3.29.32 \(\int \frac {x^2}{(b+a x^2) \sqrt [3]{x+x^3}} \, dx\)

Optimal. Leaf size=288 \[ \frac {\sqrt [3]{b} \log \left (-\sqrt [3]{b} \sqrt [3]{x^3+x} x \sqrt [3]{a-b}+x^2 (a-b)^{2/3}+b^{2/3} \left (x^3+x\right )^{2/3}\right )}{4 a \sqrt [3]{a-b}}-\frac {\sqrt [3]{b} \log \left (x \sqrt [3]{a-b}+\sqrt [3]{b} \sqrt [3]{x^3+x}\right )}{2 a \sqrt [3]{a-b}}+\frac {\sqrt {3} \sqrt [3]{b} \tan ^{-1}\left (\frac {\sqrt {3} x \sqrt [3]{a-b}}{x \sqrt [3]{a-b}-2 \sqrt [3]{b} \sqrt [3]{x^3+x}}\right )}{2 a \sqrt [3]{a-b}}-\frac {\log \left (\sqrt [3]{x^3+x}-x\right )}{2 a}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+x}+x}\right )}{2 a}+\frac {\log \left (\sqrt [3]{x^3+x} x+\left (x^3+x\right )^{2/3}+x^2\right )}{4 a} \]

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Rubi [A]  time = 0.58, antiderivative size = 445, normalized size of antiderivative = 1.55, number of steps used = 17, number of rules used = 12, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {2042, 466, 465, 494, 481, 200, 31, 634, 618, 204, 628, 617} \begin {gather*} \frac {\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{x^2+1} \log \left (\frac {x^{4/3} (a-b)^{2/3}}{\left (x^2+1\right )^{2/3}}-\frac {\sqrt [3]{b} x^{2/3} \sqrt [3]{a-b}}{\sqrt [3]{x^2+1}}+b^{2/3}\right )}{4 a \sqrt [3]{x^3+x} \sqrt [3]{a-b}}-\frac {\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{x^2+1} \log \left (\frac {x^{2/3} \sqrt [3]{a-b}}{\sqrt [3]{x^2+1}}+\sqrt [3]{b}\right )}{2 a \sqrt [3]{x^3+x} \sqrt [3]{a-b}}+\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{x^2+1} \tan ^{-1}\left (\frac {\sqrt [3]{b}-\frac {2 x^{2/3} \sqrt [3]{a-b}}{\sqrt [3]{x^2+1}}}{\sqrt {3} \sqrt [3]{b}}\right )}{2 a \sqrt [3]{x^3+x} \sqrt [3]{a-b}}-\frac {\sqrt [3]{x} \sqrt [3]{x^2+1} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{x^2+1}}\right )}{2 a \sqrt [3]{x^3+x}}+\frac {\sqrt [3]{x} \sqrt [3]{x^2+1} \log \left (\frac {x^{4/3}}{\left (x^2+1\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{x^2+1}}+1\right )}{4 a \sqrt [3]{x^3+x}}+\frac {\sqrt {3} \sqrt [3]{x} \sqrt [3]{x^2+1} \tan ^{-1}\left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{2 a \sqrt [3]{x^3+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/((b + a*x^2)*(x + x^3)^(1/3)),x]

[Out]

