\(\int \frac {(a+\frac {b}{x^4})^{5/2}}{x} \, dx\) [2076]

Optimal result

Integrand size = 15, antiderivative size = 77 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x} \, dx=-\frac {1}{2} a^2 \sqrt {a+\frac {b}{x^4}}-\frac {1}{6} a \left (a+\frac {b}{x^4}\right )^{3/2}-\frac {1}{10} \left (a+\frac {b}{x^4}\right )^{5/2}+\frac {1}{2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right ) \]

[Out]

-1/6*a*(a+b/x^4)^(3/2)-1/10*(a+b/x^4)^(5/2)+1/2*a^(5/2)*arctanh((a+b/x^4)^(1/2)/a^(1/2))-1/2*a^2*(a+b/x^4)^(1/
2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 52, 65, 214} \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x} \, dx=\frac {1}{2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right )-\frac {1}{2} a^2 \sqrt {a+\frac {b}{x^4}}-\frac {1}{6} a \left (a+\frac {b}{x^4}\right )^{3/2}-\frac {1}{10} \left (a+\frac {b}{x^4}\right )^{5/2} \]

[In]

Int[(a + b/x^4)^(5/2)/x,x]

[Out]

-1/2*(a^2*Sqrt[a + b/x^4]) - (a*(a + b/x^4)^(3/2))/6 - (a + b/x^4)^(5/2)/10 + (a^(5/2)*ArcTanh[Sqrt[a + b/x^4]
/Sqrt[a]])/2

Rule 52

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

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}& = -\left (\frac {1}{4} \text {Subst}\left (\int \frac {(a+b x)^{5/2}}{x} \, dx,x,\frac {1}{x^4}\right )\right ) \\ & = -\frac {1}{10} \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {1}{4} a \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,\frac {1}{x^4}\right ) \\ & = -\frac {1}{6} a \left (a+\frac {b}{x^4}\right )^{3/2}-\frac {1}{10} \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {1}{4} a^2 \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\frac {1}{x^4}\right ) \\ & = -\frac {1}{2} a^2 \sqrt {a+\frac {b}{x^4}}-\frac {1}{6} a \left (a+\frac {b}{x^4}\right )^{3/2}-\frac {1}{10} \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {1}{4} a^3 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x^4}\right ) \\ & = -\frac {1}{2} a^2 \sqrt {a+\frac {b}{x^4}}-\frac {1}{6} a \left (a+\frac {b}{x^4}\right )^{3/2}-\frac {1}{10} \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {a^3 \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x^4}}\right )}{2 b} \\ & = -\frac {1}{2} a^2 \sqrt {a+\frac {b}{x^4}}-\frac {1}{6} a \left (a+\frac {b}{x^4}\right )^{3/2}-\frac {1}{10} \left (a+\frac {b}{x^4}\right )^{5/2}+\frac {1}{2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x} \, dx=\frac {\sqrt {a+\frac {b}{x^4}} \left (-3 b^2-11 a b x^4-23 a^2 x^8+\frac {15 a^{5/2} x^{10} \log \left (\sqrt {a} x^2+\sqrt {b+a x^4}\right )}{\sqrt {b+a x^4}}\right )}{30 x^8} \]

[In]

Integrate[(a + b/x^4)^(5/2)/x,x]

[Out]

(Sqrt[a + b/x^4]*(-3*b^2 - 11*a*b*x^4 - 23*a^2*x^8 + (15*a^(5/2)*x^10*Log[Sqrt[a]*x^2 + Sqrt[b + a*x^4]])/Sqrt
[b + a*x^4]))/(30*x^8)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.16

method	result	size
risch	\(-\frac {\left (23 a^{2} x^{8}+11 a b \,x^{4}+3 b^{2}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{30 x^{8}}+\frac {a^{\frac {5}{2}} \ln \left (x^{2} \sqrt {a}+\sqrt {a \,x^{4}+b}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{2 \sqrt {a \,x^{4}+b}}\)	\(89\)
default	\(\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} \left (15 a^{\frac {5}{2}} \ln \left (x^{2} \sqrt {a}+\sqrt {a \,x^{4}+b}\right ) x^{10}-23 a^{2} x^{8} \sqrt {a \,x^{4}+b}-11 a b \sqrt {a \,x^{4}+b}\, x^{4}-3 b^{2} \sqrt {a \,x^{4}+b}\right )}{30 \left (a \,x^{4}+b \right )^{\frac {5}{2}}}\)	\(99\)

[In]

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

[Out]

-1/30*(23*a^2*x^8+11*a*b*x^4+3*b^2)/x^8*((a*x^4+b)/x^4)^(1/2)+1/2*a^(5/2)*ln(x^2*a^(1/2)+(a*x^4+b)^(1/2))*((a*
x^4+b)/x^4)^(1/2)*x^2/(a*x^4+b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.19 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x} \, dx=\left [\frac {15 \, a^{\frac {5}{2}} x^{8} \log \left (-2 \, a x^{4} - 2 \, \sqrt {a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}} - b\right ) - 2 \, {\left (23 \, a^{2} x^{8} + 11 \, a b x^{4} + 3 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{60 \, x^{8}}, -\frac {15 \, \sqrt {-a} a^{2} x^{8} \arctan \left (\frac {\sqrt {-a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{a x^{4} + b}\right ) + {\left (23 \, a^{2} x^{8} + 11 \, a b x^{4} + 3 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{30 \, x^{8}}\right ] \]

