3.10.19 \(\int -\frac {50 x}{-64-25 x^2-16 e^5 \log (\frac {5}{4})} \, dx\)

Optimal. Leaf size=18 \[ \log \left (4+\frac {25 x^2}{16}+e^5 \log \left (\frac {5}{4}\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 260} \begin {gather*} \log \left (25 x^2+16 \left (4+e^5 \log \left (\frac {5}{4}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50*x)/(-64 - 25*x^2 - 16*E^5*Log[5/4]),x]

[Out]

Log[25*x^2 + 16*(4 + E^5*Log[5/4])]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 260

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (50 \int \frac {x}{-64-25 x^2-16 e^5 \log \left (\frac {5}{4}\right )} \, dx\right )\\ &=\log \left (25 x^2+16 \left (4+e^5 \log \left (\frac {5}{4}\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 0.94 \begin {gather*} \log \left (64+25 x^2+16 e^5 \log \left (\frac {5}{4}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50*x)/(-64 - 25*x^2 - 16*E^5*Log[5/4]),x]

[Out]

Log[64 + 25*x^2 + 16*E^5*Log[5/4]]

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fricas [A]  time = 0.49, size = 14, normalized size = 0.78 \begin {gather*} \log \left (25 \, x^{2} - 16 \, e^{5} \log \left (\frac {4}{5}\right ) + 64\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-50*x/(16*exp(5)*log(4/5)-25*x^2-64),x, algorithm="fricas")

[Out]

log(25*x^2 - 16*e^5*log(4/5) + 64)

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giac [A]  time = 0.33, size = 14, normalized size = 0.78 \begin {gather*} \log \left (25 \, x^{2} - 16 \, e^{5} \log \left (\frac {4}{5}\right ) + 64\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-50*x/(16*exp(5)*log(4/5)-25*x^2-64),x, algorithm="giac")

[Out]

log(25*x^2 - 16*e^5*log(4/5) + 64)

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maple [A]  time = 0.14, size = 15, normalized size = 0.83

method	result	size
derivativedivides	\(\ln \left (16 \,{\mathrm e}^{5} \ln \left (\frac {4}{5}\right )-25 x^{2}-64\right )\)	\(15\)
default	\(\ln \left (16 \,{\mathrm e}^{5} \ln \left (\frac {4}{5}\right )-25 x^{2}-64\right )\)	\(15\)
norman	\(\ln \left (16 \,{\mathrm e}^{5} \ln \left (\frac {4}{5}\right )-25 x^{2}-64\right )\)	\(15\)
meijerg	\(\ln \left (1-\frac {25 x^{2}}{16 \,{\mathrm e}^{5} \ln \left (\frac {4}{5}\right )-64}\right )\)	\(19\)
risch	\(\ln \left (-32 \,{\mathrm e}^{5} \ln \relax (2)+16 \,{\mathrm e}^{5} \ln \relax (5)+25 x^{2}+64\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-50*x/(16*exp(5)*ln(4/5)-25*x^2-64),x,method=_RETURNVERBOSE)

[Out]

ln(16*exp(5)*ln(4/5)-25*x^2-64)

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maxima [A]  time = 0.35, size = 14, normalized size = 0.78 \begin {gather*} \log \left (25 \, x^{2} - 16 \, e^{5} \log \left (\frac {4}{5}\right ) + 64\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-50*x/(16*exp(5)*log(4/5)-25*x^2-64),x, algorithm="maxima")

[Out]

log(25*x^2 - 16*e^5*log(4/5) + 64)

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mupad [B]  time = 0.08, size = 14, normalized size = 0.78 \begin {gather*} \ln \left (25\,x^2-16\,{\mathrm {e}}^5\,\ln \left (\frac {4}{5}\right )+64\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x)/(25*x^2 - 16*exp(5)*log(4/5) + 64),x)

[Out]

log(25*x^2 - 16*exp(5)*log(4/5) + 64)

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sympy [A]  time = 0.14, size = 24, normalized size = 1.33 \begin {gather*} \log {\left (25 x^{2} - 32 e^{5} \log {\relax (2 )} + 64 + 16 e^{5} \log {\relax (5 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-50*x/(16*exp(5)*ln(4/5)-25*x**2-64),x)

[Out]

log(25*x**2 - 32*exp(5)*log(2) + 64 + 16*exp(5)*log(5))

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