3.89.9 \(\int \frac {(32+32 x-32 x^2-32 x^3) \log ^3(1+x)+(-6 x^5-6 x^7) \log ^3(\frac {1+x^2}{x})+(30 x^3+12 x^4+42 x^5+12 x^6+12 x^7) \log (1+x) \log ^3(\frac {1+x^2}{x})+(-60 x^2-60 x^3-60 x^4-60 x^5) \log ^2(1+x) \log ^3(\frac {1+x^2}{x})}{(x+x^2+x^3+x^4) \log ^3(1+x) \log ^3(\frac {1+x^2}{x})} \, dx\)

Optimal. Leaf size=28 \[ 5+3 \left (-5+\frac {x^2}{\log (1+x)}\right )^2+\frac {16}{\log ^2\left (\frac {1}{x}+x\right )} \]

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Rubi [B]  time = 1.18, antiderivative size = 157, normalized size of antiderivative = 5.61, number of steps used = 72, number of rules used = 16, integrand size = 162, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.099, Rules used = {6688, 2411, 2353, 2297, 2298, 2302, 30, 2306, 2309, 2178, 2418, 2389, 2400, 2399, 2390, 6686} \begin {gather*} -\frac {12 x^3 (x+1)}{\log (x+1)}+\frac {16}{\log ^2\left (\frac {x^2+1}{x}\right )}+\frac {3 (x+1)^4}{\log ^2(x+1)}-\frac {12 (x+1)^3}{\log ^2(x+1)}+\frac {18 (x+1)^2}{\log ^2(x+1)}-\frac {12 (x+1)}{\log ^2(x+1)}+\frac {3}{\log ^2(x+1)}+\frac {12 (x+1)^4}{\log (x+1)}-\frac {36 (x+1)^3}{\log (x+1)}+\frac {36 (x+1)^2}{\log (x+1)}-\frac {30 x (x+1)}{\log (x+1)}+\frac {18 (x+1)}{\log (x+1)}-\frac {30}{\log (x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((32 + 32*x - 32*x^2 - 32*x^3)*Log[1 + x]^3 + (-6*x^5 - 6*x^7)*Log[(1 + x^2)/x]^3 + (30*x^3 + 12*x^4 + 42*
x^5 + 12*x^6 + 12*x^7)*Log[1 + x]*Log[(1 + x^2)/x]^3 + (-60*x^2 - 60*x^3 - 60*x^4 - 60*x^5)*Log[1 + x]^2*Log[(
1 + x^2)/x]^3)/((x + x^2 + x^3 + x^4)*Log[1 + x]^3*Log[(1 + x^2)/x]^3),x]

[Out]

3/Log[1 + x]^2 - (12*(1 + x))/Log[1 + x]^2 + (18*(1 + x)^2)/Log[1 + x]^2 - (12*(1 + x)^3)/Log[1 + x]^2 + (3*(1
+ x)^4)/Log[1 + x]^2 - 30/Log[1 + x] + (18*(1 + x))/Log[1 + x] - (30*x*(1 + x))/Log[1 + x] - (12*x^3*(1 + x))
/Log[1 + x] + (36*(1 + x)^2)/Log[1 + x] - (36*(1 + x)^3)/Log[1 + x] + (12*(1 + x)^4)/Log[1 + x] + 16/Log[(1 +
x^2)/x]^2

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2297

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

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2302

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

Rule 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 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] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

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 2399

Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.)), x_Symbol] :> Int[ExpandIn
tegrand[(f + g*x)^q/(a + b*Log[c*(d + e*x)^n]), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g,
0] && IGtQ[q, 0]

