3.84.52 \(\int \frac {12 e^5-9 x^2+e^{-3+2 x} (-12 x^2-24 x^3)}{4 x^2} \, dx\)

Optimal. Leaf size=28 \[ 2+3 \left (-\frac {e^5}{x}-\frac {3 x}{4}-e^{-3+2 x} x\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 41, normalized size of antiderivative = 1.46, number of steps used = 7, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {12, 14, 2176, 2194} \begin {gather*} -\frac {9 x}{4}+\frac {3}{2} e^{2 x-3}-\frac {3}{2} e^{2 x-3} (2 x+1)-\frac {3 e^5}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12*E^5 - 9*x^2 + E^(-3 + 2*x)*(-12*x^2 - 24*x^3))/(4*x^2),x]

[Out]

(3*E^(-3 + 2*x))/2 - (3*E^5)/x - (9*x)/4 - (3*E^(-3 + 2*x)*(1 + 2*x))/2

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {12 e^5-9 x^2+e^{-3+2 x} \left (-12 x^2-24 x^3\right )}{x^2} \, dx\\ &=\frac {1}{4} \int \left (-12 e^{-3+2 x} (1+2 x)+\frac {3 \left (4 e^5-3 x^2\right )}{x^2}\right ) \, dx\\ &=\frac {3}{4} \int \frac {4 e^5-3 x^2}{x^2} \, dx-3 \int e^{-3+2 x} (1+2 x) \, dx\\ &=-\frac {3}{2} e^{-3+2 x} (1+2 x)+\frac {3}{4} \int \left (-3+\frac {4 e^5}{x^2}\right ) \, dx+3 \int e^{-3+2 x} \, dx\\ &=\frac {3}{2} e^{-3+2 x}-\frac {3 e^5}{x}-\frac {9 x}{4}-\frac {3}{2} e^{-3+2 x} (1+2 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 0.86 \begin {gather*} -\frac {3 e^5}{x}-\frac {9 x}{4}-3 e^{-3+2 x} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12*E^5 - 9*x^2 + E^(-3 + 2*x)*(-12*x^2 - 24*x^3))/(4*x^2),x]

[Out]

(-3*E^5)/x - (9*x)/4 - 3*E^(-3 + 2*x)*x

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fricas [A]  time = 0.75, size = 26, normalized size = 0.93 \begin {gather*} -\frac {3 \, {\left (4 \, x^{2} e^{\left (2 \, x - 3\right )} + 3 \, x^{2} + 4 \, e^{5}\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-24*x^3-12*x^2)*exp(2*x-3)+12*exp(5)-9*x^2)/x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 28, normalized size = 1.00 \begin {gather*} -\frac {3 \, {\left (3 \, x^{2} e^{3} + 4 \, x^{2} e^{\left (2 \, x\right )} + 4 \, e^{8}\right )} e^{\left (-3\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-24*x^3-12*x^2)*exp(2*x-3)+12*exp(5)-9*x^2)/x^2,x, algorithm="giac")

[Out]

-3/4*(3*x^2*e^3 + 4*x^2*e^(2*x) + 4*e^8)*e^(-3)/x

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maple [A]  time = 0.09, size = 21, normalized size = 0.75

method	result	size
risch	\(-\frac {9 x}{4}-\frac {3 \,{\mathrm e}^{5}}{x}-3 \,{\mathrm e}^{2 x -3} x\)	\(21\)
norman	\(\frac {-\frac {9 x^{2}}{4}-3 \,{\mathrm e}^{2 x -3} x^{2}-3 \,{\mathrm e}^{5}}{x}\)	\(26\)
derivativedivides	\(-\frac {9 x}{4}+\frac {27}{8}-\frac {3 \,{\mathrm e}^{5}}{x}-\frac {9 \,{\mathrm e}^{2 x -3}}{2}-\frac {3 \,{\mathrm e}^{2 x -3} \left (2 x -3\right )}{2}\)	\(34\)
default	\(-\frac {9 x}{4}+\frac {27}{8}-\frac {3 \,{\mathrm e}^{5}}{x}-\frac {9 \,{\mathrm e}^{2 x -3}}{2}-\frac {3 \,{\mathrm e}^{2 x -3} \left (2 x -3\right )}{2}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-24*x^3-12*x^2)*exp(2*x-3)+12*exp(5)-9*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

-9/4*x-3*exp(5)/x-3*exp(2*x-3)*x

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maxima [A]  time = 0.36, size = 32, normalized size = 1.14 \begin {gather*} -\frac {3}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x - 3\right )} - \frac {9}{4} \, x - \frac {3 \, e^{5}}{x} - \frac {3}{2} \, e^{\left (2 \, x - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-24*x^3-12*x^2)*exp(2*x-3)+12*exp(5)-9*x^2)/x^2,x, algorithm="maxima")

[Out]

-3/2*(2*x - 1)*e^(2*x - 3) - 9/4*x - 3*e^5/x - 3/2*e^(2*x - 3)

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mupad [B]  time = 0.09, size = 24, normalized size = 0.86 \begin {gather*} -\frac {3\,{\mathrm {e}}^5}{x}-\frac {3\,x\,{\mathrm {e}}^{-3}\,\left (4\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^3\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(2*x - 3)*(12*x^2 + 24*x^3))/4 - 3*exp(5) + (9*x^2)/4)/x^2,x)

[Out]

- (3*exp(5))/x - (3*x*exp(-3)*(4*exp(2*x) + 3*exp(3)))/4

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sympy [A]  time = 0.13, size = 22, normalized size = 0.79 \begin {gather*} - 3 x e^{2 x - 3} - \frac {9 x}{4} - \frac {3 e^{5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-24*x**3-12*x**2)*exp(2*x-3)+12*exp(5)-9*x**2)/x**2,x)

[Out]

-3*x*exp(2*x - 3) - 9*x/4 - 3*exp(5)/x

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