3.49.40 \(\int \frac {425-275 x-125 x^2-25 x^3+(300+600 x+300 x^2) \log (5-x)+(-125-225 x-75 x^2+25 x^3) \log ^2(5-x)}{-5-9 x-3 x^2+x^3} \, dx\)

Optimal. Leaf size=22 \[ 25 (1+x) \left (-1+\frac {2}{1+x}+\log (5-x)\right )^2 \]

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Rubi [B]  time = 0.20, antiderivative size = 57, normalized size of antiderivative = 2.59, number of steps used = 15, number of rules used = 9, integrand size = 72, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6742, 44, 77, 88, 2390, 2301, 2389, 2296, 2295} \begin {gather*} 25 x+\frac {100}{x+1}-25 (5-x) \log ^2(5-x)+150 \log ^2(5-x)+50 (5-x) \log (5-x)-200 \log (5-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(425 - 275*x - 125*x^2 - 25*x^3 + (300 + 600*x + 300*x^2)*Log[5 - x] + (-125 - 225*x - 75*x^2 + 25*x^3)*Lo
g[5 - x]^2)/(-5 - 9*x - 3*x^2 + x^3),x]

[Out]

25*x + 100/(1 + x) - 200*Log[5 - x] + 50*(5 - x)*Log[5 - x] + 150*Log[5 - x]^2 - 25*(5 - x)*Log[5 - x]^2

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[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 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, 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 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 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 \left (\frac {425}{(-5+x) (1+x)^2}-\frac {275 x}{(-5+x) (1+x)^2}-\frac {125 x^2}{(-5+x) (1+x)^2}-\frac {25 x^3}{(-5+x) (1+x)^2}+\frac {300 \log (5-x)}{-5+x}+25 \log ^2(5-x)\right ) \, dx\\ &=-\left (25 \int \frac {x^3}{(-5+x) (1+x)^2} \, dx\right )+25 \int \log ^2(5-x) \, dx-125 \int \frac {x^2}{(-5+x) (1+x)^2} \, dx-275 \int \frac {x}{(-5+x) (1+x)^2} \, dx+300 \int \frac {\log (5-x)}{-5+x} \, dx+425 \int \frac {1}{(-5+x) (1+x)^2} \, dx\\ &=-\left (25 \int \left (1+\frac {125}{36 (-5+x)}+\frac {1}{6 (1+x)^2}-\frac {17}{36 (1+x)}\right ) \, dx\right )-25 \operatorname {Subst}\left (\int \log ^2(x) \, dx,x,5-x\right )-125 \int \left (\frac {25}{36 (-5+x)}-\frac {1}{6 (1+x)^2}+\frac {11}{36 (1+x)}\right ) \, dx-275 \int \left (\frac {5}{36 (-5+x)}+\frac {1}{6 (1+x)^2}-\frac {5}{36 (1+x)}\right ) \, dx+300 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,5-x\right )+425 \int \left (\frac {1}{36 (-5+x)}-\frac {1}{6 (1+x)^2}-\frac {1}{36 (1+x)}\right ) \, dx\\ &=-25 x+\frac {100}{1+x}-200 \log (5-x)+150 \log ^2(5-x)-25 (5-x) \log ^2(5-x)+50 \operatorname {Subst}(\int \log (x) \, dx,x,5-x)\\ &=25 x+\frac {100}{1+x}-200 \log (5-x)+50 (5-x) \log (5-x)+150 \log ^2(5-x)-25 (5-x) \log ^2(5-x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 40, normalized size = 1.82 \begin {gather*} \frac {25 \left (4+x+x^2-2 \left (-1+x^2\right ) \log (5-x)+(1+x)^2 \log ^2(5-x)\right )}{1+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(425 - 275*x - 125*x^2 - 25*x^3 + (300 + 600*x + 300*x^2)*Log[5 - x] + (-125 - 225*x - 75*x^2 + 25*x
^3)*Log[5 - x]^2)/(-5 - 9*x - 3*x^2 + x^3),x]

[Out]

(25*(4 + x + x^2 - 2*(-1 + x^2)*Log[5 - x] + (1 + x)^2*Log[5 - x]^2))/(1 + x)

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fricas [A]  time = 0.65, size = 43, normalized size = 1.95 \begin {gather*} \frac {25 \, {\left ({\left (x^{2} + 2 \, x + 1\right )} \log \left (-x + 5\right )^{2} + x^{2} - 2 \, {\left (x^{2} - 1\right )} \log \left (-x + 5\right ) + x + 4\right )}}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x^3-75*x^2-225*x-125)*log(5-x)^2+(300*x^2+600*x+300)*log(5-x)-25*x^3-125*x^2-275*x+425)/(x^3-3*
x^2-9*x-5),x, algorithm="fricas")

[Out]

25*((x^2 + 2*x + 1)*log(-x + 5)^2 + x^2 - 2*(x^2 - 1)*log(-x + 5) + x + 4)/(x + 1)

