3.49.84 \(\int \frac {-2 e^{5/2}-3 e^5}{4 x^2+e^5 (121+66 x+9 x^2)+e^{5/2} (44 x+12 x^2)} \, dx\)

Optimal. Leaf size=25 \[ -4+\frac {1}{2 \left (5+x+\frac {x}{e^{5/2}}+\frac {1+x}{2}\right )} \]

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Rubi [A]  time = 0.03, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {12, 1981, 27, 6, 32} \begin {gather*} \frac {e^{5/2}}{\left (2+3 e^{5/2}\right ) x+11 e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^(5/2) - 3*E^5)/(4*x^2 + E^5*(121 + 66*x + 9*x^2) + E^(5/2)*(44*x + 12*x^2)),x]

[Out]

E^(5/2)/(11*E^(5/2) + (2 + 3*E^(5/2))*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 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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 1981

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (\left (e^{5/2} \left (2+3 e^{5/2}\right )\right ) \int \frac {1}{4 x^2+e^5 \left (121+66 x+9 x^2\right )+e^{5/2} \left (44 x+12 x^2\right )} \, dx\right )\\ &=-\left (\left (e^{5/2} \left (2+3 e^{5/2}\right )\right ) \int \frac {1}{121 e^5+22 e^{5/2} \left (2+3 e^{5/2}\right ) x+\left (2+3 e^{5/2}\right )^2 x^2} \, dx\right )\\ &=-\left (\left (e^{5/2} \left (2+3 e^{5/2}\right )\right ) \int \frac {1}{\left (11 e^{5/2}+2 x+3 e^{5/2} x\right )^2} \, dx\right )\\ &=-\left (\left (e^{5/2} \left (2+3 e^{5/2}\right )\right ) \int \frac {1}{\left (11 e^{5/2}+\left (2+3 e^{5/2}\right ) x\right )^2} \, dx\right )\\ &=\frac {e^{5/2}}{11 e^{5/2}+\left (2+3 e^{5/2}\right ) x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 43, normalized size = 1.72 \begin {gather*} -\frac {-2 e^{5/2}-3 e^5}{\left (2+3 e^{5/2}\right ) \left (2 x+e^{5/2} (11+3 x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^(5/2) - 3*E^5)/(4*x^2 + E^5*(121 + 66*x + 9*x^2) + E^(5/2)*(44*x + 12*x^2)),x]

[Out]

-((-2*E^(5/2) - 3*E^5)/((2 + 3*E^(5/2))*(2*x + E^(5/2)*(11 + 3*x))))

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fricas [A]  time = 0.89, size = 17, normalized size = 0.68 \begin {gather*} \frac {e^{\frac {5}{2}}}{{\left (3 \, x + 11\right )} e^{\frac {5}{2}} + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*exp(5/2)^2-2*exp(5/2))/((9*x^2+66*x+121)*exp(5/2)^2+(12*x^2+44*x)*exp(5/2)+4*x^2),x, algorithm="
fricas")

[Out]

e^(5/2)/((3*x + 11)*e^(5/2) + 2*x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*exp(5/2)^2-2*exp(5/2))/((9*x^2+66*x+121)*exp(5/2)^2+(12*x^2+44*x)*exp(5/2)+4*x^2),x, algorithm="
giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -2*(3*exp(5)+2*exp(5/2))*1/44/sqrt(-exp(
5/2)^2+exp(5))*atan((9*sageVARx*exp(5)+12*sageVARx*exp(5/2)+4*sageVARx+33*exp(5)+22*exp(5/2))*1/22/sqrt(-exp(5
/2)^2+exp(5)))

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maple [A]  time = 0.22, size = 19, normalized size = 0.76

