3.14.4 \(\int \frac {-10 x+15 x^2+10 x^4+28 x^5+7 x^6+2 x^8+14 x^9+(10 x+20 x^4+6 x^5+12 x^8) \log (\frac {1+2 x^3}{x^2})}{1+2 x^3} \, dx\)

Optimal. Leaf size=19 \[ x^2 \left (5+x^4\right ) \left (x+\log \left (\frac {1}{x^2}+2 x\right )\right ) \]

________________________________________________________________________________________

Rubi [B]  time = 0.66, antiderivative size = 129, normalized size of antiderivative = 6.79, number of steps used = 54, number of rules used = 19, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6725, 292, 31, 634, 617, 204, 628, 260, 321, 266, 43, 302, 200, 2528, 2525, 12, 459, 446, 77} \begin {gather*} x^7+5 x^3+5 x^2 \log \left (\frac {2 x^3+1}{x^2}\right )+x^6 \log \left (\frac {2 x^3+1}{x^2}\right )-\frac {5 \sqrt {3} \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{2^{2/3}}+\frac {5 \sqrt [3]{2} \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {5 \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10*x + 15*x^2 + 10*x^4 + 28*x^5 + 7*x^6 + 2*x^8 + 14*x^9 + (10*x + 20*x^4 + 6*x^5 + 12*x^8)*Log[(1 + 2*x
^3)/x^2])/(1 + 2*x^3),x]

[Out]

5*x^3 + x^7 + (5*ArcTan[(1 - 2*2^(1/3)*x)/Sqrt[3]])/(2^(2/3)*Sqrt[3]) + (5*2^(1/3)*ArcTan[(1 - 2*2^(1/3)*x)/Sq
rt[3]])/Sqrt[3] - (5*Sqrt[3]*ArcTan[(1 - 2*2^(1/3)*x)/Sqrt[3]])/2^(2/3) + 5*x^2*Log[(1 + 2*x^3)/x^2] + x^6*Log
[(1 + 2*x^3)/x^2]

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 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 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 260

