3.34.8 \(\int \frac {(-200+100 x) \log (10)+(64 x^3-32 x^4+4 x^5) \log (2 e^{\frac {x^2}{25}})}{400 x^2-200 x^3+25 x^4} \, dx\)

Optimal. Leaf size=27 \[ -\frac {2 \log (10)}{(-4+x) x}+\log ^2\left (2 e^{\frac {x^2}{25}}\right ) \]

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Rubi [A]  time = 0.17, antiderivative size = 41, normalized size of antiderivative = 1.52, number of steps used = 9, number of rules used = 7, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.123, Rules used = {1594, 27, 12, 6688, 74, 2551, 30} \begin {gather*} -\frac {x^4}{625}+\frac {2}{25} x^2 \log \left (2 e^{\frac {x^2}{25}}\right )+\frac {2 \log (10)}{(4-x) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-200 + 100*x)*Log[10] + (64*x^3 - 32*x^4 + 4*x^5)*Log[2*E^(x^2/25)])/(400*x^2 - 200*x^3 + 25*x^4),x]

[Out]

-1/625*x^4 + (2*Log[10])/((4 - x)*x) + (2*x^2*Log[2*E^(x^2/25)])/25

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(-200+100 x) \log (10)+\left (64 x^3-32 x^4+4 x^5\right ) \log \left (2 e^{\frac {x^2}{25}}\right )}{x^2 \left (400-200 x+25 x^2\right )} \, dx\\ &=\int \frac {(-200+100 x) \log (10)+\left (64 x^3-32 x^4+4 x^5\right ) \log \left (2 e^{\frac {x^2}{25}}\right )}{25 (-4+x)^2 x^2} \, dx\\ &=\frac {1}{25} \int \frac {(-200+100 x) \log (10)+\left (64 x^3-32 x^4+4 x^5\right ) \log \left (2 e^{\frac {x^2}{25}}\right )}{(-4+x)^2 x^2} \, dx\\ &=\frac {1}{25} \int \left (\frac {100 (-2+x) \log (10)}{(-4+x)^2 x^2}+4 x \log \left (2 e^{\frac {x^2}{25}}\right )\right ) \, dx\\ &=\frac {4}{25} \int x \log \left (2 e^{\frac {x^2}{25}}\right ) \, dx+(4 \log (10)) \int \frac {-2+x}{(-4+x)^2 x^2} \, dx\\ &=\frac {2 \log (10)}{(4-x) x}+\frac {2}{25} x^2 \log \left (2 e^{\frac {x^2}{25}}\right )-\frac {2}{25} \int \frac {2 x^3}{25} \, dx\\ &=\frac {2 \log (10)}{(4-x) x}+\frac {2}{25} x^2 \log \left (2 e^{\frac {x^2}{25}}\right )-\frac {4 \int x^3 \, dx}{625}\\ &=-\frac {x^4}{625}+\frac {2 \log (10)}{(4-x) x}+\frac {2}{25} x^2 \log \left (2 e^{\frac {x^2}{25}}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 47, normalized size = 1.74 \begin {gather*} -\frac {x^4}{625}-\frac {\log (10)}{2 (-4+x)}+\frac {\log (10)}{2 x}+\frac {2}{25} x^2 \log \left (2 e^{\frac {x^2}{25}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-200 + 100*x)*Log[10] + (64*x^3 - 32*x^4 + 4*x^5)*Log[2*E^(x^2/25)])/(400*x^2 - 200*x^3 + 25*x^4),
x]

[Out]

-1/625*x^4 - Log[10]/(2*(-4 + x)) + Log[10]/(2*x) + (2*x^2*Log[2*E^(x^2/25)])/25

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fricas [A]  time = 0.80, size = 37, normalized size = 1.37 \begin {gather*} \frac {x^{6} - 4 \, x^{5} + 50 \, {\left (x^{4} - 4 \, x^{3}\right )} \log \relax (2) - 1250 \, \log \left (10\right )}{625 \, {\left (x^{2} - 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^5-32*x^4+64*x^3)*log(2*exp(1/25*x^2))+(100*x-200)*log(10))/(25*x^4-200*x^3+400*x^2),x, algorit
hm="fricas")

[Out]

1/625*(x^6 - 4*x^5 + 50*(x^4 - 4*x^3)*log(2) - 1250*log(10))/(x^2 - 4*x)

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giac [A]  time = 0.19, size = 26, normalized size = 0.96 \begin {gather*} \frac {1}{625} \, x^{4} + \frac {2}{25} \, x^{2} \log \relax (2) - \frac {2 \, \log \left (10\right )}{x^{2} - 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^5-32*x^4+64*x^3)*log(2*exp(1/25*x^2))+(100*x-200)*log(10))/(25*x^4-200*x^3+400*x^2),x, algorit
hm="giac")

