3.102.94 \(\int \frac {-45+9 e^4+e^{-2+2 x} (-9-18 x)}{25 x^2-10 e^4 x^2+e^8 x^2+e^{-4+4 x} x^2+e^{-2+2 x} (10 x^2-2 e^4 x^2)} \, dx\)

Optimal. Leaf size=21 \[ \frac {9}{\left (5-e^4+e^{-2+2 x}\right ) x} \]

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Rubi [A]  time = 0.30, antiderivative size = 27, normalized size of antiderivative = 1.29, number of steps used = 5, number of rules used = 4, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {6, 6741, 12, 6687} \begin {gather*} \frac {9 e^2}{\left (e^{2 x}+e^2 \left (5-e^4\right )\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-45 + 9*E^4 + E^(-2 + 2*x)*(-9 - 18*x))/(25*x^2 - 10*E^4*x^2 + E^8*x^2 + E^(-4 + 4*x)*x^2 + E^(-2 + 2*x)*
(10*x^2 - 2*E^4*x^2)),x]

[Out]

(9*E^2)/((E^(2*x) + E^2*(5 - E^4))*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 6687

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-45+9 e^4+e^{-2+2 x} (-9-18 x)}{e^8 x^2+e^{-4+4 x} x^2+\left (25-10 e^4\right ) x^2+e^{-2+2 x} \left (10 x^2-2 e^4 x^2\right )} \, dx\\ &=\int \frac {-45+9 e^4+e^{-2+2 x} (-9-18 x)}{e^{-4+4 x} x^2+\left (25-10 e^4+e^8\right ) x^2+e^{-2+2 x} \left (10 x^2-2 e^4 x^2\right )} \, dx\\ &=\int \frac {e^4 \left (-45 \left (1-\frac {e^4}{5}\right )+e^{-2+2 x} (-9-18 x)\right )}{\left (e^{2 x}+5 e^2 \left (1-\frac {e^4}{5}\right )\right )^2 x^2} \, dx\\ &=e^4 \int \frac {-45 \left (1-\frac {e^4}{5}\right )+e^{-2+2 x} (-9-18 x)}{\left (e^{2 x}+5 e^2 \left (1-\frac {e^4}{5}\right )\right )^2 x^2} \, dx\\ &=\frac {9 e^2}{\left (e^{2 x}+e^2 \left (5-e^4\right )\right ) x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 26, normalized size = 1.24 \begin {gather*} \frac {9 e^2}{\left (5 e^2-e^6+e^{2 x}\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-45 + 9*E^4 + E^(-2 + 2*x)*(-9 - 18*x))/(25*x^2 - 10*E^4*x^2 + E^8*x^2 + E^(-4 + 4*x)*x^2 + E^(-2 +
2*x)*(10*x^2 - 2*E^4*x^2)),x]

[Out]

(9*E^2)/((5*E^2 - E^6 + E^(2*x))*x)

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fricas [A]  time = 0.50, size = 21, normalized size = 1.00 \begin {gather*} -\frac {9}{x e^{4} - x e^{\left (2 \, x - 2\right )} - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x-9)*exp(2*x-2)+9*exp(4)-45)/(x^2*exp(2*x-2)^2+(-2*x^2*exp(4)+10*x^2)*exp(2*x-2)+x^2*exp(4)^2-
10*x^2*exp(4)+25*x^2),x, algorithm="fricas")

[Out]

-9/(x*e^4 - x*e^(2*x - 2) - 5*x)

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giac [A]  time = 0.12, size = 23, normalized size = 1.10 \begin {gather*} -\frac {9 \, e^{2}}{x e^{6} - 5 \, x e^{2} - x e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x-9)*exp(2*x-2)+9*exp(4)-45)/(x^2*exp(2*x-2)^2+(-2*x^2*exp(4)+10*x^2)*exp(2*x-2)+x^2*exp(4)^2-
10*x^2*exp(4)+25*x^2),x, algorithm="giac")

[Out]

-9*e^2/(x*e^6 - 5*x*e^2 - x*e^(2*x))

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maple [A]  time = 0.34, size = 20, normalized size = 0.95

method	result	size
norman	\(-\frac {9}{x \left (-5+{\mathrm e}^{4}-{\mathrm e}^{2 x -2}\right )}\)	\(20\)
risch	\(-\frac {9}{x \left (-5+{\mathrm e}^{4}-{\mathrm e}^{2 x -2}\right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-18*x-9)*exp(2*x-2)+9*exp(4)-45)/(x^2*exp(2*x-2)^2+(-2*x^2*exp(4)+10*x^2)*exp(2*x-2)+x^2*exp(4)^2-10*x^2
*exp(4)+25*x^2),x,method=_RETURNVERBOSE)

[Out]

-9/x/(-5+exp(4)-exp(2*x-2))

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maxima [A]  time = 0.40, size = 23, normalized size = 1.10 \begin {gather*} -\frac {9 \, e^{2}}{x {\left (e^{6} - 5 \, e^{2}\right )} - x e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x-9)*exp(2*x-2)+9*exp(4)-45)/(x^2*exp(2*x-2)^2+(-2*x^2*exp(4)+10*x^2)*exp(2*x-2)+x^2*exp(4)^2-
10*x^2*exp(4)+25*x^2),x, algorithm="maxima")

[Out]

-9*e^2/(x*(e^6 - 5*e^2) - x*e^(2*x))

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mupad [B]  time = 0.36, size = 19, normalized size = 0.90 \begin {gather*} \frac {9}{x\,\left ({\mathrm {e}}^{2\,x-2}-{\mathrm {e}}^4+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x - 2)*(18*x + 9) - 9*exp(4) + 45)/(x^2*exp(8) - 10*x^2*exp(4) - exp(2*x - 2)*(2*x^2*exp(4) - 10*x
^2) + x^2*exp(4*x - 4) + 25*x^2),x)

[Out]

9/(x*(exp(2*x - 2) - exp(4) + 5))

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sympy [A]  time = 0.14, size = 17, normalized size = 0.81 \begin {gather*} \frac {9}{x e^{2 x - 2} - x e^{4} + 5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x-9)*exp(2*x-2)+9*exp(4)-45)/(x**2*exp(2*x-2)**2+(-2*x**2*exp(4)+10*x**2)*exp(2*x-2)+x**2*exp(
4)**2-10*x**2*exp(4)+25*x**2),x)

[Out]

9/(x*exp(2*x - 2) - x*exp(4) + 5*x)

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