3.42.85 \(\int \frac {-16 e^{2 e^2}-80 x^4+40 e^5 x^4-5 e^{10} x^4+5 x^5}{5 x^5} \, dx\)

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

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Rubi [A]  time = 0.02, antiderivative size = 29, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 3, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {6, 12, 14} \begin {gather*} \frac {4 e^{2 e^2}}{5 x^4}+x-\left (4-e^5\right )^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16*E^(2*E^2) - 80*x^4 + 40*E^5*x^4 - 5*E^10*x^4 + 5*x^5)/(5*x^5),x]

[Out]

(4*E^(2*E^2))/(5*x^4) + x - (4 - E^5)^2*Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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} &=\int \frac {-16 e^{2 e^2}-5 e^{10} x^4+\left (-80+40 e^5\right ) x^4+5 x^5}{5 x^5} \, dx\\ &=\int \frac {-16 e^{2 e^2}+\left (-80+40 e^5-5 e^{10}\right ) x^4+5 x^5}{5 x^5} \, dx\\ &=\frac {1}{5} \int \frac {-16 e^{2 e^2}+\left (-80+40 e^5-5 e^{10}\right ) x^4+5 x^5}{x^5} \, dx\\ &=\frac {1}{5} \int \left (5-\frac {16 e^{2 e^2}}{x^5}-\frac {5 \left (-4+e^5\right )^2}{x}\right ) \, dx\\ &=\frac {4 e^{2 e^2}}{5 x^4}+x-\left (4-e^5\right )^2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 34, normalized size = 1.13 \begin {gather*} \frac {4 e^{2 e^2}}{5 x^4}+x-16 \log (x)+8 e^5 \log (x)-e^{10} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*E^(2*E^2) - 80*x^4 + 40*E^5*x^4 - 5*E^10*x^4 + 5*x^5)/(5*x^5),x]

[Out]

(4*E^(2*E^2))/(5*x^4) + x - 16*Log[x] + 8*E^5*Log[x] - E^10*Log[x]

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fricas [A]  time = 0.59, size = 41, normalized size = 1.37 \begin {gather*} \frac {5 \, x^{5} - 5 \, {\left (x^{4} e^{10} - 8 \, x^{4} e^{5} + 16 \, x^{4}\right )} \log \relax (x) + 4 \, e^{\left (2 \, e^{2}\right )}}{5 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-16*exp(exp(2))^2-5*x^4*exp(5)^2+40*x^4*exp(5)+5*x^5-80*x^4)/x^5,x, algorithm="fricas")

[Out]

1/5*(5*x^5 - 5*(x^4*e^10 - 8*x^4*e^5 + 16*x^4)*log(x) + 4*e^(2*e^2))/x^4

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giac [A]  time = 0.13, size = 25, normalized size = 0.83 \begin {gather*} -{\left (e^{10} - 8 \, e^{5} + 16\right )} \log \left ({\left | x \right |}\right ) + x + \frac {4 \, e^{\left (2 \, e^{2}\right )}}{5 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-16*exp(exp(2))^2-5*x^4*exp(5)^2+40*x^4*exp(5)+5*x^5-80*x^4)/x^5,x, algorithm="giac")

[Out]

-(e^10 - 8*e^5 + 16)*log(abs(x)) + x + 4/5*e^(2*e^2)/x^4

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maple [A]  time = 0.06, size = 27, normalized size = 0.90

method	result	size
default	\(x +\frac {\left (40 \,{\mathrm e}^{5}-5 \,{\mathrm e}^{10}-80\right ) \ln \relax (x )}{5}+\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}}{5 x^{4}}\)	\(27\)
risch	\(x +\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}}{5 x^{4}}+8 \,{\mathrm e}^{5} \ln \relax (x )-\ln \relax (x ) {\mathrm e}^{10}-16 \ln \relax (x )\)	\(29\)
norman	\(\frac {x^{5}+\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}}{5}}{x^{4}}+\left (8 \,{\mathrm e}^{5}-{\mathrm e}^{10}-16\right ) \ln \relax (x )\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-16*exp(exp(2))^2-5*x^4*exp(5)^2+40*x^4*exp(5)+5*x^5-80*x^4)/x^5,x,method=_RETURNVERBOSE)

[Out]

x+1/5*(40*exp(5)-5*exp(10)-80)*ln(x)+4/5*exp(2*exp(2))/x^4

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maxima [A]  time = 0.36, size = 24, normalized size = 0.80 \begin {gather*} -{\left (e^{10} - 8 \, e^{5} + 16\right )} \log \relax (x) + x + \frac {4 \, e^{\left (2 \, e^{2}\right )}}{5 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-16*exp(exp(2))^2-5*x^4*exp(5)^2+40*x^4*exp(5)+5*x^5-80*x^4)/x^5,x, algorithm="maxima")

[Out]

-(e^10 - 8*e^5 + 16)*log(x) + x + 4/5*e^(2*e^2)/x^4

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mupad [B]  time = 2.89, size = 24, normalized size = 0.80 \begin {gather*} x+\frac {4\,{\mathrm {e}}^{2\,{\mathrm {e}}^2}}{5\,x^4}-\ln \relax (x)\,\left ({\mathrm {e}}^{10}-8\,{\mathrm {e}}^5+16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((16*exp(2*exp(2)))/5 - 8*x^4*exp(5) + x^4*exp(10) + 16*x^4 - x^5)/x^5,x)

[Out]

x + (4*exp(2*exp(2)))/(5*x^4) - log(x)*(exp(10) - 8*exp(5) + 16)

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sympy [A]  time = 0.32, size = 24, normalized size = 0.80 \begin {gather*} x - \left (-4 + e^{5}\right )^{2} \log {\relax (x )} + \frac {4 e^{2 e^{2}}}{5 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-16*exp(exp(2))**2-5*x**4*exp(5)**2+40*x**4*exp(5)+5*x**5-80*x**4)/x**5,x)

[Out]

x - (-4 + exp(5))**2*log(x) + 4*exp(2*exp(2))/(5*x**4)

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