3.170 \(\int \frac {e+f x}{x \sqrt {-1+x^3}} \, dx\)

Optimal. Leaf size=137 \[ \frac {2}{3} e \tan ^{-1}\left (\sqrt {x^3-1}\right )-\frac {2 \sqrt {2-\sqrt {3}} f (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}} \]

[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 = 0.05, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1832, 266, 63, 203, 12, 219} \[ \frac {2}{3} e \tan ^{-1}\left (\sqrt {x^3-1}\right )-\frac {2 \sqrt {2-\sqrt {3}} f (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}} \]

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 63

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 203

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

Rule 219

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

Rule 266

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 1832

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 \operatorname {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) \operatorname {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 [C]  time = 0.03, size = 52, normalized size = 0.38 \[ \frac {2}{3} e \tan ^{-1}\left (\sqrt {x^3-1}\right )+\frac {f x \sqrt {1-x^3} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};x^3\right )}{\sqrt {x^3-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*ArcTan[Sqrt[-1 + x^3]])/3 + (f*x*Sqrt[1 - x^3]*Hypergeometric2F1[1/3, 1/2, 4/3, x^3])/Sqrt[-1 + x^3]

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fricas [F]  time = 1.17, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{3} - 1} {\left (f x + e\right )}}{x^{4} - x}, x\right ) \]

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]

integral(sqrt(x^3 - 1)*(f*x + e)/(x^4 - x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f x + e}{\sqrt {x^{3} - 1} x}\,{d x} \]

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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maple [A]  time = 0.01, size = 129, normalized size = 0.94 \[ \frac {2 e \arctan \left (\sqrt {x^{3}-1}\right )}{3}+\frac {2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, f \EllipticF \left (\sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*f*(-3/2-1/2*I*3^(1/2))*((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2)*((x+1/2-1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2)*
((x+1/2+1/2*I*3^(1/2))/(3/2+1/2*I*3^(1/2)))^(1/2)/(x^3-1)^(1/2)*EllipticF(((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2),(
(3/2+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 \[ \int \frac {f x + e}{\sqrt {x^{3} - 1} x}\,{d x} \]

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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mupad [B]  time = 2.62, size = 207, normalized size = 1.51 \[ -\frac {\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\left (f\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )+e\,\Pi \left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\left (\sqrt {3}-3{}\mathrm {i}\right )\,1{}\mathrm {i}}{\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)/(x*(x^3 - 1)^(1/2)),x)

[Out]

-((-(x - (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 - 3/2))^(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) - 3i)*1i)/(((3^(1/2)*1i)/2 - 1/
2)*((3^(1/2)*1i)/2 + 1/2) - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) + x^3)^(1/2)

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sympy [A]  time = 3.09, size = 60, normalized size = 0.44 \[ e \left (\begin {cases} \frac {2 i \operatorname {acosh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\- \frac {2 \operatorname {asin}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} & \text {otherwise} \end {cases}\right ) - \frac {i f x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*Piecewise((2*I*acosh(x**(-3/2))/3, 1/Abs(x**3) > 1), (-2*asin(x**(-3/2))/3, True)) - I*f*x*gamma(1/3)*hyper(
(1/3, 1/2), (4/3,), x**3)/(3*gamma(4/3))

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