(Sqrt[3]*x^(1/3)*(1 + x^2)^(1/3)*ArcTan[(1 + (2*x^(2/3))/(1 + x^2)^(1/3))/Sqrt[3]])/(2*a*(x + x^3)^(1/3)) + (S
qrt[3]*b^(1/3)*x^(1/3)*(1 + x^2)^(1/3)*ArcTan[(b^(1/3) - (2*(a - b)^(1/3)*x^(2/3))/(1 + x^2)^(1/3))/(Sqrt[3]*b
^(1/3))])/(2*a*(a - b)^(1/3)*(x + x^3)^(1/3)) - (x^(1/3)*(1 + x^2)^(1/3)*Log[1 - x^(2/3)/(1 + x^2)^(1/3)])/(2*
a*(x + x^3)^(1/3)) + (x^(1/3)*(1 + x^2)^(1/3)*Log[1 + x^(4/3)/(1 + x^2)^(2/3) + x^(2/3)/(1 + x^2)^(1/3)])/(4*a
*(x + x^3)^(1/3)) - (b^(1/3)*x^(1/3)*(1 + x^2)^(1/3)*Log[b^(1/3) + ((a - b)^(1/3)*x^(2/3))/(1 + x^2)^(1/3)])/(
2*a*(a - b)^(1/3)*(x + x^3)^(1/3)) + (b^(1/3)*x^(1/3)*(1 + x^2)^(1/3)*Log[b^(2/3) + ((a - b)^(2/3)*x^(4/3))/(1
+ x^2)^(2/3) - ((a - b)^(1/3)*b^(1/3)*x^(2/3))/(1 + x^2)^(1/3)])/(4*a*(a - b)^(1/3)*(x + x^3)^(1/3))

Rule 31

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

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 465

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

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 481

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

Rule 494

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[(k*a^(p + (m + 1)/n))/n, Subst[Int[(x^((k*(m + 1))/n - 1)*(c - (b*c - a*d)*x^k)^q)/(1 - b*x^k)^(p
+ q + (m + 1)/n + 1), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 2042

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.))^(q_.), x_Symbol]
:> Dist[(e^IntPart[m]*(e*x)^FracPart[m]*(a*x^j + b*x^(j + n))^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a
+ b*x^n)^FracPart[p]), Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, j, m, n,
p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])

Rubi steps

\begin {align*} \int \frac {x^2}{\left (b+a x^2\right ) \sqrt [3]{x+x^3}} \, dx &=\frac {\left (\sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \int \frac {x^{5/3}}{\sqrt [3]{1+x^2} \left (b+a x^2\right )} \, dx}{\sqrt [3]{x+x^3}}\\ &=\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x^7}{\sqrt [3]{1+x^6} \left (b+a x^6\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x+x^3}}\\ &=\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt [3]{1+x^3} \left (b+a x^3\right )} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{x+x^3}}\\ &=\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (1-x^3\right ) \left (b-(-a+b) x^3\right )} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 \sqrt [3]{x+x^3}}\\ &=\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^3} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{x+x^3}}-\frac {\left (3 b \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{b+(a-b) x^3} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{x+x^3}}\\ &=\frac {\left (\sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{x+x^3}}+\frac {\left (\sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {2+x}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{x+x^3}}-\frac {\left (\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{b}+\sqrt [3]{a-b} x} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{x+x^3}}-\frac {\left (\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {2 \sqrt [3]{b}-\sqrt [3]{a-b} x}{b^{2/3}-\sqrt [3]{a-b} \sqrt [3]{b} x+(a-b)^{2/3} x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{x+x^3}}\\ &=-\frac {\sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{x+x^3}}-\frac {\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (\sqrt [3]{b}+\frac {\sqrt [3]{a-b} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{a-b} \sqrt [3]{x+x^3}}+\frac {\left (\sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 a \sqrt [3]{x+x^3}}+\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 a \sqrt [3]{x+x^3}}+\frac {\left (\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {-\sqrt [3]{a-b} \sqrt [3]{b}+2 (a-b)^{2/3} x}{b^{2/3}-\sqrt [3]{a-b} \sqrt [3]{b} x+(a-b)^{2/3} x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 a \sqrt [3]{a-b} \sqrt [3]{x+x^3}}-\frac {\left (3 b^{2/3} \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{b^{2/3}-\sqrt [3]{a-b} \sqrt [3]{b} x+(a-b)^{2/3} x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 a \sqrt [3]{x+x^3}}\\ &=-\frac {\sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{x+x^3}}+\frac {\sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (1+\frac {x^{4/3}}{\left (1+x^2\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 a \sqrt [3]{x+x^3}}-\frac {\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (\sqrt [3]{b}+\frac {\sqrt [3]{a-b} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{a-b} \sqrt [3]{x+x^3}}+\frac {\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (b^{2/3}+\frac {(a-b)^{2/3} x^{4/3}}{\left (1+x^2\right )^{2/3}}-\frac {\sqrt [3]{a-b} \sqrt [3]{b} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 a \sqrt [3]{a-b} \sqrt [3]{x+x^3}}-\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{x+x^3}}-\frac {\left (3 \sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{a-b} x^{2/3}}{\sqrt [3]{b} \sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{a-b} \sqrt [3]{x+x^3}}\\ &=\frac {\sqrt {3} \sqrt [3]{x} \sqrt [3]{1+x^2} \tan ^{-1}\left (\frac {1+\frac {2 x^{2/3}}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{2 a \sqrt [3]{x+x^3}}+\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{1+x^2} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{a-b} x^{2/3}}{\sqrt [3]{b} \sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{2 a \sqrt [3]{a-b} \sqrt [3]{x+x^3}}-\frac {\sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{x+x^3}}+\frac {\sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (1+\frac {x^{4/3}}{\left (1+x^2\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 a \sqrt [3]{x+x^3}}-\frac {\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (\sqrt [3]{b}+\frac {\sqrt [3]{a-b} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 a \sqrt [3]{a-b} \sqrt [3]{x+x^3}}+\frac {\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (b^{2/3}+\frac {(a-b)^{2/3} x^{4/3}}{\left (1+x^2\right )^{2/3}}-\frac {\sqrt [3]{a-b} \sqrt [3]{b} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 a \sqrt [3]{a-b} \sqrt [3]{x+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 53, normalized size = 0.18 \begin {gather*} \frac {3 x^3 \sqrt [3]{x^2+1} F_1\left (\frac {4}{3};\frac {1}{3},1;\frac {7}{3};-x^2,-\frac {a x^2}{b}\right )}{8 b \sqrt [3]{x^3+x}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2/((b + a*x^2)*(x + x^3)^(1/3)),x]

