3.84.27 \(\int \frac {e^{-e^{\frac {3}{5+x}}} (1500 x+600 x^2+90 e^{\frac {3}{5+x}} x^2+60 x^3)}{25+10 x+x^2} \, dx\)

Optimal. Leaf size=22 \[ 10 \left (-4+3 e^{-e^{\frac {3}{5+x}}} x^2\right ) \]

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Rubi [A]  time = 0.11, antiderivative size = 18, normalized size of antiderivative = 0.82, number of steps used = 2, number of rules used = 2, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {27, 2288} \begin {gather*} 30 e^{-e^{\frac {3}{x+5}}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1500*x + 600*x^2 + 90*E^(3/(5 + x))*x^2 + 60*x^3)/(E^E^(3/(5 + x))*(25 + 10*x + x^2)),x]

[Out]

(30*x^2)/E^E^(3/(5 + 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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-e^{\frac {3}{5+x}}} \left (1500 x+600 x^2+90 e^{\frac {3}{5+x}} x^2+60 x^3\right )}{(5+x)^2} \, dx\\ &=30 e^{-e^{\frac {3}{5+x}}} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 18, normalized size = 0.82 \begin {gather*} 30 e^{-e^{\frac {3}{5+x}}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1500*x + 600*x^2 + 90*E^(3/(5 + x))*x^2 + 60*x^3)/(E^E^(3/(5 + x))*(25 + 10*x + x^2)),x]

[Out]

(30*x^2)/E^E^(3/(5 + x))

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fricas [A]  time = 0.74, size = 16, normalized size = 0.73 \begin {gather*} 30 \, x^{2} e^{\left (-e^{\left (\frac {3}{x + 5}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*x^2*exp(3/(5+x))+60*x^3+600*x^2+1500*x)/(x^2+10*x+25)/exp(exp(3/(5+x))),x, algorithm="fricas")

[Out]

30*x^2*e^(-e^(3/(x + 5)))

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giac [A]  time = 0.23, size = 16, normalized size = 0.73 \begin {gather*} 30 \, x^{2} e^{\left (-e^{\left (\frac {3}{x + 5}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*x^2*exp(3/(5+x))+60*x^3+600*x^2+1500*x)/(x^2+10*x+25)/exp(exp(3/(5+x))),x, algorithm="giac")

[Out]

30*x^2*e^(-e^(3/(x + 5)))

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maple [A]  time = 0.37, size = 17, normalized size = 0.77

method	result	size
risch	\(30 x^{2} {\mathrm e}^{-{\mathrm e}^{\frac {3}{5+x}}}\)	\(17\)
norman	\(\frac {\left (30 x^{3}+150 x^{2}\right ) {\mathrm e}^{-{\mathrm e}^{\frac {3}{5+x}}}}{5+x}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((90*x^2*exp(3/(5+x))+60*x^3+600*x^2+1500*x)/(x^2+10*x+25)/exp(exp(3/(5+x))),x,method=_RETURNVERBOSE)

[Out]

30*x^2*exp(-exp(3/(5+x)))

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maxima [A]  time = 0.40, size = 16, normalized size = 0.73 \begin {gather*} 30 \, x^{2} e^{\left (-e^{\left (\frac {3}{x + 5}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*x^2*exp(3/(5+x))+60*x^3+600*x^2+1500*x)/(x^2+10*x+25)/exp(exp(3/(5+x))),x, algorithm="maxima")

[Out]

30*x^2*e^(-e^(3/(x + 5)))

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mupad [B]  time = 5.29, size = 16, normalized size = 0.73 \begin {gather*} 30\,x^2\,{\mathrm {e}}^{-{\mathrm {e}}^{\frac {3}{x+5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-exp(3/(x + 5)))*(1500*x + 90*x^2*exp(3/(x + 5)) + 600*x^2 + 60*x^3))/(10*x + x^2 + 25),x)

[Out]

30*x^2*exp(-exp(3/(x + 5)))

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sympy [A]  time = 6.77, size = 12, normalized size = 0.55 \begin {gather*} 30 x^{2} e^{- e^{\frac {3}{x + 5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*x**2*exp(3/(5+x))+60*x**3+600*x**2+1500*x)/(x**2+10*x+25)/exp(exp(3/(5+x))),x)

[Out]

30*x**2*exp(-exp(3/(x + 5)))

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