Rule 2400

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((
d + e*x)*(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1))/(b*e*n*(p + 1)), x] + (-Dist[(q + 1)/(b*n*(p + 1)), I
nt[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x] + Dist[(q*(e*f - d*g))/(b*e*n*(p + 1)), Int[(f + g*x
)^(q - 1)*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x]) /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g,
0] && LtQ[p, -1] && GtQ[q, 0]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {6 x^4}{(1+x) \log ^3(1+x)}+\frac {6 x^2 \left (5+2 x+2 x^2\right )}{(1+x) \log ^2(1+x)}-\frac {60 x}{\log (1+x)}-\frac {32 \left (-1+x^2\right )}{x \left (1+x^2\right ) \log ^3\left (\frac {1+x^2}{x}\right )}\right ) \, dx\\ &=-\left (6 \int \frac {x^4}{(1+x) \log ^3(1+x)} \, dx\right )+6 \int \frac {x^2 \left (5+2 x+2 x^2\right )}{(1+x) \log ^2(1+x)} \, dx-32 \int \frac {-1+x^2}{x \left (1+x^2\right ) \log ^3\left (\frac {1+x^2}{x}\right )} \, dx-60 \int \frac {x}{\log (1+x)} \, dx\\ &=\frac {16}{\log ^2\left (\frac {1+x^2}{x}\right )}+6 \int \left (-\frac {5}{\log ^2(1+x)}+\frac {5 x}{\log ^2(1+x)}+\frac {2 x^3}{\log ^2(1+x)}+\frac {5}{(1+x) \log ^2(1+x)}\right ) \, dx-6 \operatorname {Subst}\left (\int \frac {(-1+x)^4}{x \log ^3(x)} \, dx,x,1+x\right )-60 \int \left (-\frac {1}{\log (1+x)}+\frac {1+x}{\log (1+x)}\right ) \, dx\\ &=\frac {16}{\log ^2\left (\frac {1+x^2}{x}\right )}-6 \operatorname {Subst}\left (\int \left (-\frac {4}{\log ^3(x)}+\frac {1}{x \log ^3(x)}+\frac {6 x}{\log ^3(x)}-\frac {4 x^2}{\log ^3(x)}+\frac {x^3}{\log ^3(x)}\right ) \, dx,x,1+x\right )+12 \int \frac {x^3}{\log ^2(1+x)} \, dx-30 \int \frac {1}{\log ^2(1+x)} \, dx+30 \int \frac {x}{\log ^2(1+x)} \, dx+30 \int \frac {1}{(1+x) \log ^2(1+x)} \, dx+60 \int \frac {1}{\log (1+x)} \, dx-60 \int \frac {1+x}{\log (1+x)} \, dx\\ &=-\frac {30 x (1+x)}{\log (1+x)}-\frac {12 x^3 (1+x)}{\log (1+x)}+\frac {16}{\log ^2\left (\frac {1+x^2}{x}\right )}-6 \operatorname {Subst}\left (\int \frac {1}{x \log ^3(x)} \, dx,x,1+x\right )-6 \operatorname {Subst}\left (\int \frac {x^3}{\log ^3(x)} \, dx,x,1+x\right )+24 \operatorname {Subst}\left (\int \frac {1}{\log ^3(x)} \, dx,x,1+x\right )+24 \operatorname {Subst}\left (\int \frac {x^2}{\log ^3(x)} \, dx,x,1+x\right )+30 \int \frac {1}{\log (1+x)} \, dx-30 \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,1+x\right )+30 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,1+x\right )+36 \int \frac {x^2}{\log (1+x)} \, dx-36 \operatorname {Subst}\left (\int \frac {x}{\log ^3(x)} \, dx,x,1+x\right )+48 \int \frac {x^3}{\log (1+x)} \, dx+60 \int \frac {x}{\log (1+x)} \, dx+60 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,1+x\right )-60 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,1+x\right )\\ &=-\frac {12 (1+x)}{\log ^2(1+x)}+\frac {18 (1+x)^2}{\log ^2(1+x)}-\frac {12 (1+x)^3}{\log ^2(1+x)}+\frac {3 (1+x)^4}{\log ^2(1+x)}+\frac {30 (1+x)}{\log (1+x)}-\frac {30 x (1+x)}{\log (1+x)}-\frac {12 x^3 (1+x)}{\log (1+x)}+\frac {16}{\log ^2\left (\frac {1+x^2}{x}\right )}+60 \text {li}(1+x)-6 \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (1+x)\right )+12 \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,1+x\right )-12 \operatorname {Subst}\left (\int \frac {x^3}{\log ^2(x)} \, dx,x,1+x\right )+30 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (1+x)\right )+36 \int \left (\frac {1}{\log (1+x)}-\frac {2 (1+x)}{\log (1+x)}+\frac {(1+x)^2}{\log (1+x)}\right ) \, dx-36 \operatorname {Subst}\left (\int \frac {x}{\log ^2(x)} \, dx,x,1+x\right )+36 \operatorname {Subst}\left (\int \frac {x^2}{\log ^2(x)} \, dx,x,1+x\right )+48 \int \left (-\frac {1}{\log (1+x)}+\frac {3 (1+x)}{\log (1+x)}-\frac {3 (1+x)^2}{\log (1+x)}+\frac {(1+x)^3}{\log (1+x)}\right ) \, dx+60 \int \left (-\frac {1}{\log (1+x)}+\frac {1+x}{\log (1+x)}\right ) \, dx-60 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (1+x)\right )\\ &=-60 \text {Ei}(2 \log (1+x))+\frac {3}{\log ^2(1+x)}-\frac {12 (1+x)}{\log ^2(1+x)}+\frac {18 (1+x)^2}{\log ^2(1+x)}-\frac {12 (1+x)^3}{\log ^2(1+x)}+\frac {3 (1+x)^4}{\log ^2(1+x)}-\frac {30}{\log (1+x)}+\frac {18 (1+x)}{\log (1+x)}-\frac {30 x (1+x)}{\log (1+x)}-\frac {12 x^3 (1+x)}{\log (1+x)}+\frac {36 (1+x)^2}{\log (1+x)}-\frac {36 (1+x)^3}{\log (1+x)}+\frac {12 (1+x)^4}{\log (1+x)}+\frac {16}{\log ^2\left (\frac {1+x^2}{x}\right )}+60 \text {li}(1+x)+12 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,1+x\right )+36 \int \frac {1}{\log (1+x)} \, dx+36 \int \frac {(1+x)^2}{\log (1+x)} \, dx-48 \int \frac {1}{\log (1+x)} \, dx+48 \int \frac {(1+x)^3}{\log (1+x)} \, dx-48 \operatorname {Subst}\left (\int \frac {x^3}{\log (x)} \, dx,x,1+x\right )-60 \int \frac {1}{\log (1+x)} \, dx+60 \int \frac {1+x}{\log (1+x)} \, dx-72 \int \frac {1+x}{\log (1+x)} \, dx-72 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,1+x\right )+108 \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,1+x\right )+144 \int \frac {1+x}{\log (1+x)} \, dx-144 \int \frac {(1+x)^2}{\log (1+x)} \, dx\\ &=-60 \text {Ei}(2 \log (1+x))+\frac {3}{\log ^2(1+x)}-\frac {12 (1+x)}{\log ^2(1+x)}+\frac {18 (1+x)^2}{\log ^2(1+x)}-\frac {12 (1+x)^3}{\log ^2(1+x)}+\frac {3 (1+x)^4}{\log ^2(1+x)}-\frac {30}{\log (1+x)}+\frac {18 (1+x)}{\log (1+x)}-\frac {30 x (1+x)}{\log (1+x)}-\frac {12 x^3 (1+x)}{\log (1+x)}+\frac {36 (1+x)^2}{\log (1+x)}-\frac {36 (1+x)^3}{\log (1+x)}+\frac {12 (1+x)^4}{\log (1+x)}+\frac {16}{\log ^2\left (\frac {1+x^2}{x}\right )}+72 \text {li}(1+x)+36 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,1+x\right )+36 \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,1+x\right )-48 \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (1+x)\right )-48 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,1+x\right )+48 \operatorname {Subst}\left (\int \frac {x^3}{\log (x)} \, dx,x,1+x\right )-60 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,1+x\right )+60 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,1+x\right )-72 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (1+x)\right )-72 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,1+x\right )+108 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (1+x)\right )+144 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,1+x\right )-144 \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,1+x\right )\\ &=-132 \text {Ei}(2 \log (1+x))+108 \text {Ei}(3 \log (1+x))-48 \text {Ei}(4 \log (1+x))+\frac {3}{\log ^2(1+x)}-\frac {12 (1+x)}{\log ^2(1+x)}+\frac {18 (1+x)^2}{\log ^2(1+x)}-\frac {12 (1+x)^3}{\log ^2(1+x)}+\frac {3 (1+x)^4}{\log ^2(1+x)}-\frac {30}{\log (1+x)}+\frac {18 (1+x)}{\log (1+x)}-\frac {30 x (1+x)}{\log (1+x)}-\frac {12 x^3 (1+x)}{\log (1+x)}+\frac {36 (1+x)^2}{\log (1+x)}-\frac {36 (1+x)^3}{\log (1+x)}+\frac {12 (1+x)^4}{\log (1+x)}+\frac {16}{\log ^2\left (\frac {1+x^2}{x}\right )}+36 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (1+x)\right )+48 \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (1+x)\right )+60 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (1+x)\right )-72 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (1+x)\right )+144 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (1+x)\right )-144 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (1+x)\right )\\ &=\frac {3}{\log ^2(1+x)}-\frac {12 (1+x)}{\log ^2(1+x)}+\frac {18 (1+x)^2}{\log ^2(1+x)}-\frac {12 (1+x)^3}{\log ^2(1+x)}+\frac {3 (1+x)^4}{\log ^2(1+x)}-\frac {30}{\log (1+x)}+\frac {18 (1+x)}{\log (1+x)}-\frac {30 x (1+x)}{\log (1+x)}-\frac {12 x^3 (1+x)}{\log (1+x)}+\frac {36 (1+x)^2}{\log (1+x)}-\frac {36 (1+x)^3}{\log (1+x)}+\frac {12 (1+x)^4}{\log (1+x)}+\frac {16}{\log ^2\left (\frac {1+x^2}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.26, size = 46, normalized size = 1.64 \begin {gather*} 60 \text {Ei}(\log (1+x))+\frac {3 x^4}{\log ^2(1+x)}-\frac {30 x^2}{\log (1+x)}+\frac {16}{\log ^2\left (\frac {1}{x}+x\right )}-60 \text {li}(1+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((32 + 32*x - 32*x^2 - 32*x^3)*Log[1 + x]^3 + (-6*x^5 - 6*x^7)*Log[(1 + x^2)/x]^3 + (30*x^3 + 12*x^4
+ 42*x^5 + 12*x^6 + 12*x^7)*Log[1 + x]*Log[(1 + x^2)/x]^3 + (-60*x^2 - 60*x^3 - 60*x^4 - 60*x^5)*Log[1 + x]^2
*Log[(1 + x^2)/x]^3)/((x + x^2 + x^3 + x^4)*Log[1 + x]^3*Log[(1 + x^2)/x]^3),x]

