3.65.3 \(\int \frac {-96-160 x-32 x^2+32 x^3+(-96+256 x+192 x^2-128 x^3+32 x^4) \log (x)+(288 x-288 x^2-32 x^3+32 x^4) \log ^2(x)}{1+3 x+3 x^2+x^3} \, dx\)

Optimal. Leaf size=20 \[ -4+\left (4+4 x \left (1-\frac {4}{1+x}\right ) \log (x)\right )^2 \]

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Rubi [B]  time = 0.54, antiderivative size = 61, normalized size of antiderivative = 3.05, number of steps used = 31, number of rules used = 18, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6688, 12, 6742, 43, 2357, 2295, 2304, 2314, 31, 2317, 2391, 2296, 2305, 2319, 2347, 2344, 2301, 2318} \begin {gather*} 16 x^2 \log ^2(x)-128 x \log ^2(x)+\frac {640 x \log ^2(x)}{x+1}+\frac {256 \log ^2(x)}{(x+1)^2}-256 \log ^2(x)+32 x \log (x)-\frac {128 x \log (x)}{x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-96 - 160*x - 32*x^2 + 32*x^3 + (-96 + 256*x + 192*x^2 - 128*x^3 + 32*x^4)*Log[x] + (288*x - 288*x^2 - 32
*x^3 + 32*x^4)*Log[x]^2)/(1 + 3*x + 3*x^2 + x^3),x]

[Out]

32*x*Log[x] - (128*x*Log[x])/(1 + x) - 256*Log[x]^2 - 128*x*Log[x]^2 + 16*x^2*Log[x]^2 + (256*Log[x]^2)/(1 + x
)^2 + (640*x*Log[x]^2)/(1 + 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 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 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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 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 2317

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

Rule 2318

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

Rule 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2344

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

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 \frac {32 \left ((-3+x) (1+x)^2+\left (-3+8 x+6 x^2-4 x^3+x^4\right ) \log (x)+x \left (9-9 x-x^2+x^3\right ) \log ^2(x)\right )}{(1+x)^3} \, dx\\ &=32 \int \frac {(-3+x) (1+x)^2+\left (-3+8 x+6 x^2-4 x^3+x^4\right ) \log (x)+x \left (9-9 x-x^2+x^3\right ) \log ^2(x)}{(1+x)^3} \, dx\\ &=32 \int \left (\frac {-3+x}{1+x}+\frac {\left (-3+11 x-5 x^2+x^3\right ) \log (x)}{(1+x)^2}+\frac {(-3+x) (-1+x) x (3+x) \log ^2(x)}{(1+x)^3}\right ) \, dx\\ &=32 \int \frac {-3+x}{1+x} \, dx+32 \int \frac {\left (-3+11 x-5 x^2+x^3\right ) \log (x)}{(1+x)^2} \, dx+32 \int \frac {(-3+x) (-1+x) x (3+x) \log ^2(x)}{(1+x)^3} \, dx\\ &=32 \int \left (1-\frac {4}{1+x}\right ) \, dx+32 \int \left (-7 \log (x)+x \log (x)-\frac {20 \log (x)}{(1+x)^2}+\frac {24 \log (x)}{1+x}\right ) \, dx+32 \int \left (-4 \log ^2(x)+x \log ^2(x)-\frac {16 \log ^2(x)}{(1+x)^3}+\frac {20 \log ^2(x)}{(1+x)^2}\right ) \, dx\\ &=32 x-128 \log (1+x)+32 \int x \log (x) \, dx+32 \int x \log ^2(x) \, dx-128 \int \log ^2(x) \, dx-224 \int \log (x) \, dx-512 \int \frac {\log ^2(x)}{(1+x)^3} \, dx-640 \int \frac {\log (x)}{(1+x)^2} \, dx+640 \int \frac {\log ^2(x)}{(1+x)^2} \, dx+768 \int \frac {\log (x)}{1+x} \, dx\\ &=256 x-8 x^2-224 x \log (x)+16 x^2 \log (x)-\frac {640 x \log (x)}{1+x}-128 x \log ^2(x)+16 x^2 \log ^2(x)+\frac {256 \log ^2(x)}{(1+x)^2}+\frac {640 x \log ^2(x)}{1+x}-128 \log (1+x)+768 \log (x) \log (1+x)-32 \int x \log (x) \, dx+256 \int \log (x) \, dx-512 \int \frac {\log (x)}{x (1+x)^2} \, dx+640 \int \frac {1}{1+x} \, dx-768 \int \frac {\log (1+x)}{x} \, dx-1280 \int \frac {\log (x)}{1+x} \, dx\\ &=32 x \log (x)-\frac {640 x \log (x)}{1+x}-128 x \log ^2(x)+16 x^2 \log ^2(x)+\frac {256 \log ^2(x)}{(1+x)^2}+\frac {640 x \log ^2(x)}{1+x}+512 \log (1+x)-512 \log (x) \log (1+x)+768 \text {Li}_2(-x)+512 \int \frac {\log (x)}{(1+x)^2} \, dx-512 \int \frac {\log (x)}{x (1+x)} \, dx+1280 \int \frac {\log (1+x)}{x} \, dx\\ &=32 x \log (x)-\frac {128 x \log (x)}{1+x}-128 x \log ^2(x)+16 x^2 \log ^2(x)+\frac {256 \log ^2(x)}{(1+x)^2}+\frac {640 x \log ^2(x)}{1+x}+512 \log (1+x)-512 \log (x) \log (1+x)-512 \text {Li}_2(-x)-512 \int \frac {1}{1+x} \, dx-512 \int \frac {\log (x)}{x} \, dx+512 \int \frac {\log (x)}{1+x} \, dx\\ &=32 x \log (x)-\frac {128 x \log (x)}{1+x}-256 \log ^2(x)-128 x \log ^2(x)+16 x^2 \log ^2(x)+\frac {256 \log ^2(x)}{(1+x)^2}+\frac {640 x \log ^2(x)}{1+x}-512 \text {Li}_2(-x)-512 \int \frac {\log (1+x)}{x} \, dx\\ &=32 x \log (x)-\frac {128 x \log (x)}{1+x}-256 \log ^2(x)-128 x \log ^2(x)+16 x^2 \log ^2(x)+\frac {256 \log ^2(x)}{(1+x)^2}+\frac {640 x \log ^2(x)}{1+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 26, normalized size = 1.30 \begin {gather*} \frac {16 (-3+x) x \log (x) (2 (1+x)+(-3+x) x \log (x))}{(1+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-96 - 160*x - 32*x^2 + 32*x^3 + (-96 + 256*x + 192*x^2 - 128*x^3 + 32*x^4)*Log[x] + (288*x - 288*x^
2 - 32*x^3 + 32*x^4)*Log[x]^2)/(1 + 3*x + 3*x^2 + x^3),x]

