3.42.16 \(\int \frac {9 e+9 e^5+9 x^2}{e^2 (e+e^5-x) (-x-2 x^2-x^3+e (1+2 x+x^2)+e^5 (1+2 x+x^2))} \, dx\)

Optimal. Leaf size=23 \[ -25+\frac {9 x}{e^2 \left (e+e^5-x\right ) (1+x)} \]

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Rubi [B]  time = 0.21, antiderivative size = 50, normalized size of antiderivative = 2.17, number of steps used = 6, number of rules used = 4, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6, 12, 6688, 894} \begin {gather*} \frac {9 \left (1+e^4\right )}{e \left (1+e+e^5\right ) \left (e \left (1+e^4\right )-x\right )}-\frac {9}{e^2 \left (1+e+e^5\right ) (x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9*E + 9*E^5 + 9*x^2)/(E^2*(E + E^5 - x)*(-x - 2*x^2 - x^3 + E*(1 + 2*x + x^2) + E^5*(1 + 2*x + x^2))),x]

[Out]

(9*(1 + E^4))/(E*(1 + E + E^5)*(E*(1 + E^4) - x)) - 9/(E^2*(1 + E + E^5)*(1 + x))

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 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9 e+9 e^5+9 x^2}{e^2 \left (e+e^5-x\right ) \left (-x-2 x^2-x^3+\left (e+e^5\right ) \left (1+2 x+x^2\right )\right )} \, dx\\ &=\frac {\int \frac {9 e+9 e^5+9 x^2}{\left (e+e^5-x\right ) \left (-x-2 x^2-x^3+\left (e+e^5\right ) \left (1+2 x+x^2\right )\right )} \, dx}{e^2}\\ &=\frac {\int \frac {9 \left (e+e^5+x^2\right )}{\left (e+e^5-x\right )^2 (1+x)^2} \, dx}{e^2}\\ &=\frac {9 \int \frac {e+e^5+x^2}{\left (e+e^5-x\right )^2 (1+x)^2} \, dx}{e^2}\\ &=\frac {9 \int \left (\frac {e+e^5}{\left (1+e+e^5\right ) \left (e+e^5-x\right )^2}+\frac {1}{\left (1+e+e^5\right ) (1+x)^2}\right ) \, dx}{e^2}\\ &=\frac {9 \left (1+e^4\right )}{e \left (1+e+e^5\right ) \left (e \left (1+e^4\right )-x\right )}-\frac {9}{e^2 \left (1+e+e^5\right ) (1+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 21, normalized size = 0.91 \begin {gather*} \frac {9 x}{e^2 \left (e+e^5-x\right ) (1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9*E + 9*E^5 + 9*x^2)/(E^2*(E + E^5 - x)*(-x - 2*x^2 - x^3 + E*(1 + 2*x + x^2) + E^5*(1 + 2*x + x^2)
)),x]

[Out]

(9*x)/(E^2*(E + E^5 - x)*(1 + x))

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fricas [A]  time = 0.47, size = 27, normalized size = 1.17 \begin {gather*} \frac {9 \, x}{{\left (x + 1\right )} e^{7} + {\left (x + 1\right )} e^{3} - {\left (x^{2} + x\right )} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*exp(5)+9*exp(1)+9*x^2)/((x^2+2*x+1)*exp(5)+(x^2+2*x+1)*exp(1)-x^3-2*x^2-x)/exp(log(exp(5)+exp(1)-
x)+2),x, algorithm="fricas")

[Out]

9*x/((x + 1)*e^7 + (x + 1)*e^3 - (x^2 + x)*e^2)

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giac [A]  time = 0.12, size = 30, normalized size = 1.30 \begin {gather*} -\frac {9 \, x e^{\left (-2\right )}}{x^{2} - x e^{5} - x e + x - e^{5} - e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*exp(5)+9*exp(1)+9*x^2)/((x^2+2*x+1)*exp(5)+(x^2+2*x+1)*exp(1)-x^3-2*x^2-x)/exp(log(exp(5)+exp(1)-
x)+2),x, algorithm="giac")

[Out]

-9*x*e^(-2)/(x^2 - x*e^5 - x*e + x - e^5 - e)

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maple [A]  time = 0.13, size = 23, normalized size = 1.00

method	result	size
gosper	\(\frac {9 x \,{\mathrm e}^{-2}}{\left ({\mathrm e}^{5}+{\mathrm e}-x \right ) \left (x +1\right )}\)	\(23\)
norman	\(\frac {9 x \,{\mathrm e}^{-2}}{\left ({\mathrm e}^{5}+{\mathrm e}-x \right ) \left (x +1\right )}\)	\(23\)
risch	\(\frac {9 \,{\mathrm e}^{-2} x}{x \,{\mathrm e}^{5}+{\mathrm e}^{5}+x \,{\mathrm e}-x^{2}+{\mathrm e}-x}\)	\(29\)
default	\(9 \,{\mathrm e}^{-2} \left (-\frac {{\mathrm e}^{5}+{\mathrm e}}{\left ({\mathrm e}^{5}+{\mathrm e}+1\right ) \left (-{\mathrm e}^{5}-{\mathrm e}+x \right )}-\frac {1}{\left ({\mathrm e}^{5}+{\mathrm e}+1\right ) \left (x +1\right )}\right )\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*exp(5)+9*exp(1)+9*x^2)/((x^2+2*x+1)*exp(5)+(x^2+2*x+1)*exp(1)-x^3-2*x^2-x)/exp(ln(exp(5)+exp(1)-x)+2),x
,method=_RETURNVERBOSE)

[Out]

9*x/exp(ln(exp(5)+exp(1)-x)+2)/(x+1)

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maxima [A]  time = 0.35, size = 28, normalized size = 1.22 \begin {gather*} -\frac {9 \, x e^{\left (-2\right )}}{x^{2} - x {\left (e^{5} + e - 1\right )} - e^{5} - e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*exp(5)+9*exp(1)+9*x^2)/((x^2+2*x+1)*exp(5)+(x^2+2*x+1)*exp(1)-x^3-2*x^2-x)/exp(log(exp(5)+exp(1)-
x)+2),x, algorithm="maxima")

[Out]

-9*x*e^(-2)/(x^2 - x*(e^5 + e - 1) - e^5 - e)

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mupad [B]  time = 4.00, size = 20, normalized size = 0.87 \begin {gather*} \frac {9\,x\,{\mathrm {e}}^{-2}}{\left (x+1\right )\,\left (\mathrm {e}-x+{\mathrm {e}}^5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- log(exp(1) - x + exp(5)) - 2)*(9*exp(1) + 9*exp(5) + 9*x^2))/(x - exp(1)*(2*x + x^2 + 1) - exp(5)*
(2*x + x^2 + 1) + 2*x^2 + x^3),x)

[Out]

(9*x*exp(-2))/((x + 1)*(exp(1) - x + exp(5)))

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sympy [A]  time = 1.14, size = 29, normalized size = 1.26 \begin {gather*} - \frac {9 x}{x^{2} e^{2} + x \left (- e^{7} - e^{3} + e^{2}\right ) - e^{7} - e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*exp(5)+9*exp(1)+9*x**2)/((x**2+2*x+1)*exp(5)+(x**2+2*x+1)*exp(1)-x**3-2*x**2-x)/exp(ln(exp(5)+exp
(1)-x)+2),x)

[Out]

-9*x/(x**2*exp(2) + x*(-exp(7) - exp(3) + exp(2)) - exp(7) - exp(3))

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