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

Optimal. Leaf size=25 \[ -e^{-2-\frac {e^4}{x}+\frac {5 x}{4}}+\frac {x}{3} \]

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Rubi [A]  time = 0.20, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 14, 6706} \begin {gather*} \frac {x}{3}-e^{\frac {5 x}{4}-\frac {e^4}{x}-2} \end {gather*}

Antiderivative was successfully verified.

[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 6706

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 25, normalized size = 1.00 \begin {gather*} -e^{-2-\frac {e^4}{x}+\frac {5 x}{4}}+\frac {x}{3} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.52, size = 25, normalized size = 1.00 \begin {gather*} \frac {1}{3} \, x - e^{\left (\frac {5 \, x^{2} - 8 \, x - 4 \, e^{4}}{4 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.44, size = 30, normalized size = 1.20 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.06, size = 26, normalized size = 1.04

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 1.26, size = 19, normalized size = 0.76 \begin {gather*} \frac {1}{3} \, x - e^{\left (\frac {5}{4} \, x - \frac {e^{4}}{x} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.95, size = 19, normalized size = 0.76 \begin {gather*} \frac {x}{3}-{\mathrm {e}}^{\frac {5\,x}{4}-\frac {{\mathrm {e}}^4}{x}-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.20, size = 19, normalized size = 0.76 \begin {gather*} \frac {x}{3} - e^{\frac {\frac {5 x^{2}}{4} - 2 x - e^{4}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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