[Out]

(16*(-3 + x)*x*Log[x]*(2*(1 + x) + (-3 + x)*x*Log[x]))/(1 + x)^2

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fricas [B]  time = 0.69, size = 48, normalized size = 2.40 \begin {gather*} \frac {16 \, {\left ({\left (x^{4} - 6 \, x^{3} + 9 \, x^{2}\right )} \log \relax (x)^{2} + 2 \, {\left (x^{3} - 2 \, x^{2} - 3 \, x\right )} \log \relax (x)\right )}}{x^{2} + 2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^4-32*x^3-288*x^2+288*x)*log(x)^2+(32*x^4-128*x^3+192*x^2+256*x-96)*log(x)+32*x^3-32*x^2-160*x
-96)/(x^3+3*x^2+3*x+1),x, algorithm="fricas")

[Out]

16*((x^4 - 6*x^3 + 9*x^2)*log(x)^2 + 2*(x^3 - 2*x^2 - 3*x)*log(x))/(x^2 + 2*x + 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {32 \, {\left (x^{3} + {\left (x^{4} - x^{3} - 9 \, x^{2} + 9 \, x\right )} \log \relax (x)^{2} - x^{2} + {\left (x^{4} - 4 \, x^{3} + 6 \, x^{2} + 8 \, x - 3\right )} \log \relax (x) - 5 \, x - 3\right )}}{x^{3} + 3 \, x^{2} + 3 \, x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^4-32*x^3-288*x^2+288*x)*log(x)^2+(32*x^4-128*x^3+192*x^2+256*x-96)*log(x)+32*x^3-32*x^2-160*x
-96)/(x^3+3*x^2+3*x+1),x, algorithm="giac")

[Out]

integrate(32*(x^3 + (x^4 - x^3 - 9*x^2 + 9*x)*log(x)^2 - x^2 + (x^4 - 4*x^3 + 6*x^2 + 8*x - 3)*log(x) - 5*x -
3)/(x^3 + 3*x^2 + 3*x + 1), x)

