3.23.89 \(\int \frac {39-9 \log (4+e^2)}{169-78 e^2 x+9 e^4 x^2+(-78+18 e^2 x) \log (4+e^2)+9 \log ^2(4+e^2)} \, dx\)

Optimal. Leaf size=22 \[ \frac {x}{\frac {13}{3}-e^2 x-\log \left (4+e^2\right )} \]

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Rubi [A]  time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.45, number of steps used = 4, number of rules used = 4, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 1981, 27, 32} \begin {gather*} \frac {13-3 \log \left (4+e^2\right )}{e^2 \left (-3 e^2 x+13-3 \log \left (4+e^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(39 - 9*Log[4 + E^2])/(169 - 78*E^2*x + 9*E^4*x^2 + (-78 + 18*E^2*x)*Log[4 + E^2] + 9*Log[4 + E^2]^2),x]

[Out]

(13 - 3*Log[4 + E^2])/(E^2*(13 - 3*E^2*x - 3*Log[4 + E^2]))

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 (3 \left (13-3 \log \left (4+e^2\right )\right )\right ) \int \frac {1}{169-78 e^2 x+9 e^4 x^2+\left (-78+18 e^2 x\right ) \log \left (4+e^2\right )+9 \log ^2\left (4+e^2\right )} \, dx\\ &=\left (3 \left (13-3 \log \left (4+e^2\right )\right )\right ) \int \frac {1}{9 e^4 x^2-6 e^2 x \left (13-3 \log \left (4+e^2\right )\right )+\left (13-3 \log \left (4+e^2\right )\right )^2} \, dx\\ &=\left (3 \left (13-3 \log \left (4+e^2\right )\right )\right ) \int \frac {1}{\left (-13+3 e^2 x+3 \log \left (4+e^2\right )\right )^2} \, dx\\ &=\frac {13-3 \log \left (4+e^2\right )}{e^2 \left (13-3 e^2 x-3 \log \left (4+e^2\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 1.59 \begin {gather*} -\frac {39-9 \log \left (4+e^2\right )}{3 e^2 \left (-13+3 e^2 x+3 \log \left (4+e^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(39 - 9*Log[4 + E^2])/(169 - 78*E^2*x + 9*E^4*x^2 + (-78 + 18*E^2*x)*Log[4 + E^2] + 9*Log[4 + E^2]^2
),x]

[Out]

-1/3*(39 - 9*Log[4 + E^2])/(E^2*(-13 + 3*E^2*x + 3*Log[4 + E^2]))

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fricas [A]  time = 0.78, size = 31, normalized size = 1.41 \begin {gather*} \frac {3 \, \log \left (e^{2} + 4\right ) - 13}{3 \, x e^{4} + 3 \, e^{2} \log \left (e^{2} + 4\right ) - 13 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*log(4+exp(2))+39)/(9*log(4+exp(2))^2+(18*exp(2)*x-78)*log(4+exp(2))+9*x^2*exp(2)^2-78*exp(2)*x+1
69),x, algorithm="fricas")

[Out]

(3*log(e^2 + 4) - 13)/(3*x*e^4 + 3*e^2*log(e^2 + 4) - 13*e^2)

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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((-9*log(4+exp(2))+39)/(9*log(4+exp(2))^2+(18*exp(2)*x-78)*log(4+exp(2))+9*x^2*exp(2)^2-78*exp(2)*x+1
69),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -3*(3*ln(exp(2)+4)-13)*1/6/sqrt(exp(2)^2
-exp(4))/(-3*ln(exp(2)+4)+13)*ln(sqrt((18*sageVARx*exp(4)+18*exp(2)*ln(exp(2)+4)-78*exp(2))^2+(-6*sqrt(-exp(2)
^2+exp(4))*(-3*ln(exp

