\(\int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx\) [1745]

Optimal result

Integrand size = 17, antiderivative size = 118 \[ \int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx=\frac {1}{2} a \left (x+x^3\right )^{2/3}+\frac {1}{6} \left (-\sqrt {3} a+3 \sqrt {3} b\right ) \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )+\frac {1}{6} (a-3 b) \log \left (-x+\sqrt [3]{x+x^3}\right )+\frac {1}{12} (-a+3 b) \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right ) \]

[Out]

1/2*a*(x^3+x)^(2/3)+1/6*(-3^(1/2)*a+3*3^(1/2)*b)*arctan(3^(1/2)*x/(x+2*(x^3+x)^(1/3)))+1/6*(a-3*b)*ln(-x+(x^3+
x)^(1/3))+1/12*(-a+3*b)*ln(x^2+x*(x^3+x)^(1/3)+(x^3+x)^(2/3))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.89, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2078, 2036, 335, 281, 245, 2049} \[ \int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx=-\frac {a \sqrt [3]{x} \sqrt [3]{x^2+1} \arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{2 \sqrt {3} \sqrt [3]{x^3+x}}+\frac {1}{2} a \left (x^3+x\right )^{2/3}+\frac {a \sqrt [3]{x} \sqrt [3]{x^2+1} \log \left (x^{2/3}-\sqrt [3]{x^2+1}\right )}{4 \sqrt [3]{x^3+x}}+\frac {\sqrt {3} b \sqrt [3]{x} \sqrt [3]{x^2+1} \arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{x^3+x}}-\frac {3 b \sqrt [3]{x} \sqrt [3]{x^2+1} \log \left (x^{2/3}-\sqrt [3]{x^2+1}\right )}{4 \sqrt [3]{x^3+x}} \]

[In]

Int[(b + a*x^2)/(x + x^3)^(1/3),x]

[Out]

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

Rule 245

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

Rule 281

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

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2049

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

Rule 2078

Int[(Pq_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[Pq*(a*x^j + b*x^n)^p, x]
, x] /; FreeQ[{a, b, j, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !IntegerQ[p] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {b}{\sqrt [3]{x+x^3}}+\frac {a x^2}{\sqrt [3]{x+x^3}}\right ) \, dx \\ & = a \int \frac {x^2}{\sqrt [3]{x+x^3}} \, dx+b \int \frac {1}{\sqrt [3]{x+x^3}} \, dx \\ & = \frac {1}{2} a \left (x+x^3\right )^{2/3}-\frac {1}{3} a \int \frac {1}{\sqrt [3]{x+x^3}} \, dx+\frac {\left (b \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{1+x^2}} \, dx}{\sqrt [3]{x+x^3}} \\ & = \frac {1}{2} a \left (x+x^3\right )^{2/3}-\frac {\left (a \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{1+x^2}} \, dx}{3 \sqrt [3]{x+x^3}}+\frac {\left (3 b \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {x}{\sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x+x^3}} \\ & = \frac {1}{2} a \left (x+x^3\right )^{2/3}-\frac {\left (a \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {x}{\sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x+x^3}}+\frac {\left (3 b \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{1+x^3}} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{x+x^3}} \\ & = \frac {1}{2} a \left (x+x^3\right )^{2/3}+\frac {\sqrt {3} b \sqrt [3]{x} \sqrt [3]{1+x^2} \arctan \left (\frac {1+\frac {2 x^{2/3}}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{2 \sqrt [3]{x+x^3}}-\frac {3 b \sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (x^{2/3}-\sqrt [3]{1+x^2}\right )}{4 \sqrt [3]{x+x^3}}-\frac {\left (a \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{1+x^3}} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{x+x^3}} \\ & = \frac {1}{2} a \left (x+x^3\right )^{2/3}-\frac {a \sqrt [3]{x} \sqrt [3]{1+x^2} \arctan \left (\frac {1+\frac {2 x^{2/3}}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{2 \sqrt {3} \sqrt [3]{x+x^3}}+\frac {\sqrt {3} b \sqrt [3]{x} \sqrt [3]{1+x^2} \arctan \left (\frac {1+\frac {2 x^{2/3}}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{2 \sqrt [3]{x+x^3}}+\frac {a \sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (x^{2/3}-\sqrt [3]{1+x^2}\right )}{4 \sqrt [3]{x+x^3}}-\frac {3 b \sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (x^{2/3}-\sqrt [3]{1+x^2}\right )}{4 \sqrt [3]{x+x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.28 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.73 \[ \int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx=\frac {a x \left (1+x^2\right )}{2 \sqrt [3]{x+x^3}}-\frac {(a-3 b) \sqrt [3]{x} \sqrt [3]{1+x^2} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{1+x^2}}\right )}{2 \sqrt {3} \sqrt [3]{x+x^3}}+\frac {(a-3 b) \sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (-x^{2/3}+\sqrt [3]{1+x^2}\right )}{6 \sqrt [3]{x+x^3}}+\frac {(-a+3 b) \sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (x^{4/3}+x^{2/3} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )}{12 \sqrt [3]{x+x^3}} \]

