3.20.38 \(\int \frac {b+a x^6}{x^6 \sqrt [3]{x+x^3}} \, dx\)

Optimal. Leaf size=135 \[ \frac {1}{4} \left (a-i \sqrt {3} a\right ) \log \left (-2 i \sqrt [3]{x^3+x}+\sqrt {3} x-i x\right )+\frac {1}{4} \left (a+i \sqrt {3} a\right ) \log \left (2 i \sqrt [3]{x^3+x}+\sqrt {3} x+i x\right )-\frac {1}{2} a \log \left (\sqrt [3]{x^3+x}-x\right )-\frac {3 b \left (x^3+x\right )^{2/3} \left (9 x^4-6 x^2+5\right )}{80 x^6} \]

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Rubi [A]  time = 0.18, antiderivative size = 156, normalized size of antiderivative = 1.16, number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {2052, 2011, 329, 275, 239, 2016, 2014} \begin {gather*} -\frac {3 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} a \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 \sqrt [3]{x^3+x}}-\frac {3 b \left (x^3+x\right )^{2/3}}{16 x^6}+\frac {9 b \left (x^3+x\right )^{2/3}}{40 x^4}-\frac {27 b \left (x^3+x\right )^{2/3}}{80 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b + a*x^6)/(x^6*(x + x^3)^(1/3)),x]

[Out]

(-3*b*(x + x^3)^(2/3))/(16*x^6) + (9*b*(x + x^3)^(2/3))/(40*x^4) - (27*b*(x + x^3)^(2/3))/(80*x^2) + (Sqrt[3]*
a*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)) - (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))

Rule 239

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 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 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 2011

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 2014

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

Rule 2016

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

Rule 2052

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

Rubi steps

\begin {align*} \int \frac {b+a x^6}{x^6 \sqrt [3]{x+x^3}} \, dx &=\int \left (\frac {a}{\sqrt [3]{x+x^3}}+\frac {b}{x^6 \sqrt [3]{x+x^3}}\right ) \, dx\\ &=a \int \frac {1}{\sqrt [3]{x+x^3}} \, dx+b \int \frac {1}{x^6 \sqrt [3]{x+x^3}} \, dx\\ &=-\frac {3 b \left (x+x^3\right )^{2/3}}{16 x^6}-\frac {1}{4} (3 b) \int \frac {1}{x^4 \sqrt [3]{x+x^3}} \, dx+\frac {\left (a \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 {3 b \left (x+x^3\right )^{2/3}}{16 x^6}+\frac {9 b \left (x+x^3\right )^{2/3}}{40 x^4}+\frac {1}{20} (9 b) \int \frac {1}{x^2 \sqrt [3]{x+x^3}} \, dx+\frac {\left (3 a \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x+x^3}}\\ &=-\frac {3 b \left (x+x^3\right )^{2/3}}{16 x^6}+\frac {9 b \left (x+x^3\right )^{2/3}}{40 x^4}-\frac {27 b \left (x+x^3\right )^{2/3}}{80 x^2}+\frac {\left (3 a \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1+x^3}} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{x+x^3}}\\ &=-\frac {3 b \left (x+x^3\right )^{2/3}}{16 x^6}+\frac {9 b \left (x+x^3\right )^{2/3}}{40 x^4}-\frac {27 b \left (x+x^3\right )^{2/3}}{80 x^2}+\frac {\sqrt {3} a \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 \sqrt [3]{x+x^3}}-\frac {3 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}}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 149, normalized size = 1.10 \begin {gather*} \frac {3 \sqrt [3]{x} \sqrt [3]{x^2+1} \left (\frac {1}{12} a \left (-2 \log \left (1-\frac {x^{2/3}}{\sqrt [3]{x^2+1}}\right )+\log \left (\frac {x^{4/3}}{\left (x^2+1\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{x^2+1}}+1\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )\right )-\frac {b \left (x^2+1\right )^{2/3} \left (9 x^4-6 x^2+5\right )}{80 x^{16/3}}\right )}{\sqrt [3]{x^3+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b + a*x^6)/(x^6*(x + x^3)^(1/3)),x]

[Out]

(3*x^(1/3)*(1 + x^2)^(1/3)*(-1/80*(b*(1 + x^2)^(2/3)*(5 - 6*x^2 + 9*x^4))/x^(16/3) + (a*(2*Sqrt[3]*ArcTan[(1 +
(2*x^(2/3))/(1 + x^2)^(1/3))/Sqrt[3]] - 2*Log[1 - x^(2/3)/(1 + x^2)^(1/3)] + Log[1 + x^(4/3)/(1 + x^2)^(2/3)
+ x^(2/3)/(1 + x^2)^(1/3)]))/12))/(x + x^3)^(1/3)

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IntegrateAlgebraic [A]  time = 0.30, size = 112, normalized size = 0.83 \begin {gather*} -\frac {3 b \left (x+x^3\right )^{2/3} \left (5-6 x^2+9 x^4\right )}{80 x^6}+\frac {1}{2} \sqrt {3} a \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )-\frac {1}{2} a \log \left (-x+\sqrt [3]{x+x^3}\right )+\frac {1}{4} a \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^6)/(x^6*(x + x^3)^(1/3)),x]

[Out]

