3.20.52 \(\int \frac {32 x+16 e^3 x+2 e^6 x}{1+e^5-2 x^2+x^4+e^{5/2} (-2+2 x^2)} \, dx\)

Optimal. Leaf size=24 \[ \frac {\left (4+e^3\right )^2}{1-e^{5/2}-x^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6, 12, 1989, 28, 261} \begin {gather*} \frac {\left (4+e^3\right )^2}{-x^2-e^{5/2}+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32*x + 16*E^3*x + 2*E^6*x)/(1 + E^5 - 2*x^2 + x^4 + E^(5/2)*(-2 + 2*x^2)),x]

[Out]

(4 + E^3)^2/(1 - E^(5/2) - x^2)

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 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 1989

Int[(u_)^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[(d*x)^m*ExpandToSum[u, x]^p, x] /; FreeQ[{d, m, p}, x] &&
TrinomialQ[u, x] &&  !TrinomialMatchQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^6 x+\left (32+16 e^3\right ) x}{1+e^5-2 x^2+x^4+e^{5/2} \left (-2+2 x^2\right )} \, dx\\ &=\int \frac {\left (32+16 e^3+2 e^6\right ) x}{1+e^5-2 x^2+x^4+e^{5/2} \left (-2+2 x^2\right )} \, dx\\ &=\left (2 \left (4+e^3\right )^2\right ) \int \frac {x}{1+e^5-2 x^2+x^4+e^{5/2} \left (-2+2 x^2\right )} \, dx\\ &=\left (2 \left (4+e^3\right )^2\right ) \int \frac {x}{\left (-1+e^{5/2}\right )^2-2 \left (1-e^{5/2}\right ) x^2+x^4} \, dx\\ &=\left (2 \left (4+e^3\right )^2\right ) \int \frac {x}{\left (-1+e^{5/2}+x^2\right )^2} \, dx\\ &=\frac {\left (4+e^3\right )^2}{1-e^{5/2}-x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.88 \begin {gather*} -\frac {\left (4+e^3\right )^2}{-1+e^{5/2}+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32*x + 16*E^3*x + 2*E^6*x)/(1 + E^5 - 2*x^2 + x^4 + E^(5/2)*(-2 + 2*x^2)),x]

[Out]

-((4 + E^3)^2/(-1 + E^(5/2) + x^2))

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fricas [A]  time = 0.71, size = 19, normalized size = 0.79 \begin {gather*} -\frac {e^{6} + 8 \, e^{3} + 16}{x^{2} + e^{\frac {5}{2}} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(3)^2+16*x*exp(3)+32*x)/(exp(5/4)^4+(2*x^2-2)*exp(5/4)^2+x^4-2*x^2+1),x, algorithm="fricas")

[Out]

-(e^6 + 8*e^3 + 16)/(x^2 + e^(5/2) - 1)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(3)^2+16*x*exp(3)+32*x)/(exp(5/4)^4+(2*x^2-2)*exp(5/4)^2+x^4-2*x^2+1),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.10, size = 24, normalized size = 1.00

method	result	size
gosper	\(-\frac {{\mathrm e}^{6}+8 \,{\mathrm e}^{3}+16}{{\mathrm e}^{\frac {5}{2}}+x^{2}-1}\)	\(24\)
norman	\(\frac {-{\mathrm e}^{6}-8 \,{\mathrm e}^{3}-16}{{\mathrm e}^{\frac {5}{2}}+x^{2}-1}\)	\(25\)
risch	\(-\frac {{\mathrm e}^{6}}{{\mathrm e}^{\frac {5}{2}}+x^{2}-1}-\frac {8 \,{\mathrm e}^{3}}{{\mathrm e}^{\frac {5}{2}}+x^{2}-1}-\frac {16}{{\mathrm e}^{\frac {5}{2}}+x^{2}-1}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*exp(3)^2+16*x*exp(3)+32*x)/(exp(5/4)^4+(2*x^2-2)*exp(5/4)^2+x^4-2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

-(exp(3)^2+8*exp(3)+16)/(exp(5/4)^2+x^2-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, \int \frac {x e^{6} + 8 \, x e^{3} + 16 \, x}{x^{4} - 2 \, x^{2} + 2 \, {\left (x^{2} - 1\right )} e^{\frac {5}{2}} + e^{5} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(3)^2+16*x*exp(3)+32*x)/(exp(5/4)^4+(2*x^2-2)*exp(5/4)^2+x^4-2*x^2+1),x, algorithm="maxima")

[Out]

2*integrate((x*e^6 + 8*x*e^3 + 16*x)/(x^4 - 2*x^2 + 2*(x^2 - 1)*e^(5/2) + e^5 + 1), x)

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mupad [B]  time = 1.16, size = 17, normalized size = 0.71 \begin {gather*} -\frac {{\left ({\mathrm {e}}^3+4\right )}^2}{x^2+{\mathrm {e}}^{5/2}-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x + 16*x*exp(3) + 2*x*exp(6))/(exp(5) + exp(5/2)*(2*x^2 - 2) - 2*x^2 + x^4 + 1),x)

[Out]

-(exp(3) + 4)^2/(exp(5/2) + x^2 - 1)

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sympy [B]  time = 0.25, size = 42, normalized size = 1.75 \begin {gather*} \frac {\left (1 - e^{\frac {5}{2}}\right ) \left (32 + 16 e^{3} + 2 e^{6}\right )}{x^{2} \left (-2 + 2 e^{\frac {5}{2}}\right ) - 4 e^{\frac {5}{2}} + 2 + 2 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(3)**2+16*x*exp(3)+32*x)/(exp(5/4)**4+(2*x**2-2)*exp(5/4)**2+x**4-2*x**2+1),x)

[Out]

(1 - exp(5/2))*(32 + 16*exp(3) + 2*exp(6))/(x**2*(-2 + 2*exp(5/2)) - 4*exp(5/2) + 2 + 2*exp(5))

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