[In]

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

[Out]

(a*x*(1 + x^2))/(2*(x + x^3)^(1/3)) - ((a - 3*b)*x^(1/3)*(1 + x^2)^(1/3)*ArcTan[(Sqrt[3]*x^(2/3))/(x^(2/3) + 2
*(1 + x^2)^(1/3))])/(2*Sqrt[3]*(x + x^3)^(1/3)) + ((a - 3*b)*x^(1/3)*(1 + x^2)^(1/3)*Log[-x^(2/3) + (1 + x^2)^
(1/3)])/(6*(x + x^3)^(1/3)) + ((-a + 3*b)*x^(1/3)*(1 + x^2)^(1/3)*Log[x^(4/3) + x^(2/3)*(1 + x^2)^(1/3) + (1 +
x^2)^(2/3)])/(12*(x + x^3)^(1/3))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 0.95 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.31

method	result	size
meijerg	\(\frac {3 a \,x^{\frac {8}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {4}{3}\right ], \left [\frac {7}{3}\right ], -x^{2}\right )}{8}+\frac {3 b \,x^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{2}\right )}{2}\)	\(36\)
risch	\(\frac {a x \left (x^{2}+1\right )}{2 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}}+\frac {3 b \,x^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{2}\right )}{2}-\frac {a \,x^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{2}\right )}{2}\)	\(54\)
pseudoelliptic	\(-\frac {x \left (\frac {\left (-a +3 b \right ) \ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}+{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )}{2}+\sqrt {3}\, \left (a -3 b \right ) \arctan \left (\frac {\left (2 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )+\left (a -3 b \right ) \ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}-x}{x}\right )+3 {\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}} a \right )}{6 \left (-{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}+x \right ) \left ({\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}+x \left (x +{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}\right )\right )}\)	\(150\)
trager	\(\frac {a \left (x^{3}+x \right )^{\frac {2}{3}}}{2}+\frac {\left (a -3 b \right ) \left (6 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \ln \left (180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}+144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x +174 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+15 x \left (x^{3}+x \right )^{\frac {1}{3}}-180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2}+20 x^{2}+36 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )+8\right )-6 \ln \left (180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}-144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x -114 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-9 x \left (x^{3}+x \right )^{\frac {1}{3}}-180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2}-4 x^{2}-96 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )-3\right ) \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )-\ln \left (180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}-144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x -114 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-9 x \left (x^{3}+x \right )^{\frac {1}{3}}-180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2}-4 x^{2}-96 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )-3\right )\right )}{6}\)	\(438\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 47.61 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.85 \[ \int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx=-\frac {1}{6} \, \sqrt {3} {\left (a - 3 \, b\right )} \arctan \left (-\frac {196 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {1}{3}} x - \sqrt {3} {\left (539 \, x^{2} + 507\right )} - 1274 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{2205 \, x^{2} + 2197}\right ) + \frac {1}{12} \, {\left (a - 3 \, b\right )} \log \left (3 \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - 3 \, {\left (x^{3} + x\right )}^{\frac {2}{3}} + 1\right ) + \frac {1}{2} \, {\left (x^{3} + x\right )}^{\frac {2}{3}} a \]

[In]

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

[Out]

-1/6*sqrt(3)*(a - 3*b)*arctan(-(196*sqrt(3)*(x^3 + x)^(1/3)*x - sqrt(3)*(539*x^2 + 507) - 1274*sqrt(3)*(x^3 +
x)^(2/3))/(2205*x^2 + 2197)) + 1/12*(a - 3*b)*log(3*(x^3 + x)^(1/3)*x - 3*(x^3 + x)^(2/3) + 1) + 1/2*(x^3 + x)
^(2/3)*a

Sympy [F]

\[ \int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx=\int \frac {a x^{2} + b}{\sqrt [3]{x \left (x^{2} + 1\right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx=\int { \frac {a x^{2} + b}{{\left (x^{3} + x\right )}^{\frac {1}{3}}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.70 \[ \int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx=\frac {1}{2} \, a x^{2} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + \frac {1}{6} \, \sqrt {3} {\left (a - 3 \, b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{12} \, {\left (a - 3 \, b\right )} \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{6} \, {\left (a - 3 \, b\right )} \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]

[In]

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

[Out]

1/2*a*x^2*(1/x^2 + 1)^(2/3) + 1/6*sqrt(3)*(a - 3*b)*arctan(1/3*sqrt(3)*(2*(1/x^2 + 1)^(1/3) + 1)) - 1/12*(a -
3*b)*log((1/x^2 + 1)^(2/3) + (1/x^2 + 1)^(1/3) + 1) + 1/6*(a - 3*b)*log(abs((1/x^2 + 1)^(1/3) - 1))

Mupad [F(-1)]

Timed out. \[ \int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx=\int \frac {a\,x^2+b}{{\left (x^3+x\right )}^{1/3}} \,d x \]

[In]

int((b + a*x^2)/(x + x^3)^(1/3),x)

[Out]

int((b + a*x^2)/(x + x^3)^(1/3), x)