3.2076 \(\int \frac{(a+\frac{b}{x^4})^{5/2}}{x} \, dx\)

Optimal. Leaf size=77 \[ -\frac{1}{2} a^2 \sqrt{a+\frac{b}{x^4}}+\frac{1}{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^4}}}{\sqrt{a}}\right )-\frac{1}{6} a \left (a+\frac{b}{x^4}\right )^{3/2}-\frac{1}{10} \left (a+\frac{b}{x^4}\right )^{5/2} \]

[Out]

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

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Rubi [A]  time = 0.0405486, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 50, 63, 208} \[ -\frac{1}{2} a^2 \sqrt{a+\frac{b}{x^4}}+\frac{1}{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^4}}}{\sqrt{a}}\right )-\frac{1}{6} a \left (a+\frac{b}{x^4}\right )^{3/2}-\frac{1}{10} \left (a+\frac{b}{x^4}\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(5/2)/x,x]

[Out]

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

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]]

Rule 50

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 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 208

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

Rubi steps

\begin{align*} \int \frac{\left (a+\frac{b}{x^4}\right )^{5/2}}{x} \, dx &=-\left (\frac{1}{4} \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 [C]  time = 0.0176662, size = 54, normalized size = 0.7 \[ -\frac{b^2 \sqrt{a+\frac{b}{x^4}} \, _2F_1\left (-\frac{5}{2},-\frac{5}{2};-\frac{3}{2};-\frac{a x^4}{b}\right )}{10 x^8 \sqrt{\frac{a x^4}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(5/2)/x,x]

[Out]

-(b^2*Sqrt[a + b/x^4]*Hypergeometric2F1[-5/2, -5/2, -3/2, -((a*x^4)/b)])/(10*x^8*Sqrt[1 + (a*x^4)/b])

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Maple [A]  time = 0.017, size = 99, normalized size = 1.3 \begin{align*}{\frac{1}{30} \left ({\frac{a{x}^{4}+b}{{x}^{4}}} \right ) ^{{\frac{5}{2}}} \left ( 15\,{a}^{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\,ba\sqrt{a{x}^{4}+b}{x}^{4}-3\,{b}^{2}\sqrt{a{x}^{4}+b} \right ) \left ( a{x}^{4}+b \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*((a*x^4+b)/x^4)^(5/2)*(15*a^(5/2)*ln(x^2*a^(1/2)+(a*x^4+b)^(1/2))*x^10-23*a^2*x^8*(a*x^4+b)^(1/2)-11*b*a*
(a*x^4+b)^(1/2)*x^4-3*b^2*(a*x^4+b)^(1/2))/(a*x^4+b)^(5/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59527, size = 396, normalized size = 5.14 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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Sympy [A]  time = 4.40094, size = 107, normalized size = 1.39 \begin{align*} - \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.48088, size = 86, normalized size = 1.12 \begin{align*} -\frac{a^{3} \arctan \left (\frac{\sqrt{a + \frac{b}{x^{4}}}}{\sqrt{-a}}\right )}{2 \, \sqrt{-a}} - \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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