\(\int \frac {1}{x^5 \sqrt {a-b x^4}} \, dx\) [835]

Optimal result

Integrand size = 16, antiderivative size = 52 \[ \int \frac {1}{x^5 \sqrt {a-b x^4}} \, dx=-\frac {\sqrt {a-b x^4}}{4 a x^4}-\frac {b \text {arctanh}\left (\frac {\sqrt {a-b x^4}}{\sqrt {a}}\right )}{4 a^{3/2}} \]

[Out]

-1/4*b*arctanh((-b*x^4+a)^(1/2)/a^(1/2))/a^(3/2)-1/4*(-b*x^4+a)^(1/2)/a/x^4

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {272, 44, 65, 214} \[ \int \frac {1}{x^5 \sqrt {a-b x^4}} \, dx=-\frac {b \text {arctanh}\left (\frac {\sqrt {a-b x^4}}{\sqrt {a}}\right )}{4 a^{3/2}}-\frac {\sqrt {a-b x^4}}{4 a x^4} \]

[In]

Int[1/(x^5*Sqrt[a - b*x^4]),x]

[Out]

-1/4*Sqrt[a - b*x^4]/(a*x^4) - (b*ArcTanh[Sqrt[a - b*x^4]/Sqrt[a]])/(4*a^(3/2))

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 214

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

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}{4} \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a-b x}} \, dx,x,x^4\right ) \\ & = -\frac {\sqrt {a-b x^4}}{4 a x^4}+\frac {b \text {Subst}\left (\int \frac {1}{x \sqrt {a-b x}} \, dx,x,x^4\right )}{8 a} \\ & = -\frac {\sqrt {a-b x^4}}{4 a x^4}-\frac {\text {Subst}\left (\int \frac {1}{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a-b x^4}\right )}{4 a} \\ & = -\frac {\sqrt {a-b x^4}}{4 a x^4}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {a-b x^4}}{\sqrt {a}}\right )}{4 a^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \sqrt {a-b x^4}} \, dx=-\frac {\sqrt {a-b x^4}}{4 a x^4}-\frac {b \text {arctanh}\left (\frac {\sqrt {a-b x^4}}{\sqrt {a}}\right )}{4 a^{3/2}} \]

[In]

Integrate[1/(x^5*Sqrt[a - b*x^4]),x]

[Out]

-1/4*Sqrt[a - b*x^4]/(a*x^4) - (b*ArcTanh[Sqrt[a - b*x^4]/Sqrt[a]])/(4*a^(3/2))

Maple [A] (verified)

Time = 4.32 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85

method	result	size
pseudoelliptic	\(-\frac {\operatorname {arctanh}\left (\frac {\sqrt {-b \,x^{4}+a}}{\sqrt {a}}\right ) b \,x^{4}+\sqrt {a}\, \sqrt {-b \,x^{4}+a}}{4 a^{\frac {3}{2}} x^{4}}\)	\(44\)
default	\(-\frac {\sqrt {-b \,x^{4}+a}}{4 a \,x^{4}}-\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {-b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}\)	\(50\)
risch	\(-\frac {\sqrt {-b \,x^{4}+a}}{4 a \,x^{4}}-\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {-b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}\)	\(50\)
elliptic	\(-\frac {\sqrt {-b \,x^{4}+a}}{4 a \,x^{4}}-\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {-b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}\)	\(50\)

[In]

int(1/x^5/(-b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/a^(3/2)*(arctanh((-b*x^4+a)^(1/2)/a^(1/2))*b*x^4+a^(1/2)*(-b*x^4+a)^(1/2))/x^4

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 2.15 \[ \int \frac {1}{x^5 \sqrt {a-b x^4}} \, dx=\left [\frac {\sqrt {a} b x^{4} \log \left (\frac {b x^{4} + 2 \, \sqrt {-b x^{4} + a} \sqrt {a} - 2 \, a}{x^{4}}\right ) - 2 \, \sqrt {-b x^{4} + a} a}{8 \, a^{2} x^{4}}, \frac {\sqrt {-a} b x^{4} \arctan \left (\frac {\sqrt {-b x^{4} + a} \sqrt {-a}}{a}\right ) - \sqrt {-b x^{4} + a} a}{4 \, a^{2} x^{4}}\right ] \]

