3.25.35 \(\int \frac {1}{3} e^{2 x} (x+x^2) \log (5) \, dx\)

Optimal. Leaf size=14 \[ \frac {1}{6} e^{2 x} x^2 \log (5) \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} \frac {1}{6} e^{2 x} x^2 \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*x)*(x + x^2)*Log[5])/3,x]

[Out]

(E^(2*x)*x^2*Log[5])/6

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}{3} \log (5) \int e^{2 x} \left (x+x^2\right ) \, dx\\ &=\frac {1}{3} \log (5) \int e^{2 x} x (1+x) \, dx\\ &=\frac {1}{3} \log (5) \int \left (e^{2 x} x+e^{2 x} x^2\right ) \, dx\\ &=\frac {1}{3} \log (5) \int e^{2 x} x \, dx+\frac {1}{3} \log (5) \int e^{2 x} x^2 \, dx\\ &=\frac {1}{6} e^{2 x} x \log (5)+\frac {1}{6} e^{2 x} x^2 \log (5)-\frac {1}{6} \log (5) \int e^{2 x} \, dx-\frac {1}{3} \log (5) \int e^{2 x} x \, dx\\ &=-\frac {1}{12} e^{2 x} \log (5)+\frac {1}{6} e^{2 x} x^2 \log (5)+\frac {1}{6} \log (5) \int e^{2 x} \, dx\\ &=\frac {1}{6} e^{2 x} x^2 \log (5)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} \frac {1}{6} e^{2 x} x^2 \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(2*x)*x^2*Log[5])/6

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 11, normalized size = 0.79 \begin {gather*} \frac {1}{6} \, x^{2} e^{\left (2 \, x\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^2*e^(2*x)*log(5)

________________________________________________________________________________________

giac [A]  time = 0.25, size = 11, normalized size = 0.79 \begin {gather*} \frac {1}{6} \, x^{2} e^{\left (2 \, x\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^2*e^(2*x)*log(5)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 12, normalized size = 0.86

method	result	size
gosper	\(\frac {x^{2} \ln \relax (5) {\mathrm e}^{2 x}}{6}\)	\(12\)
default	\(\frac {x^{2} \ln \relax (5) {\mathrm e}^{2 x}}{6}\)	\(12\)
norman	\(\frac {x^{2} \ln \relax (5) {\mathrm e}^{2 x}}{6}\)	\(12\)
risch	\(\frac {x^{2} \ln \relax (5) {\mathrm e}^{2 x}}{6}\)	\(12\)
meijerg	\(-\frac {\ln \relax (5) \left (2-\frac {\left (12 x^{2}-12 x +6\right ) {\mathrm e}^{2 x}}{3}\right )}{24}+\frac {\ln \relax (5) \left (1-\frac {\left (-4 x +2\right ) {\mathrm e}^{2 x}}{2}\right )}{12}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^2*ln(5)*exp(x)^2

________________________________________________________________________________________

maxima [B]  time = 0.52, size = 30, normalized size = 2.14 \begin {gather*} \frac {1}{12} \, {\left ({\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )}\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.36, size = 11, normalized size = 0.79 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{2\,x}\,\ln \relax (5)}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*log(5)*(x + x^2))/3,x)

[Out]

(x^2*exp(2*x)*log(5))/6

________________________________________________________________________________________

sympy [A]  time = 0.10, size = 12, normalized size = 0.86 \begin {gather*} \frac {x^{2} e^{2 x} \log {\relax (5 )}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*exp(2*x)*log(5)/6

________________________________________________________________________________________