3.62.75 \(\int \frac {-27 x+42 x^2-9 x^3+(12-3 x) \log (20-5 x)}{-4+x} \, dx\)

Optimal. Leaf size=20 \[ 3 x \left (x-x^2-\log (5 (4-x))\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 34, normalized size of antiderivative = 1.70, number of steps used = 6, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6742, 771, 2389, 2295} \begin {gather*} -3 x^3+3 x^2-12 \log (4-x)+3 (4-x) \log (5 (4-x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-27*x + 42*x^2 - 9*x^3 + (12 - 3*x)*Log[20 - 5*x])/(-4 + x),x]

[Out]

3*x^2 - 3*x^3 - 12*Log[4 - x] + 3*(4 - x)*Log[5*(4 - x)]

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

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 \left (-\frac {3 x \left (9-14 x+3 x^2\right )}{-4+x}-3 \log (20-5 x)\right ) \, dx\\ &=-\left (3 \int \frac {x \left (9-14 x+3 x^2\right )}{-4+x} \, dx\right )-3 \int \log (20-5 x) \, dx\\ &=\frac {3}{5} \operatorname {Subst}(\int \log (x) \, dx,x,20-5 x)-3 \int \left (1+\frac {4}{-4+x}-2 x+3 x^2\right ) \, dx\\ &=3 x^2-3 x^3-12 \log (4-x)+3 (4-x) \log (5 (4-x))\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 40, normalized size = 2.00 \begin {gather*} -123 (-4+x)-33 (-4+x)^2-3 (-4+x)^3+3 x-3 (-4+x) \log (-5 (-4+x))-12 \log (-4+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-27*x + 42*x^2 - 9*x^3 + (12 - 3*x)*Log[20 - 5*x])/(-4 + x),x]

[Out]

-123*(-4 + x) - 33*(-4 + x)^2 - 3*(-4 + x)^3 + 3*x - 3*(-4 + x)*Log[-5*(-4 + x)] - 12*Log[-4 + x]

________________________________________________________________________________________

fricas [A]  time = 0.73, size = 20, normalized size = 1.00 \begin {gather*} -3 \, x^{3} + 3 \, x^{2} - 3 \, x \log \left (-5 \, x + 20\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x+12)*log(-5*x+20)-9*x^3+42*x^2-27*x)/(x-4),x, algorithm="fricas")

[Out]

-3*x^3 + 3*x^2 - 3*x*log(-5*x + 20)

________________________________________________________________________________________

giac [A]  time = 0.34, size = 20, normalized size = 1.00 \begin {gather*} -3 \, x^{3} + 3 \, x^{2} - 3 \, x \log \left (-5 \, x + 20\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x+12)*log(-5*x+20)-9*x^3+42*x^2-27*x)/(x-4),x, algorithm="giac")

[Out]

-3*x^3 + 3*x^2 - 3*x*log(-5*x + 20)

________________________________________________________________________________________

maple [A]  time = 0.39, size = 21, normalized size = 1.05

method	result	size
norman	\(3 x^{2}-3 x^{3}-3 \ln \left (-5 x +20\right ) x\)	\(21\)
risch	\(3 x^{2}-3 x^{3}-3 \ln \left (-5 x +20\right ) x\)	\(21\)
derivativedivides	\(\frac {3 \left (-5 x +20\right )^{3}}{125}+\frac {3 \left (-5 x +20\right ) \ln \left (-5 x +20\right )}{5}-120 x +480-\frac {33 \left (-5 x +20\right )^{2}}{25}-12 \ln \left (-5 x +20\right )\)	\(45\)
default	\(\frac {3 \left (-5 x +20\right )^{3}}{125}+\frac {3 \left (-5 x +20\right ) \ln \left (-5 x +20\right )}{5}-120 x +480-\frac {33 \left (-5 x +20\right )^{2}}{25}-12 \ln \left (-5 x +20\right )\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x+12)*ln(-5*x+20)-9*x^3+42*x^2-27*x)/(x-4),x,method=_RETURNVERBOSE)

[Out]

3*x^2-3*x^3-3*ln(-5*x+20)*x

________________________________________________________________________________________

maxima [C]  time = 0.43, size = 49, normalized size = 2.45 \begin {gather*} -3 \, x^{3} + 3 \, x^{2} - 12 \, {\left (-i \, \pi - \log \relax (5)\right )} \log \left (x - 4\right ) + 12 \, \log \left (x - 4\right )^{2} - 3 \, {\left (x + 4 \, \log \left (x - 4\right )\right )} \log \left (-5 \, x + 20\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x+12)*log(-5*x+20)-9*x^3+42*x^2-27*x)/(x-4),x, algorithm="maxima")

[Out]

-3*x^3 + 3*x^2 - 12*(-I*pi - log(5))*log(x - 4) + 12*log(x - 4)^2 - 3*(x + 4*log(x - 4))*log(-5*x + 20)

________________________________________________________________________________________

mupad [B]  time = 0.11, size = 16, normalized size = 0.80 \begin {gather*} -3\,x\,\left (\ln \left (20-5\,x\right )-x+x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(27*x + log(20 - 5*x)*(3*x - 12) - 42*x^2 + 9*x^3)/(x - 4),x)

[Out]

-3*x*(log(20 - 5*x) - x + x^2)

________________________________________________________________________________________

sympy [A]  time = 0.11, size = 19, normalized size = 0.95 \begin {gather*} - 3 x^{3} + 3 x^{2} - 3 x \log {\left (20 - 5 x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x+12)*ln(-5*x+20)-9*x**3+42*x**2-27*x)/(x-4),x)

[Out]

-3*x**3 + 3*x**2 - 3*x*log(20 - 5*x)

________________________________________________________________________________________