\(\int \frac {(-b+a x^2) \sqrt [3]{x+x^3}}{x^2} \, dx\) [1871]

Optimal result

Integrand size = 22, antiderivative size = 129 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [3]{x+x^3}}{x^2} \, dx=\frac {\left (3 b+a x^2\right ) \sqrt [3]{x+x^3}}{2 x}+\frac {1}{6} \left (-\sqrt {3} a+3 \sqrt {3} b\right ) \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )+\frac {1}{6} (-a+3 b) \log \left (-x+\sqrt [3]{x+x^3}\right )+\frac {1}{12} (a-3 b) \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right ) \]

[Out]

1/2*(a*x^2+3*b)*(x^3+x)^(1/3)/x+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 = 0.09 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2063, 2029, 2057, 335, 281, 337} \[ \int \frac {\left (-b+a x^2\right ) \sqrt [3]{x+x^3}}{x^2} \, dx=-\frac {x^{2/3} \left (x^2+1\right )^{2/3} (a-3 b) \arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{2 \sqrt {3} \left (x^3+x\right )^{2/3}}+\frac {1}{2} x \sqrt [3]{x^3+x} (a-3 b)-\frac {x^{2/3} \left (x^2+1\right )^{2/3} (a-3 b) \log \left (x^{2/3}-\sqrt [3]{x^2+1}\right )}{4 \left (x^3+x\right )^{2/3}}+\frac {3 b \left (x^3+x\right )^{4/3}}{2 x^2} \]

[In]

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

[Out]

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

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 337

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

Rule 2029

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

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 2063

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

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

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.62 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [3]{x+x^3}}{x^2} \, dx=\frac {\left (3 b+a x^2\right ) \sqrt [3]{x+x^3}}{2 x}-\frac {(a-3 b) \sqrt [3]{x+x^3} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{1+x^2}}\right )}{2 \sqrt {3} \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {(-a+3 b) \sqrt [3]{x+x^3} \log \left (-x^{2/3}+\sqrt [3]{1+x^2}\right )}{6 \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {(a-3 b) \sqrt [3]{x+x^3} \log \left (x^{4/3}+x^{2/3} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )}{12 \sqrt [3]{x} \sqrt [3]{1+x^2}} \]

[In]

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

[Out]

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

Maple [C] (verified)

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

Time = 2.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.28

method	result	size
meijerg	\(\frac {3 a \,x^{\frac {4}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -x^{2}\right )}{4}+\frac {3 b \operatorname {hypergeom}\left (\left [-\frac {1}{3}, -\frac {1}{3}\right ], \left [\frac {2}{3}\right ], -x^{2}\right )}{2 x^{\frac {2}{3}}}\)	\(36\)
pseudoelliptic	\(\frac {3 \left (-a \,x^{2}-3 b \right ) {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}+\left (a -3 b \right ) \left (-\sqrt {3}\, \arctan \left (\frac {\left (2 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )+\ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}-x}{x}\right )-\frac {\ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}+{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )}{2}\right ) x}{6 \left (-{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}+x \right ) \left ({\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}+{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +x^{2}\right )}\)	\(150\)
trager	\(\frac {\left (a \,x^{2}+3 b \right ) \left (x^{3}+x \right )^{\frac {1}{3}}}{2 x}+\frac {\left (a -3 b \right ) \left (3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \ln \left (45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}+72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+72 \left (x^{3}+x \right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x +57 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}-45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-9 x \left (x^{3}+x \right )^{\frac {1}{3}}-4 x^{2}+48 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )-3\right )-3 \ln \left (45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}-72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-72 \left (x^{3}+x \right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x -87 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}-45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+15 x \left (x^{3}+x \right )^{\frac {1}{3}}+20 x^{2}-18 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+8\right ) \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+\ln \left (45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}-72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-72 \left (x^{3}+x \right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x -87 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}-45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+15 x \left (x^{3}+x \right )^{\frac {1}{3}}+20 x^{2}-18 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+8\right )\right )}{6}\)	\(447\)
risch	\(\text {Expression too large to display}\)	\(745\)

[In]

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

[Out]

3/4*a*x^(4/3)*hypergeom([-1/3,2/3],[5/3],-x^2)+3/2*b/x^(2/3)*hypergeom([-1/3,-1/3],[2/3],-x^2)

Fricas [A] (verification not implemented)

none

Time = 40.32 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.88 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [3]{x+x^3}}{x^2} \, dx=-\frac {2 \, \sqrt {3} {\left (a - 3 \, b\right )} x \arctan \left (-\frac {196 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {1}{3}} x - \sqrt {3} {\left (539 \, x^{2} + 507\right )} - 1274 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{2205 \, x^{2} + 2197}\right ) + {\left (a - 3 \, b\right )} x \log \left (3 \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - 3 \, {\left (x^{3} + x\right )}^{\frac {2}{3}} + 1\right ) - 6 \, {\left (a x^{2} + 3 \, b\right )} {\left (x^{3} + x\right )}^{\frac {1}{3}}}{12 \, x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.72 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [3]{x+x^3}}{x^2} \, dx=\frac {1}{2} \, a x^{2} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + \frac {1}{6} \, \sqrt {3} {\left (a - 3 \, b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{12} \, {\left (a - 3 \, b\right )} \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{6} \, {\left (a - 3 \, b\right )} \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) + \frac {3}{2} \, b {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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