(-3*b*(x + x^3)^(2/3)*(5 - 6*x^2 + 9*x^4))/(80*x^6) + (Sqrt[3]*a*ArcTan[(Sqrt[3]*x)/(x + 2*(x + x^3)^(1/3))])/
2 - (a*Log[-x + (x + x^3)^(1/3)])/2 + (a*Log[x^2 + x*(x + x^3)^(1/3) + (x + x^3)^(2/3)])/4

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fricas [A]  time = 1.36, size = 118, normalized size = 0.87 \begin {gather*} \frac {40 \, \sqrt {3} a x^{6} \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 ) - 20 \, a x^{6} \log \left (3 \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - 3 \, {\left (x^{3} + x\right )}^{\frac {2}{3}} + 1\right ) - 3 \, {\left (9 \, b x^{4} - 6 \, b x^{2} + 5 \, b\right )} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{80 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^6+b)/x^6/(x^3+x)^(1/3),x, algorithm="fricas")

[Out]

1/80*(40*sqrt(3)*a*x^6*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)) - 20*a*x^6*log(3*(x^3 + x)^(1/3)*x - 3*(x^3 + x)^(2/3) + 1) - 3*(9*b*x^4 - 6*b*x^
2 + 5*b)*(x^3 + x)^(2/3))/x^6

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giac [A]  time = 0.22, size = 88, normalized size = 0.65 \begin {gather*} -\frac {3}{16} \, b {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {8}{3}} + \frac {3}{5} \, b {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {5}{3}} - \frac {1}{2} \, \sqrt {3} a \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{4} \, a \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}{2} \, a \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) - \frac {3}{4} \, b {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^6+b)/x^6/(x^3+x)^(1/3),x, algorithm="giac")

[Out]

-3/16*b*(1/x^2 + 1)^(8/3) + 3/5*b*(1/x^2 + 1)^(5/3) - 1/2*sqrt(3)*a*arctan(1/3*sqrt(3)*(2*(1/x^2 + 1)^(1/3) +
1)) + 1/4*a*log((1/x^2 + 1)^(2/3) + (1/x^2 + 1)^(1/3) + 1) - 1/2*a*log(abs((1/x^2 + 1)^(1/3) - 1)) - 3/4*b*(1/
x^2 + 1)^(2/3)

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maple [C]  time = 1.55, size = 44, normalized size = 0.33

method	result	size
meijerg	\(\frac {3 a \,x^{\frac {2}{3}} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{2}\right )}{2}-\frac {3 b \left (\frac {9}{5} x^{4}-\frac {6}{5} x^{2}+1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}}{16 x^{\frac {16}{3}}}\)	\(44\)
risch	\(-\frac {3 b \left (x^{2}+1\right ) \left (9 x^{4}-6 x^{2}+5\right )}{80 x^{5} \left (x \left (x^{2}+1\right )\right )^{\frac {1}{3}}}+\frac {3 a \,x^{\frac {2}{3}} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{2}\right )}{2}\)	\(51\)
trager	\(-\frac {3 b \left (x^{3}+x \right )^{\frac {2}{3}} \left (9 x^{4}-6 x^{2}+5\right )}{80 x^{6}}-\frac {a \left (6 \ln \left (-36 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}-144 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+90 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x +60 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}+36 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+9 x \left (x^{3}+x \right )^{\frac {1}{3}}-25 x^{2}-18 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )-10\right ) \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )+\ln \left (-144 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}+144 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-54 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x -66 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}+144 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-15 x \left (x^{3}+x \right )^{\frac {1}{3}}+20 x^{2}-102 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )+15\right )-\ln \left (-36 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}-144 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+90 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x +60 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}+36 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+9 x \left (x^{3}+x \right )^{\frac {1}{3}}-25 x^{2}-18 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )-10\right )\right )}{2}\)	\(436\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^6+b)/x^6/(x^3+x)^(1/3),x,method=_RETURNVERBOSE)

[Out]

3/2*a*x^(2/3)*hypergeom([1/3,1/3],[4/3],-x^2)-3/16*b/x^(16/3)*(9/5*x^4-6/5*x^2+1)*(x^2+1)^(2/3)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{6} + b}{{\left (x^{3} + x\right )}^{\frac {1}{3}} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^6+b)/x^6/(x^3+x)^(1/3),x, algorithm="maxima")

[Out]

integrate((a*x^6 + b)/((x^3 + x)^(1/3)*x^6), x)

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mupad [B]  time = 1.30, size = 54, normalized size = 0.40 \begin {gather*} \frac {3\,a\,x\,{\left (x^2+1\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ -x^2\right )}{2\,{\left (x^3+x\right )}^{1/3}}-\frac {3\,b\,{\left (x^3+x\right )}^{2/3}\,\left (9\,x^4-6\,x^2+5\right )}{80\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b + a*x^6)/(x^6*(x + x^3)^(1/3)),x)

[Out]

(3*a*x*(x^2 + 1)^(1/3)*hypergeom([1/3, 1/3], 4/3, -x^2))/(2*(x + x^3)^(1/3)) - (3*b*(x + x^3)^(2/3)*(9*x^4 - 6
*x^2 + 5))/(80*x^6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{6} + b}{x^{6} \sqrt [3]{x \left (x^{2} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**6+b)/x**6/(x**3+x)**(1/3),x)

[Out]

Integral((a*x**6 + b)/(x**6*(x*(x**2 + 1))**(1/3)), x)

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