3.50.34 \(\int \frac {48+e^{4 x} (24-48 x)+e^{e^2} (-16+e^{4 x} (-8+16 x))}{x^3} \, dx\)

Optimal. Leaf size=23 \[ 4-\frac {4 \left (-3+e^{e^2}\right ) \left (-2-e^{4 x}\right )}{x^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 34, normalized size of antiderivative = 1.48, number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {14, 2197} \begin {gather*} -\frac {4 \left (3-e^{e^2}\right ) e^{4 x}}{x^2}-\frac {8 \left (3-e^{e^2}\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(48 + E^(4*x)*(24 - 48*x) + E^E^2*(-16 + E^(4*x)*(-8 + 16*x)))/x^3,x]

[Out]

(-8*(3 - E^E^2))/x^2 - (4*E^(4*x)*(3 - E^E^2))/x^2

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 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 \left (-\frac {16 \left (-3+e^{e^2}\right )}{x^3}+\frac {8 e^{4 x} \left (-3+e^{e^2}\right ) (-1+2 x)}{x^3}\right ) \, dx\\ &=-\frac {8 \left (3-e^{e^2}\right )}{x^2}-\left (8 \left (3-e^{e^2}\right )\right ) \int \frac {e^{4 x} (-1+2 x)}{x^3} \, dx\\ &=-\frac {8 \left (3-e^{e^2}\right )}{x^2}-\frac {4 e^{4 x} \left (3-e^{e^2}\right )}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 25, normalized size = 1.09 \begin {gather*} 8 \left (-3+e^{e^2}\right ) \left (\frac {1}{x^2}+\frac {e^{4 x}}{2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(48 + E^(4*x)*(24 - 48*x) + E^E^2*(-16 + E^(4*x)*(-8 + 16*x)))/x^3,x]

[Out]

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

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fricas [A]  time = 0.67, size = 23, normalized size = 1.00 \begin {gather*} \frac {4 \, {\left ({\left (e^{\left (4 \, x\right )} + 2\right )} e^{\left (e^{2}\right )} - 3 \, e^{\left (4 \, x\right )} - 6\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x-8)*exp(x)^4-16)*exp(exp(2))+(-48*x+24)*exp(x)^4+48)/x^3,x, algorithm="fricas")

[Out]

4*((e^(4*x) + 2)*e^(e^2) - 3*e^(4*x) - 6)/x^2

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giac [A]  time = 0.21, size = 27, normalized size = 1.17 \begin {gather*} -\frac {4 \, {\left (3 \, e^{\left (4 \, x\right )} - e^{\left (4 \, x + e^{2}\right )} - 2 \, e^{\left (e^{2}\right )} + 6\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x-8)*exp(x)^4-16)*exp(exp(2))+(-48*x+24)*exp(x)^4+48)/x^3,x, algorithm="giac")

[Out]

-4*(3*e^(4*x) - e^(4*x + e^2) - 2*e^(e^2) + 6)/x^2

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maple [A]  time = 0.05, size = 24, normalized size = 1.04

method	result	size
norman	\(\frac {\left (4 \,{\mathrm e}^{{\mathrm e}^{2}}-12\right ) {\mathrm e}^{4 x}+8 \,{\mathrm e}^{{\mathrm e}^{2}}-24}{x^{2}}\)	\(24\)
risch	\(\frac {8 \,{\mathrm e}^{{\mathrm e}^{2}}}{x^{2}}-\frac {24}{x^{2}}+\frac {4 \left ({\mathrm e}^{{\mathrm e}^{2}}-3\right ) {\mathrm e}^{4 x}}{x^{2}}\)	\(29\)
default	\(-\frac {24}{x^{2}}-\frac {12 \,{\mathrm e}^{4 x}}{x^{2}}+\frac {8 \,{\mathrm e}^{{\mathrm e}^{2}}}{x^{2}}-8 \,{\mathrm e}^{{\mathrm e}^{2}} \left (-\frac {{\mathrm e}^{4 x}}{2 x^{2}}-\frac {2 \,{\mathrm e}^{4 x}}{x}-8 \expIntegralEi \left (1, -4 x \right )\right )+16 \,{\mathrm e}^{{\mathrm e}^{2}} \left (-\frac {{\mathrm e}^{4 x}}{x}-4 \expIntegralEi \left (1, -4 x \right )\right )\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x-8)*exp(x)^4-16)*exp(exp(2))+(-48*x+24)*exp(x)^4+48)/x^3,x,method=_RETURNVERBOSE)

[Out]

((4*exp(exp(2))-12)*exp(x)^4+8*exp(exp(2))-24)/x^2

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maxima [C]  time = 0.40, size = 48, normalized size = 2.09 \begin {gather*} 64 \, e^{\left (e^{2}\right )} \Gamma \left (-1, -4 \, x\right ) + 128 \, e^{\left (e^{2}\right )} \Gamma \left (-2, -4 \, x\right ) + \frac {8 \, e^{\left (e^{2}\right )}}{x^{2}} - \frac {24}{x^{2}} - 192 \, \Gamma \left (-1, -4 \, x\right ) - 384 \, \Gamma \left (-2, -4 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x-8)*exp(x)^4-16)*exp(exp(2))+(-48*x+24)*exp(x)^4+48)/x^3,x, algorithm="maxima")

[Out]

64*e^(e^2)*gamma(-1, -4*x) + 128*e^(e^2)*gamma(-2, -4*x) + 8*e^(e^2)/x^2 - 24/x^2 - 192*gamma(-1, -4*x) - 384*
gamma(-2, -4*x)

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mupad [B]  time = 0.14, size = 16, normalized size = 0.70 \begin {gather*} \frac {4\,\left ({\mathrm {e}}^{{\mathrm {e}}^2}-3\right )\,\left ({\mathrm {e}}^{4\,x}+2\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(2))*(exp(4*x)*(16*x - 8) - 16) - exp(4*x)*(48*x - 24) + 48)/x^3,x)

[Out]

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

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sympy [A]  time = 0.15, size = 29, normalized size = 1.26 \begin {gather*} \frac {\left (-12 + 4 e^{e^{2}}\right ) e^{4 x}}{x^{2}} - \frac {48 - 16 e^{e^{2}}}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x-8)*exp(x)**4-16)*exp(exp(2))+(-48*x+24)*exp(x)**4+48)/x**3,x)

[Out]

(-12 + 4*exp(exp(2)))*exp(4*x)/x**2 - (48 - 16*exp(exp(2)))/(2*x**2)

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