method	result	size
gosper	\(\frac {{\mathrm e}^{\frac {5}{2}}}{3 \,{\mathrm e}^{\frac {5}{2}} x +11 \,{\mathrm e}^{\frac {5}{2}}+2 x}\)	\(19\)
norman	\(\frac {\left (-\frac {3 \,{\mathrm e}^{\frac {5}{2}}}{11}-\frac {2}{11}\right ) x}{3 \,{\mathrm e}^{\frac {5}{2}} x +11 \,{\mathrm e}^{\frac {5}{2}}+2 x}\)	\(24\)
risch	\(\frac {{\mathrm e}^{5}}{\left (3 \,{\mathrm e}^{\frac {5}{2}}+2\right ) \left ({\mathrm e}^{\frac {5}{2}} x +\frac {11 \,{\mathrm e}^{\frac {5}{2}}}{3}+\frac {2 x}{3}\right )}+\frac {2 \,{\mathrm e}^{\frac {5}{2}}}{3 \left (3 \,{\mathrm e}^{\frac {5}{2}}+2\right ) \left ({\mathrm e}^{\frac {5}{2}} x +\frac {11 \,{\mathrm e}^{\frac {5}{2}}}{3}+\frac {2 x}{3}\right )}\)	\(53\)
meijerg	\(-\frac {3 \left (3 \,{\mathrm e}^{\frac {5}{2}}+2\right )^{2} x}{121 \left (9 \,{\mathrm e}^{5}+12 \,{\mathrm e}^{\frac {5}{2}}+4\right ) \left (1+\frac {x \,{\mathrm e}^{-\frac {5}{2}} \left (3 \,{\mathrm e}^{\frac {5}{2}}+2\right )}{11}\right )}-\frac {2 \,{\mathrm e}^{-\frac {5}{2}} \left (3 \,{\mathrm e}^{\frac {5}{2}}+2\right )^{2} x}{121 \left (9 \,{\mathrm e}^{5}+12 \,{\mathrm e}^{\frac {5}{2}}+4\right ) \left (1+\frac {x \,{\mathrm e}^{-\frac {5}{2}} \left (3 \,{\mathrm e}^{\frac {5}{2}}+2\right )}{11}\right )}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*exp(5/2)^2-2*exp(5/2))/((9*x^2+66*x+121)*exp(5/2)^2+(12*x^2+44*x)*exp(5/2)+4*x^2),x,method=_RETURNVERB
OSE)

[Out]

exp(5/2)/(3*exp(5/2)*x+11*exp(5/2)+2*x)

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maxima [B]  time = 0.36, size = 33, normalized size = 1.32 \begin {gather*} \frac {3 \, e^{5} + 2 \, e^{\frac {5}{2}}}{x {\left (9 \, e^{5} + 12 \, e^{\frac {5}{2}} + 4\right )} + 33 \, e^{5} + 22 \, e^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*exp(5/2)^2-2*exp(5/2))/((9*x^2+66*x+121)*exp(5/2)^2+(12*x^2+44*x)*exp(5/2)+4*x^2),x, algorithm="
maxima")

[Out]

(3*e^5 + 2*e^(5/2))/(x*(9*e^5 + 12*e^(5/2) + 4) + 33*e^5 + 22*e^(5/2))

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mupad [B]  time = 0.14, size = 18, normalized size = 0.72 \begin {gather*} \frac {{\mathrm {e}}^{5/2}}{11\,{\mathrm {e}}^{5/2}+x\,\left (3\,{\mathrm {e}}^{5/2}+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*exp(5) + 2*exp(5/2))/(exp(5/2)*(44*x + 12*x^2) + exp(5)*(66*x + 9*x^2 + 121) + 4*x^2),x)

[Out]

exp(5/2)/(11*exp(5/2) + x*(3*exp(5/2) + 2))

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sympy [B]  time = 0.36, size = 114, normalized size = 4.56 \begin {gather*} \frac {\left (- 3 e^{5} - 2 e^{\frac {5}{2}}\right ) \left (- 81 e^{10} - 216 e^{\frac {15}{2}} - 216 e^{5} - 96 e^{\frac {5}{2}} - 16\right )}{x \left (64 + 576 e^{\frac {5}{2}} + 2160 e^{5} + 4320 e^{\frac {15}{2}} + 4860 e^{10} + 2916 e^{\frac {25}{2}} + 729 e^{15}\right ) + 352 e^{\frac {5}{2}} + 2640 e^{5} + 7920 e^{\frac {15}{2}} + 11880 e^{10} + 8910 e^{\frac {25}{2}} + 2673 e^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*exp(5/2)**2-2*exp(5/2))/((9*x**2+66*x+121)*exp(5/2)**2+(12*x**2+44*x)*exp(5/2)+4*x**2),x)

[Out]

(-3*exp(5) - 2*exp(5/2))*(-81*exp(10) - 216*exp(15/2) - 216*exp(5) - 96*exp(5/2) - 16)/(x*(64 + 576*exp(5/2) +
2160*exp(5) + 4320*exp(15/2) + 4860*exp(10) + 2916*exp(25/2) + 729*exp(15)) + 352*exp(5/2) + 2640*exp(5) + 79
20*exp(15/2) + 11880*exp(10) + 8910*exp(25/2) + 2673*exp(15))

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