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maple [B]  time = 0.12, size = 48, normalized size = 2.40

method	result	size
risch	\(\frac {16 x^{2} \left (x^{2}-6 x +9\right ) \ln \relax (x )^{2}}{x^{2}+2 x +1}+\frac {32 \left (x^{2}+x +4\right ) \ln \relax (x )}{x +1}-128 \ln \relax (x )\)	\(48\)
norman	\(\frac {-96 x \ln \relax (x )-64 x^{2} \ln \relax (x )+144 x^{2} \ln \relax (x )^{2}+32 x^{3} \ln \relax (x )-96 x^{3} \ln \relax (x )^{2}+16 x^{4} \ln \relax (x )^{2}}{\left (x +1\right )^{2}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((32*x^4-32*x^3-288*x^2+288*x)*ln(x)^2+(32*x^4-128*x^3+192*x^2+256*x-96)*ln(x)+32*x^3-32*x^2-160*x-96)/(x^
3+3*x^2+3*x+1),x,method=_RETURNVERBOSE)

[Out]

16*x^2*(x^2-6*x+9)/(x^2+2*x+1)*ln(x)^2+32*(x^2+x+4)/(x+1)*ln(x)-128*ln(x)

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maxima [B]  time = 0.42, size = 173, normalized size = 8.65 \begin {gather*} 32 \, x - \frac {128 \, {\left (2 \, x + 1\right )} \log \relax (x)}{x^{2} + 2 \, x + 1} - \frac {16 \, {\left (2 \, x^{3} - {\left (x^{4} - 6 \, x^{3} + 9 \, x^{2}\right )} \log \relax (x)^{2} + 4 \, x^{2} - {\left (2 \, x^{3} - 9 \, x^{2}\right )} \log \relax (x) - 9 \, x - 11\right )}}{x^{2} + 2 \, x + 1} - \frac {16 \, {\left (6 \, x + 5\right )}}{x^{2} + 2 \, x + 1} - \frac {16 \, {\left (4 \, x + 3\right )}}{x^{2} + 2 \, x + 1} + \frac {80 \, {\left (2 \, x + 1\right )}}{x^{2} + 2 \, x + 1} + \frac {48 \, \log \relax (x)}{x^{2} + 2 \, x + 1} + \frac {48}{x^{2} + 2 \, x + 1} - \frac {176}{x + 1} + 80 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^4-32*x^3-288*x^2+288*x)*log(x)^2+(32*x^4-128*x^3+192*x^2+256*x-96)*log(x)+32*x^3-32*x^2-160*x
-96)/(x^3+3*x^2+3*x+1),x, algorithm="maxima")

[Out]

32*x - 128*(2*x + 1)*log(x)/(x^2 + 2*x + 1) - 16*(2*x^3 - (x^4 - 6*x^3 + 9*x^2)*log(x)^2 + 4*x^2 - (2*x^3 - 9*
x^2)*log(x) - 9*x - 11)/(x^2 + 2*x + 1) - 16*(6*x + 5)/(x^2 + 2*x + 1) - 16*(4*x + 3)/(x^2 + 2*x + 1) + 80*(2*
x + 1)/(x^2 + 2*x + 1) + 48*log(x)/(x^2 + 2*x + 1) + 48/(x^2 + 2*x + 1) - 176/(x + 1) + 80*log(x)

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mupad [B]  time = 4.50, size = 29, normalized size = 1.45 \begin {gather*} \frac {16\,x\,\ln \relax (x)\,\left (x-3\right )\,\left (2\,x+x^2\,\ln \relax (x)-3\,x\,\ln \relax (x)+2\right )}{{\left (x+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(160*x - log(x)*(256*x + 192*x^2 - 128*x^3 + 32*x^4 - 96) - log(x)^2*(288*x - 288*x^2 - 32*x^3 + 32*x^4)
+ 32*x^2 - 32*x^3 + 96)/(3*x + 3*x^2 + x^3 + 1),x)

[Out]

(16*x*log(x)*(x - 3)*(2*x + x^2*log(x) - 3*x*log(x) + 2))/(x + 1)^2

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sympy [B]  time = 0.26, size = 49, normalized size = 2.45 \begin {gather*} - 128 \log {\relax (x )} + \frac {\left (16 x^{4} - 96 x^{3} + 144 x^{2}\right ) \log {\relax (x )}^{2}}{x^{2} + 2 x + 1} + \frac {\left (32 x^{2} + 32 x + 128\right ) \log {\relax (x )}}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x**4-32*x**3-288*x**2+288*x)*ln(x)**2+(32*x**4-128*x**3+192*x**2+256*x-96)*ln(x)+32*x**3-32*x**
2-160*x-96)/(x**3+3*x**2+3*x+1),x)

[Out]

-128*log(x) + (16*x**4 - 96*x**3 + 144*x**2)*log(x)**2/(x**2 + 2*x + 1) + (32*x**2 + 32*x + 128)*log(x)/(x + 1
)

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