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

Rule 266

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

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 459

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

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

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 2528

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

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {10 x}{1+2 x^3}+\frac {15 x^2}{1+2 x^3}+\frac {10 x^4}{1+2 x^3}+\frac {28 x^5}{1+2 x^3}+\frac {7 x^6}{1+2 x^3}+\frac {2 x^8}{1+2 x^3}+\frac {14 x^9}{1+2 x^3}+2 x \left (5+3 x^4\right ) \log \left (\frac {1+2 x^3}{x^2}\right )\right ) \, dx\\ &=2 \int \frac {x^8}{1+2 x^3} \, dx+2 \int x \left (5+3 x^4\right ) \log \left (\frac {1+2 x^3}{x^2}\right ) \, dx+7 \int \frac {x^6}{1+2 x^3} \, dx-10 \int \frac {x}{1+2 x^3} \, dx+10 \int \frac {x^4}{1+2 x^3} \, dx+14 \int \frac {x^9}{1+2 x^3} \, dx+15 \int \frac {x^2}{1+2 x^3} \, dx+28 \int \frac {x^5}{1+2 x^3} \, dx\\ &=\frac {5 x^2}{2}+\frac {5}{2} \log \left (1+2 x^3\right )+\frac {2}{3} \operatorname {Subst}\left (\int \frac {x^2}{1+2 x} \, dx,x,x^3\right )+2 \int \left (5 x \log \left (\frac {1+2 x^3}{x^2}\right )+3 x^5 \log \left (\frac {1+2 x^3}{x^2}\right )\right ) \, dx-5 \int \frac {x}{1+2 x^3} \, dx+7 \int \left (-\frac {1}{4}+\frac {x^3}{2}+\frac {1}{4 \left (1+2 x^3\right )}\right ) \, dx+\frac {28}{3} \operatorname {Subst}\left (\int \frac {x}{1+2 x} \, dx,x,x^3\right )+14 \int \left (\frac {1}{8}-\frac {x^3}{4}+\frac {x^6}{2}-\frac {1}{8 \left (1+2 x^3\right )}\right ) \, dx+\frac {1}{3} \left (5\ 2^{2/3}\right ) \int \frac {1}{1+\sqrt [3]{2} x} \, dx-\frac {1}{3} \left (5\ 2^{2/3}\right ) \int \frac {1+\sqrt [3]{2} x}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx\\ &=\frac {5 x^2}{2}+x^7+\frac {5}{3} \sqrt [3]{2} \log \left (1+\sqrt [3]{2} x\right )+\frac {5}{2} \log \left (1+2 x^3\right )+\frac {2}{3} \operatorname {Subst}\left (\int \left (-\frac {1}{4}+\frac {x}{2}+\frac {1}{4 (1+2 x)}\right ) \, dx,x,x^3\right )+6 \int x^5 \log \left (\frac {1+2 x^3}{x^2}\right ) \, dx+\frac {28}{3} \operatorname {Subst}\left (\int \left (\frac {1}{2}-\frac {1}{2 (1+2 x)}\right ) \, dx,x,x^3\right )+10 \int x \log \left (\frac {1+2 x^3}{x^2}\right ) \, dx-\frac {5 \int \frac {-\sqrt [3]{2}+2\ 2^{2/3} x}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx}{3\ 2^{2/3}}+\frac {5 \int \frac {1}{1+\sqrt [3]{2} x} \, dx}{3 \sqrt [3]{2}}-\frac {5 \int \frac {1+\sqrt [3]{2} x}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx}{3 \sqrt [3]{2}}-\frac {5 \int \frac {1}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx}{\sqrt [3]{2}}\\ &=\frac {5 x^2}{2}+\frac {9 x^3}{2}+\frac {x^6}{6}+x^7+\frac {5 \log \left (1+\sqrt [3]{2} x\right )}{3\ 2^{2/3}}+\frac {5}{3} \sqrt [3]{2} \log \left (1+\sqrt [3]{2} x\right )-\frac {5 \log \left (1-\sqrt [3]{2} x+2^{2/3} x^2\right )}{3\ 2^{2/3}}+\frac {1}{4} \log \left (1+2 x^3\right )+5 x^2 \log \left (\frac {1+2 x^3}{x^2}\right )+x^6 \log \left (\frac {1+2 x^3}{x^2}\right )-5 \int \frac {2 x \left (-1+x^3\right )}{1+2 x^3} \, dx-\frac {5 \int \frac {-\sqrt [3]{2}+2\ 2^{2/3} x}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx}{6\ 2^{2/3}}-\frac {5 \int \frac {1}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx}{2 \sqrt [3]{2}}-\left (5 \sqrt [3]{2}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2 \sqrt [3]{2} x\right )-\int \frac {2 x^5 \left (-1+x^3\right )}{1+2 x^3} \, dx\\ &=\frac {5 x^2}{2}+\frac {9 x^3}{2}+\frac {x^6}{6}+x^7+\frac {5 \sqrt [3]{2} \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {5 \log \left (1+\sqrt [3]{2} x\right )}{3\ 2^{2/3}}+\frac {5}{3} \sqrt [3]{2} \log \left (1+\sqrt [3]{2} x\right )-\frac {5 \log \left (1-\sqrt [3]{2} x+2^{2/3} x^2\right )}{2\ 2^{2/3}}+\frac {1}{4} \log \left (1+2 x^3\right )+5 x^2 \log \left (\frac {1+2 x^3}{x^2}\right )+x^6 \log \left (\frac {1+2 x^3}{x^2}\right )-2 \int \frac {x^5 \left (-1+x^3\right )}{1+2 x^3} \, dx-10 \int \frac {x \left (-1+x^3\right )}{1+2 x^3} \, dx-\frac {5 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2 \sqrt [3]{2} x\right )}{2^{2/3}}\\ &=\frac {9 x^3}{2}+\frac {x^6}{6}+x^7+\frac {5 \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {5 \sqrt [3]{2} \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {5 \log \left (1+\sqrt [3]{2} x\right )}{3\ 2^{2/3}}+\frac {5}{3} \sqrt [3]{2} \log \left (1+\sqrt [3]{2} x\right )-\frac {5 \log \left (1-\sqrt [3]{2} x+2^{2/3} x^2\right )}{2\ 2^{2/3}}+\frac {1}{4} \log \left (1+2 x^3\right )+5 x^2 \log \left (\frac {1+2 x^3}{x^2}\right )+x^6 \log \left (\frac {1+2 x^3}{x^2}\right )-\frac {2}{3} \operatorname {Subst}\left (\int \frac {(-1+x) x}{1+2 x} \, dx,x,x^3\right )+15 \int \frac {x}{1+2 x^3} \, dx\\ &=\frac {9 x^3}{2}+\frac {x^6}{6}+x^7+\frac {5 \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {5 \sqrt [3]{2} \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {5 \log \left (1+\sqrt [3]{2} x\right )}{3\ 2^{2/3}}+\frac {5}{3} \sqrt [3]{2} \log \left (1+\sqrt [3]{2} x\right )-\frac {5 \log \left (1-\sqrt [3]{2} x+2^{2/3} x^2\right )}{2\ 2^{2/3}}+\frac {1}{4} \log \left (1+2 x^3\right )+5 x^2 \log \left (\frac {1+2 x^3}{x^2}\right )+x^6 \log \left (\frac {1+2 x^3}{x^2}\right )-\frac {2}{3} \operatorname {Subst}\left (\int \left (-\frac {3}{4}+\frac {x}{2}+\frac {3}{4 (1+2 x)}\right ) \, dx,x,x^3\right )-\frac {5 \int \frac {1}{1+\sqrt [3]{2} x} \, dx}{\sqrt [3]{2}}+\frac {5 \int \frac {1+\sqrt [3]{2} x}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx}{\sqrt [3]{2}}\\ &=5 x^3+x^7+\frac {5 \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {5 \sqrt [3]{2} \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {5 \log \left (1-\sqrt [3]{2} x+2^{2/3} x^2\right )}{2\ 2^{2/3}}+5 x^2 \log \left (\frac {1+2 x^3}{x^2}\right )+x^6 \log \left (\frac {1+2 x^3}{x^2}\right )+\frac {5 \int \frac {-\sqrt [3]{2}+2\ 2^{2/3} x}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx}{2\ 2^{2/3}}+\frac {15 \int \frac {1}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx}{2 \sqrt [3]{2}}\\ &=5 x^3+x^7+\frac {5 \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {5 \sqrt [3]{2} \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{\sqrt {3}}+5 x^2 \log \left (\frac {1+2 x^3}{x^2}\right )+x^6 \log \left (\frac {1+2 x^3}{x^2}\right )+\frac {15 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2 \sqrt [3]{2} x\right )}{2^{2/3}}\\ &=5 x^3+x^7+\frac {5 \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {5 \sqrt [3]{2} \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {5 \sqrt {3} \tan ^{-1}\left (\frac {1-2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{2^{2/3}}+5 x^2 \log \left (\frac {1+2 x^3}{x^2}\right )+x^6 \log \left (\frac {1+2 x^3}{x^2}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 34, normalized size = 1.79 \begin {gather*} 5 x^3+x^7+5 x^2 \log \left (\frac {1}{x^2}+2 x\right )+x^6 \log \left (\frac {1}{x^2}+2 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x + 15*x^2 + 10*x^4 + 28*x^5 + 7*x^6 + 2*x^8 + 14*x^9 + (10*x + 20*x^4 + 6*x^5 + 12*x^8)*Log[(1
+ 2*x^3)/x^2])/(1 + 2*x^3),x]

