\(\int \frac {4 x^2+e^{\frac {-4 e^4-8 x+5 x^2}{4 x}} (-12 e^4-15 x^2)}{12 x^2} \, dx\) [3208]

Optimal result

Integrand size = 48, antiderivative size = 25 \[ \int \frac {4 x^2+e^{\frac {-4 e^4-8 x+5 x^2}{4 x}} \left (-12 e^4-15 x^2\right )}{12 x^2} \, dx=-e^{-2-\frac {e^4}{x}+\frac {5 x}{4}}+\frac {x}{3} \]

[Out]

1/3*x-exp(5/4*x-exp(4)/x-2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 14, 6838} \[ \int \frac {4 x^2+e^{\frac {-4 e^4-8 x+5 x^2}{4 x}} \left (-12 e^4-15 x^2\right )}{12 x^2} \, dx=\frac {x}{3}-e^{\frac {5 x}{4}-\frac {e^4}{x}-2} \]

[In]

Int[(4*x^2 + E^((-4*E^4 - 8*x + 5*x^2)/(4*x))*(-12*E^4 - 15*x^2))/(12*x^2),x]

[Out]

-E^(-2 - E^4/x + (5*x)/4) + x/3

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 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{12} \int \frac {4 x^2+e^{\frac {-4 e^4-8 x+5 x^2}{4 x}} \left (-12 e^4-15 x^2\right )}{x^2} \, dx \\ & = \frac {1}{12} \int \left (4-\frac {3 e^{-2-\frac {e^4}{x}+\frac {5 x}{4}} \left (4 e^4+5 x^2\right )}{x^2}\right ) \, dx \\ & = \frac {x}{3}-\frac {1}{4} \int \frac {e^{-2-\frac {e^4}{x}+\frac {5 x}{4}} \left (4 e^4+5 x^2\right )}{x^2} \, dx \\ & = -e^{-2-\frac {e^4}{x}+\frac {5 x}{4}}+\frac {x}{3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^2+e^{\frac {-4 e^4-8 x+5 x^2}{4 x}} \left (-12 e^4-15 x^2\right )}{12 x^2} \, dx=-e^{-2-\frac {e^4}{x}+\frac {5 x}{4}}+\frac {x}{3} \]

[In]

Integrate[(4*x^2 + E^((-4*E^4 - 8*x + 5*x^2)/(4*x))*(-12*E^4 - 15*x^2))/(12*x^2),x]

[Out]

-E^(-2 - E^4/x + (5*x)/4) + x/3

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04

method	result	size
risch	\(\frac {x}{3}-{\mathrm e}^{-\frac {-5 x^{2}+4 \,{\mathrm e}^{4}+8 x}{4 x}}\)	\(26\)
parallelrisch	\(\frac {x}{3}-{\mathrm e}^{-\frac {-5 x^{2}+4 \,{\mathrm e}^{4}+8 x}{4 x}}\)	\(26\)
parts	\(\frac {x}{3}-{\mathrm e}^{\frac {-4 \,{\mathrm e}^{4}+5 x^{2}-8 x}{4 x}}\)	\(26\)
norman	\(\frac {\frac {x^{2}}{3}-x \,{\mathrm e}^{\frac {-4 \,{\mathrm e}^{4}+5 x^{2}-8 x}{4 x}}}{x}\)	\(33\)

[In]

int(1/12*((-12*exp(4)-15*x^2)*exp(1/4*(-4*exp(4)+5*x^2-8*x)/x)+4*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/3*x-exp(-1/4*(-5*x^2+4*exp(4)+8*x)/x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^2+e^{\frac {-4 e^4-8 x+5 x^2}{4 x}} \left (-12 e^4-15 x^2\right )}{12 x^2} \, dx=\frac {1}{3} \, x - e^{\left (\frac {5 \, x^{2} - 8 \, x - 4 \, e^{4}}{4 \, x}\right )} \]

[In]

integrate(1/12*((-12*exp(4)-15*x^2)*exp(1/4*(-4*exp(4)+5*x^2-8*x)/x)+4*x^2)/x^2,x, algorithm="fricas")

[Out]

1/3*x - e^(1/4*(5*x^2 - 8*x - 4*e^4)/x)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {4 x^2+e^{\frac {-4 e^4-8 x+5 x^2}{4 x}} \left (-12 e^4-15 x^2\right )}{12 x^2} \, dx=\frac {x}{3} - e^{\frac {\frac {5 x^{2}}{4} - 2 x - e^{4}}{x}} \]

[In]

integrate(1/12*((-12*exp(4)-15*x**2)*exp(1/4*(-4*exp(4)+5*x**2-8*x)/x)+4*x**2)/x**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {4 x^2+e^{\frac {-4 e^4-8 x+5 x^2}{4 x}} \left (-12 e^4-15 x^2\right )}{12 x^2} \, dx=\frac {1}{3} \, x - e^{\left (\frac {5}{4} \, x - \frac {e^{4}}{x} - 2\right )} \]

[In]

integrate(1/12*((-12*exp(4)-15*x^2)*exp(1/4*(-4*exp(4)+5*x^2-8*x)/x)+4*x^2)/x^2,x, algorithm="maxima")

[Out]

1/3*x - e^(5/4*x - e^4/x - 2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {4 x^2+e^{\frac {-4 e^4-8 x+5 x^2}{4 x}} \left (-12 e^4-15 x^2\right )}{12 x^2} \, dx=\frac {1}{3} \, {\left (x e^{4} - 3 \, e^{\left (\frac {5 \, x^{2} + 8 \, x - 4 \, e^{4}}{4 \, x}\right )}\right )} e^{\left (-4\right )} \]

[In]

integrate(1/12*((-12*exp(4)-15*x^2)*exp(1/4*(-4*exp(4)+5*x^2-8*x)/x)+4*x^2)/x^2,x, algorithm="giac")

[Out]

1/3*(x*e^4 - 3*e^(1/4*(5*x^2 + 8*x - 4*e^4)/x))*e^(-4)

Mupad [B] (verification not implemented)

Time = 9.37 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {4 x^2+e^{\frac {-4 e^4-8 x+5 x^2}{4 x}} \left (-12 e^4-15 x^2\right )}{12 x^2} \, dx=\frac {x}{3}-{\mathrm {e}}^{\frac {5\,x}{4}-\frac {{\mathrm {e}}^4}{x}-2} \]

[In]

int(-((exp(-(2*x + exp(4) - (5*x^2)/4)/x)*(12*exp(4) + 15*x^2))/12 - x^2/3)/x^2,x)

[Out]

x/3 - exp((5*x)/4 - exp(4)/x - 2)