3.2.71 \(\int \frac {e+f x}{x \sqrt {-1-x^3}} \, dx\) [171]

Optimal. Leaf size=131 \[ \frac {2}{3} e \tan ^{-1}\left (\sqrt {-1-x^3}\right )+\frac {2 \sqrt {2-\sqrt {3}} f (1+x) \sqrt {\frac {1-x+x^2}{\left (1-\sqrt {3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac {1+\sqrt {3}+x}{1-\sqrt {3}+x}\right )|-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1+x}{\left (1-\sqrt {3}+x\right )^2}} \sqrt {-1-x^3}} \]

[Out]

2/3*e*arctan((-x^3-1)^(1/2))+2/3*f*(1+x)*EllipticF((1+x+3^(1/2))/(1+x-3^(1/2)),2*I-I*3^(1/2))*(1/2*6^(1/2)-1/2
*2^(1/2))*((x^2-x+1)/(1+x-3^(1/2))^2)^(1/2)*3^(3/4)/(-x^3-1)^(1/2)/((-1-x)/(1+x-3^(1/2))^2)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1846, 272, 65, 210, 12, 225} \begin {gather*} \frac {2 \sqrt {2-\sqrt {3}} f (x+1) \sqrt {\frac {x^2-x+1}{\left (x-\sqrt {3}+1\right )^2}} F\left (\text {ArcSin}\left (\frac {x+\sqrt {3}+1}{x-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {x+1}{\left (x-\sqrt {3}+1\right )^2}} \sqrt {-x^3-1}}+\frac {2}{3} e \text {ArcTan}\left (\sqrt {-x^3-1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e + f*x)/(x*Sqrt[-1 - x^3]),x]

[Out]

(2*e*ArcTan[Sqrt[-1 - x^3]])/3 + (2*Sqrt[2 - Sqrt[3]]*f*(1 + x)*Sqrt[(1 - x + x^2)/(1 - Sqrt[3] + x)^2]*Ellipt
icF[ArcSin[(1 + Sqrt[3] + x)/(1 - Sqrt[3] + x)], -7 + 4*Sqrt[3]])/(3^(1/4)*Sqrt[-((1 + x)/(1 - Sqrt[3] + x)^2)
]*Sqrt[-1 - x^3])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt
[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sq
rt[(-s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3])*s + r*x)/((1 - Sqrt[3])*s + r
*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 272

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

Rule 1846

Int[(Pq_)/((x_)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[Coeff[Pq, x, 0], Int[1/(x*Sqrt[a + b*x^n]), x
], x] + Int[ExpandToSum[(Pq - Coeff[Pq, x, 0])/x, x]/Sqrt[a + b*x^n], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] &
& IGtQ[n, 0] && NeQ[Coeff[Pq, x, 0], 0]

Rubi steps

\begin {align*} \int \frac {e+f x}{x \sqrt {-1-x^3}} \, dx &=e \int \frac {1}{x \sqrt {-1-x^3}} \, dx+\int \frac {f}{\sqrt {-1-x^3}} \, dx\\ &=\frac {1}{3} e \text {Subst}\left (\int \frac {1}{\sqrt {-1-x} x} \, dx,x,x^3\right )+f \int \frac {1}{\sqrt {-1-x^3}} \, dx\\ &=\frac {2 \sqrt {2-\sqrt {3}} f (1+x) \sqrt {\frac {1-x+x^2}{\left (1-\sqrt {3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac {1+\sqrt {3}+x}{1-\sqrt {3}+x}\right )|-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1+x}{\left (1-\sqrt {3}+x\right )^2}} \sqrt {-1-x^3}}-\frac {1}{3} (2 e) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {-1-x^3}\right )\\ &=\frac {2}{3} e \tan ^{-1}\left (\sqrt {-1-x^3}\right )+\frac {2 \sqrt {2-\sqrt {3}} f (1+x) \sqrt {\frac {1-x+x^2}{\left (1-\sqrt {3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac {1+\sqrt {3}+x}{1-\sqrt {3}+x}\right )|-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1+x}{\left (1-\sqrt {3}+x\right )^2}} \sqrt {-1-x^3}}\\ \end {align*}