[Out]

60*ExpIntegralEi[Log[1 + x]] + (3*x^4)/Log[1 + x]^2 - (30*x^2)/Log[1 + x] + 16/Log[x^(-1) + x]^2 - 60*LogInteg
ral[1 + x]

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fricas [B]  time = 0.47, size = 66, normalized size = 2.36 \begin {gather*} \frac {3 \, x^{4} \log \left (\frac {x^{2} + 1}{x}\right )^{2} - 30 \, x^{2} \log \left (x + 1\right ) \log \left (\frac {x^{2} + 1}{x}\right )^{2} + 16 \, \log \left (x + 1\right )^{2}}{\log \left (x + 1\right )^{2} \log \left (\frac {x^{2} + 1}{x}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^3-32*x^2+32*x+32)*log(x+1)^3+(-60*x^5-60*x^4-60*x^3-60*x^2)*log((x^2+1)/x)^3*log(x+1)^2+(12*
x^7+12*x^6+42*x^5+12*x^4+30*x^3)*log((x^2+1)/x)^3*log(x+1)+(-6*x^7-6*x^5)*log((x^2+1)/x)^3)/(x^4+x^3+x^2+x)/lo
g((x^2+1)/x)^3/log(x+1)^3,x, algorithm="fricas")

[Out]

