3.1.37 \(\int \frac {1}{(a-b x^3) (a+b x^3)^{2/3}} \, dx\) [37]

Optimal. Leaf size=452 \[ -\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3} a^{4/3} \sqrt [3]{b}}-\frac {\tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{2\ 2^{2/3} \sqrt {3} a^{4/3} \sqrt [3]{b}}+\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 a \left (a+b x^3\right )^{2/3}}-\frac {\log \left (2^{2/3}-\frac {\sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{6\ 2^{2/3} a^{4/3} \sqrt [3]{b}}+\frac {\log \left (1+\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}-\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}\right )}{6\ 2^{2/3} a^{4/3} \sqrt [3]{b}}-\frac {\log \left (1+\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}\right )}{3\ 2^{2/3} a^{4/3} \sqrt [3]{b}}+\frac {\log \left (2 \sqrt [3]{2}+\frac {\left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}+\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}\right )}{12\ 2^{2/3} a^{4/3} \sqrt [3]{b}} \]

[Out]

1/2*x*(1+b*x^3/a)^(2/3)*hypergeom([1/3, 2/3],[4/3],-b*x^3/a)/a/(b*x^3+a)^(2/3)-1/12*ln(2^(2/3)+(-a^(1/3)-b^(1/
3)*x)/(b*x^3+a)^(1/3))*2^(1/3)/a^(4/3)/b^(1/3)+1/12*ln(1+2^(2/3)*(a^(1/3)+b^(1/3)*x)^2/(b*x^3+a)^(2/3)-2^(1/3)
*(a^(1/3)+b^(1/3)*x)/(b*x^3+a)^(1/3))*2^(1/3)/a^(4/3)/b^(1/3)-1/6*ln(1+2^(1/3)*(a^(1/3)+b^(1/3)*x)/(b*x^3+a)^(
1/3))*2^(1/3)/a^(4/3)/b^(1/3)+1/24*ln(2*2^(1/3)+(a^(1/3)+b^(1/3)*x)^2/(b*x^3+a)^(2/3)+2^(2/3)*(a^(1/3)+b^(1/3)
*x)/(b*x^3+a)^(1/3))*2^(1/3)/a^(4/3)/b^(1/3)-1/6*arctan(1/3*(1-2*2^(1/3)*(a^(1/3)+b^(1/3)*x)/(b*x^3+a)^(1/3))*
3^(1/2))*2^(1/3)/a^(4/3)/b^(1/3)*3^(1/2)-1/12*arctan(1/3*(1+2^(1/3)*(a^(1/3)+b^(1/3)*x)/(b*x^3+a)^(1/3))*3^(1/
2))*2^(1/3)/a^(4/3)/b^(1/3)*3^(1/2)