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Mathematica [A]
time = 10.18, size = 138, normalized size = 1.05 \begin {gather*} \frac {2}{3} e \tan ^{-1}\left (\sqrt {-1-x^3}\right )-\frac {2 f \left (\sqrt [3]{-1}-x\right ) \sqrt {\frac {1+x}{1+\sqrt [3]{-1}}} \sqrt {-\frac {(-1)^{2/3} \left ((-1)^{2/3}+x\right )}{1+\sqrt [3]{-1}}} F\left (\sin ^{-1}\left (\sqrt {\frac {1+(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right )|\sqrt [3]{-1}\right )}{\sqrt {\frac {1+(-1)^{2/3} x}{1+\sqrt [3]{-1}}} \sqrt {-1-x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)/(x*Sqrt[-1 - x^3]),x]

[Out]

(2*e*ArcTan[Sqrt[-1 - x^3]])/3 - (2*f*((-1)^(1/3) - x)*Sqrt[(1 + x)/(1 + (-1)^(1/3))]*Sqrt[-(((-1)^(2/3)*((-1)
^(2/3) + x))/(1 + (-1)^(1/3)))]*EllipticF[ArcSin[Sqrt[(1 + (-1)^(2/3)*x)/(1 + (-1)^(1/3))]], (-1)^(1/3)])/(Sqr
t[(1 + (-1)^(2/3)*x)/(1 + (-1)^(1/3))]*Sqrt[-1 - x^3])

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Maple [A]
time = 0.24, size = 122, normalized size = 0.93

method	result	size
meijerg	\(-i f x \hypergeom \left (\left [\frac {1}{3}, \frac {1}{2}\right ], \left [\frac {4}{3}\right ], -x^{3}\right )-\frac {i e \left (-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{3}+1}}{2}\right )+\left (-2 \ln \left (2\right )+3 \ln \left (x \right )\right ) \sqrt {\pi }\right )}{3 \sqrt {\pi }}\)	\(56\)
default	\(-\frac {2 i f \sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {1+x}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x -\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \EllipticF \left (\frac {\sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}-1}}+\frac {2 e \arctan \left (\sqrt {-x^{3}-1}\right )}{3}\)	\(122\)
elliptic	\(-\frac {2 i f \sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {1+x}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x -\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \EllipticF \left (\frac {\sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}-1}}+\frac {2 e \arctan \left (\sqrt {-x^{3}-1}\right )}{3}\)	\(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*I*f*3^(1/2)*(I*(x-1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*((1+x)/(3/2+1/2*I*3^(1/2)))^(1/2)*(-I*(x-1/2+1/2*I*3^
(1/2))*3^(1/2))^(1/2)/(-x^3-1)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x-1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2),(I*3^(1/2)/
(3/2+1/2*I*3^(1/2)))^(1/2))+2/3*e*arctan((-x^3-1)^(1/2))

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

[Out]

integrate((f*x + e)/(sqrt(-x^3 - 1)*x), x)

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Fricas [A]
time = 0.10, size = 28, normalized size = 0.21 \begin {gather*} \frac {1}{3} \, \arctan \left (\frac {{\left (x^{3} + 2\right )} \sqrt {-x^{3} - 1}}{2 \, {\left (x^{3} + 1\right )}}\right ) e \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*arctan(1/2*(x^3 + 2)*sqrt(-x^3 - 1)/(x^3 + 1))*e

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Sympy [A]
time = 1.61, size = 46, normalized size = 0.35 \begin {gather*} \frac {2 i e \operatorname {asinh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} - \frac {i f x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)/x/(-x**3-1)**(1/2),x)

[Out]

2*I*e*asinh(x**(-3/2))/3 - I*f*x*gamma(1/3)*hyper((1/3, 1/2), (4/3,), x**3*exp_polar(I*pi))/(3*gamma(4/3))

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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((f*x+e)/x/(-x^3-1)^(1/2),x, algorithm="giac")

[Out]

integrate((f*x + e)/(sqrt(-x^3 - 1)*x), x)

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Mupad [B]
time = 2.66, size = 223, normalized size = 1.70 \begin {gather*} \frac {\left (3+\sqrt {3}\,1{}\mathrm {i}\right )\,\sqrt {x^3+1}\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\left (f\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )-e\,\Pi \left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}}{\sqrt {-x^3-1}\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)/(x*(- x^3 - 1)^(1/2)),x)

[Out]

((3^(1/2)*1i + 3)*(x^3 + 1)^(1/2)*((x + (3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*(f*ellipticF(asin(
((x + 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)) - e*ellipticPi((3^(1/
2)*1i)/2 + 3/2, asin(((x + 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))
*((x + 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*(((3^(1/2)*1i)/2 - x + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/2))/((- x^3 - 1
)^(1/2)*(x^3 - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) - ((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2
+ 1/2))^(1/2))

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