\(\int \frac {1}{x^4 \sqrt {1-x^3}} \, dx\) [461]

Optimal result

Integrand size = 15, antiderivative size = 35 \[ \int \frac {1}{x^4 \sqrt {1-x^3}} \, dx=-\frac {\sqrt {1-x^3}}{3 x^3}-\frac {1}{3} \text {arctanh}\left (\sqrt {1-x^3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 44, 65, 212} \[ \int \frac {1}{x^4 \sqrt {1-x^3}} \, dx=-\frac {1}{3} \text {arctanh}\left (\sqrt {1-x^3}\right )-\frac {\sqrt {1-x^3}}{3 x^3} \]

[In]

Int[1/(x^4*Sqrt[1 - x^3]),x]

[Out]

-1/3*Sqrt[1 - x^3]/x^3 - ArcTanh[Sqrt[1 - x^3]]/3

Rule 44

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

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 212

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

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]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x^2} \, dx,x,x^3\right ) \\ & = -\frac {\sqrt {1-x^3}}{3 x^3}+\frac {1}{6} \text {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} \text {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} \tanh ^{-1}\left (\sqrt {1-x^3}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \sqrt {1-x^3}} \, dx=-\frac {\sqrt {1-x^3}}{3 x^3}-\frac {1}{3} \text {arctanh}\left (\sqrt {1-x^3}\right ) \]

[In]

Integrate[1/(x^4*Sqrt[1 - x^3]),x]

[Out]

-1/3*Sqrt[1 - x^3]/x^3 - ArcTanh[Sqrt[1 - x^3]]/3

Maple [A] (verified)

Time = 4.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80

method	result	size
default	\(-\frac {\operatorname {arctanh}\left (\sqrt {-x^{3}+1}\right )}{3}-\frac {\sqrt {-x^{3}+1}}{3 x^{3}}\)	\(28\)
elliptic	\(-\frac {\operatorname {arctanh}\left (\sqrt {-x^{3}+1}\right )}{3}-\frac {\sqrt {-x^{3}+1}}{3 x^{3}}\)	\(28\)
risch	\(\frac {x^{3}-1}{3 x^{3} \sqrt {-x^{3}+1}}-\frac {\operatorname {arctanh}\left (\sqrt {-x^{3}+1}\right )}{3}\)	\(33\)
trager	\(-\frac {\sqrt {-x^{3}+1}}{3 x^{3}}-\frac {\ln \left (-\frac {-x^{3}+2 \sqrt {-x^{3}+1}+2}{x^{3}}\right )}{6}\)	\(42\)
pseudoelliptic	\(\frac {\ln \left (-1+\sqrt {-x^{3}+1}\right ) x^{3}-\ln \left (1+\sqrt {-x^{3}+1}\right ) x^{3}-2 \sqrt {-x^{3}+1}}{6 x^{3}}\)	\(51\)
meijerg	\(-\frac {\frac {\sqrt {\pi }}{x^{3}}-\frac {\left (1-2 \ln \left (2\right )+3 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{2}-\frac {\sqrt {\pi }\, \left (-4 x^{3}+8\right )}{8 x^{3}}+\frac {\sqrt {\pi }\, \sqrt {-x^{3}+1}}{x^{3}}+\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{3}+1}}{2}\right )}{3 \sqrt {\pi }}\)	\(82\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.43 \[ \int \frac {1}{x^4 \sqrt {1-x^3}} \, dx=-\frac {x^{3} \log \left (\sqrt {-x^{3} + 1} + 1\right ) - x^{3} \log \left (\sqrt {-x^{3} + 1} - 1\right ) + 2 \, \sqrt {-x^{3} + 1}}{6 \, x^{3}} \]

[In]

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

[Out]

-1/6*(x^3*log(sqrt(-x^3 + 1) + 1) - x^3*log(sqrt(-x^3 + 1) - 1) + 2*sqrt(-x^3 + 1))/x^3

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.06 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.34 \[ \int \frac {1}{x^4 \sqrt {1-x^3}} \, dx=\begin {cases} - \frac {\operatorname {acosh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} + \frac {1}{3 x^{\frac {3}{2}} \sqrt {-1 + \frac {1}{x^{3}}}} - \frac {1}{3 x^{\frac {9}{2}} \sqrt {-1 + \frac {1}{x^{3}}}} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\\frac {i \operatorname {asin}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} - \frac {i \sqrt {1 - \frac {1}{x^{3}}}}{3 x^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^4 \sqrt {1-x^3}} \, dx=-\frac {\sqrt {-x^{3} + 1}}{3 \, x^{3}} - \frac {1}{6} \, \log \left (\sqrt {-x^{3} + 1} + 1\right ) + \frac {1}{6} \, \log \left (\sqrt {-x^{3} + 1} - 1\right ) \]

[In]

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

[Out]

-1/3*sqrt(-x^3 + 1)/x^3 - 1/6*log(sqrt(-x^3 + 1) + 1) + 1/6*log(sqrt(-x^3 + 1) - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {1}{x^4 \sqrt {1-x^3}} \, dx=-\frac {\sqrt {-x^{3} + 1}}{3 \, x^{3}} - \frac {1}{6} \, \log \left (\sqrt {-x^{3} + 1} + 1\right ) + \frac {1}{6} \, \log \left ({\left | \sqrt {-x^{3} + 1} - 1 \right |}\right ) \]

[In]

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

[Out]

-1/3*sqrt(-x^3 + 1)/x^3 - 1/6*log(sqrt(-x^3 + 1) + 1) + 1/6*log(abs(sqrt(-x^3 + 1) - 1))

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 195, normalized size of antiderivative = 5.57 \[ \int \frac {1}{x^4 \sqrt {1-x^3}} \, dx=-\frac {\sqrt {1-x^3}}{3\,x^3}-\frac {\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {x^3-1}\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\Pi \left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{\sqrt {1-x^3}\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

[In]

int(1/(x^4*(1 - x^3)^(1/2)),x)

[Out]

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