3.34.88 \(\int \frac {e^8 (-16+\frac {320}{e^4}+\frac {4 (-400-32 x^3+3 x^4)}{e^8})}{24 x^2} \, dx\)

Optimal. Leaf size=30 \[ \frac {2}{3} \left (-4+\frac {x}{4}\right ) x \left (-\frac {\left (5-\frac {e^4}{2}\right )^2}{x^2}+x\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 14} \begin {gather*} \frac {x^3}{6}-\frac {8 x^2}{3}+\frac {2 \left (10-e^4\right )^2}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^8*(-16 + 320/E^4 + (4*(-400 - 32*x^3 + 3*x^4))/E^8))/(24*x^2),x]

[Out]

(2*(10 - E^4)^2)/(3*x) - (8*x^2)/3 + x^3/6

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{24} e^8 \int \frac {-16+\frac {320}{e^4}+\frac {4 \left (-400-32 x^3+3 x^4\right )}{e^8}}{x^2} \, dx\\ &=\frac {1}{24} e^8 \int \left (-\frac {16 \left (-10+e^4\right )^2}{e^8 x^2}-\frac {128 x}{e^8}+\frac {12 x^2}{e^8}\right ) \, dx\\ &=\frac {2 \left (10-e^4\right )^2}{3 x}-\frac {8 x^2}{3}+\frac {x^3}{6}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.93 \begin {gather*} \frac {1}{6} \left (\frac {4 \left (100-20 e^4+e^8\right )}{x}-16 x^2+x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^8*(-16 + 320/E^4 + (4*(-400 - 32*x^3 + 3*x^4))/E^8))/(24*x^2),x]

[Out]

((4*(100 - 20*E^4 + E^8))/x - 16*x^2 + x^3)/6

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fricas [A]  time = 0.92, size = 39, normalized size = 1.30 \begin {gather*} \frac {{\left ({\left (x^{4} - 16 \, x^{3} + 400\right )} e^{\left (2 \, \log \relax (2) - 8\right )} - 160 \, e^{\left (\log \relax (2) - 4\right )} + 16\right )} e^{\left (-2 \, \log \relax (2) + 8\right )}}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((3*x^4-32*x^3-400)*exp(log(2)-4)^2+160*exp(log(2)-4)-16)/x^2/exp(log(2)-4)^2,x, algorithm="fric
as")

[Out]

1/6*((x^4 - 16*x^3 + 400)*e^(2*log(2) - 8) - 160*e^(log(2) - 4) + 16)*e^(-2*log(2) + 8)/x

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giac [B]  time = 0.15, size = 56, normalized size = 1.87 \begin {gather*} \frac {1}{6} \, {\left (x^{3} e^{\left (2 \, \log \relax (2) - 8\right )} - 16 \, x^{2} e^{\left (2 \, \log \relax (2) - 8\right )} + \frac {16 \, {\left (25 \, e^{\left (2 \, \log \relax (2) - 8\right )} - 10 \, e^{\left (\log \relax (2) - 4\right )} + 1\right )}}{x}\right )} e^{\left (-2 \, \log \relax (2) + 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((3*x^4-32*x^3-400)*exp(log(2)-4)^2+160*exp(log(2)-4)-16)/x^2/exp(log(2)-4)^2,x, algorithm="giac
")

[Out]

1/6*(x^3*e^(2*log(2) - 8) - 16*x^2*e^(2*log(2) - 8) + 16*(25*e^(2*log(2) - 8) - 10*e^(log(2) - 4) + 1)/x)*e^(-
2*log(2) + 8)

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maple [A]  time = 0.05, size = 37, normalized size = 1.23

method	result	size
risch	\(\frac {x^{3}}{6}-\frac {8 x^{2}}{3}-\frac {40 \,{\mathrm e}^{8} {\mathrm e}^{-4}}{3 x}+\frac {200 \,{\mathrm e}^{8} {\mathrm e}^{-8}}{3 x}+\frac {2 \,{\mathrm e}^{8}}{3 x}\)	\(37\)
norman	\(\frac {\left (-\frac {8 x^{3} {\mathrm e}^{4}}{3}+\frac {x^{4} {\mathrm e}^{4}}{6}+\frac {2 \,{\mathrm e}^{4} \left ({\mathrm e}^{8}-20 \,{\mathrm e}^{4}+100\right )}{3}\right ) {\mathrm e}^{-4}}{x}\)	\(38\)
gosper	\(\frac {\left (4 \,{\mathrm e}^{-8} x^{4}-64 \,{\mathrm e}^{-8} x^{3}+16+1600 \,{\mathrm e}^{-8}-160 \,{\mathrm e}^{\ln \relax (2)-4}\right ) {\mathrm e}^{8}}{24 x}\)	\(54\)
default	\(\frac {{\mathrm e}^{8} \left (x^{3} {\mathrm e}^{2 \ln \relax (2)-8}-16 \,{\mathrm e}^{2 \ln \relax (2)-8} x^{2}-\frac {160 \,{\mathrm e}^{\ln \relax (2)-4}-400 \,{\mathrm e}^{2 \ln \relax (2)-8}-16}{x}\right )}{24}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*((3*x^4-32*x^3-400)*exp(ln(2)-4)^2+160*exp(ln(2)-4)-16)/x^2/exp(ln(2)-4)^2,x,method=_RETURNVERBOSE)

[Out]

1/6*x^3-8/3*x^2-40/3*exp(8)/x*exp(-4)+200/3*exp(8)/x*exp(-8)+2/3*exp(8)/x

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maxima [A]  time = 0.35, size = 32, normalized size = 1.07 \begin {gather*} \frac {1}{6} \, {\left ({\left (x^{3} - 16 \, x^{2}\right )} e^{\left (-8\right )} + \frac {4 \, {\left (e^{8} - 20 \, e^{4} + 100\right )} e^{\left (-8\right )}}{x}\right )} e^{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((3*x^4-32*x^3-400)*exp(log(2)-4)^2+160*exp(log(2)-4)-16)/x^2/exp(log(2)-4)^2,x, algorithm="maxi
ma")

[Out]

1/6*((x^3 - 16*x^2)*e^(-8) + 4*(e^8 - 20*e^4 + 100)*e^(-8)/x)*e^8

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mupad [B]  time = 0.06, size = 23, normalized size = 0.77 \begin {gather*} \frac {x^4-16\,x^3-80\,{\mathrm {e}}^4+4\,{\mathrm {e}}^8+400}{6\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(8 - 2*log(2))*((exp(2*log(2) - 8)*(32*x^3 - 3*x^4 + 400))/6 - (80*exp(log(2) - 4))/3 + 8/3))/x^2,x)

[Out]

(4*exp(8) - 80*exp(4) - 16*x^3 + x^4 + 400)/(6*x)

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sympy [A]  time = 0.14, size = 26, normalized size = 0.87 \begin {gather*} \frac {x^{3}}{6} - \frac {8 x^{2}}{3} + \frac {- 80 e^{4} + 400 + 4 e^{8}}{6 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((3*x**4-32*x**3-400)*exp(ln(2)-4)**2+160*exp(ln(2)-4)-16)/x**2/exp(ln(2)-4)**2,x)

[Out]

x**3/6 - 8*x**2/3 + (-80*exp(4) + 400 + 4*exp(8))/(6*x)

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