[In]

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

[Out]

[1/60*(15*a^(5/2)*x^8*log(-2*a*x^4 - 2*sqrt(a)*x^4*sqrt((a*x^4 + b)/x^4) - b) - 2*(23*a^2*x^8 + 11*a*b*x^4 + 3
*b^2)*sqrt((a*x^4 + b)/x^4))/x^8, -1/30*(15*sqrt(-a)*a^2*x^8*arctan(sqrt(-a)*x^4*sqrt((a*x^4 + b)/x^4)/(a*x^4
+ b)) + (23*a^2*x^8 + 11*a*b*x^4 + 3*b^2)*sqrt((a*x^4 + b)/x^4))/x^8]

Sympy [A] (verification not implemented)

Time = 2.46 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x} \, dx=- \frac {23 a^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x^{4}}}}{30} - \frac {a^{\frac {5}{2}} \log {\left (\frac {b}{a x^{4}} \right )}}{4} + \frac {a^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {b}{a x^{4}}} + 1 \right )}}{2} - \frac {11 a^{\frac {3}{2}} b \sqrt {1 + \frac {b}{a x^{4}}}}{30 x^{4}} - \frac {\sqrt {a} b^{2} \sqrt {1 + \frac {b}{a x^{4}}}}{10 x^{8}} \]

[In]

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

[Out]

-23*a**(5/2)*sqrt(1 + b/(a*x**4))/30 - a**(5/2)*log(b/(a*x**4))/4 + a**(5/2)*log(sqrt(1 + b/(a*x**4)) + 1)/2 -
11*a**(3/2)*b*sqrt(1 + b/(a*x**4))/(30*x**4) - sqrt(a)*b**2*sqrt(1 + b/(a*x**4))/(10*x**8)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x} \, dx=-\frac {1}{4} \, a^{\frac {5}{2}} \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{4}}} + \sqrt {a}}\right ) - \frac {1}{10} \, {\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}} - \frac {1}{6} \, {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} a - \frac {1}{2} \, \sqrt {a + \frac {b}{x^{4}}} a^{2} \]

[In]

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

[Out]

-1/4*a^(5/2)*log((sqrt(a + b/x^4) - sqrt(a))/(sqrt(a + b/x^4) + sqrt(a))) - 1/10*(a + b/x^4)^(5/2) - 1/6*(a +
b/x^4)^(3/2)*a - 1/2*sqrt(a + b/x^4)*a^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (57) = 114\).

Time = 0.31 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.34 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x} \, dx=-\frac {1}{4} \, a^{\frac {5}{2}} \log \left ({\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{2}\right ) + \frac {45 \, {\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{8} a^{\frac {5}{2}} b - 90 \, {\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{6} a^{\frac {5}{2}} b^{2} + 140 \, {\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{4} a^{\frac {5}{2}} b^{3} - 70 \, {\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{2} a^{\frac {5}{2}} b^{4} + 23 \, a^{\frac {5}{2}} b^{5}}{15 \, {\left ({\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{2} - b\right )}^{5}} \]

[In]

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

[Out]

-1/4*a^(5/2)*log((sqrt(a)*x^2 - sqrt(a*x^4 + b))^2) + 1/15*(45*(sqrt(a)*x^2 - sqrt(a*x^4 + b))^8*a^(5/2)*b - 9
0*(sqrt(a)*x^2 - sqrt(a*x^4 + b))^6*a^(5/2)*b^2 + 140*(sqrt(a)*x^2 - sqrt(a*x^4 + b))^4*a^(5/2)*b^3 - 70*(sqrt
(a)*x^2 - sqrt(a*x^4 + b))^2*a^(5/2)*b^4 + 23*a^(5/2)*b^5)/((sqrt(a)*x^2 - sqrt(a*x^4 + b))^2 - b)^5

Mupad [B] (verification not implemented)

Time = 6.64 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x} \, dx=-\frac {a\,{\left (a+\frac {b}{x^4}\right )}^{3/2}}{6}-\frac {{\left (a+\frac {b}{x^4}\right )}^{5/2}}{10}-\frac {a^2\,\sqrt {a+\frac {b}{x^4}}}{2}-\frac {a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x^4}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{2} \]

[In]

int((a + b/x^4)^(5/2)/x,x)

[Out]

- (a^(5/2)*atan(((a + b/x^4)^(1/2)*1i)/a^(1/2))*1i)/2 - (a*(a + b/x^4)^(3/2))/6 - (a + b/x^4)^(5/2)/10 - (a^2*
(a + b/x^4)^(1/2))/2