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giac [A]  time = 0.14, size = 39, normalized size = 1.77 \begin {gather*} 25 \, {\left (x + 1\right )} \log \left (-x + 5\right )^{2} - 50 \, x \log \left (-x + 5\right ) + 25 \, x + \frac {100}{x + 1} + 50 \, \log \left (x - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x^3-75*x^2-225*x-125)*log(5-x)^2+(300*x^2+600*x+300)*log(5-x)-25*x^3-125*x^2-275*x+425)/(x^3-3*
x^2-9*x-5),x, algorithm="giac")

[Out]

25*(x + 1)*log(-x + 5)^2 - 50*x*log(-x + 5) + 25*x + 100/(x + 1) + 50*log(x - 5)

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maple [B]  time = 0.08, size = 51, normalized size = 2.32

method	result	size
risch	\(\left (25 x +25\right ) \ln \left (5-x \right )^{2}-50 \ln \left (5-x \right ) x +\frac {50 x \ln \left (x -5\right )+25 x^{2}+50 \ln \left (x -5\right )+25 x +100}{x +1}\)	\(51\)
derivativedivides	\(-25 \ln \left (5-x \right )^{2} \left (5-x \right )+50 \left (5-x \right ) \ln \left (5-x \right )-125+25 x +150 \ln \left (5-x \right )^{2}-\frac {100}{-x -1}-200 \ln \left (5-x \right )\)	\(61\)
default	\(-25 \ln \left (5-x \right )^{2} \left (5-x \right )+50 \left (5-x \right ) \ln \left (5-x \right )-125+25 x +150 \ln \left (5-x \right )^{2}-\frac {100}{-x -1}-200 \ln \left (5-x \right )\)	\(61\)
norman	\(\frac {-75 x +50 \ln \left (5-x \right )+25 x^{2}+25 \ln \left (5-x \right )^{2}-50 x^{2} \ln \left (5-x \right )+50 \ln \left (5-x \right )^{2} x +25 \ln \left (5-x \right )^{2} x^{2}}{x +1}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((25*x^3-75*x^2-225*x-125)*ln(5-x)^2+(300*x^2+600*x+300)*ln(5-x)-25*x^3-125*x^2-275*x+425)/(x^3-3*x^2-9*x-
5),x,method=_RETURNVERBOSE)

[Out]

(25*x+25)*ln(5-x)^2-50*ln(5-x)*x+25*(2*x*ln(x-5)+x^2+2*ln(x-5)+x+4)/(x+1)

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maxima [B]  time = 0.41, size = 67, normalized size = 3.05 \begin {gather*} \frac {25 \, {\left (36 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (-x + 5\right )^{2} + 36 \, x^{2} - {\left (72 \, x^{2} + 17 \, x - 55\right )} \log \left (-x + 5\right ) + 36 \, x + 42\right )}}{36 \, {\left (x + 1\right )}} + \frac {425}{6 \, {\left (x + 1\right )}} + \frac {425}{36} \, \log \left (x - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x^3-75*x^2-225*x-125)*log(5-x)^2+(300*x^2+600*x+300)*log(5-x)-25*x^3-125*x^2-275*x+425)/(x^3-3*
x^2-9*x-5),x, algorithm="maxima")

[Out]

25/36*(36*(x^2 + 2*x + 1)*log(-x + 5)^2 + 36*x^2 - (72*x^2 + 17*x - 55)*log(-x + 5) + 36*x + 42)/(x + 1) + 425
/6/(x + 1) + 425/36*log(x - 5)

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mupad [B]  time = 0.17, size = 46, normalized size = 2.09 \begin {gather*} 50\,\ln \left (x-5\right )+25\,{\ln \left (5-x\right )}^2+\frac {100}{x+1}+x\,\left (25\,{\ln \left (5-x\right )}^2-50\,\ln \left (5-x\right )+25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((275*x + log(5 - x)^2*(225*x + 75*x^2 - 25*x^3 + 125) - log(5 - x)*(600*x + 300*x^2 + 300) + 125*x^2 + 25*
x^3 - 425)/(9*x + 3*x^2 - x^3 + 5),x)

[Out]

50*log(x - 5) + 25*log(5 - x)^2 + 100/(x + 1) + x*(25*log(5 - x)^2 - 50*log(5 - x) + 25)

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sympy [A]  time = 0.19, size = 34, normalized size = 1.55 \begin {gather*} - 50 x \log {\left (5 - x \right )} + 25 x + \left (25 x + 25\right ) \log {\left (5 - x \right )}^{2} + 50 \log {\left (x - 5 \right )} + \frac {100}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x**3-75*x**2-225*x-125)*ln(5-x)**2+(300*x**2+600*x+300)*ln(5-x)-25*x**3-125*x**2-275*x+425)/(x*
*3-3*x**2-9*x-5),x)

[Out]

-50*x*log(5 - x) + 25*x + (25*x + 25)*log(5 - x)**2 + 50*log(x - 5) + 100/(x + 1)

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