[In]

integrate(1/x^5/(-b*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(sqrt(a)*b*x^4*log((b*x^4 + 2*sqrt(-b*x^4 + a)*sqrt(a) - 2*a)/x^4) - 2*sqrt(-b*x^4 + a)*a)/(a^2*x^4), 1/4
*(sqrt(-a)*b*x^4*arctan(sqrt(-b*x^4 + a)*sqrt(-a)/a) - sqrt(-b*x^4 + a)*a)/(a^2*x^4)]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.48 \[ \int \frac {1}{x^5 \sqrt {a-b x^4}} \, dx=\begin {cases} - \frac {\sqrt {b} \sqrt {\frac {a}{b x^{4}} - 1}}{4 a x^{2}} - \frac {b \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4 a^{\frac {3}{2}}} & \text {for}\: \left |{\frac {a}{b x^{4}}}\right | > 1 \\\frac {i}{4 \sqrt {b} x^{6} \sqrt {- \frac {a}{b x^{4}} + 1}} - \frac {i \sqrt {b}}{4 a x^{2} \sqrt {- \frac {a}{b x^{4}} + 1}} + \frac {i b \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/x**5/(-b*x**4+a)**(1/2),x)

[Out]

Piecewise((-sqrt(b)*sqrt(a/(b*x**4) - 1)/(4*a*x**2) - b*acosh(sqrt(a)/(sqrt(b)*x**2))/(4*a**(3/2)), Abs(a/(b*x
**4)) > 1), (I/(4*sqrt(b)*x**6*sqrt(-a/(b*x**4) + 1)) - I*sqrt(b)/(4*a*x**2*sqrt(-a/(b*x**4) + 1)) + I*b*asin(
sqrt(a)/(sqrt(b)*x**2))/(4*a**(3/2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.37 \[ \int \frac {1}{x^5 \sqrt {a-b x^4}} \, dx=-\frac {\sqrt {-b x^{4} + a} b}{4 \, {\left ({\left (b x^{4} - a\right )} a + a^{2}\right )}} + \frac {b \log \left (\frac {\sqrt {-b x^{4} + a} - \sqrt {a}}{\sqrt {-b x^{4} + a} + \sqrt {a}}\right )}{8 \, a^{\frac {3}{2}}} \]

[In]

integrate(1/x^5/(-b*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

-1/4*sqrt(-b*x^4 + a)*b/((b*x^4 - a)*a + a^2) + 1/8*b*log((sqrt(-b*x^4 + a) - sqrt(a))/(sqrt(-b*x^4 + a) + sqr
t(a)))/a^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^5 \sqrt {a-b x^4}} \, dx=\frac {\frac {b^{2} \arctan \left (\frac {\sqrt {-b x^{4} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {\sqrt {-b x^{4} + a} b}{a x^{4}}}{4 \, b} \]

[In]

integrate(1/x^5/(-b*x^4+a)^(1/2),x, algorithm="giac")

[Out]

1/4*(b^2*arctan(sqrt(-b*x^4 + a)/sqrt(-a))/(sqrt(-a)*a) - sqrt(-b*x^4 + a)*b/(a*x^4))/b

Mupad [B] (verification not implemented)

Time = 5.96 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^5 \sqrt {a-b x^4}} \, dx=-\frac {\sqrt {a-b\,x^4}}{4\,a\,x^4}-\frac {b\,\mathrm {atanh}\left (\frac {\sqrt {a-b\,x^4}}{\sqrt {a}}\right )}{4\,a^{3/2}} \]

[In]

int(1/(x^5*(a - b*x^4)^(1/2)),x)

[Out]

- (a - b*x^4)^(1/2)/(4*a*x^4) - (b*atanh((a - b*x^4)^(1/2)/a^(1/2)))/(4*a^(3/2))