(3*x^4*log((x^2 + 1)/x)^2 - 30*x^2*log(x + 1)*log((x^2 + 1)/x)^2 + 16*log(x + 1)^2)/(log(x + 1)^2*log((x^2 + 1
)/x)^2)

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giac [B]  time = 0.90, size = 91, normalized size = 3.25 \begin {gather*} \frac {16 \, {\left (x^{2} - 1\right )}}{x^{2} \log \left (x^{2} + 1\right )^{2} - 2 \, x^{2} \log \left (x^{2} + 1\right ) \log \relax (x) + x^{2} \log \relax (x)^{2} - \log \left (x^{2} + 1\right )^{2} + 2 \, \log \left (x^{2} + 1\right ) \log \relax (x) - \log \relax (x)^{2}} + \frac {3 \, {\left (x^{4} - 10 \, x^{2} \log \left (x + 1\right )\right )}}{\log \left (x + 1\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^3-32*x^2+32*x+32)*log(x+1)^3+(-60*x^5-60*x^4-60*x^3-60*x^2)*log((x^2+1)/x)^3*log(x+1)^2+(12*
x^7+12*x^6+42*x^5+12*x^4+30*x^3)*log((x^2+1)/x)^3*log(x+1)+(-6*x^7-6*x^5)*log((x^2+1)/x)^3)/(x^4+x^3+x^2+x)/lo
g((x^2+1)/x)^3/log(x+1)^3,x, algorithm="giac")

[Out]

16*(x^2 - 1)/(x^2*log(x^2 + 1)^2 - 2*x^2*log(x^2 + 1)*log(x) + x^2*log(x)^2 - log(x^2 + 1)^2 + 2*log(x^2 + 1)*
log(x) - log(x)^2) + 3*(x^4 - 10*x^2*log(x + 1))/log(x + 1)^2

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maple [B]  time = 0.27, size = 106, normalized size = 3.79

method	result	size
default	\(\frac {3 \left (x +1\right )^{4}}{\ln \left (x +1\right )^{2}}-\frac {12 \left (x +1\right )^{3}}{\ln \left (x +1\right )^{2}}+\frac {18 \left (x +1\right )^{2}}{\ln \left (x +1\right )^{2}}-\frac {30 \left (x +1\right )^{2}}{\ln \left (x +1\right )}-\frac {12 \left (x +1\right )}{\ln \left (x +1\right )^{2}}+\frac {60 x +60}{\ln \left (x +1\right )}+\frac {3}{\ln \left (x +1\right )^{2}}+\frac {16}{\ln \left (\frac {x^{2}+1}{x}\right )^{2}}-\frac {30}{\ln \left (x +1\right )}\)	\(106\)
risch	\(\frac {3 x^{2} \left (x^{2}-10 \ln \left (x +1\right )\right )}{\ln \left (x +1\right )^{2}}-\frac {64}{\left (\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x^{2}+1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+1\right )}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+1\right )}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x^{2}+1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+1\right )}{x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (x^{2}+1\right )}{x}\right )^{3}-2 i \ln \relax (x )+2 i \ln \left (x^{2}+1\right )\right )^{2}}\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-32*x^3-32*x^2+32*x+32)*ln(x+1)^3+(-60*x^5-60*x^4-60*x^3-60*x^2)*ln((x^2+1)/x)^3*ln(x+1)^2+(12*x^7+12*x^
6+42*x^5+12*x^4+30*x^3)*ln((x^2+1)/x)^3*ln(x+1)+(-6*x^7-6*x^5)*ln((x^2+1)/x)^3)/(x^4+x^3+x^2+x)/ln((x^2+1)/x)^
3/ln(x+1)^3,x,method=_RETURNVERBOSE)

[Out]

3*(x+1)^4/ln(x+1)^2-12*(x+1)^3/ln(x+1)^2+18*(x+1)^2/ln(x+1)^2-30*(x+1)^2/ln(x+1)-12*(x+1)/ln(x+1)^2+60/ln(x+1)
*(x+1)+3/ln(x+1)^2+16/ln((x^2+1)/x)^2-30/ln(x+1)

