3.73.4 \(\int \frac {e^{-\frac {25 x^4}{4}} (e^{6+\frac {25 x^4}{4}}-25 x^5+e^6 (1-25 x^4))}{2 e^{12}+4 e^6 x+2 x^2} \, dx\)

Optimal. Leaf size=24 \[ \frac {x+e^{-\frac {25 x^4}{4}} x}{2 \left (e^6+x\right )} \]

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Rubi [B]  time = 0.34, antiderivative size = 49, normalized size of antiderivative = 2.04, number of steps used = 5, number of rules used = 4, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {27, 12, 6742, 2288} \begin {gather*} \frac {e^{-\frac {25 x^4}{4}} \left (x^5+e^6 x^4\right )}{2 x^3 \left (x+e^6\right )^2}-\frac {e^6}{2 \left (x+e^6\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(6 + (25*x^4)/4) - 25*x^5 + E^6*(1 - 25*x^4))/(E^((25*x^4)/4)*(2*E^12 + 4*E^6*x + 2*x^2)),x]

[Out]

-1/2*E^6/(E^6 + x) + (E^6*x^4 + x^5)/(2*E^((25*x^4)/4)*x^3*(E^6 + x)^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

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 {e^{-\frac {25 x^4}{4}} \left (e^{6+\frac {25 x^4}{4}}-25 x^5+e^6 \left (1-25 x^4\right )\right )}{2 \left (e^6+x\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{-\frac {25 x^4}{4}} \left (e^{6+\frac {25 x^4}{4}}-25 x^5+e^6 \left (1-25 x^4\right )\right )}{\left (e^6+x\right )^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {e^6}{\left (e^6+x\right )^2}+\frac {e^{-\frac {25 x^4}{4}} \left (e^6-25 e^6 x^4-25 x^5\right )}{\left (e^6+x\right )^2}\right ) \, dx\\ &=-\frac {e^6}{2 \left (e^6+x\right )}+\frac {1}{2} \int \frac {e^{-\frac {25 x^4}{4}} \left (e^6-25 e^6 x^4-25 x^5\right )}{\left (e^6+x\right )^2} \, dx\\ &=-\frac {e^6}{2 \left (e^6+x\right )}+\frac {e^{-\frac {25 x^4}{4}} \left (e^6 x^4+x^5\right )}{2 x^3 \left (e^6+x\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 28, normalized size = 1.17 \begin {gather*} \frac {-e^6+e^{-\frac {25 x^4}{4}} x}{2 \left (e^6+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(6 + (25*x^4)/4) - 25*x^5 + E^6*(1 - 25*x^4))/(E^((25*x^4)/4)*(2*E^12 + 4*E^6*x + 2*x^2)),x]

[Out]

(-E^6 + x/E^((25*x^4)/4))/(2*(E^6 + x))

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fricas [A]  time = 0.62, size = 31, normalized size = 1.29 \begin {gather*} \frac {{\left (x e^{6} - e^{\left (\frac {25}{4} \, x^{4} + 12\right )}\right )} e^{\left (-\frac {25}{4} \, x^{4} - 6\right )}}{2 \, {\left (x + e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(3)^2*exp(25/4*x^4)+(-25*x^4+1)*exp(3)^2-25*x^5)/(2*exp(3)^4+4*x*exp(3)^2+2*x^2)/exp(25/4*x^4),x
, algorithm="fricas")

[Out]

1/2*(x*e^6 - e^(25/4*x^4 + 12))*e^(-25/4*x^4 - 6)/(x + e^6)

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giac [A]  time = 0.20, size = 33, normalized size = 1.38 \begin {gather*} \frac {x - e^{\left (\frac {25}{4} \, x^{4} + 6\right )}}{2 \, {\left (x e^{\left (\frac {25}{4} \, x^{4}\right )} + e^{\left (\frac {25}{4} \, x^{4} + 6\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(3)^2*exp(25/4*x^4)+(-25*x^4+1)*exp(3)^2-25*x^5)/(2*exp(3)^4+4*x*exp(3)^2+2*x^2)/exp(25/4*x^4),x
, algorithm="giac")

[Out]

1/2*(x - e^(25/4*x^4 + 6))/(x*e^(25/4*x^4) + e^(25/4*x^4 + 6))

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maple [A]  time = 0.17, size = 27, normalized size = 1.12

method	result	size
risch	\(-\frac {{\mathrm e}^{6}}{2 \left ({\mathrm e}^{6}+x \right )}+\frac {x \,{\mathrm e}^{-\frac {25 x^{4}}{4}}}{2 \,{\mathrm e}^{6}+2 x}\)	\(27\)
norman	\(\frac {\left (-\frac {{\mathrm e}^{6} {\mathrm e}^{\frac {25 x^{4}}{4}}}{2}+\frac {x}{2}\right ) {\mathrm e}^{-\frac {25 x^{4}}{4}}}{{\mathrm e}^{6}+x}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3)^2*exp(25/4*x^4)+(-25*x^4+1)*exp(3)^2-25*x^5)/(2*exp(3)^4+4*x*exp(3)^2+2*x^2)/exp(25/4*x^4),x,metho
d=_RETURNVERBOSE)

[Out]

-1/2*exp(6)/(exp(6)+x)+1/2*x/(exp(6)+x)*exp(-25/4*x^4)

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maxima [A]  time = 0.41, size = 21, normalized size = 0.88 \begin {gather*} \frac {x e^{\left (-\frac {25}{4} \, x^{4}\right )} - e^{6}}{2 \, {\left (x + e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(3)^2*exp(25/4*x^4)+(-25*x^4+1)*exp(3)^2-25*x^5)/(2*exp(3)^4+4*x*exp(3)^2+2*x^2)/exp(25/4*x^4),x
, algorithm="maxima")

[Out]

1/2*(x*e^(-25/4*x^4) - e^6)/(x + e^6)

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mupad [B]  time = 0.18, size = 21, normalized size = 0.88 \begin {gather*} \frac {x+x\,{\mathrm {e}}^{-\frac {25\,x^4}{4}}}{2\,x+2\,{\mathrm {e}}^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(25*x^4)/4)*(exp(6)*(25*x^4 - 1) - exp(6)*exp((25*x^4)/4) + 25*x^5))/(2*exp(12) + 4*x*exp(6) + 2*x^
2),x)

[Out]

(x + x*exp(-(25*x^4)/4))/(2*x + 2*exp(6))

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sympy [A]  time = 0.23, size = 29, normalized size = 1.21 \begin {gather*} \frac {x e^{- \frac {25 x^{4}}{4}}}{2 x + 2 e^{6}} - \frac {e^{6}}{2 x + 2 e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(3)**2*exp(25/4*x**4)+(-25*x**4+1)*exp(3)**2-25*x**5)/(2*exp(3)**4+4*x*exp(3)**2+2*x**2)/exp(25/
4*x**4),x)

[Out]

x*exp(-25*x**4/4)/(2*x + 2*exp(6)) - exp(6)/(2*x + 2*exp(6))

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