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maple [A]  time = 1.68, size = 31, normalized size = 1.41

method	result	size
gosper	\(\frac {\left (3 \ln \left (4+{\mathrm e}^{2}\right )-13\right ) {\mathrm e}^{-2}}{3 \,{\mathrm e}^{2} x +3 \ln \left (4+{\mathrm e}^{2}\right )-13}\)	\(31\)
norman	\(\frac {\left (3 \ln \left (4+{\mathrm e}^{2}\right )-13\right ) {\mathrm e}^{-2}}{3 \,{\mathrm e}^{2} x +3 \ln \left (4+{\mathrm e}^{2}\right )-13}\)	\(31\)
risch	\(\frac {{\mathrm e}^{-2} \ln \left (4+{\mathrm e}^{2}\right )}{{\mathrm e}^{2} x +\ln \left (4+{\mathrm e}^{2}\right )-\frac {13}{3}}-\frac {13 \,{\mathrm e}^{-2}}{3 \left ({\mathrm e}^{2} x +\ln \left (4+{\mathrm e}^{2}\right )-\frac {13}{3}\right )}\)	\(40\)
meijerg	\(\frac {9 \ln \left (4+{\mathrm e}^{2}\right ) \left (-3 \ln \left (4+{\mathrm e}^{2}\right )+13\right ) x}{\left (3 \ln \left (4+{\mathrm e}^{2}\right )-13\right )^{3} \left (1+\frac {3 x \,{\mathrm e}^{2}}{3 \ln \left (4+{\mathrm e}^{2}\right )-13}\right )}-\frac {39 \left (-3 \ln \left (4+{\mathrm e}^{2}\right )+13\right ) x}{\left (3 \ln \left (4+{\mathrm e}^{2}\right )-13\right )^{3} \left (1+\frac {3 x \,{\mathrm e}^{2}}{3 \ln \left (4+{\mathrm e}^{2}\right )-13}\right )}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-9*ln(4+exp(2))+39)/(9*ln(4+exp(2))^2+(18*exp(2)*x-78)*ln(4+exp(2))+9*x^2*exp(2)^2-78*exp(2)*x+169),x,met
hod=_RETURNVERBOSE)

[Out]

(3*ln(4+exp(2))-13)/exp(2)/(3*exp(2)*x+3*ln(4+exp(2))-13)

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maxima [A]  time = 0.34, size = 31, normalized size = 1.41 \begin {gather*} \frac {3 \, \log \left (e^{2} + 4\right ) - 13}{3 \, x e^{4} + 3 \, e^{2} \log \left (e^{2} + 4\right ) - 13 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*log(4+exp(2))+39)/(9*log(4+exp(2))^2+(18*exp(2)*x-78)*log(4+exp(2))+9*x^2*exp(2)^2-78*exp(2)*x+1
69),x, algorithm="maxima")

[Out]

(3*log(e^2 + 4) - 13)/(3*x*e^4 + 3*e^2*log(e^2 + 4) - 13*e^2)

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mupad [B]  time = 0.18, size = 28, normalized size = 1.27 \begin {gather*} \frac {{\mathrm {e}}^{-2}\,\left (3\,\ln \left ({\mathrm {e}}^2+4\right )-13\right )}{3\,\ln \left ({\mathrm {e}}^2+4\right )+3\,x\,{\mathrm {e}}^2-13} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*log(exp(2) + 4) - 39)/(log(exp(2) + 4)*(18*x*exp(2) - 78) - 78*x*exp(2) + 9*log(exp(2) + 4)^2 + 9*x^2*
exp(4) + 169),x)

[Out]

(exp(-2)*(3*log(exp(2) + 4) - 13))/(3*log(exp(2) + 4) + 3*x*exp(2) - 13)

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sympy [B]  time = 0.26, size = 34, normalized size = 1.55 \begin {gather*} - \frac {39 - 9 \log {\left (4 + e^{2} \right )}}{9 x e^{4} - 39 e^{2} + 9 e^{2} \log {\left (4 + e^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*ln(4+exp(2))+39)/(9*ln(4+exp(2))**2+(18*exp(2)*x-78)*ln(4+exp(2))+9*x**2*exp(2)**2-78*exp(2)*x+1
69),x)

[Out]

-(39 - 9*log(4 + exp(2)))/(9*x*exp(4) - 39*exp(2) + 9*exp(2)*log(4 + exp(2)))

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