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maxima [B]  time = 0.51, size = 126, normalized size = 4.50 \begin {gather*} \frac {3 \, x^{4} \log \relax (x)^{2} - 30 \, x^{2} \log \left (x + 1\right ) \log \relax (x)^{2} + 3 \, {\left (x^{4} - 10 \, x^{2} \log \left (x + 1\right )\right )} \log \left (x^{2} + 1\right )^{2} - 6 \, {\left (x^{4} \log \relax (x) - 10 \, x^{2} \log \left (x + 1\right ) \log \relax (x)\right )} \log \left (x^{2} + 1\right ) + 16 \, \log \left (x + 1\right )^{2}}{\log \left (x^{2} + 1\right )^{2} \log \left (x + 1\right )^{2} - 2 \, \log \left (x^{2} + 1\right ) \log \left (x + 1\right )^{2} \log \relax (x) + \log \left (x + 1\right )^{2} \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^3-32*x^2+32*x+32)*log(x+1)^3+(-60*x^5-60*x^4-60*x^3-60*x^2)*log((x^2+1)/x)^3*log(x+1)^2+(12*
x^7+12*x^6+42*x^5+12*x^4+30*x^3)*log((x^2+1)/x)^3*log(x+1)+(-6*x^7-6*x^5)*log((x^2+1)/x)^3)/(x^4+x^3+x^2+x)/lo
g((x^2+1)/x)^3/log(x+1)^3,x, algorithm="maxima")

[Out]

(3*x^4*log(x)^2 - 30*x^2*log(x + 1)*log(x)^2 + 3*(x^4 - 10*x^2*log(x + 1))*log(x^2 + 1)^2 - 6*(x^4*log(x) - 10
*x^2*log(x + 1)*log(x))*log(x^2 + 1) + 16*log(x + 1)^2)/(log(x^2 + 1)^2*log(x + 1)^2 - 2*log(x^2 + 1)*log(x +
1)^2*log(x) + log(x + 1)^2*log(x)^2)

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mupad [B]  time = 5.97, size = 54, normalized size = 1.93 \begin {gather*} \frac {16}{{\ln \left (\frac {1}{x}\right )}^2+2\,\ln \left (\frac {1}{x}\right )\,\ln \left (x^2+1\right )+{\ln \left (x^2+1\right )}^2}-\frac {30\,x^2}{\ln \left (x+1\right )}+\frac {3\,x^4}{{\ln \left (x+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((x^2 + 1)/x)^3*(6*x^5 + 6*x^7) - log(x + 1)^3*(32*x - 32*x^2 - 32*x^3 + 32) - log(x + 1)*log((x^2 +
1)/x)^3*(30*x^3 + 12*x^4 + 42*x^5 + 12*x^6 + 12*x^7) + log(x + 1)^2*log((x^2 + 1)/x)^3*(60*x^2 + 60*x^3 + 60*x
^4 + 60*x^5))/(log(x + 1)^3*log((x^2 + 1)/x)^3*(x + x^2 + x^3 + x^4)),x)

[Out]

16/(log(x^2 + 1)^2 + 2*log(1/x)*log(x^2 + 1) + log(1/x)^2) - (30*x^2)/log(x + 1) + (3*x^4)/log(x + 1)^2

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sympy [A]  time = 0.39, size = 32, normalized size = 1.14 \begin {gather*} \frac {3 x^{4} - 30 x^{2} \log {\left (x + 1 \right )}}{\log {\left (x + 1 \right )}^{2}} + \frac {16}{\log {\left (\frac {x^{2} + 1}{x} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x**3-32*x**2+32*x+32)*ln(x+1)**3+(-60*x**5-60*x**4-60*x**3-60*x**2)*ln((x**2+1)/x)**3*ln(x+1)*
*2+(12*x**7+12*x**6+42*x**5+12*x**4+30*x**3)*ln((x**2+1)/x)**3*ln(x+1)+(-6*x**7-6*x**5)*ln((x**2+1)/x)**3)/(x*
*4+x**3+x**2+x)/ln((x**2+1)/x)**3/ln(x+1)**3,x)

[Out]

(3*x**4 - 30*x**2*log(x + 1))/log(x + 1)**2 + 16/log((x**2 + 1)/x)**2

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