3.21.45 \(\int \frac {20-4 x+225 x^2-90 x^3+9 x^4+20 \log (2 x)}{75-30 x+3 x^2} \, dx\)

Optimal. Leaf size=19 \[ 1+x^3-\frac {4 x \log (2 x)}{3 (-5+x)} \]

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Rubi [A]  time = 0.18, antiderivative size = 20, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {27, 12, 6742, 43, 2314, 31} \begin {gather*} x^3+\frac {4 x \log (2 x)}{3 (5-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20 - 4*x + 225*x^2 - 90*x^3 + 9*x^4 + 20*Log[2*x])/(75 - 30*x + 3*x^2),x]

[Out]

x^3 + (4*x*Log[2*x])/(3*(5 - 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2314

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {20-4 x+225 x^2-90 x^3+9 x^4+20 \log (2 x)}{3 (-5+x)^2} \, dx\\ &=\frac {1}{3} \int \frac {20-4 x+225 x^2-90 x^3+9 x^4+20 \log (2 x)}{(-5+x)^2} \, dx\\ &=\frac {1}{3} \int \left (\frac {20}{(-5+x)^2}-\frac {4 x}{(-5+x)^2}+\frac {225 x^2}{(-5+x)^2}-\frac {90 x^3}{(-5+x)^2}+\frac {9 x^4}{(-5+x)^2}+\frac {20 \log (2 x)}{(-5+x)^2}\right ) \, dx\\ &=\frac {20}{3 (5-x)}-\frac {4}{3} \int \frac {x}{(-5+x)^2} \, dx+3 \int \frac {x^4}{(-5+x)^2} \, dx+\frac {20}{3} \int \frac {\log (2 x)}{(-5+x)^2} \, dx-30 \int \frac {x^3}{(-5+x)^2} \, dx+75 \int \frac {x^2}{(-5+x)^2} \, dx\\ &=\frac {20}{3 (5-x)}+\frac {4 x \log (2 x)}{3 (5-x)}-\frac {4}{3} \int \left (\frac {5}{(-5+x)^2}+\frac {1}{-5+x}\right ) \, dx+\frac {4}{3} \int \frac {1}{-5+x} \, dx+3 \int \left (75+\frac {625}{(-5+x)^2}+\frac {500}{-5+x}+10 x+x^2\right ) \, dx-30 \int \left (10+\frac {125}{(-5+x)^2}+\frac {75}{-5+x}+x\right ) \, dx+75 \int \left (1+\frac {25}{(-5+x)^2}+\frac {10}{-5+x}\right ) \, dx\\ &=x^3+\frac {4 x \log (2 x)}{3 (5-x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 27, normalized size = 1.42 \begin {gather*} \frac {1}{3} \left (3 x^3-4 \log (x)+\frac {20 \log (2 x)}{5-x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 - 4*x + 225*x^2 - 90*x^3 + 9*x^4 + 20*Log[2*x])/(75 - 30*x + 3*x^2),x]

[Out]

(3*x^3 - 4*Log[x] + (20*Log[2*x])/(5 - x))/3

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fricas [A]  time = 0.99, size = 25, normalized size = 1.32 \begin {gather*} \frac {3 \, x^{4} - 15 \, x^{3} - 4 \, x \log \left (2 \, x\right )}{3 \, {\left (x - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*log(2*x)+9*x^4-90*x^3+225*x^2-4*x+20)/(3*x^2-30*x+75),x, algorithm="fricas")

[Out]

1/3*(3*x^4 - 15*x^3 - 4*x*log(2*x))/(x - 5)

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giac [A]  time = 0.19, size = 19, normalized size = 1.00 \begin {gather*} x^{3} - \frac {20 \, \log \left (2 \, x\right )}{3 \, {\left (x - 5\right )}} - \frac {4}{3} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*log(2*x)+9*x^4-90*x^3+225*x^2-4*x+20)/(3*x^2-30*x+75),x, algorithm="giac")

[Out]

x^3 - 20/3*log(2*x)/(x - 5) - 4/3*log(x)

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maple [A]  time = 0.41, size = 19, normalized size = 1.00

method	result	size
derivativedivides	\(x^{3}-\frac {8 \ln \left (2 x \right ) x}{3 \left (2 x -10\right )}\)	\(19\)
default	\(x^{3}-\frac {8 \ln \left (2 x \right ) x}{3 \left (2 x -10\right )}\)	\(19\)
risch	\(-\frac {20 \ln \left (2 x \right )}{3 \left (x -5\right )}+x^{3}-\frac {4 \ln \relax (x )}{3}\)	\(20\)
norman	\(\frac {x^{4}-5 x^{3}-\frac {4 x \ln \left (2 x \right )}{3}}{x -5}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*ln(2*x)+9*x^4-90*x^3+225*x^2-4*x+20)/(3*x^2-30*x+75),x,method=_RETURNVERBOSE)

[Out]

x^3-8/3*ln(2*x)*x/(2*x-10)

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maxima [A]  time = 0.39, size = 19, normalized size = 1.00 \begin {gather*} x^{3} - \frac {20 \, \log \left (2 \, x\right )}{3 \, {\left (x - 5\right )}} - \frac {4}{3} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*log(2*x)+9*x^4-90*x^3+225*x^2-4*x+20)/(3*x^2-30*x+75),x, algorithm="maxima")

[Out]

x^3 - 20/3*log(2*x)/(x - 5) - 4/3*log(x)

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mupad [B]  time = 1.21, size = 27, normalized size = 1.42 \begin {gather*} -\frac {x\,\left (4\,\ln \left (2\,x\right )+15\,x^2-3\,x^3\right )}{3\,\left (x-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*log(2*x) - 4*x + 225*x^2 - 90*x^3 + 9*x^4 + 20)/(3*x^2 - 30*x + 75),x)

[Out]

-(x*(4*log(2*x) + 15*x^2 - 3*x^3))/(3*(x - 5))

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sympy [A]  time = 0.14, size = 20, normalized size = 1.05 \begin {gather*} x^{3} - \frac {4 \log {\relax (x )}}{3} - \frac {20 \log {\left (2 x \right )}}{3 x - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*ln(2*x)+9*x**4-90*x**3+225*x**2-4*x+20)/(3*x**2-30*x+75),x)

[Out]

x**3 - 4*log(x)/3 - 20*log(2*x)/(3*x - 15)

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