[Out]

5*x^3 + x^7 + 5*x^2*Log[x^(-2) + 2*x] + x^6*Log[x^(-2) + 2*x]

________________________________________________________________________________________

fricas [A]  time = 0.90, size = 31, normalized size = 1.63 \begin {gather*} x^{7} + 5 \, x^{3} + {\left (x^{6} + 5 \, x^{2}\right )} \log \left (\frac {2 \, x^{3} + 1}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^8+6*x^5+20*x^4+10*x)*log((2*x^3+1)/x^2)+14*x^9+2*x^8+7*x^6+28*x^5+10*x^4+15*x^2-10*x)/(2*x^3+
1),x, algorithm="fricas")

[Out]

x^7 + 5*x^3 + (x^6 + 5*x^2)*log((2*x^3 + 1)/x^2)

________________________________________________________________________________________

giac [A]  time = 0.37, size = 31, normalized size = 1.63 \begin {gather*} x^{7} + 5 \, x^{3} + {\left (x^{6} + 5 \, x^{2}\right )} \log \left (\frac {2 \, x^{3} + 1}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^8+6*x^5+20*x^4+10*x)*log((2*x^3+1)/x^2)+14*x^9+2*x^8+7*x^6+28*x^5+10*x^4+15*x^2-10*x)/(2*x^3+
1),x, algorithm="giac")

[Out]

x^7 + 5*x^3 + (x^6 + 5*x^2)*log((2*x^3 + 1)/x^2)