[Out]

(3*x^3*(1 + x^2)^(1/3)*AppellF1[4/3, 1/3, 1, 7/3, -x^2, -((a*x^2)/b)])/(8*b*(x + x^3)^(1/3))

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IntegrateAlgebraic [A]  time = 0.71, size = 291, normalized size = 1.01 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )}{2 a}+\frac {\sqrt {3} \sqrt [3]{b} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{a-b} x}{\sqrt [3]{a-b} x-2 \sqrt [3]{b} \sqrt [3]{x+x^3}}\right )}{2 a \sqrt [3]{a-b}}-\frac {\log \left (-a x+a \sqrt [3]{x+x^3}\right )}{2 a}-\frac {\sqrt [3]{b} \log \left (\sqrt [3]{a-b} x+\sqrt [3]{b} \sqrt [3]{x+x^3}\right )}{2 a \sqrt [3]{a-b}}+\frac {\log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right )}{4 a}+\frac {\sqrt [3]{b} \log \left ((a-b)^{2/3} x^2-\sqrt [3]{a-b} \sqrt [3]{b} x \sqrt [3]{x+x^3}+b^{2/3} \left (x+x^3\right )^{2/3}\right )}{4 a \sqrt [3]{a-b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2/((b + a*x^2)*(x + x^3)^(1/3)),x]

