3.29.48 \(\int \frac {e^{3+x} (40-30 x+10 x^2)}{3 x^3+9 x^3 \log ^2(4)} \, dx\)

Optimal. Leaf size=25 \[ \frac {5 e^{3+x} (-4+2 x)}{x^2 \left (3+9 \log ^2(4)\right )} \]

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Rubi [A]  time = 0.13, antiderivative size = 45, normalized size of antiderivative = 1.80, number of steps used = 10, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {6, 12, 2199, 2177, 2178} \begin {gather*} \frac {10 e^{x+3}}{3 x \left (1+3 \log ^2(4)\right )}-\frac {20 e^{x+3}}{3 x^2 \left (1+3 \log ^2(4)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3 + x)*(40 - 30*x + 10*x^2))/(3*x^3 + 9*x^3*Log[4]^2),x]

[Out]

(-20*E^(3 + x))/(3*x^2*(1 + 3*Log[4]^2)) + (10*E^(3 + x))/(3*x*(1 + 3*Log[4]^2))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), 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] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2199

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{3+x} \left (40-30 x+10 x^2\right )}{x^3 \left (3+9 \log ^2(4)\right )} \, dx\\ &=\frac {\int \frac {e^{3+x} \left (40-30 x+10 x^2\right )}{x^3} \, dx}{3 \left (1+3 \log ^2(4)\right )}\\ &=\frac {\int \left (\frac {40 e^{3+x}}{x^3}-\frac {30 e^{3+x}}{x^2}+\frac {10 e^{3+x}}{x}\right ) \, dx}{3 \left (1+3 \log ^2(4)\right )}\\ &=\frac {10 \int \frac {e^{3+x}}{x} \, dx}{3 \left (1+3 \log ^2(4)\right )}-\frac {10 \int \frac {e^{3+x}}{x^2} \, dx}{1+3 \log ^2(4)}+\frac {40 \int \frac {e^{3+x}}{x^3} \, dx}{3 \left (1+3 \log ^2(4)\right )}\\ &=-\frac {20 e^{3+x}}{3 x^2 \left (1+3 \log ^2(4)\right )}+\frac {10 e^{3+x}}{x \left (1+3 \log ^2(4)\right )}+\frac {10 e^3 \text {Ei}(x)}{3 \left (1+3 \log ^2(4)\right )}+\frac {20 \int \frac {e^{3+x}}{x^2} \, dx}{3 \left (1+3 \log ^2(4)\right )}-\frac {10 \int \frac {e^{3+x}}{x} \, dx}{1+3 \log ^2(4)}\\ &=-\frac {20 e^{3+x}}{3 x^2 \left (1+3 \log ^2(4)\right )}+\frac {10 e^{3+x}}{3 x \left (1+3 \log ^2(4)\right )}-\frac {20 e^3 \text {Ei}(x)}{3 \left (1+3 \log ^2(4)\right )}+\frac {20 \int \frac {e^{3+x}}{x} \, dx}{3 \left (1+3 \log ^2(4)\right )}\\ &=-\frac {20 e^{3+x}}{3 x^2 \left (1+3 \log ^2(4)\right )}+\frac {10 e^{3+x}}{3 x \left (1+3 \log ^2(4)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 25, normalized size = 1.00 \begin {gather*} \frac {10 e^{3+x} (-2+x)}{3 x^2 \left (1+3 \log ^2(4)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 + x)*(40 - 30*x + 10*x^2))/(3*x^3 + 9*x^3*Log[4]^2),x]

[Out]

(10*E^(3 + x)*(-2 + x))/(3*x^2*(1 + 3*Log[4]^2))

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fricas [A]  time = 0.49, size = 24, normalized size = 0.96 \begin {gather*} \frac {10 \, {\left (x - 2\right )} e^{\left (x + 3\right )}}{3 \, {\left (12 \, x^{2} \log \relax (2)^{2} + x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x^2-30*x+40)*exp(3)*exp(x)/(36*x^3*log(2)^2+3*x^3),x, algorithm="fricas")

[Out]

