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

Optimal. Leaf size=129 \[ \frac {1}{6} (3 b-a) \log \left (\sqrt [3]{x^3+x}-x\right )+\frac {1}{6} \left (3 \sqrt {3} b-\sqrt {3} a\right ) \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+x}+x}\right )+\frac {\sqrt [3]{x^3+x} \left (a x^2+3 b\right )}{2 x}+\frac {1}{12} (a-3 b) \log \left (\sqrt [3]{x^3+x} x+\left (x^3+x\right )^{2/3}+x^2\right ) \]

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Rubi [A]  time = 0.21, antiderivative size = 215, normalized size of antiderivative = 1.67, number of steps used = 12, number of rules used = 12, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {2038, 2004, 2032, 329, 275, 331, 292, 31, 634, 618, 204, 628} \begin {gather*} \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 (1-\frac {x^{2/3}}{\sqrt [3]{x^2+1}}\right )}{6 \left (x^3+x\right )^{2/3}}+\frac {x^{2/3} \left (x^2+1\right )^{2/3} (a-3 b) \log \left (\frac {x^{4/3}}{\left (x^2+1\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{x^2+1}}+1\right )}{12 \left (x^3+x\right )^{2/3}}-\frac {x^{2/3} \left (x^2+1\right )^{2/3} (a-3 b) \tan ^{-1}\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 {3 b \left (x^3+x\right )^{4/3}}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, 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 275

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 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 329

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

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 2004

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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[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 2038

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*} \int \frac {\left (-b+a x^2\right ) \sqrt [3]{x+x^3}}{x^2} \, dx &=\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 ) \operatorname {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 ) \operatorname {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 {\left ((-a+3 b) x^{2/3} \left (1+x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x}{1-x^3} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\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 {\left ((-a+3 b) x^{2/3} \left (1+x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 \left (x+x^3\right )^{2/3}}+\frac {\left ((-a+3 b) x^{2/3} \left (1+x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1-x}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 \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} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 \left (x+x^3\right )^{2/3}}-\frac {\left ((-a+3 b) x^{2/3} \left (1+x^2\right )^{2/3}\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 )}{12 \left (x+x^3\right )^{2/3}}+\frac {\left ((-a+3 b) x^{2/3} \left (1+x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 \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} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 \left (x+x^3\right )^{2/3}}+\frac {(a-3 b) x^{2/3} \left (1+x^2\right )^{2/3} \log \left (1+\frac {x^{4/3}}{\left (1+x^2\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{12 \left (x+x^3\right )^{2/3}}-\frac {\left ((-a+3 b) x^{2/3} \left (1+x^2\right )^{2/3}\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 \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} \tan ^{-1}\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 (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 \left (x+x^3\right )^{2/3}}+\frac {(a-3 b) x^{2/3} \left (1+x^2\right )^{2/3} \log \left (1+\frac {x^{4/3}}{\left (1+x^2\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{12 \left (x+x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 62, normalized size = 0.48 \begin {gather*} \frac {3 \sqrt [3]{x^3+x} \left (x^2 (a-3 b) \, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};-x^2\right )+2 b \left (x^2+1\right )^{4/3}\right )}{4 x \sqrt [3]{x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(x + x^3)^(1/3)*(2*b*(1 + x^2)^(4/3) + (a - 3*b)*x^2*Hypergeometric2F1[-1/3, 2/3, 5/3, -x^2]))/(4*x*(1 + x^
2)^(1/3))

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IntegrateAlgebraic [A]  time = 0.32, size = 129, normalized size = 1.00 \begin {gather*} \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 ) \tan ^{-1}\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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 71.03, size = 114, normalized size = 0.88 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.48, size = 93, normalized size = 0.72 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 3.32, size = 36, normalized size = 0.28

method	result	size
meijerg	\(\frac {3 a \,x^{\frac {4}{3}} \hypergeom \left (\left [-\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -x^{2}\right )}{4}+\frac {3 b \hypergeom \left (\left [-\frac {1}{3}, -\frac {1}{3}\right ], \left [\frac {2}{3}\right ], -x^{2}\right )}{2 x^{\frac {2}{3}}}\)	\(36\)
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 \ln \left (45 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}-72 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-72 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x -87 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}-45 \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 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+8\right ) \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )-3 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \ln \left (45 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}+72 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+72 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x +57 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}-45 \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 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )-3\right )-\ln \left (45 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}-72 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-72 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x -87 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}-45 \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 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+8\right )\right )}{6}\)	\(449\)
risch	\(\frac {\left (a \,x^{2}+3 b \right ) \left (x \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{2 x}-\frac {\left (a -3 b \right ) \left (-\RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{4}-20 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{4}+48 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}+100 x^{4}-30 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}-14 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}-60 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} x^{2}-\RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2}+48 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}-36 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}+140 x^{2}+6 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )-60 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}+40}{x^{2}+1}\right )+2 \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{4}+38 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{4}-18 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}-16 x^{4}-30 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}+70 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}-60 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} x^{2}-5 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2}-18 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}+96 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}-28 x^{2}+32 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )-60 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}-12}{x^{2}+1}\right )\right ) \left (x \left (x^{2}+1\right )\right )^{\frac {1}{3}} \left (x^{2} \left (x^{2}+1\right )^{2}\right )^{\frac {1}{3}}}{12 x \left (x^{2}+1\right )}\)	\(514\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {\left (b-a\,x^2\right )\,{\left (x^3+x\right )}^{1/3}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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