3.38.68 \(\int \frac {5-640 x+500 x^2+105 x \log (x)}{x-50 x^2+625 x^3} \, dx\) [3768]

Optimal. Leaf size=20 \[ \frac {-4+x+\log (x)-4 x \log (x)}{\frac {1}{5}-5 x} \]

[Out]

(ln(x)-4*x*ln(x)+x-4)/(1/5-5*x)

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Rubi [A]
time = 0.35, antiderivative size = 28, normalized size of antiderivative = 1.40, number of steps used = 10, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1608, 27, 6873, 12, 6874, 907, 2351, 31} \begin {gather*} -\frac {99}{5 (1-25 x)}+\frac {105 x \log (x)}{1-25 x}+5 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - 640*x + 500*x^2 + 105*x*Log[x])/(x - 50*x^2 + 625*x^3),x]

[Out]

-99/(5*(1 - 25*x)) + 5*Log[x] + (105*x*Log[x])/(1 - 25*x)

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 31

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

Rule 907

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1608

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 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[x*(d + e*x^r)^(q +
1)*((a + b*Log[c*x^n])/d), x] - Dist[b*(n/d), Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 6873

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

Rule 6874

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5-640 x+500 x^2+105 x \log (x)}{x \left (1-50 x+625 x^2\right )} \, dx\\ &=\int \frac {5-640 x+500 x^2+105 x \log (x)}{x (-1+25 x)^2} \, dx\\ &=\int \frac {5 \left (1-128 x+100 x^2+21 x \log (x)\right )}{(1-25 x)^2 x} \, dx\\ &=5 \int \frac {1-128 x+100 x^2+21 x \log (x)}{(1-25 x)^2 x} \, dx\\ &=5 \int \left (\frac {1-128 x+100 x^2}{x (-1+25 x)^2}+\frac {21 \log (x)}{(-1+25 x)^2}\right ) \, dx\\ &=5 \int \frac {1-128 x+100 x^2}{x (-1+25 x)^2} \, dx+105 \int \frac {\log (x)}{(-1+25 x)^2} \, dx\\ &=\frac {105 x \log (x)}{1-25 x}+5 \int \left (\frac {1}{x}-\frac {99}{(-1+25 x)^2}-\frac {21}{-1+25 x}\right ) \, dx+105 \int \frac {1}{-1+25 x} \, dx\\ &=-\frac {99}{5 (1-25 x)}+5 \log (x)+\frac {105 x \log (x)}{1-25 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.03, size = 20, normalized size = 1.00 \begin {gather*} \frac {5 (99+25 (-1+4 x) \log (x))}{-25+625 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - 640*x + 500*x^2 + 105*x*Log[x])/(x - 50*x^2 + 625*x^3),x]

[Out]

(5*(99 + 25*(-1 + 4*x)*Log[x]))/(-25 + 625*x)

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Maple [A]
time = 0.26, size = 27, normalized size = 1.35

method	result	size
norman	\(\frac {-5 \ln \left (x \right )+495 x +20 x \ln \left (x \right )}{-1+25 x}\)	\(22\)
default	\(5 \ln \left (x \right )+\frac {99}{5 \left (-1+25 x \right )}-\frac {105 \ln \left (x \right ) x}{-1+25 x}\)	\(27\)
risch	\(-\frac {21 \ln \left (x \right )}{5 \left (-1+25 x \right )}+\frac {100 x \ln \left (x \right )-4 \ln \left (x \right )+99}{-5+125 x}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((105*x*ln(x)+500*x^2-640*x+5)/(625*x^3-50*x^2+x),x,method=_RETURNVERBOSE)

[Out]

5*ln(x)+99/5/(-1+25*x)-105*ln(x)*x/(-1+25*x)

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Maxima [A]
time = 0.27, size = 25, normalized size = 1.25 \begin {gather*} -\frac {21 \, \log \left (x\right )}{5 \, {\left (25 \, x - 1\right )}} + \frac {99}{5 \, {\left (25 \, x - 1\right )}} + \frac {4}{5} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((105*x*log(x)+500*x^2-640*x+5)/(625*x^3-50*x^2+x),x, algorithm="maxima")

[Out]

-21/5*log(x)/(25*x - 1) + 99/5/(25*x - 1) + 4/5*log(x)

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Fricas [A]
time = 0.35, size = 20, normalized size = 1.00 \begin {gather*} \frac {25 \, {\left (4 \, x - 1\right )} \log \left (x\right ) + 99}{5 \, {\left (25 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((105*x*log(x)+500*x^2-640*x+5)/(625*x^3-50*x^2+x),x, algorithm="fricas")

[Out]

1/5*(25*(4*x - 1)*log(x) + 99)/(25*x - 1)

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Sympy [A]
time = 0.07, size = 22, normalized size = 1.10 \begin {gather*} \frac {4 \log {\left (x \right )}}{5} - \frac {21 \log {\left (x \right )}}{125 x - 5} + \frac {99}{125 x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((105*x*ln(x)+500*x**2-640*x+5)/(625*x**3-50*x**2+x),x)

[Out]

4*log(x)/5 - 21*log(x)/(125*x - 5) + 99/(125*x - 5)

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Giac [A]
time = 0.39, size = 25, normalized size = 1.25 \begin {gather*} -\frac {21 \, \log \left (x\right )}{5 \, {\left (25 \, x - 1\right )}} + \frac {99}{5 \, {\left (25 \, x - 1\right )}} + \frac {4}{5} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((105*x*log(x)+500*x^2-640*x+5)/(625*x^3-50*x^2+x),x, algorithm="giac")

[Out]

-21/5*log(x)/(25*x - 1) + 99/5/(25*x - 1) + 4/5*log(x)

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Mupad [B]
time = 2.14, size = 20, normalized size = 1.00 \begin {gather*} \frac {4\,\ln \left (x\right )}{5}-\frac {\frac {21\,\ln \left (x\right )}{5}-\frac {99}{5}}{25\,x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((105*x*log(x) - 640*x + 500*x^2 + 5)/(x - 50*x^2 + 625*x^3),x)

[Out]

(4*log(x))/5 - ((21*log(x))/5 - 99/5)/(25*x - 1)

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