10/3*(x - 2)*e^(x + 3)/(12*x^2*log(2)^2 + x^2)

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giac [A]  time = 0.27, size = 30, normalized size = 1.20 \begin {gather*} \frac {10 \, {\left (x e^{\left (x + 3\right )} - 2 \, e^{\left (x + 3\right )}\right )}}{3 \, {\left (12 \, x^{2} \log \relax (2)^{2} + x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x^2-30*x+40)*exp(3)*exp(x)/(36*x^3*log(2)^2+3*x^3),x, algorithm="giac")

[Out]

10/3*(x*e^(x + 3) - 2*e^(x + 3))/(12*x^2*log(2)^2 + x^2)

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maple [A]  time = 0.06, size = 23, normalized size = 0.92

method	result	size
gosper	\(\frac {10 \,{\mathrm e}^{3+x} \left (x -2\right )}{3 x^{2} \left (12 \ln \relax (2)^{2}+1\right )}\)	\(23\)
risch	\(\frac {10 \,{\mathrm e}^{3+x} \left (x -2\right )}{3 x^{2} \left (12 \ln \relax (2)^{2}+1\right )}\)	\(23\)
default	\(\frac {10 \,{\mathrm e}^{3} \left (-\frac {2 \,{\mathrm e}^{x}}{\left (12 \ln \relax (2)^{2}+1\right ) x^{2}}+\frac {{\mathrm e}^{x}}{x \left (12 \ln \relax (2)^{2}+1\right )}\right )}{3}\)	\(39\)
norman	\(\frac {-\frac {20 \,{\mathrm e}^{3} {\mathrm e}^{x}}{3 \left (12 \ln \relax (2)^{2}+1\right )}+\frac {10 \,{\mathrm e}^{3} x \,{\mathrm e}^{x}}{3 \left (12 \ln \relax (2)^{2}+1\right )}}{x^{2}}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x^2-30*x+40)*exp(3)*exp(x)/(36*x^3*ln(2)^2+3*x^3),x,method=_RETURNVERBOSE)

[Out]

10/3*exp(3+x)*(x-2)/x^2/(12*ln(2)^2+1)

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maxima [C]  time = 0.64, size = 55, normalized size = 2.20 \begin {gather*} \frac {10 \, {\rm Ei}\relax (x) e^{3}}{3 \, {\left (12 \, \log \relax (2)^{2} + 1\right )}} - \frac {10 \, e^{3} \Gamma \left (-1, -x\right )}{12 \, \log \relax (2)^{2} + 1} - \frac {40 \, e^{3} \Gamma \left (-2, -x\right )}{3 \, {\left (12 \, \log \relax (2)^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x^2-30*x+40)*exp(3)*exp(x)/(36*x^3*log(2)^2+3*x^3),x, algorithm="maxima")

[Out]

10/3*Ei(x)*e^3/(12*log(2)^2 + 1) - 10*e^3*gamma(-1, -x)/(12*log(2)^2 + 1) - 40/3*e^3*gamma(-2, -x)/(12*log(2)^
2 + 1)

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mupad [B]  time = 0.09, size = 29, normalized size = 1.16 \begin {gather*} -\frac {20\,{\mathrm {e}}^{x+3}-10\,x\,{\mathrm {e}}^{x+3}}{3\,x^2\,\left (12\,{\ln \relax (2)}^2+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3)*exp(x)*(10*x^2 - 30*x + 40))/(36*x^3*log(2)^2 + 3*x^3),x)

[Out]

-(20*exp(x + 3) - 10*x*exp(x + 3))/(3*x^2*(12*log(2)^2 + 1))

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sympy [A]  time = 0.13, size = 29, normalized size = 1.16 \begin {gather*} \frac {\left (10 x e^{3} - 20 e^{3}\right ) e^{x}}{3 x^{2} + 36 x^{2} \log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x**2-30*x+40)*exp(3)*exp(x)/(36*x**3*ln(2)**2+3*x**3),x)

[Out]

(10*x*exp(3) - 20*exp(3))*exp(x)/(3*x**2 + 36*x**2*log(2)**2)

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