\(\int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx\) [994]

Optimal result

Integrand size = 15, antiderivative size = 75 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=-\frac {\sqrt [4]{a+b x^4}}{4 x^4}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}} \]

[Out]

-1/4*(b*x^4+a)^(1/4)/x^4-1/8*b*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(3/4)-1/8*b*arctanh((b*x^4+a)^(1/4)/a^(1/4))/
a^(3/4)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {272, 43, 65, 218, 212, 209} \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{4 x^4} \]

[In]

Int[(a + b*x^4)^(1/4)/x^5,x]

[Out]

-1/4*(a + b*x^4)^(1/4)/x^4 - (b*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(8*a^(3/4)) - (b*ArcTanh[(a + b*x^4)^(1/4)/
a^(1/4)])/(8*a^(3/4))

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[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 209

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

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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/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}{4} \text {Subst}\left (\int \frac {\sqrt [4]{a+b x}}{x^2} \, dx,x,x^4\right ) \\ & = -\frac {\sqrt [4]{a+b x^4}}{4 x^4}+\frac {1}{16} b \text {Subst}\left (\int \frac {1}{x (a+b x)^{3/4}} \, dx,x,x^4\right ) \\ & = -\frac {\sqrt [4]{a+b x^4}}{4 x^4}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^4}{b}} \, dx,x,\sqrt [4]{a+b x^4}\right ) \\ & = -\frac {\sqrt [4]{a+b x^4}}{4 x^4}-\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )}{8 \sqrt {a}}-\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )}{8 \sqrt {a}} \\ & = -\frac {\sqrt [4]{a+b x^4}}{4 x^4}-\frac {b \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=-\frac {\sqrt [4]{a+b x^4}}{4 x^4}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}} \]

[In]

Integrate[(a + b*x^4)^(1/4)/x^5,x]

[Out]

-1/4*(a + b*x^4)^(1/4)/x^4 - (b*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(8*a^(3/4)) - (b*ArcTanh[(a + b*x^4)^(1/4)/
a^(1/4)])/(8*a^(3/4))

Maple [A] (verified)

Time = 4.46 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.13

method	result	size
pseudoelliptic	\(-\frac {\ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right ) b \,x^{4}+2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) b \,x^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{\frac {3}{4}}}{16 x^{4} a^{\frac {3}{4}}}\)	\(85\)

[In]

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

[Out]

-1/16*(ln((-(b*x^4+a)^(1/4)-a^(1/4))/(-(b*x^4+a)^(1/4)+a^(1/4)))*b*x^4+2*arctan((b*x^4+a)^(1/4)/a^(1/4))*b*x^4
+4*(b*x^4+a)^(1/4)*a^(3/4))/x^4/a^(3/4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 171, normalized size of antiderivative = 2.28 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=-\frac {\left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x^{4} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} b + a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) + i \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x^{4} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} b + i \, a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) - i \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x^{4} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} b - i \, a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) - \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x^{4} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} b - a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) + 4 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{16 \, x^{4}} \]

[In]

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

[Out]

-1/16*((b^4/a^3)^(1/4)*x^4*log((b*x^4 + a)^(1/4)*b + a*(b^4/a^3)^(1/4)) + I*(b^4/a^3)^(1/4)*x^4*log((b*x^4 + a
)^(1/4)*b + I*a*(b^4/a^3)^(1/4)) - I*(b^4/a^3)^(1/4)*x^4*log((b*x^4 + a)^(1/4)*b - I*a*(b^4/a^3)^(1/4)) - (b^4
/a^3)^(1/4)*x^4*log((b*x^4 + a)^(1/4)*b - a*(b^4/a^3)^(1/4)) + 4*(b*x^4 + a)^(1/4))/x^4

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.73 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=- \frac {\sqrt [4]{b} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 x^{3} \Gamma \left (\frac {7}{4}\right )} \]

[In]

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

[Out]

-b**(1/4)*gamma(3/4)*hyper((-1/4, 3/4), (7/4,), a*exp_polar(I*pi)/(b*x**4))/(4*x**3*gamma(7/4))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=-\frac {b \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{8 \, a^{\frac {3}{4}}} + \frac {b \log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{16 \, a^{\frac {3}{4}}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{4 \, x^{4}} \]

[In]

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

[Out]

-1/8*b*arctan((b*x^4 + a)^(1/4)/a^(1/4))/a^(3/4) + 1/16*b*log(((b*x^4 + a)^(1/4) - a^(1/4))/((b*x^4 + a)^(1/4)
+ a^(1/4)))/a^(3/4) - 1/4*(b*x^4 + a)^(1/4)/x^4

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (55) = 110\).

Time = 0.29 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.76 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=\frac {\frac {2 \, \sqrt {2} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} b^{2} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{a} - \frac {8 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b}{x^{4}}}{32 \, b} \]

[In]

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

[Out]

1/32*(2*sqrt(2)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 2*s
qrt(2)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + sqrt(2)*b^2
*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/(-a)^(3/4) + sqrt(2)*(-a)^(1/4)*b^2*lo
g(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a - 8*(b*x^4 + a)^(1/4)*b/x^4)/b

Mupad [B] (verification not implemented)

Time = 5.88 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=-\frac {{\left (b\,x^4+a\right )}^{1/4}}{4\,x^4}-\frac {b\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{3/4}}-\frac {b\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{3/4}} \]

[In]

int((a + b*x^4)^(1/4)/x^5,x)

[Out]

- (a + b*x^4)^(1/4)/(4*x^4) - (b*atan((a + b*x^4)^(1/4)/a^(1/4)))/(8*a^(3/4)) - (b*atanh((a + b*x^4)^(1/4)/a^(
1/4)))/(8*a^(3/4))