3.43.36 \(\int \frac {-25 x-10 x^2-x^3+e^{\frac {e^4}{x}} (x^2+e^4 (5+x))}{25 x^2+10 x^3+x^4} \, dx\)

Optimal. Leaf size=22 \[ 2-\frac {e^{\frac {e^4}{x}}}{5+x}-\log (x) \]

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Rubi [A]  time = 0.41, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1594, 27, 6742, 2288} \begin {gather*} -\frac {e^{\frac {e^4}{x}}}{x+5}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25*x - 10*x^2 - x^3 + E^(E^4/x)*(x^2 + E^4*(5 + x)))/(25*x^2 + 10*x^3 + x^4),x]

[Out]

-(E^(E^4/x)/(5 + x)) - Log[x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-25 x-10 x^2-x^3+e^{\frac {e^4}{x}} \left (x^2+e^4 (5+x)\right )}{x^2 \left (25+10 x+x^2\right )} \, dx\\ &=\int \frac {-25 x-10 x^2-x^3+e^{\frac {e^4}{x}} \left (x^2+e^4 (5+x)\right )}{x^2 (5+x)^2} \, dx\\ &=\int \left (-\frac {1}{x}+\frac {e^{\frac {e^4}{x}} \left (5 e^4+e^4 x+x^2\right )}{x^2 (5+x)^2}\right ) \, dx\\ &=-\log (x)+\int \frac {e^{\frac {e^4}{x}} \left (5 e^4+e^4 x+x^2\right )}{x^2 (5+x)^2} \, dx\\ &=-\frac {e^{\frac {e^4}{x}}}{5+x}-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 21, normalized size = 0.95 \begin {gather*} -\frac {e^{\frac {e^4}{x}}}{5+x}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25*x - 10*x^2 - x^3 + E^(E^4/x)*(x^2 + E^4*(5 + x)))/(25*x^2 + 10*x^3 + x^4),x]

[Out]

-(E^(E^4/x)/(5 + x)) - Log[x]

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fricas [A]  time = 0.69, size = 21, normalized size = 0.95 \begin {gather*} -\frac {{\left (x + 5\right )} \log \relax (x) + e^{\left (\frac {e^{4}}{x}\right )}}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5+x)*exp(4)+x^2)*exp(exp(4)/x)-x^3-10*x^2-25*x)/(x^4+10*x^3+25*x^2),x, algorithm="fricas")

[Out]

-((x + 5)*log(x) + e^(e^4/x))/(x + 5)

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giac [B]  time = 0.24, size = 54, normalized size = 2.45 \begin {gather*} \frac {{\left (e^{8} \log \left (\frac {e^{4}}{x}\right ) + \frac {5 \, e^{8} \log \left (\frac {e^{4}}{x}\right )}{x} - \frac {e^{\left (\frac {e^{4}}{x} + 8\right )}}{x}\right )} e^{\left (-4\right )}}{\frac {5 \, e^{4}}{x} + e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5+x)*exp(4)+x^2)*exp(exp(4)/x)-x^3-10*x^2-25*x)/(x^4+10*x^3+25*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.21, size = 20, normalized size = 0.91

