3.83.58 \(\int \frac {e^{14+2 x} (-3+6 x)}{2 e^4 x^2+5 x^2 \log (3)} \, dx\)

Optimal. Leaf size=25 \[ 3+\frac {3 e^{10+2 x}}{x \left (2+\frac {5 \log (3)}{e^4}\right )} \]

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Rubi [A]  time = 0.04, antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {6, 12, 2197} \begin {gather*} \frac {3 e^{2 x+14}}{x \left (2 e^4+\log (243)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(14 + 2*x)*(-3 + 6*x))/(2*E^4*x^2 + 5*x^2*Log[3]),x]

[Out]

(3*E^(14 + 2*x))/(x*(2*E^4 + Log[243]))

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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{14+2 x} (-3+6 x)}{x^2 \left (2 e^4+5 \log (3)\right )} \, dx\\ &=\frac {\int \frac {e^{14+2 x} (-3+6 x)}{x^2} \, dx}{2 e^4+\log (243)}\\ &=\frac {3 e^{14+2 x}}{x \left (2 e^4+\log (243)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.88 \begin {gather*} \frac {3 e^{2 (7+x)}}{x \left (2 e^4+\log (243)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(14 + 2*x)*(-3 + 6*x))/(2*E^4*x^2 + 5*x^2*Log[3]),x]

[Out]

(3*E^(2*(7 + x)))/(x*(2*E^4 + Log[243]))

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fricas [A]  time = 0.70, size = 21, normalized size = 0.84 \begin {gather*} \frac {3 \, e^{\left (2 \, x + 14\right )}}{2 \, x e^{4} + 5 \, x \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x-3)*exp(1)^4*exp(5+x)^2/(5*x^2*log(3)+2*x^2*exp(1)^4),x, algorithm="fricas")

[Out]

3*e^(2*x + 14)/(2*x*e^4 + 5*x*log(3))

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giac [A]  time = 1.00, size = 21, normalized size = 0.84 \begin {gather*} \frac {3 \, e^{\left (2 \, x + 14\right )}}{2 \, x e^{4} + 5 \, x \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x-3)*exp(1)^4*exp(5+x)^2/(5*x^2*log(3)+2*x^2*exp(1)^4),x, algorithm="giac")

[Out]

3*e^(2*x + 14)/(2*x*e^4 + 5*x*log(3))

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maple [A]  time = 0.37, size = 23, normalized size = 0.92

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{14+2 x}}{\left (2 \,{\mathrm e}^{4}+5 \ln \relax (3)\right ) x}\)	\(23\)
derivativedivides	\(\frac {3 \,{\mathrm e}^{2 x +10} {\mathrm e}^{4}}{x \left (2 \,{\mathrm e}^{4}+5 \ln \relax (3)\right )}\)	\(27\)
default	\(\frac {3 \,{\mathrm e}^{2 x +10} {\mathrm e}^{4}}{x \left (2 \,{\mathrm e}^{4}+5 \ln \relax (3)\right )}\)	\(27\)
gosper	\(\frac {3 \,{\mathrm e}^{2 x +10} {\mathrm e}^{4}}{x \left (2 \,{\mathrm e}^{4}+5 \ln \relax (3)\right )}\)	\(29\)
norman	\(\frac {3 \,{\mathrm e}^{2 x +10} {\mathrm e}^{4}}{x \left (2 \,{\mathrm e}^{4}+5 \ln \relax (3)\right )}\)	\(29\)
meijerg	\(\frac {6 \,{\mathrm e}^{-2 x \,{\mathrm e}^{10}+14+2 x} \left (\ln \relax (x )+\ln \relax (2)+10+i \pi -\ln \left (-2 x \,{\mathrm e}^{10}\right )-\expIntegralEi \left (1, -2 x \,{\mathrm e}^{10}\right )\right )}{2 \,{\mathrm e}^{4}+5 \ln \relax (3)}+\frac {6 \,{\mathrm e}^{24-2 x \,{\mathrm e}^{10}+2 x} \left (\frac {{\mathrm e}^{-10}}{2 x}-9-\ln \relax (x )-\ln \relax (2)-i \pi -\frac {{\mathrm e}^{-10} \left (4 x \,{\mathrm e}^{10}+2\right )}{4 x}+\frac {{\mathrm e}^{-10+2 x \,{\mathrm e}^{10}}}{2 x}+\ln \left (-2 x \,{\mathrm e}^{10}\right )+\expIntegralEi \left (1, -2 x \,{\mathrm e}^{10}\right )\right )}{2 \,{\mathrm e}^{4}+5 \ln \relax (3)}\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x-3)*exp(1)^4*exp(5+x)^2/(5*x^2*ln(3)+2*x^2*exp(1)^4),x,method=_RETURNVERBOSE)

[Out]

3/(2*exp(4)+5*ln(3))/x*exp(14+2*x)

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maxima [C]  time = 0.91, size = 40, normalized size = 1.60 \begin {gather*} \frac {6 \, {\rm Ei}\left (2 \, x\right ) e^{14}}{2 \, e^{4} + 5 \, \log \relax (3)} - \frac {6 \, e^{14} \Gamma \left (-1, -2 \, x\right )}{2 \, e^{4} + 5 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x-3)*exp(1)^4*exp(5+x)^2/(5*x^2*log(3)+2*x^2*exp(1)^4),x, algorithm="maxima")

[Out]

6*Ei(2*x)*e^14/(2*e^4 + 5*log(3)) - 6*e^14*gamma(-1, -2*x)/(2*e^4 + 5*log(3))

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mupad [B]  time = 0.09, size = 20, normalized size = 0.80 \begin {gather*} \frac {3\,{\mathrm {e}}^{2\,x+14}}{x\,\left (2\,{\mathrm {e}}^4+\ln \left (243\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4)*exp(2*x + 10)*(6*x - 3))/(2*x^2*exp(4) + 5*x^2*log(3)),x)

[Out]

(3*exp(2*x + 14))/(x*(2*exp(4) + log(243)))

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sympy [A]  time = 0.13, size = 24, normalized size = 0.96 \begin {gather*} \frac {3 e^{4} e^{2 x + 10}}{5 x \log {\relax (3 )} + 2 x e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x-3)*exp(1)**4*exp(5+x)**2/(5*x**2*ln(3)+2*x**2*exp(1)**4),x)

[Out]

3*exp(4)*exp(2*x + 10)/(5*x*log(3) + 2*x*exp(4))

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