3.4.75 \(\int \frac {\sqrt {-1+x^3}}{x^4} \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{3} \tan ^{-1}\left (\sqrt {x^3-1}\right )-\frac {\sqrt {x^3-1}}{3 x^3} \]

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {266, 47, 63, 203} \begin {gather*} \frac {1}{3} \tan ^{-1}\left (\sqrt {x^3-1}\right )-\frac {\sqrt {x^3-1}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + x^3]/x^4,x]

[Out]

-1/3*Sqrt[-1 + x^3]/x^3 + ArcTan[Sqrt[-1 + x^3]]/3

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, 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 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]]

Rubi steps

\begin {align*} \int \frac {\sqrt {-1+x^3}}{x^4} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x}}{x^2} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {-1+x^3}}{3 x^3}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} x} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {-1+x^3}}{3 x^3}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {-1+x^3}\right )\\ &=-\frac {\sqrt {-1+x^3}}{3 x^3}+\frac {1}{3} \tan ^{-1}\left (\sqrt {-1+x^3}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 48, normalized size = 1.55 \begin {gather*} -\frac {x^3+\sqrt {1-x^3} x^3 \tanh ^{-1}\left (\sqrt {1-x^3}\right )-1}{3 x^3 \sqrt {x^3-1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 + x^3]/x^4,x]

[Out]

-1/3*(-1 + x^3 + x^3*Sqrt[1 - x^3]*ArcTanh[Sqrt[1 - x^3]])/(x^3*Sqrt[-1 + x^3])

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IntegrateAlgebraic [A]  time = 0.02, size = 31, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {-1+x^3}}{3 x^3}+\frac {1}{3} \tan ^{-1}\left (\sqrt {-1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[-1 + x^3]/x^4,x]

[Out]

-1/3*Sqrt[-1 + x^3]/x^3 + ArcTan[Sqrt[-1 + x^3]]/3

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fricas [A]  time = 0.47, size = 27, normalized size = 0.87 \begin {gather*} \frac {x^{3} \arctan \left (\sqrt {x^{3} - 1}\right ) - \sqrt {x^{3} - 1}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)^(1/2)/x^4,x, algorithm="fricas")

[Out]

1/3*(x^3*arctan(sqrt(x^3 - 1)) - sqrt(x^3 - 1))/x^3

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giac [A]  time = 0.23, size = 23, normalized size = 0.74 \begin {gather*} -\frac {\sqrt {x^{3} - 1}}{3 \, x^{3}} + \frac {1}{3} \, \arctan \left (\sqrt {x^{3} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)^(1/2)/x^4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.33, size = 24, normalized size = 0.77

method	result	size
default	\(-\frac {\sqrt {x^{3}-1}}{3 x^{3}}+\frac {\arctan \left (\sqrt {x^{3}-1}\right )}{3}\)	\(24\)
risch	\(-\frac {\sqrt {x^{3}-1}}{3 x^{3}}+\frac {\arctan \left (\sqrt {x^{3}-1}\right )}{3}\)	\(24\)
elliptic	\(-\frac {\sqrt {x^{3}-1}}{3 x^{3}}+\frac {\arctan \left (\sqrt {x^{3}-1}\right )}{3}\)	\(24\)
trager	\(-\frac {\sqrt {x^{3}-1}}{3 x^{3}}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \sqrt {x^{3}-1}+2 \RootOf \left (\textit {\_Z}^{2}+1\right )}{x^{3}}\right )}{6}\)	\(57\)
meijerg	\(\frac {\sqrt {\mathrm {signum}\left (x^{3}-1\right )}\, \left (-\frac {2 \sqrt {\pi }}{x^{3}}-\left (-2 \ln \relax (2)-1+3 \ln \relax (x )+i \pi \right ) \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (-4 x^{3}+8\right )}{4 x^{3}}-\frac {2 \sqrt {\pi }\, \sqrt {-x^{3}+1}}{x^{3}}+2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{3}+1}}{2}\right )\right )}{6 \sqrt {-\mathrm {signum}\left (x^{3}-1\right )}\, \sqrt {\pi }}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3-1)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.73, size = 23, normalized size = 0.74 \begin {gather*} -\frac {\sqrt {x^{3} - 1}}{3 \, x^{3}} + \frac {1}{3} \, \arctan \left (\sqrt {x^{3} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)^(1/2)/x^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.17, size = 177, normalized size = 5.71 \begin {gather*} -\frac {\sqrt {x^3-1}}{3\,x^3}-\frac {\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\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}}}\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\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 )}{\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((x^3 - 1)^(1/2)/x^4,x)

[Out]

- (x^3 - 1)^(1/2)/(3*x^3) - (((3^(1/2)*1i)/2 + 3/2)*(-(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)*(-(x - 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*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)))/(((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 = 1.29, size = 82, normalized size = 2.65 \begin {gather*} \begin {cases} \frac {i \operatorname {acosh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} + \frac {i}{3 x^{\frac {3}{2}} \sqrt {-1 + \frac {1}{x^{3}}}} - \frac {i}{3 x^{\frac {9}{2}} \sqrt {-1 + \frac {1}{x^{3}}}} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\- \frac {\operatorname {asin}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} - \frac {\sqrt {1 - \frac {1}{x^{3}}}}{3 x^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3-1)**(1/2)/x**4,x)

[Out]

Piecewise((I*acosh(x**(-3/2))/3 + I/(3*x**(3/2)*sqrt(-1 + x**(-3))) - I/(3*x**(9/2)*sqrt(-1 + x**(-3))), 1/Abs
(x**3) > 1), (-asin(x**(-3/2))/3 - sqrt(1 - 1/x**3)/(3*x**(3/2)), True))

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