3.17.56 \(\int \frac {4-4 e^3+2 x^2+e^{x^4} (-2+8 x^4)}{x^2+2 e^5 x^2} \, dx\)

Optimal. Leaf size=27 \[ \frac {2 \left (-2+2 e^3+e^{x^4}+x^2\right )}{x+2 e^5 x} \]

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Rubi [A]  time = 0.08, antiderivative size = 53, normalized size of antiderivative = 1.96, number of steps used = 7, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6, 12, 14, 2288} \begin {gather*} \frac {2 e^{x^4}}{\left (1+2 e^5\right ) x}+\frac {2 x}{1+2 e^5}-\frac {4 \left (1-e^3\right )}{\left (1+2 e^5\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 - 4*E^3 + 2*x^2 + E^x^4*(-2 + 8*x^4))/(x^2 + 2*E^5*x^2),x]

[Out]

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

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]]

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 {4-4 e^3+2 x^2+e^{x^4} \left (-2+8 x^4\right )}{\left (1+2 e^5\right ) x^2} \, dx\\ &=\frac {\int \frac {4-4 e^3+2 x^2+e^{x^4} \left (-2+8 x^4\right )}{x^2} \, dx}{1+2 e^5}\\ &=\frac {\int \left (-\frac {2 \left (-2+2 e^3-x^2\right )}{x^2}+\frac {2 e^{x^4} \left (-1+4 x^4\right )}{x^2}\right ) \, dx}{1+2 e^5}\\ &=-\frac {2 \int \frac {-2+2 e^3-x^2}{x^2} \, dx}{1+2 e^5}+\frac {2 \int \frac {e^{x^4} \left (-1+4 x^4\right )}{x^2} \, dx}{1+2 e^5}\\ &=\frac {2 e^{x^4}}{\left (1+2 e^5\right ) x}-\frac {2 \int \left (-1+\frac {2 \left (-1+e^3\right )}{x^2}\right ) \, dx}{1+2 e^5}\\ &=\frac {2 e^{x^4}}{\left (1+2 e^5\right ) x}-\frac {4 \left (1-e^3\right )}{\left (1+2 e^5\right ) x}+\frac {2 x}{1+2 e^5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 29, normalized size = 1.07 \begin {gather*} \frac {2 \left (-2+2 e^3+e^{x^4}+x^2\right )}{\left (1+2 e^5\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 4*E^3 + 2*x^2 + E^x^4*(-2 + 8*x^4))/(x^2 + 2*E^5*x^2),x]

[Out]

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

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fricas [A]  time = 0.88, size = 24, normalized size = 0.89 \begin {gather*} \frac {2 \, {\left (x^{2} + 2 \, e^{3} + e^{\left (x^{4}\right )} - 2\right )}}{2 \, x e^{5} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4-2)*exp(x^4)-4*exp(3)+2*x^2+4)/(2*x^2*exp(5)+x^2),x, algorithm="fricas")

[Out]

2*(x^2 + 2*e^3 + e^(x^4) - 2)/(2*x*e^5 + x)

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giac [B]  time = 0.23, size = 52, normalized size = 1.93 \begin {gather*} \frac {2 \, {\left (2 \, x^{2} e^{5} + x^{2} + 4 \, e^{8} - 4 \, e^{5} + 2 \, e^{3} + 2 \, e^{\left (x^{4} + 5\right )} + e^{\left (x^{4}\right )} - 2\right )}}{4 \, x e^{10} + 4 \, x e^{5} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4-2)*exp(x^4)-4*exp(3)+2*x^2+4)/(2*x^2*exp(5)+x^2),x, algorithm="giac")

[Out]

2*(2*x^2*e^5 + x^2 + 4*e^8 - 4*e^5 + 2*e^3 + 2*e^(x^4 + 5) + e^(x^4) - 2)/(4*x*e^10 + 4*x*e^5 + x)

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maple [A]  time = 0.31, size = 47, normalized size = 1.74

method	result	size
norman	\(\frac {\frac {4 \,{\mathrm e}^{3}-4}{2 \,{\mathrm e}^{5}+1}+\frac {2 x^{2}}{2 \,{\mathrm e}^{5}+1}+\frac {2 \,{\mathrm e}^{x^{4}}}{2 \,{\mathrm e}^{5}+1}}{x}\)	\(47\)
risch	\(\frac {2 x}{2 \,{\mathrm e}^{5}+1}+\frac {8 \,{\mathrm e}^{8}}{\left (2 \,{\mathrm e}^{5}+1\right )^{2} x}-\frac {8 \,{\mathrm e}^{5}}{\left (2 \,{\mathrm e}^{5}+1\right )^{2} x}+\frac {4 \,{\mathrm e}^{3}}{\left (2 \,{\mathrm e}^{5}+1\right )^{2} x}-\frac {4}{\left (2 \,{\mathrm e}^{5}+1\right )^{2} x}+\frac {2 \,{\mathrm e}^{x^{4}}}{\left (2 \,{\mathrm e}^{5}+1\right ) x}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^4-2)*exp(x^4)-4*exp(3)+2*x^2+4)/(2*x^2*exp(5)+x^2),x,method=_RETURNVERBOSE)

[Out]

(4*(exp(3)-1)/(2*exp(5)+1)+2/(2*exp(5)+1)*x^2+2/(2*exp(5)+1)*exp(x^4))/x

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maxima [C]  time = 0.45, size = 94, normalized size = 3.48 \begin {gather*} -\frac {2 \, x^{3} \Gamma \left (\frac {3}{4}, -x^{4}\right )}{\left (-x^{4}\right )^{\frac {3}{4}} {\left (2 \, e^{5} + 1\right )}} + \frac {2 \, x}{2 \, e^{5} + 1} + \frac {\left (-x^{4}\right )^{\frac {1}{4}} \Gamma \left (-\frac {1}{4}, -x^{4}\right )}{2 \, x {\left (2 \, e^{5} + 1\right )}} + \frac {4 \, e^{3}}{x {\left (2 \, e^{5} + 1\right )}} - \frac {4}{x {\left (2 \, e^{5} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4-2)*exp(x^4)-4*exp(3)+2*x^2+4)/(2*x^2*exp(5)+x^2),x, algorithm="maxima")

[Out]

-2*x^3*gamma(3/4, -x^4)/((-x^4)^(3/4)*(2*e^5 + 1)) + 2*x/(2*e^5 + 1) + 1/2*(-x^4)^(1/4)*gamma(-1/4, -x^4)/(x*(
2*e^5 + 1)) + 4*e^3/(x*(2*e^5 + 1)) - 4/(x*(2*e^5 + 1))

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mupad [B]  time = 1.09, size = 26, normalized size = 0.96 \begin {gather*} \frac {2\,\left ({\mathrm {e}}^{x^4}+2\,{\mathrm {e}}^3+x^2-2\right )}{x\,\left (2\,{\mathrm {e}}^5+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^4)*(8*x^4 - 2) - 4*exp(3) + 2*x^2 + 4)/(2*x^2*exp(5) + x^2),x)

[Out]

(2*(exp(x^4) + 2*exp(3) + x^2 - 2))/(x*(2*exp(5) + 1))

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sympy [A]  time = 0.16, size = 32, normalized size = 1.19 \begin {gather*} \frac {2 x + \frac {-4 + 4 e^{3}}{x}}{1 + 2 e^{5}} + \frac {2 e^{x^{4}}}{x + 2 x e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**4-2)*exp(x**4)-4*exp(3)+2*x**2+4)/(2*x**2*exp(5)+x**2),x)

[Out]

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

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