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Rubi [A]
time = 0.20, antiderivative size = 452, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 11, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {421, 252, 251, 420, 493, 298, 31, 648, 631, 210, 642} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {1-\frac {2 \sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3} a^{4/3} \sqrt [3]{b}}-\frac {\text {ArcTan}\left (\frac {\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{2\ 2^{2/3} \sqrt {3} a^{4/3} \sqrt [3]{b}}-\frac {\log \left (2^{2/3}-\frac {\sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{6\ 2^{2/3} a^{4/3} \sqrt [3]{b}}+\frac {\log \left (\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}-\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+1\right )}{6\ 2^{2/3} a^{4/3} \sqrt [3]{b}}-\frac {\log \left (\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+1\right )}{3\ 2^{2/3} a^{4/3} \sqrt [3]{b}}+\frac {\log \left (\frac {\left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}+\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+2 \sqrt [3]{2}\right )}{12\ 2^{2/3} a^{4/3} \sqrt [3]{b}}+\frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 a \left (a+b x^3\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[(1 - (2*2^(1/3)*(a^(1/3) + b^(1/3)*x))/(a + b*x^3)^(1/3))/Sqrt[3]]/(2^(2/3)*Sqrt[3]*a^(4/3)*b^(1/3)))
- ArcTan[(1 + (2^(1/3)*(a^(1/3) + b^(1/3)*x))/(a + b*x^3)^(1/3))/Sqrt[3]]/(2*2^(2/3)*Sqrt[3]*a^(4/3)*b^(1/3))
+ (x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(2*a*(a + b*x^3)^(2/3)) - Log[2^(2
/3) - (a^(1/3) + b^(1/3)*x)/(a + b*x^3)^(1/3)]/(6*2^(2/3)*a^(4/3)*b^(1/3)) + Log[1 + (2^(2/3)*(a^(1/3) + b^(1/
3)*x)^2)/(a + b*x^3)^(2/3) - (2^(1/3)*(a^(1/3) + b^(1/3)*x))/(a + b*x^3)^(1/3)]/(6*2^(2/3)*a^(4/3)*b^(1/3)) -
Log[1 + (2^(1/3)*(a^(1/3) + b^(1/3)*x))/(a + b*x^3)^(1/3)]/(3*2^(2/3)*a^(4/3)*b^(1/3)) + Log[2*2^(1/3) + (a^(1
/3) + b^(1/3)*x)^2/(a + b*x^3)^(2/3) + (2^(2/3)*(a^(1/3) + b^(1/3)*x))/(a + b*x^3)^(1/3)]/(12*2^(2/3)*a^(4/3)*
b^(1/3))

Rule 31

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

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 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 298

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 420

Int[((a_) + (b_.)*(x_)^3)^(1/3)/((c_) + (d_.)*(x_)^3), x_Symbol] :> With[{q = Rt[b/a, 3]}, Dist[9*(a/(c*q)), S
ubst[Int[x/((4 - a*x^3)*(1 + 2*a*x^3)), x], x, (1 + q*x)/(a + b*x^3)^(1/3)], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[b*c - a*d, 0] && EqQ[b*c + a*d, 0]

Rule 421

Int[1/(((a_) + (b_.)*(x_)^3)^(2/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x^3)^
(2/3), x], x] - Dist[d/(b*c - a*d), Int[(a + b*x^3)^(1/3)/(c + d*x^3), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ
[b*c - a*d, 0] && EqQ[b*c + a*d, 0]

Rule 493

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[b/(b*c - a*d), I
nt[(e*x)^m/(a + b*x^n), x], x] - Dist[d/(b*c - a*d), Int[(e*x)^m/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 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 642

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 648

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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a-b x^3\right ) \left (a+b x^3\right )^{2/3}} \, dx &=\frac {\left (1+\frac {b x^3}{a}\right )^{2/3} \int \frac {1}{\left (a-b x^3\right ) \left (1+\frac {b x^3}{a}\right )^{2/3}} \, dx}{\left (a+b x^3\right )^{2/3}}\\ &=\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} F_1\left (\frac {1}{3};1,\frac {2}{3};\frac {4}{3};\frac {b x^3}{a},-\frac {b x^3}{a}\right )}{a \left (a+b x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 10.05, size = 153, normalized size = 0.34 \begin {gather*} \frac {4 a x F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )}{\left (a-b x^3\right ) \left (a+b x^3\right )^{2/3} \left (4 a F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )+b x^3 \left (3 F_1\left (\frac {4}{3};\frac {2}{3},2;\frac {7}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )-2 F_1\left (\frac {4}{3};\frac {5}{3},1;\frac {7}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*a*x*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), (b*x^3)/a])/((a - b*x^3)*(a + b*x^3)^(2/3)*(4*a*AppellF1[1/3,
2/3, 1, 4/3, -((b*x^3)/a), (b*x^3)/a] + b*x^3*(3*AppellF1[4/3, 2/3, 2, 7/3, -((b*x^3)/a), (b*x^3)/a] - 2*Appel
lF1[4/3, 5/3, 1, 7/3, -((b*x^3)/a), (b*x^3)/a])))

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (-b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
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(1/(-b*x^3+a)/(b*x^3+a)^(2/3),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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