3.30.1 \(\int \frac {-9-17 x-98 x^2-25 x^3+(1800 x+475 x^2-75 x^3) \log (-9+x)+(9-x) \log (x)}{-9+x} \, dx\)

Optimal. Leaf size=19 \[ x (2+x-25 x (4+x) \log (-9+x)-\log (x)) \]

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Rubi [A]  time = 0.25, antiderivative size = 30, normalized size of antiderivative = 1.58, number of steps used = 10, number of rules used = 6, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6742, 1850, 43, 2414, 77, 2295} \begin {gather*} -25 x^3 \log (x-9)+x^2-100 x^2 \log (x-9)+2 x-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9 - 17*x - 98*x^2 - 25*x^3 + (1800*x + 475*x^2 - 75*x^3)*Log[-9 + x] + (9 - x)*Log[x])/(-9 + x),x]

[Out]

2*x + x^2 - 100*x^2*Log[-9 + x] - 25*x^3*Log[-9 + x] - x*Log[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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

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 2414

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*(x_)^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = IntHide[x^m*(f + g*x^r)^q, x]}, Dist[a + b*Log[c*(d + e*x)^n], u, x] - Dist[b*e*n, Int[SimplifyI
ntegrand[u/(d + e*x), x], x], x] /; InverseFunctionFreeQ[u, x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, q, r}, x]
&& IntegerQ[m] && IntegerQ[q] && IntegerQ[r]

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 {-9-17 x-98 x^2-25 x^3+1800 x \log (-9+x)+475 x^2 \log (-9+x)-75 x^3 \log (-9+x)}{-9+x}-\log (x)\right ) \, dx\\ &=\int \frac {-9-17 x-98 x^2-25 x^3+1800 x \log (-9+x)+475 x^2 \log (-9+x)-75 x^3 \log (-9+x)}{-9+x} \, dx-\int \log (x) \, dx\\ &=x-x \log (x)+\int \left (\frac {-9-17 x-98 x^2-25 x^3}{-9+x}-25 x (8+3 x) \log (-9+x)\right ) \, dx\\ &=x-x \log (x)-25 \int x (8+3 x) \log (-9+x) \, dx+\int \frac {-9-17 x-98 x^2-25 x^3}{-9+x} \, dx\\ &=x-100 x^2 \log (-9+x)-25 x^3 \log (-9+x)-x \log (x)+25 \int \frac {(-4-x) x^2}{9-x} \, dx+\int \left (-2924-\frac {26325}{-9+x}-323 x-25 x^2\right ) \, dx\\ &=-2923 x-\frac {323 x^2}{2}-\frac {25 x^3}{3}-26325 \log (9-x)-100 x^2 \log (-9+x)-25 x^3 \log (-9+x)-x \log (x)+25 \int \left (117+\frac {1053}{-9+x}+13 x+x^2\right ) \, dx\\ &=2 x+x^2-100 x^2 \log (-9+x)-25 x^3 \log (-9+x)-x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 30, normalized size = 1.58 \begin {gather*} 2 x+x^2-100 x^2 \log (-9+x)-25 x^3 \log (-9+x)-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9 - 17*x - 98*x^2 - 25*x^3 + (1800*x + 475*x^2 - 75*x^3)*Log[-9 + x] + (9 - x)*Log[x])/(-9 + x),x]

[Out]

2*x + x^2 - 100*x^2*Log[-9 + x] - 25*x^3*Log[-9 + x] - x*Log[x]

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fricas [A]  time = 0.57, size = 27, normalized size = 1.42 \begin {gather*} x^{2} - 25 \, {\left (x^{3} + 4 \, x^{2}\right )} \log \left (x - 9\right ) - x \log \relax (x) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9-x)*log(x)+(-75*x^3+475*x^2+1800*x)*log(x-9)-25*x^3-98*x^2-17*x-9)/(x-9),x, algorithm="fricas")

