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

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

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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} $$\begin{array}{c}{x}^7+5x^3+5x^2\log \left(\frac{2x^3+1}{x^2}\right)+x^6\log \left(\frac{2x^3+1}{x^2}\right)-\frac{5\sqrt{3}{\tan }^{-1}\left(\frac{1-2\sqrt[3]{2}x}{\sqrt{3}}\right)}{2^{2/3}}+\frac{5\sqrt[3]{2}{\tan }^{-1}\left(\frac{1-2\sqrt[3]{2}x}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{5{\tan }^{-1}\left(\frac{1-2\sqrt[3]{2}x}{\sqrt{3}}\right)}{2^{2/3}\sqrt{3}}\end{array}$$

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{array}{c}\begin{array}{rl}\text{integral}& =\int (-\frac{10x}{1+2x^3}+\frac{15x^2}{1+2x^3}+\frac{10x^4}{1+2x^3}+\frac{28x^5}{1+2x^3}+\frac{7x^6}{1+2x^3}+\frac{2x^8}{1+2x^3}+\frac{14x^9}{1+2x^3}+2x(5+3x^4)\log \left(\frac{1+2x^3}{x^2}\right))dx\\ & =2\int \frac{x^8}{1+2x^3}dx+2\int x(5+3x^4)\log \left(\frac{1+2x^3}{x^2}\right)dx+7\int \frac{x^6}{1+2x^3}dx-10\int \frac{x}{1+2x^3}dx+10\int \frac{x^4}{1+2x^3}dx+14\int \frac{x^9}{1+2x^3}dx+15\int \frac{x^2}{1+2x^3}dx+28\int \frac{x^5}{1+2x^3}dx\\ & =\frac{5x^2}{2}+\frac{5}{2}\log (1+2x^3)+\frac{2}{3}\mathrm{Subst}(\int \frac{x^2}{1+2x}dx,x,x^3)+2\int (5x\log \left(\frac{1+2x^3}{x^2}\right)+3x^5\log \left(\frac{1+2x^3}{x^2}\right))dx-5\int \frac{x}{1+2x^3}dx+7\int (-\frac{1}{4}+\frac{x^3}{2}+\frac{1}{4(1+2x^3)})dx+\frac{28}{3}\mathrm{Subst}(\int \frac{x}{1+2x}dx,x,x^3)+14\int (\frac{1}{8}-\frac{x^3}{4}+\frac{x^6}{2}-\frac{1}{8(1+2x^3)})dx+\frac{1}{3}\left(52^{2/3}\right)\int \frac{1}{1+\sqrt[3]{2}x}dx-\frac{1}{3}\left(52^{2/3}\right)\int \frac{1+\sqrt[3]{2}x}{1-\sqrt[3]{2}x+2^{2/3}{x}^2}dx\\ & =\frac{5x^2}{2}+x^7+\frac{5}{3}\sqrt[3]{2}\log (1+\sqrt[3]{2}x)+\frac{5}{2}\log (1+2x^3)+\frac{2}{3}\mathrm{Subst}(\int (-\frac{1}{4}+\frac{x}{2}+\frac{1}{4(1+2x)})dx,x,x^3)+6\int x^5\log \left(\frac{1+2x^3}{x^2}\right)dx+\frac{28}{3}\mathrm{Subst}(\int (\frac{1}{2}-\frac{1}{2(1+2x)})dx,x,x^3)+10\int x\log \left(\frac{1+2x^3}{x^2}\right)dx-\frac{5\int \frac{-\sqrt[3]{2}+22^{2/3}x}{1-\sqrt[3]{2}x+2^{2/3}{x}^2}dx}{32^{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{5x^2}{2}+\frac{9x^3}{2}+\frac{x^6}{6}+x^7+\frac{5\log (1+\sqrt[3]{2}x)}{32^{2/3}}+\frac{5}{3}\sqrt[3]{2}\log (1+\sqrt[3]{2}x)-\frac{5\log (1-\sqrt[3]{2}x+2^{2/3}{x}^2)}{32^{2/3}}+\frac{1}{4}\log (1+2x^3)+5x^2\log \left(\frac{1+2x^3}{x^2}\right)+x^6\log \left(\frac{1+2x^3}{x^2}\right)-5\int \frac{2x(-1+x^3)}{1+2x^3}dx-\frac{5\int \frac{-\sqrt[3]{2}+22^{2/3}x}{1-\sqrt[3]{2}x+2^{2/3}{x}^2}dx}{62^{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)\mathrm{Subst}(\int \frac{1}{-3-x^2}dx,x,1-2\sqrt[3]{2}x)-\int \frac{2x^5(-1+x^3)}{1+2x^3}dx\\ & =\frac{5x^2}{2}+\frac{9x^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 (1+\sqrt[3]{2}x)}{32^{2/3}}+\frac{5}{3}\sqrt[3]{2}\log (1+\sqrt[3]{2}x)-\frac{5\log (1-\sqrt[3]{2}x+2^{2/3}{x}^2)}{22^{2/3}}+\frac{1}{4}\log (1+2x^3)+5x^2\log \left(\frac{1+2x^3}{x^2}\right)+x^6\log \left(\frac{1+2x^3}{x^2}\right)-2\int \frac{x^5(-1+x^3)}{1+2x^3}dx-10\int \frac{x(-1+x^3)}{1+2x^3}dx-\frac{5\mathrm{Subst}(\int \frac{1}{-3-x^2}dx,x,1-2\sqrt[3]{2}x)}{2^{2/3}}\\ & =\frac{9x^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 (1+\sqrt[3]{2}x)}{32^{2/3}}+\frac{5}{3}\sqrt[3]{2}\log (1+\sqrt[3]{2}x)-\frac{5\log (1-\sqrt[3]{2}x+2^{2/3}{x}^2)}{22^{2/3}}+\frac{1}{4}\log (1+2x^3)+5x^2\log \left(\frac{1+2x^3}{x^2}\right)+x^6\log \left(\frac{1+2x^3}{x^2}\right)-\frac{2}{3}\mathrm{Subst}(\int \frac{(-1+x)x}{1+2x}dx,x,x^3)+15\int \frac{x}{1+2x^3}dx\\ & =\frac{9x^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 (1+\sqrt[3]{2}x)}{32^{2/3}}+\frac{5}{3}\sqrt[3]{2}\log (1+\sqrt[3]{2}x)-\frac{5\log (1-\sqrt[3]{2}x+2^{2/3}{x}^2)}{22^{2/3}}+\frac{1}{4}\log (1+2x^3)+5x^2\log \left(\frac{1+2x^3}{x^2}\right)+x^6\log \left(\frac{1+2x^3}{x^2}\right)-\frac{2}{3}\mathrm{Subst}(\int (-\frac{3}{4}+\frac{x}{2}+\frac{3}{4(1+2x)})dx,x,x^3)-\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}}\\ & =5x^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 (1-\sqrt[3]{2}x+2^{2/3}{x}^2)}{22^{2/3}}+5x^2\log \left(\frac{1+2x^3}{x^2}\right)+x^6\log \left(\frac{1+2x^3}{x^2}\right)+\frac{5\int \frac{-\sqrt[3]{2}+22^{2/3}x}{1-\sqrt[3]{2}x+2^{2/3}{x}^2}dx}{22^{2/3}}+\frac{15\int \frac{1}{1-\sqrt[3]{2}x+2^{2/3}{x}^2}dx}{2\sqrt[3]{2}}\\ & =5x^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}}+5x^2\log \left(\frac{1+2x^3}{x^2}\right)+x^6\log \left(\frac{1+2x^3}{x^2}\right)+\frac{15\mathrm{Subst}(\int \frac{1}{-3-x^2}dx,x,1-2\sqrt[3]{2}x)}{2^{2/3}}\\ & =5x^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}}+5x^2\log \left(\frac{1+2x^3}{x^2}\right)+x^6\log \left(\frac{1+2x^3}{x^2}\right)\end{array}\end{array}$$

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

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]

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

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)

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

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)

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

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

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

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

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)

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

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)

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

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)

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