relax (5)+4+x \right )}{5}+\frac {48 \,{\mathrm e}^{-2 \ln \relax (5)+4+x}}{5}-\frac {48 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \ln \relax (5)}{5}+\frac {12 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \ln \relax (5)^{2}}{5}+\frac {12 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \left (-2 \ln \relax (5)+4+x \right ) \ln \relax (5)}{5}\)	\(102\)
derivativedivides	\(-\frac {72 \ln \relax (5)^{3} {\mathrm e}^{2}}{25}-\frac {108 \ln \relax (5)^{2} {\mathrm e}^{2} \left (-2 \ln \relax (5)+4+x \right )}{25}-\frac {54 \ln \relax (5) {\mathrm e}^{2} \left (-2 \ln \relax (5)+4+x \right )^{2}}{25}-\frac {9 \,{\mathrm e}^{2} \left (-2 \ln \relax (5)+4+x \right )^{3}}{25}+\frac {432 \ln \relax (5)^{2} {\mathrm e}^{2}}{25}+\frac {432 \ln \relax (5) {\mathrm e}^{2} \left (-2 \ln \relax (5)+4+x \right )}{25}+\frac {108 \,{\mathrm e}^{2} \left (-2 \ln \relax (5)+4+x \right )^{2}}{25}-\frac {864 \,{\mathrm e}^{2} \ln \relax (5)}{25}-\frac {432 \,{\mathrm e}^{2} \left (-2 \ln \relax (5)+4+x \right )}{25}+\frac {576 \,{\mathrm e}^{2}}{25}+\frac {3 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \left (-2 \ln \relax (5)+4+x \right )^{2}}{5}-\frac {24 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \left (-2 \ln \relax (5)+4+x \right )}{5}+\frac {48 \,{\mathrm e}^{-2 \ln \relax (5)+4+x}}{5}-\frac {48 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \ln \relax (5)}{5}+\frac {12 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \ln \relax (5)^{2}}{5}+\frac {12 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \left (-2 \ln \relax (5)+4+x \right ) \ln \relax (5)}{5}\)	\(219\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(15*x^2+30*x)*exp(-2*ln(5)+4+x)-27/25*x^2*exp(1)^2,x,method=_RETURNVERBOSE)

[Out]

-9/25*x^3*exp(2)+3/125*x^2*exp(4+x)

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maxima [A]  time = 0.36, size = 41, normalized size = 1.64 \begin {gather*} -\frac {9}{25} \, x^{3} e^{2} + \frac {3}{125} \, {\left (x^{2} e^{4} - 2 \, x e^{4} + 2 \, e^{4}\right )} e^{x} + \frac {6}{125} \, {\left (x e^{4} - e^{4}\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(15*x^2+30*x)*exp(-2*log(5)+4+x)-27/25*x^2*exp(1)^2,x, algorithm="maxima")

[Out]

-9/25*x^3*e^2 + 3/125*(x^2*e^4 - 2*x*e^4 + 2*e^4)*e^x + 6/125*(x*e^4 - e^4)*e^x

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mupad [B]  time = 0.08, size = 15, normalized size = 0.60 \begin {gather*} \frac {3\,x^2\,\left ({\mathrm {e}}^{x+4}-15\,x\,{\mathrm {e}}^2\right )}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x - 2*log(5) + 4)*(30*x + 15*x^2))/25 - (27*x^2*exp(2))/25,x)

[Out]

(3*x^2*(exp(x + 4) - 15*x*exp(2)))/125

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sympy [A]  time = 0.10, size = 20, normalized size = 0.80 \begin {gather*} - \frac {9 x^{3} e^{2}}{25} + \frac {3 x^{2} e^{x + 4}}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(15*x**2+30*x)*exp(-2*ln(5)+4+x)-27/25*x**2*exp(1)**2,x)

[Out]

-9*x**3*exp(2)/25 + 3*x**2*exp(x + 4)/125

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