[Out]

x^2 - 25*(x^3 + 4*x^2)*log(x - 9) - x*log(x) + 2*x

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giac [A]  time = 0.22, size = 27, normalized size = 1.42 \begin {gather*} x^{2} - 25 \, {\left (x^{3} + 4 \, x^{2}\right )} \log \left (x - 9\right ) - x \log \relax (x) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9-x)*log(x)+(-75*x^3+475*x^2+1800*x)*log(x-9)-25*x^3-98*x^2-17*x-9)/(x-9),x, algorithm="giac")

[Out]

x^2 - 25*(x^3 + 4*x^2)*log(x - 9) - x*log(x) + 2*x

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maple [A]  time = 0.44, size = 29, normalized size = 1.53

method	result	size
risch	\(\left (-25 x^{3}-100 x^{2}\right ) \ln \left (x -9\right )+x^{2}-x \ln \relax (x )+2 x\)	\(29\)
default	\(x^{2}+2 x -26325 \ln \left (x -9\right )-25 \ln \left (x -9\right ) \left (x -9\right )^{3}-\frac {91125}{2}-775 \ln \left (x -9\right ) \left (x -9\right )^{2}-7875 \ln \left (x -9\right ) \left (x -9\right )-x \ln \relax (x )\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((9-x)*ln(x)+(-75*x^3+475*x^2+1800*x)*ln(x-9)-25*x^3-98*x^2-17*x-9)/(x-9),x,method=_RETURNVERBOSE)

[Out]

(-25*x^3-100*x^2)*ln(x-9)+x^2-x*ln(x)+2*x

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maxima [B]  time = 0.48, size = 71, normalized size = 3.74 \begin {gather*} x^{2} - \frac {25}{2} \, {\left (2 \, x^{3} + 27 \, x^{2} + 486 \, x + 4374 \, \log \left (x - 9\right )\right )} \log \left (x - 9\right ) + \frac {475}{2} \, {\left (x^{2} + 18 \, x + 162 \, \log \left (x - 9\right )\right )} \log \left (x - 9\right ) + 1800 \, {\left (x + 9 \, \log \left (x - 9\right )\right )} \log \left (x - 9\right ) - x \log \relax (x) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9-x)*log(x)+(-75*x^3+475*x^2+1800*x)*log(x-9)-25*x^3-98*x^2-17*x-9)/(x-9),x, algorithm="maxima")

[Out]

x^2 - 25/2*(2*x^3 + 27*x^2 + 486*x + 4374*log(x - 9))*log(x - 9) + 475/2*(x^2 + 18*x + 162*log(x - 9))*log(x -
9) + 1800*(x + 9*log(x - 9))*log(x - 9) - x*log(x) + 2*x

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mupad [B]  time = 1.88, size = 29, normalized size = 1.53 \begin {gather*} 2\,x-\ln \left (x-9\right )\,\left (25\,x^3+100\,x^2\right )-x\,\ln \relax (x)+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(17*x + log(x)*(x - 9) - log(x - 9)*(1800*x + 475*x^2 - 75*x^3) + 98*x^2 + 25*x^3 + 9)/(x - 9),x)

[Out]

2*x - log(x - 9)*(100*x^2 + 25*x^3) - x*log(x) + x^2

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sympy [A]  time = 0.54, size = 37, normalized size = 1.95 \begin {gather*} x^{2} - x \log {\relax (x )} + 2 x + \left (- 25 x^{3} - 100 x^{2} + \frac {29025}{4}\right ) \log {\left (x - 9 \right )} - \frac {29025 \log {\left (x - 9 \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9-x)*ln(x)+(-75*x**3+475*x**2+1800*x)*ln(x-9)-25*x**3-98*x**2-17*x-9)/(x-9),x)

[Out]

x**2 - x*log(x) + 2*x + (-25*x**3 - 100*x**2 + 29025/4)*log(x - 9) - 29025*log(x - 9)/4

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