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x + 2*(x + x^3)^(1/3))])/(2*a) + (Sqrt[3]*b^(1/3)*ArcTan[(Sqrt[3]*(a - b)^(1/3)*x
)/((a - b)^(1/3)*x - 2*b^(1/3)*(x + x^3)^(1/3))])/(2*a*(a - b)^(1/3)) - Log[-(a*x) + a*(x + x^3)^(1/3)]/(2*a)
- (b^(1/3)*Log[(a - b)^(1/3)*x + b^(1/3)*(x + x^3)^(1/3)])/(2*a*(a - b)^(1/3)) + Log[x^2 + x*(x + x^3)^(1/3) +
(x + x^3)^(2/3)]/(4*a) + (b^(1/3)*Log[(a - b)^(2/3)*x^2 - (a - b)^(1/3)*b^(1/3)*x*(x + x^3)^(1/3) + b^(2/3)*(
x + x^3)^(2/3)])/(4*a*(a - b)^(1/3))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x^2+b)/(x^3+x)^(1/3),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.41, size = 256, normalized size = 0.89 \begin {gather*} -\frac {b \left (-\frac {a - b}{b}\right )^{\frac {2}{3}} \log \left ({\left | -\left (-\frac {a - b}{b}\right )^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} \right |}\right )}{2 \, {\left (a^{2} - a b\right )}} - \frac {3 \, {\left (-a b^{2} + b^{3}\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a - b}{b}\right )^{\frac {1}{3}} + 2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a - b}{b}\right )^{\frac {1}{3}}}\right )}{2 \, {\left (\sqrt {3} a^{2} b - \sqrt {3} a b^{2}\right )}} + \frac {{\left (-a b^{2} + b^{3}\right )}^{\frac {2}{3}} \log \left (\left (-\frac {a - b}{b}\right )^{\frac {2}{3}} + \left (-\frac {a - b}{b}\right )^{\frac {1}{3}} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}}\right )}{4 \, {\left (a^{2} b - a b^{2}\right )}} - \frac {\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right )}{2 \, a} + \frac {\log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}{4 \, a} - \frac {\log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right )}{2 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x^2+b)/(x^3+x)^(1/3),x, algorithm="giac")

[Out]

-1/2*b*(-(a - b)/b)^(2/3)*log(abs(-(-(a - b)/b)^(1/3) + (1/x^2 + 1)^(1/3)))/(a^2 - a*b) - 3/2*(-a*b^2 + b^3)^(
2/3)*arctan(1/3*sqrt(3)*((-(a - b)/b)^(1/3) + 2*(1/x^2 + 1)^(1/3))/(-(a - b)/b)^(1/3))/(sqrt(3)*a^2*b - sqrt(3
)*a*b^2) + 1/4*(-a*b^2 + b^3)^(2/3)*log((-(a - b)/b)^(2/3) + (-(a - b)/b)^(1/3)*(1/x^2 + 1)^(1/3) + (1/x^2 + 1
)^(2/3))/(a^2*b - a*b^2) - 1/2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(1/x^2 + 1)^(1/3) + 1))/a + 1/4*log((1/x^2 + 1)^(
2/3) + (1/x^2 + 1)^(1/3) + 1)/a - 1/2*log(abs((1/x^2 + 1)^(1/3) - 1))/a

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {x^{2}}{\left (a \,x^{2}+b \right ) \left (x^{3}+x \right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x^2+b)/(x^3+x)^(1/3),x)

[Out]

int(x^2/(a*x^2+b)/(x^3+x)^(1/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{{\left (a x^{2} + b\right )} {\left (x^{3} + x\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x^2+b)/(x^3+x)^(1/3),x, algorithm="maxima")

[Out]

integrate(x^2/((a*x^2 + b)*(x^3 + x)^(1/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2}{\left (a\,x^2+b\right )\,{\left (x^3+x\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((b + a*x^2)*(x + x^3)^(1/3)),x)

[Out]

int(x^2/((b + a*x^2)*(x + x^3)^(1/3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\sqrt [3]{x \left (x^{2} + 1\right )} \left (a x^{2} + b\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a*x**2+b)/(x**3+x)**(1/3),x)

[Out]

Integral(x**2/((x*(x**2 + 1))**(1/3)*(a*x**2 + b)), x)

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