3.41.1 \(\int -\frac {1568 e^{\frac {1}{16} e^{4+\log ^2(2)}} x}{(49+16 x^2) (196+64 x^2)} \, dx\)

Optimal. Leaf size=22 \[ \frac {e^{e^{(-2+\log (2))^2}}}{4+\frac {64 x^2}{49}} \]

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.23, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 21, 261} \begin {gather*} \frac {49 e^{\frac {1}{16} e^{4+\log ^2(2)}}}{4 \left (16 x^2+49\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1568*E^(E^(4 + Log[2]^2)/16)*x)/((49 + 16*x^2)*(196 + 64*x^2)),x]

[Out]

(49*E^(E^(4 + Log[2]^2)/16))/(4*(49 + 16*x^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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (\left (1568 e^{\frac {1}{16} e^{4+\log ^2(2)}}\right ) \int \frac {x}{\left (49+16 x^2\right ) \left (196+64 x^2\right )} \, dx\right )\\ &=-\left (\left (392 e^{\frac {1}{16} e^{4+\log ^2(2)}}\right ) \int \frac {x}{\left (49+16 x^2\right )^2} \, dx\right )\\ &=\frac {49 e^{\frac {1}{16} e^{4+\log ^2(2)}}}{4 \left (49+16 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.23 \begin {gather*} \frac {49 e^{\frac {1}{16} e^{4+\log ^2(2)}}}{4 \left (49+16 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1568*E^(E^(4 + Log[2]^2)/16)*x)/((49 + 16*x^2)*(196 + 64*x^2)),x]

[Out]

(49*E^(E^(4 + Log[2]^2)/16))/(4*(49 + 16*x^2))

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fricas [A]  time = 0.53, size = 23, normalized size = 1.05 \begin {gather*} \frac {49 \, e^{\left (e^{\left (\log \relax (2)^{2} - 4 \, \log \relax (2) + 4\right )}\right )}}{4 \, {\left (16 \, x^{2} + 49\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-32*x*exp(-log(64/49*x^2+4)+exp(log(2)^2-4*log(2)+4))/(16*x^2+49),x, algorithm="fricas")

[Out]

49/4*e^(e^(log(2)^2 - 4*log(2) + 4))/(16*x^2 + 49)

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giac [A]  time = 0.20, size = 21, normalized size = 0.95 \begin {gather*} \frac {49 \, e^{\left (\frac {1}{16} \, e^{\left (\log \relax (2)^{2} + 4\right )}\right )}}{4 \, {\left (16 \, x^{2} + 49\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-32*x*exp(-log(64/49*x^2+4)+exp(log(2)^2-4*log(2)+4))/(16*x^2+49),x, algorithm="giac")

[Out]

49/4*e^(1/16*e^(log(2)^2 + 4))/(16*x^2 + 49)

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maple [A]  time = 0.07, size = 20, normalized size = 0.91

method	result	size
risch	\(\frac {49 \,{\mathrm e}^{\frac {{\mathrm e}^{\ln \relax (2)^{2}+4}}{16}}}{64 \left (x^{2}+\frac {49}{16}\right )}\)	\(20\)
default	\(\frac {49 \,{\mathrm e}^{\frac {{\mathrm e}^{\ln \relax (2)^{2}+4}}{16}}}{4 \left (16 x^{2}+49\right )}\)	\(22\)
norman	\(\frac {49 \,{\mathrm e}^{\frac {{\mathrm e}^{\ln \relax (2)^{2}+4}}{16}}}{4 \left (16 x^{2}+49\right )}\)	\(23\)
gosper	\({\mathrm e}^{-\ln \left (\frac {64 x^{2}}{49}+4\right )+{\mathrm e}^{\ln \relax (2)^{2}-4 \ln \relax (2)+4}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-32*x*exp(-ln(64/49*x^2+4)+exp(ln(2)^2-4*ln(2)+4))/(16*x^2+49),x,method=_RETURNVERBOSE)

[Out]

49/64*exp(1/16*exp(ln(2)^2+4))/(x^2+49/16)

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maxima [A]  time = 0.37, size = 21, normalized size = 0.95 \begin {gather*} \frac {49 \, e^{\left (\frac {1}{16} \, e^{\left (\log \relax (2)^{2} + 4\right )}\right )}}{4 \, {\left (16 \, x^{2} + 49\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-32*x*exp(-log(64/49*x^2+4)+exp(log(2)^2-4*log(2)+4))/(16*x^2+49),x, algorithm="maxima")

[Out]

49/4*e^(1/16*e^(log(2)^2 + 4))/(16*x^2 + 49)

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mupad [B]  time = 0.09, size = 21, normalized size = 0.95 \begin {gather*} \frac {49\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\ln \relax (2)}^2}\,{\mathrm {e}}^4}{16}}}{4\,\left (16\,x^2+49\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(32*x*exp(exp(log(2)^2 - 4*log(2) + 4) - log((64*x^2)/49 + 4)))/(16*x^2 + 49),x)

[Out]

(49*exp((exp(log(2)^2)*exp(4))/16))/(4*(16*x^2 + 49))

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sympy [A]  time = 0.10, size = 20, normalized size = 0.91 \begin {gather*} \frac {392 e^{\frac {e^{4} e^{\log {\relax (2 )}^{2}}}{16}}}{512 x^{2} + 1568} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-32*x*exp(-ln(64/49*x**2+4)+exp(ln(2)**2-4*ln(2)+4))/(16*x**2+49),x)

[Out]

392*exp(exp(4)*exp(log(2)**2)/16)/(512*x**2 + 1568)

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