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

Optimal. Leaf size=17 \[ \frac {\left (e^3+e^{2 x}\right ) x^3}{\log (5)} \]

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Rubi [A]  time = 0.10, antiderivative size = 25, normalized size of antiderivative = 1.47, number of steps used = 12, 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 {e^{2 x} x^3}{\log (5)}+\frac {e^3 x^3}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 {\int \left (3 e^3 x^2+e^{2 x} \left (3 x^2+2 x^3\right )\right ) \, dx}{\log (5)}\\ &=\frac {e^3 x^3}{\log (5)}+\frac {\int e^{2 x} \left (3 x^2+2 x^3\right ) \, dx}{\log (5)}\\ &=\frac {e^3 x^3}{\log (5)}+\frac {\int e^{2 x} x^2 (3+2 x) \, dx}{\log (5)}\\ &=\frac {e^3 x^3}{\log (5)}+\frac {\int \left (3 e^{2 x} x^2+2 e^{2 x} x^3\right ) \, dx}{\log (5)}\\ &=\frac {e^3 x^3}{\log (5)}+\frac {2 \int e^{2 x} x^3 \, dx}{\log (5)}+\frac {3 \int e^{2 x} x^2 \, dx}{\log (5)}\\ &=\frac {3 e^{2 x} x^2}{2 \log (5)}+\frac {e^3 x^3}{\log (5)}+\frac {e^{2 x} x^3}{\log (5)}-\frac {3 \int e^{2 x} x \, dx}{\log (5)}-\frac {3 \int e^{2 x} x^2 \, dx}{\log (5)}\\ &=-\frac {3 e^{2 x} x}{2 \log (5)}+\frac {e^3 x^3}{\log (5)}+\frac {e^{2 x} x^3}{\log (5)}+\frac {3 \int e^{2 x} \, dx}{2 \log (5)}+\frac {3 \int e^{2 x} x \, dx}{\log (5)}\\ &=\frac {3 e^{2 x}}{4 \log (5)}+\frac {e^3 x^3}{\log (5)}+\frac {e^{2 x} x^3}{\log (5)}-\frac {3 \int e^{2 x} \, dx}{2 \log (5)}\\ &=\frac {e^3 x^3}{\log (5)}+\frac {e^{2 x} x^3}{\log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 1.00 \begin {gather*} \frac {\left (e^3+e^{2 x}\right ) x^3}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((E^3 + E^(2*x))*x^3)/Log[5]

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fricas [A]  time = 0.67, size = 20, normalized size = 1.18 \begin {gather*} \frac {x^{3} e^{3} + x^{3} e^{\left (2 \, x\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^3*e^3 + x^3*e^(2*x))/log(5)

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giac [A]  time = 0.17, size = 20, normalized size = 1.18 \begin {gather*} \frac {x^{3} e^{3} + x^{3} e^{\left (2 \, x\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^3*e^3 + x^3*e^(2*x))/log(5)

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maple [A]  time = 0.04, size = 21, normalized size = 1.24

method	result	size
default	\(\frac {{\mathrm e}^{2 x} x^{3}+x^{3} {\mathrm e}^{3}}{\ln \relax (5)}\)	\(21\)
derivativedivides	\(\frac {8 x^{3} {\mathrm e}^{3}+8 \,{\mathrm e}^{2 x} x^{3}}{8 \ln \relax (5)}\)	\(24\)
norman	\(\frac {x^{3} {\mathrm e}^{2 x}}{\ln \relax (5)}+\frac {{\mathrm e}^{3} x^{3}}{\ln \relax (5)}\)	\(24\)
risch	\(\frac {x^{3} {\mathrm e}^{2 x}}{\ln \relax (5)}+\frac {{\mathrm e}^{3} x^{3}}{\ln \relax (5)}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.35, size = 50, normalized size = 2.94 \begin {gather*} \frac {4 \, x^{3} e^{3} + {\left (4 \, x^{3} - 6 \, x^{2} + 6 \, x - 3\right )} e^{\left (2 \, x\right )} + 3 \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )}}{4 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*x^3*e^3 + (4*x^3 - 6*x^2 + 6*x - 3)*e^(2*x) + 3*(2*x^2 - 2*x + 1)*e^(2*x))/log(5)

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mupad [B]  time = 3.08, size = 15, normalized size = 0.88 \begin {gather*} \frac {x^3\,\left ({\mathrm {e}}^{2\,x}+{\mathrm {e}}^3\right )}{\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^3*(exp(2*x) + exp(3)))/log(5)

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sympy [A]  time = 0.11, size = 20, normalized size = 1.18 \begin {gather*} \frac {x^{3} e^{2 x}}{\log {\relax (5 )}} + \frac {x^{3} e^{3}}{\log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*exp(2*x)/log(5) + x**3*exp(3)/log(5)

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