3.79.40 \(\int \frac {64 x}{1344+368 x^2+25 x^4+(2944+400 x^2) \log (5)+1600 \log ^2(5)} \, dx\)

Optimal. Leaf size=22 \[ \log \left (-\frac {25}{8}+\frac {2}{4+\frac {x^2}{2}+4 \log (5)}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 29, normalized size of antiderivative = 1.32, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 1989, 1107, 616, 31} \begin {gather*} \log \left (25 x^2+8 (21+25 \log (5))\right )-\log \left (x^2+8 (1+\log (5))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(64*x)/(1344 + 368*x^2 + 25*x^4 + (2944 + 400*x^2)*Log[5] + 1600*Log[5]^2),x]

[Out]

-Log[x^2 + 8*(1 + Log[5])] + Log[25*x^2 + 8*(21 + 25*Log[5])]

Rule 12

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

Rule 31

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

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 1107

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

Rule 1989

Int[(u_)^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[(d*x)^m*ExpandToSum[u, x]^p, x] /; FreeQ[{d, m, p}, x] &&
TrinomialQ[u, x] &&  !TrinomialMatchQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=64 \int \frac {x}{1344+368 x^2+25 x^4+\left (2944+400 x^2\right ) \log (5)+1600 \log ^2(5)} \, dx\\ &=64 \int \frac {x}{25 x^4+64 (1+\log (5)) (21+25 \log (5))+16 x^2 (23+25 \log (5))} \, dx\\ &=32 \operatorname {Subst}\left (\int \frac {1}{25 x^2+64 (1+\log (5)) (21+25 \log (5))+16 x (23+25 \log (5))} \, dx,x,x^2\right )\\ &=-\left (25 \operatorname {Subst}\left (\int \frac {1}{25 x+200 (1+\log (5))} \, dx,x,x^2\right )\right )+25 \operatorname {Subst}\left (\int \frac {1}{25 x+8 (21+25 \log (5))} \, dx,x,x^2\right )\\ &=-\log \left (x^2+8 (1+\log (5))\right )+\log \left (25 x^2+8 (21+25 \log (5))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 1.50 \begin {gather*} 64 \left (-\frac {1}{64} \log \left (8+x^2+8 \log (5)\right )+\frac {1}{64} \log \left (168+25 x^2+200 \log (5)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(64*x)/(1344 + 368*x^2 + 25*x^4 + (2944 + 400*x^2)*Log[5] + 1600*Log[5]^2),x]

[Out]

64*(-1/64*Log[8 + x^2 + 8*Log[5]] + Log[168 + 25*x^2 + 200*Log[5]]/64)

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fricas [A]  time = 0.54, size = 25, normalized size = 1.14 \begin {gather*} \log \left (25 \, x^{2} + 200 \, \log \relax (5) + 168\right ) - \log \left (x^{2} + 8 \, \log \relax (5) + 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(64*x/(1600*log(5)^2+(400*x^2+2944)*log(5)+25*x^4+368*x^2+1344),x, algorithm="fricas")

[Out]

log(25*x^2 + 200*log(5) + 168) - log(x^2 + 8*log(5) + 8)

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giac [A]  time = 0.17, size = 25, normalized size = 1.14 \begin {gather*} \log \left (25 \, x^{2} + 200 \, \log \relax (5) + 168\right ) - \log \left (x^{2} + 8 \, \log \relax (5) + 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(64*x/(1600*log(5)^2+(400*x^2+2944)*log(5)+25*x^4+368*x^2+1344),x, algorithm="giac")

[Out]

log(25*x^2 + 200*log(5) + 168) - log(x^2 + 8*log(5) + 8)

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maple [A]  time = 0.04, size = 26, normalized size = 1.18

method	result	size
default	\(\ln \left (25 x^{2}+200 \ln \relax (5)+168\right )-\ln \left (x^{2}+8 \ln \relax (5)+8\right )\)	\(26\)
norman	\(\ln \left (25 x^{2}+200 \ln \relax (5)+168\right )-\ln \left (x^{2}+8 \ln \relax (5)+8\right )\)	\(26\)
risch	\(\ln \left (25 x^{2}+200 \ln \relax (5)+168\right )-\ln \left (-x^{2}-8 \ln \relax (5)-8\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(64*x/(1600*ln(5)^2+(400*x^2+2944)*ln(5)+25*x^4+368*x^2+1344),x,method=_RETURNVERBOSE)

[Out]

ln(25*x^2+200*ln(5)+168)-ln(x^2+8*ln(5)+8)

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maxima [A]  time = 0.35, size = 25, normalized size = 1.14 \begin {gather*} \log \left (25 \, x^{2} + 200 \, \log \relax (5) + 168\right ) - \log \left (x^{2} + 8 \, \log \relax (5) + 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(64*x/(1600*log(5)^2+(400*x^2+2944)*log(5)+25*x^4+368*x^2+1344),x, algorithm="maxima")

[Out]

log(25*x^2 + 200*log(5) + 168) - log(x^2 + 8*log(5) + 8)

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mupad [B]  time = 0.16, size = 33, normalized size = 1.50 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {320000\,x^2}{58880000\,\ln \relax (5)+2\,x^2\,\left (2000000\,\ln \relax (5)+1840000\right )[phone]\,{\ln \relax (5)}^2+26880000}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((64*x)/(log(5)*(400*x^2 + 2944) + 1600*log(5)^2 + 368*x^2 + 25*x^4 + 1344),x)

[Out]

2*atanh((320000*x^2)/(58880000*log(5) + 2*x^2*(2000000*log(5) + 1840000) + 32000000*log(5)^2 + 26880000))

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sympy [A]  time = 0.41, size = 24, normalized size = 1.09 \begin {gather*} \log {\left (x^{2} + \frac {168}{25} + 8 \log {\relax (5 )} \right )} - \log {\left (x^{2} + 8 + 8 \log {\relax (5 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(64*x/(1600*ln(5)**2+(400*x**2+2944)*ln(5)+25*x**4+368*x**2+1344),x)

[Out]

log(x**2 + 168/25 + 8*log(5)) - log(x**2 + 8 + 8*log(5))

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