\(\int \frac {12 e^5-9 x^2+e^{-3+2 x} (-12 x^2-24 x^3)}{4 x^2} \, dx\) [8352]

Optimal result

Integrand size = 37, antiderivative size = 28 \[ \int \frac {12 e^5-9 x^2+e^{-3+2 x} \left (-12 x^2-24 x^3\right )}{4 x^2} \, dx=2+3 \left (-\frac {e^5}{x}-\frac {3 x}{4}-e^{-3+2 x} x\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {12, 14, 2207, 2225} \[ \int \frac {12 e^5-9 x^2+e^{-3+2 x} \left (-12 x^2-24 x^3\right )}{4 x^2} \, dx=-\frac {9 x}{4}+\frac {3}{2} e^{2 x-3}-\frac {3}{2} e^{2 x-3} (2 x+1)-\frac {3 e^5}{x} \]

[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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {12 e^5-9 x^2+e^{-3+2 x} \left (-12 x^2-24 x^3\right )}{4 x^2} \, dx=-\frac {3 e^5}{x}-\frac {9 x}{4}-3 e^{-3+2 x} x \]

[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

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {9 x}{4}-\frac {3 \,{\mathrm e}^{5}}{x}-3 \,{\mathrm e}^{-3+2 x} x\)	\(21\)
norman	\(\frac {-\frac {9 x^{2}}{4}-3 \,{\mathrm e}^{-3+2 x} x^{2}-3 \,{\mathrm e}^{5}}{x}\)	\(26\)
parallelrisch	\(-\frac {12 \,{\mathrm e}^{-3+2 x} x^{2}+9 x^{2}+12 \,{\mathrm e}^{5}}{4 x}\)	\(27\)
parts	\(-\frac {3 \,{\mathrm e}^{-3+2 x} \left (-3+2 x \right )}{2}-\frac {9 \,{\mathrm e}^{-3+2 x}}{2}-\frac {9 x}{4}-\frac {3 \,{\mathrm e}^{5}}{x}\)	\(33\)
derivativedivides	\(\frac {27}{8}-\frac {9 x}{4}-\frac {3 \,{\mathrm e}^{5}}{x}-\frac {9 \,{\mathrm e}^{-3+2 x}}{2}-\frac {3 \,{\mathrm e}^{-3+2 x} \left (-3+2 x \right )}{2}\)	\(34\)
default	\(\frac {27}{8}-\frac {9 x}{4}-\frac {3 \,{\mathrm e}^{5}}{x}-\frac {9 \,{\mathrm e}^{-3+2 x}}{2}-\frac {3 \,{\mathrm e}^{-3+2 x} \left (-3+2 x \right )}{2}\)	\(34\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {12 e^5-9 x^2+e^{-3+2 x} \left (-12 x^2-24 x^3\right )}{4 x^2} \, dx=-\frac {3 \, {\left (4 \, x^{2} e^{\left (2 \, x - 3\right )} + 3 \, x^{2} + 4 \, e^{5}\right )}}{4 \, x} \]

[In]

integrate(1/4*((-24*x^3-12*x^2)*exp(-3+2*x)+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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {12 e^5-9 x^2+e^{-3+2 x} \left (-12 x^2-24 x^3\right )}{4 x^2} \, dx=- 3 x e^{2 x - 3} - \frac {9 x}{4} - \frac {3 e^{5}}{x} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {12 e^5-9 x^2+e^{-3+2 x} \left (-12 x^2-24 x^3\right )}{4 x^2} \, dx=-\frac {3}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x - 3\right )} - \frac {9}{4} \, x - \frac {3 \, e^{5}}{x} - \frac {3}{2} \, e^{\left (2 \, x - 3\right )} \]

[In]

integrate(1/4*((-24*x^3-12*x^2)*exp(-3+2*x)+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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {12 e^5-9 x^2+e^{-3+2 x} \left (-12 x^2-24 x^3\right )}{4 x^2} \, dx=-\frac {3 \, {\left (3 \, x^{2} e^{3} + 4 \, x^{2} e^{\left (2 \, x\right )} + 4 \, e^{8}\right )} e^{\left (-3\right )}}{4 \, x} \]

[In]

integrate(1/4*((-24*x^3-12*x^2)*exp(-3+2*x)+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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {12 e^5-9 x^2+e^{-3+2 x} \left (-12 x^2-24 x^3\right )}{4 x^2} \, dx=-\frac {3\,{\mathrm {e}}^5}{x}-\frac {3\,x\,{\mathrm {e}}^{-3}\,\left (4\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^3\right )}{4} \]

[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