method	result	size
norman	\(-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}}}{5+x}-\ln \relax (x )\)	\(20\)
risch	\(-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}}}{5+x}-\ln \relax (x )\)	\(20\)
derivativedivides	\(-{\mathrm e}^{4} \left ({\mathrm e}^{8} \left (-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}} {\mathrm e}^{-8}}{5 \left ({\mathrm e}^{4}+\frac {5 \,{\mathrm e}^{4}}{x}\right )}-\frac {{\mathrm e}^{-8} {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \expIntegralEi \left (1, -\frac {{\mathrm e}^{4}}{5}-\frac {{\mathrm e}^{4}}{x}\right )}{25}\right )+{\mathrm e}^{8} \left (\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}} {\mathrm e}^{-4}}{25 \,{\mathrm e}^{4}+\frac {125 \,{\mathrm e}^{4}}{x}}+\frac {\left ({\mathrm e}^{4}-5\right ) {\mathrm e}^{-8} {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \expIntegralEi \left (1, -\frac {{\mathrm e}^{4}}{5}-\frac {{\mathrm e}^{4}}{x}\right )}{125}\right )-{\mathrm e}^{-4} \ln \left (\frac {{\mathrm e}^{4}}{x}\right )+5 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}} {\mathrm e}^{-8}}{25}-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}}}{125 \left ({\mathrm e}^{4}+\frac {5 \,{\mathrm e}^{4}}{x}\right )}-\frac {{\mathrm e}^{-4} \left ({\mathrm e}^{4}-10\right ) {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \expIntegralEi \left (1, -\frac {{\mathrm e}^{4}}{5}-\frac {{\mathrm e}^{4}}{x}\right )}{625}\right )\right )\)	\(199\)
default	\(-{\mathrm e}^{4} \left ({\mathrm e}^{8} \left (-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}} {\mathrm e}^{-8}}{5 \left ({\mathrm e}^{4}+\frac {5 \,{\mathrm e}^{4}}{x}\right )}-\frac {{\mathrm e}^{-8} {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \expIntegralEi \left (1, -\frac {{\mathrm e}^{4}}{5}-\frac {{\mathrm e}^{4}}{x}\right )}{25}\right )+{\mathrm e}^{8} \left (\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}} {\mathrm e}^{-4}}{25 \,{\mathrm e}^{4}+\frac {125 \,{\mathrm e}^{4}}{x}}+\frac {\left ({\mathrm e}^{4}-5\right ) {\mathrm e}^{-8} {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \expIntegralEi \left (1, -\frac {{\mathrm e}^{4}}{5}-\frac {{\mathrm e}^{4}}{x}\right )}{125}\right )-{\mathrm e}^{-4} \ln \left (\frac {{\mathrm e}^{4}}{x}\right )+5 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}} {\mathrm e}^{-8}}{25}-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}}}{125 \left ({\mathrm e}^{4}+\frac {5 \,{\mathrm e}^{4}}{x}\right )}-\frac {{\mathrm e}^{-4} \left ({\mathrm e}^{4}-10\right ) {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \expIntegralEi \left (1, -\frac {{\mathrm e}^{4}}{5}-\frac {{\mathrm e}^{4}}{x}\right )}{625}\right )\right )\)	\(199\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5+x)*exp(4)+x^2)*exp(exp(4)/x)-x^3-10*x^2-25*x)/(x^4+10*x^3+25*x^2),x,method=_RETURNVERBOSE)

[Out]

-exp(exp(4)/x)/(5+x)-ln(x)

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maxima [A]  time = 0.39, size = 19, normalized size = 0.86 \begin {gather*} -\frac {e^{\left (\frac {e^{4}}{x}\right )}}{x + 5} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5+x)*exp(4)+x^2)*exp(exp(4)/x)-x^3-10*x^2-25*x)/(x^4+10*x^3+25*x^2),x, algorithm="maxima")

[Out]

-e^(e^4/x)/(x + 5) - log(x)

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mupad [B]  time = 2.92, size = 19, normalized size = 0.86 \begin {gather*} -\ln \relax (x)-\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^4}{x}}}{x+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(25*x - exp(exp(4)/x)*(exp(4)*(x + 5) + x^2) + 10*x^2 + x^3)/(25*x^2 + 10*x^3 + x^4),x)

[Out]

- log(x) - exp(exp(4)/x)/(x + 5)

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sympy [A]  time = 0.18, size = 14, normalized size = 0.64 \begin {gather*} - \log {\relax (x )} - \frac {e^{\frac {e^{4}}{x}}}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5+x)*exp(4)+x**2)*exp(exp(4)/x)-x**3-10*x**2-25*x)/(x**4+10*x**3+25*x**2),x)

[Out]

-log(x) - exp(exp(4)/x)/(x + 5)

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