3.2.8 \(\int \frac {(1-x^3)^{2/3}}{a+b x} \, dx\) [108]

Optimal. Leaf size=384 \[ \frac {\left (1-x^3\right )^{2/3}}{2 b}-\frac {\left (a^3+b^3\right ) x^2 F_1\left (\frac {2}{3};\frac {1}{3},1;\frac {5}{3};x^3,-\frac {b^3 x^3}{a^3}\right )}{2 a^2 b^2}+\frac {a^2 \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^3}-\frac {\left (a^3+b^3\right )^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{a^3+b^3} x}{a \sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^3}+\frac {\left (a^3+b^3\right )^{2/3} \tan ^{-1}\left (\frac {1+\frac {2 b \sqrt [3]{1-x^3}}{\sqrt [3]{a^3+b^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^3}+\frac {a x^2 \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};x^3\right )}{2 b^2}-\frac {\left (a^3+b^3\right )^{2/3} \log \left (a^3+b^3 x^3\right )}{3 b^3}+\frac {\left (a^3+b^3\right )^{2/3} \log \left (-\frac {\sqrt [3]{a^3+b^3} x}{a}-\sqrt [3]{1-x^3}\right )}{2 b^3}-\frac {a^2 \log \left (x+\sqrt [3]{1-x^3}\right )}{2 b^3}+\frac {\left (a^3+b^3\right )^{2/3} \log \left (\sqrt [3]{a^3+b^3}-b \sqrt [3]{1-x^3}\right )}{2 b^3} \]

[Out]

1/2*(-x^3+1)^(2/3)/b-1/2*(a^3+b^3)*x^2*AppellF1(2/3,1/3,1,5/3,x^3,-b^3*x^3/a^3)/a^2/b^2+1/2*a*x^2*hypergeom([1
/3, 2/3],[5/3],x^3)/b^2-1/3*(a^3+b^3)^(2/3)*ln(b^3*x^3+a^3)/b^3+1/2*(a^3+b^3)^(2/3)*ln(-(a^3+b^3)^(1/3)*x/a-(-
x^3+1)^(1/3))/b^3-1/2*a^2*ln(x+(-x^3+1)^(1/3))/b^3+1/2*(a^3+b^3)^(2/3)*ln((a^3+b^3)^(1/3)-b*(-x^3+1)^(1/3))/b^
3+1/3*a^2*arctan(1/3*(1-2*x/(-x^3+1)^(1/3))*3^(1/2))/b^3*3^(1/2)-1/3*(a^3+b^3)^(2/3)*arctan(1/3*(1-2*(a^3+b^3)
^(1/3)*x/a/(-x^3+1)^(1/3))*3^(1/2))/b^3*3^(1/2)+1/3*(a^3+b^3)^(2/3)*arctan(1/3*(1+2*b*(-x^3+1)^(1/3)/(a^3+b^3)
^(1/3))*3^(1/2))/b^3*3^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.29, antiderivative size = 384, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {2178, 2177, 245, 2181, 384, 524, 455, 57, 631, 210, 31, 371} \begin {gather*} -\frac {\left (a^3+b^3\right )^{2/3} \log \left (a^3+b^3 x^3\right )}{3 b^3}+\frac {\left (a^3+b^3\right )^{2/3} \log \left (-\frac {x \sqrt [3]{a^3+b^3}}{a}-\sqrt [3]{1-x^3}\right )}{2 b^3}+\frac {\left (a^3+b^3\right )^{2/3} \log \left (\sqrt [3]{a^3+b^3}-b \sqrt [3]{1-x^3}\right )}{2 b^3}-\frac {\left (a^3+b^3\right )^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 x \sqrt [3]{a^3+b^3}}{a \sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^3}+\frac {\left (a^3+b^3\right )^{2/3} \tan ^{-1}\left (\frac {\frac {2 b \sqrt [3]{1-x^3}}{\sqrt [3]{a^3+b^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^3}-\frac {a^2 \log \left (\sqrt [3]{1-x^3}+x\right )}{2 b^3}+\frac {a^2 \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^3}-\frac {x^2 \left (a^3+b^3\right ) F_1\left (\frac {2}{3};\frac {1}{3},1;\frac {5}{3};x^3,-\frac {b^3 x^3}{a^3}\right )}{2 a^2 b^2}+\frac {a x^2 \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};x^3\right )}{2 b^2}+\frac {\left (1-x^3\right )^{2/3}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x^3)^(2/3)/(a + b*x),x]

[Out]

(1 - x^3)^(2/3)/(2*b) - ((a^3 + b^3)*x^2*AppellF1[2/3, 1/3, 1, 5/3, x^3, -((b^3*x^3)/a^3)])/(2*a^2*b^2) + (a^2
*ArcTan[(1 - (2*x)/(1 - x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^3) - ((a^3 + b^3)^(2/3)*ArcTan[(1 - (2*(a^3 + b^3)^(1
/3)*x)/(a*(1 - x^3)^(1/3)))/Sqrt[3]])/(Sqrt[3]*b^3) + ((a^3 + b^3)^(2/3)*ArcTan[(1 + (2*b*(1 - x^3)^(1/3))/(a^
3 + b^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^3) + (a*x^2*Hypergeometric2F1[1/3, 2/3, 5/3, x^3])/(2*b^2) - ((a^3 + b^3)
^(2/3)*Log[a^3 + b^3*x^3])/(3*b^3) + ((a^3 + b^3)^(2/3)*Log[-(((a^3 + b^3)^(1/3)*x)/a) - (1 - x^3)^(1/3)])/(2*
b^3) - (a^2*Log[x + (1 - x^3)^(1/3)])/(2*b^3) + ((a^3 + b^3)^(2/3)*Log[(a^3 + b^3)^(1/3) - b*(1 - x^3)^(1/3)])
/(2*b^3)

Rule 31

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

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 455

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

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 631

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 2177

Int[((e_.) + (f_.)*(x_))/(((c_.) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Dist[f/d, Int[1/(a +
b*x^3)^(1/3), x], x] + Dist[(d*e - c*f)/d, Int[1/((c + d*x)*(a + b*x^3)^(1/3)), x], x] /; FreeQ[{a, b, c, d,
e, f}, x]

Rule 2178

Int[((a_) + (b_.)*(x_)^3)^(2/3)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[(a + b*x^3)^(2/3)/(2*d), x] + (Dist[1/d
^2, Int[(a*d^2 + b*c^2*x)/((c + d*x)*(a + b*x^3)^(1/3)), x], x] - Dist[b*(c/d^2), Int[x/(a + b*x^3)^(1/3), x],
x]) /; FreeQ[{a, b, c, d}, x]

Rule 2181

Int[(Px_.)*((c_) + (d_.)*(x_))^(q_)*((a_) + (b_.)*(x_)^3)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c^3 + d^3*x
^3)^q*(a + b*x^3)^p, Px/(c^2 - c*d*x + d^2*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p}, x] && PolyQ[Px, x] && ILtQ
[q, 0] && RationalQ[p] && EqQ[Denominator[p], 3]

Rubi steps

\begin {align*} \int \frac {\left (1-x^3\right )^{2/3}}{a+b x} \, dx &=\int \frac {\left (1-x^3\right )^{2/3}}{a+b x} \, dx\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]
time = 14.24, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1-x^3\right )^{2/3}}{a+b x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F(-1)] Timed out
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^3+1)^(2/3)/(b*x+a),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F] N/A
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Could not integrate

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - x^3)^(2/3)/(a + b*x),x)

[Out]

int((1 - x^3)^(2/3)/(a + b*x), x)

________________________________________________________________________________________