[Out]

1/625*x^4 + 2/25*x^2*log(2) - 2*log(10)/(x^2 - 4*x)

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maple [A]  time = 0.06, size = 29, normalized size = 1.07

method	result	size
default	\(\ln \left (2 \,{\mathrm e}^{\frac {x^{2}}{25}}\right )^{2}+\frac {\ln \left (10\right )}{2 x}-\frac {\ln \left (10\right )}{2 \left (x -4\right )}\)	\(29\)
risch	\(\frac {2 x^{2} \ln \left ({\mathrm e}^{\frac {x^{2}}{25}}\right )}{25}-\frac {x^{6}-50 x^{4} \ln \relax (2)-4 x^{5}+200 x^{3} \ln \relax (2)+1250 \ln \relax (5)+1250 \ln \relax (2)}{625 \left (x -4\right ) x}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^5-32*x^4+64*x^3)*ln(2*exp(1/25*x^2))+(100*x-200)*ln(10))/(25*x^4-200*x^3+400*x^2),x,method=_RETURNVE
RBOSE)

[Out]

ln(2*exp(1/25*x^2))^2+1/2*ln(10)/x-1/2*ln(10)/(x-4)

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maxima [B]  time = 0.37, size = 170, normalized size = 6.30 \begin {gather*} -\frac {1}{625} \, x^{4} + \frac {1}{4} \, {\left (\frac {4 \, {\left (x - 2\right )}}{x^{2} - 4 \, x} + \log \left (x - 4\right ) - \log \relax (x)\right )} \log \left (10\right ) - \frac {1}{4} \, {\left (\frac {4}{x - 4} + \log \left (x - 4\right ) - \log \relax (x)\right )} \log \left (10\right ) - \frac {64}{625} \, {\left (3 \, x^{2} - 80\right )} \log \left (x - 4\right ) + \frac {256}{625} \, {\left (x^{2} - 32\right )} \log \left (x - 4\right ) - \frac {64}{625} \, {\left (x^{2} - 48\right )} \log \left (x - 4\right ) + \frac {2}{25} \, {\left (x^{2} + 16 \, x - \frac {128}{x - 4} + 96 \, \log \left (x - 4\right )\right )} \log \left (2 \, e^{\left (\frac {1}{25} \, x^{2}\right )}\right ) - \frac {32}{25} \, {\left (x - \frac {16}{x - 4} + 8 \, \log \left (x - 4\right )\right )} \log \left (2 \, e^{\left (\frac {1}{25} \, x^{2}\right )}\right ) - \frac {64}{25} \, {\left (\frac {4}{x - 4} - \log \left (x - 4\right )\right )} \log \left (2 \, e^{\left (\frac {1}{25} \, x^{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^5-32*x^4+64*x^3)*log(2*exp(1/25*x^2))+(100*x-200)*log(10))/(25*x^4-200*x^3+400*x^2),x, algorit
hm="maxima")

[Out]

-1/625*x^4 + 1/4*(4*(x - 2)/(x^2 - 4*x) + log(x - 4) - log(x))*log(10) - 1/4*(4/(x - 4) + log(x - 4) - log(x))
*log(10) - 64/625*(3*x^2 - 80)*log(x - 4) + 256/625*(x^2 - 32)*log(x - 4) - 64/625*(x^2 - 48)*log(x - 4) + 2/2
5*(x^2 + 16*x - 128/(x - 4) + 96*log(x - 4))*log(2*e^(1/25*x^2)) - 32/25*(x - 16/(x - 4) + 8*log(x - 4))*log(2
*e^(1/25*x^2)) - 64/25*(4/(x - 4) - log(x - 4))*log(2*e^(1/25*x^2))

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mupad [B]  time = 2.12, size = 25, normalized size = 0.93 \begin {gather*} \frac {2\,x^2\,\ln \relax (2)}{25}+\frac {x^4}{625}-\frac {2\,\ln \left (10\right )}{x\,\left (x-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(10)*(100*x - 200) + log(2*exp(x^2/25))*(64*x^3 - 32*x^4 + 4*x^5))/(400*x^2 - 200*x^3 + 25*x^4),x)

[Out]

(2*x^2*log(2))/25 + x^4/625 - (2*log(10))/(x*(x - 4))

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sympy [A]  time = 0.33, size = 26, normalized size = 0.96 \begin {gather*} \frac {x^{4}}{625} + \frac {2 x^{2} \log {\relax (2 )}}{25} - \frac {2 \log {\left (10 \right )}}{x^{2} - 4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**5-32*x**4+64*x**3)*ln(2*exp(1/25*x**2))+(100*x-200)*ln(10))/(25*x**4-200*x**3+400*x**2),x)

[Out]

x**4/625 + 2*x**2*log(2)/25 - 2*log(10)/(x**2 - 4*x)

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