________________________________________________________________________________________

maple [A]  time = 0.21, size = 32, normalized size = 1.68

method	result	size
risch	\(\left (x^{6}+5 x^{2}\right ) \ln \left (\frac {2 x^{3}+1}{x^{2}}\right )+x^{7}+5 x^{3}\)	\(32\)
default	\(x^{7}+5 x^{3}+5 x^{2} \ln \left (\frac {2 x^{3}+1}{x^{2}}\right )+\ln \left (\frac {2 x^{3}+1}{x^{2}}\right ) x^{6}\)	\(43\)
norman	\(x^{7}+5 x^{3}+5 x^{2} \ln \left (\frac {2 x^{3}+1}{x^{2}}\right )+\ln \left (\frac {2 x^{3}+1}{x^{2}}\right ) x^{6}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((12*x^8+6*x^5+20*x^4+10*x)*ln((2*x^3+1)/x^2)+14*x^9+2*x^8+7*x^6+28*x^5+10*x^4+15*x^2-10*x)/(2*x^3+1),x,me
thod=_RETURNVERBOSE)

[Out]

(x^6+5*x^2)*ln((2*x^3+1)/x^2)+x^7+5*x^3

________________________________________________________________________________________

maxima [B]  time = 1.15, size = 136, normalized size = 7.16 \begin {gather*} x^{7} + 5 \, x^{3} + {\left (x^{6} + 5 \, x^{2}\right )} \log \left (2 \, x^{3} + 1\right ) + \frac {1}{4} \, {\left (5 \cdot 2^{\frac {1}{3}} - 1\right )} \log \left (2^{\frac {2}{3}} x^{2} - 2^{\frac {1}{3}} x + 1\right ) - \frac {1}{4} \, {\left (10 \cdot 2^{\frac {1}{3}} + 1\right )} \log \left (\frac {1}{2} \cdot 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} x + 1\right )}\right ) - 2 \, {\left (x^{6} + 5 \, x^{2}\right )} \log \relax (x) - \frac {5}{4} \cdot 2^{\frac {1}{3}} \log \left (2^{\frac {2}{3}} x^{2} - 2^{\frac {1}{3}} x + 1\right ) + \frac {5}{2} \cdot 2^{\frac {1}{3}} \log \left (\frac {1}{2} \cdot 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} x + 1\right )}\right ) + \frac {1}{4} \, \log \left (2 \, x^{3} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^8+6*x^5+20*x^4+10*x)*log((2*x^3+1)/x^2)+14*x^9+2*x^8+7*x^6+28*x^5+10*x^4+15*x^2-10*x)/(2*x^3+
1),x, algorithm="maxima")

[Out]

x^7 + 5*x^3 + (x^6 + 5*x^2)*log(2*x^3 + 1) + 1/4*(5*2^(1/3) - 1)*log(2^(2/3)*x^2 - 2^(1/3)*x + 1) - 1/4*(10*2^
(1/3) + 1)*log(1/2*2^(2/3)*(2^(1/3)*x + 1)) - 2*(x^6 + 5*x^2)*log(x) - 5/4*2^(1/3)*log(2^(2/3)*x^2 - 2^(1/3)*x
+ 1) + 5/2*2^(1/3)*log(1/2*2^(2/3)*(2^(1/3)*x + 1)) + 1/4*log(2*x^3 + 1)

________________________________________________________________________________________

mupad [B]  time = 1.13, size = 23, normalized size = 1.21 \begin {gather*} x^2\,\left (x+\ln \left (\frac {2\,x^3+1}{x^2}\right )\right )\,\left (x^4+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((2*x^3 + 1)/x^2)*(10*x + 20*x^4 + 6*x^5 + 12*x^8) - 10*x + 15*x^2 + 10*x^4 + 28*x^5 + 7*x^6 + 2*x^8 +
14*x^9)/(2*x^3 + 1),x)

[Out]

x^2*(x + log((2*x^3 + 1)/x^2))*(x^4 + 5)

________________________________________________________________________________________

sympy [A]  time = 0.17, size = 27, normalized size = 1.42 \begin {gather*} x^{7} + 5 x^{3} + \left (x^{6} + 5 x^{2}\right ) \log {\left (\frac {2 x^{3} + 1}{x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x**8+6*x**5+20*x**4+10*x)*ln((2*x**3+1)/x**2)+14*x**9+2*x**8+7*x**6+28*x**5+10*x**4+15*x**2-10*
x)/(2*x**3+1),x)

[Out]

x**7 + 5*x**3 + (x**6 + 5*x**2)*log((2*x**3 + 1)/x**2)

________________________________________________________________________________________