3.2100 \(\int \frac{1}{(a+\frac{b}{x^4})^{5/2} x} \, dx\)

Optimal. Leaf size=64 \[ -\frac{1}{2 a^2 \sqrt{a+\frac{b}{x^4}}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^4}}}{\sqrt{a}}\right )}{2 a^{5/2}}-\frac{1}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0332934, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ -\frac{1}{2 a^2 \sqrt{a+\frac{b}{x^4}}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^4}}}{\sqrt{a}}\right )}{2 a^{5/2}}-\frac{1}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^4)^(5/2)*x),x]

[Out]

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

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 51

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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{1}{\left (a+\frac{b}{x^4}\right )^{5/2} x} \, dx &=-\left (\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{5/2}} \, dx,x,\frac{1}{x^4}\right )\right )\\ &=-\frac{1}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{3/2}} \, dx,x,\frac{1}{x^4}\right )}{4 a}\\ &=-\frac{1}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{1}{2 a^2 \sqrt{a+\frac{b}{x^4}}}-\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x^4}\right )}{4 a^2}\\ &=-\frac{1}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{1}{2 a^2 \sqrt{a+\frac{b}{x^4}}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x^4}}\right )}{2 a^2 b}\\ &=-\frac{1}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{1}{2 a^2 \sqrt{a+\frac{b}{x^4}}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^4}}}{\sqrt{a}}\right )}{2 a^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.161347, size = 93, normalized size = 1.45 \[ \frac{\frac{3 \sqrt{b} \left (a x^4+b\right ) \sqrt{\frac{a x^4}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{a} x^2}{\sqrt{b}}\right )}{x^2}-\sqrt{a} \left (4 a x^4+3 b\right )}{6 a^{5/2} \sqrt{a+\frac{b}{x^4}} \left (a x^4+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^4)^(5/2)*x),x]

[Out]

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

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Maple [B]  time = 0.02, size = 221, normalized size = 3.5 \begin{align*} -{\frac{1}{6\,{x}^{10}} \left ( a{x}^{4}+b \right ) ^{{\frac{5}{2}}} \left ( 4\,{a}^{9/2}\sqrt{-{\frac{ \left ( -a{x}^{2}+\sqrt{-ab} \right ) \left ( a{x}^{2}+\sqrt{-ab} \right ) }{a}}}{x}^{6}-3\,\ln \left ({x}^{2}\sqrt{a}+\sqrt{a{x}^{4}+b} \right ){x}^{8}{a}^{5}+3\,{a}^{7/2}\sqrt{-{\frac{ \left ( -a{x}^{2}+\sqrt{-ab} \right ) \left ( a{x}^{2}+\sqrt{-ab} \right ) }{a}}}b{x}^{2}-6\,\ln \left ({x}^{2}\sqrt{a}+\sqrt{a{x}^{4}+b} \right ){a}^{4}b{x}^{4}-3\,\ln \left ({x}^{2}\sqrt{a}+\sqrt{a{x}^{4}+b} \right ){a}^{3}{b}^{2} \right ){a}^{-{\frac{7}{2}}} \left ({\frac{a{x}^{4}+b}{{x}^{4}}} \right ) ^{-{\frac{5}{2}}} \left ( -a{x}^{2}+\sqrt{-ab} \right ) ^{-2} \left ( a{x}^{2}+\sqrt{-ab} \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x^4)^(5/2)/x,x)

[Out]

-1/6*(a*x^4+b)^(5/2)*(4*a^(9/2)*(-1/a*(-a*x^2+(-a*b)^(1/2))*(a*x^2+(-a*b)^(1/2)))^(1/2)*x^6-3*ln(x^2*a^(1/2)+(
a*x^4+b)^(1/2))*x^8*a^5+3*a^(7/2)*(-1/a*(-a*x^2+(-a*b)^(1/2))*(a*x^2+(-a*b)^(1/2)))^(1/2)*b*x^2-6*ln(x^2*a^(1/
2)+(a*x^4+b)^(1/2))*a^4*b*x^4-3*ln(x^2*a^(1/2)+(a*x^4+b)^(1/2))*a^3*b^2)/a^(7/2)/((a*x^4+b)/x^4)^(5/2)/x^10/(-
a*x^2+(-a*b)^(1/2))^2/(a*x^2+(-a*b)^(1/2))^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(1/(a+b/x^4)^(5/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.63123, size = 505, normalized size = 7.89 \begin{align*} \left [\frac{3 \,{\left (a^{2} x^{8} + 2 \, a b x^{4} + b^{2}\right )} \sqrt{a} \log \left (-2 \, a x^{4} - 2 \, \sqrt{a} x^{4} \sqrt{\frac{a x^{4} + b}{x^{4}}} - b\right ) - 2 \,{\left (4 \, a^{2} x^{8} + 3 \, a b x^{4}\right )} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{12 \,{\left (a^{5} x^{8} + 2 \, a^{4} b x^{4} + a^{3} b^{2}\right )}}, -\frac{3 \,{\left (a^{2} x^{8} + 2 \, a b x^{4} + b^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} x^{4} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{a x^{4} + b}\right ) +{\left (4 \, a^{2} x^{8} + 3 \, a b x^{4}\right )} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{6 \,{\left (a^{5} x^{8} + 2 \, a^{4} b x^{4} + a^{3} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*(a^2*x^8 + 2*a*b*x^4 + b^2)*sqrt(a)*log(-2*a*x^4 - 2*sqrt(a)*x^4*sqrt((a*x^4 + b)/x^4) - b) - 2*(4*a^
2*x^8 + 3*a*b*x^4)*sqrt((a*x^4 + b)/x^4))/(a^5*x^8 + 2*a^4*b*x^4 + a^3*b^2), -1/6*(3*(a^2*x^8 + 2*a*b*x^4 + b^
2)*sqrt(-a)*arctan(sqrt(-a)*x^4*sqrt((a*x^4 + b)/x^4)/(a*x^4 + b)) + (4*a^2*x^8 + 3*a*b*x^4)*sqrt((a*x^4 + b)/
x^4))/(a^5*x^8 + 2*a^4*b*x^4 + a^3*b^2)]

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Sympy [B]  time = 4.74391, size = 743, normalized size = 11.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*a**7*x**12*sqrt(1 + b/(a*x**4))/(12*a**(19/2)*x**12 + 36*a**(17/2)*b*x**8 + 36*a**(15/2)*b**2*x**4 + 12*a**
(13/2)*b**3) - 3*a**7*x**12*log(b/(a*x**4))/(12*a**(19/2)*x**12 + 36*a**(17/2)*b*x**8 + 36*a**(15/2)*b**2*x**4
+ 12*a**(13/2)*b**3) + 6*a**7*x**12*log(sqrt(1 + b/(a*x**4)) + 1)/(12*a**(19/2)*x**12 + 36*a**(17/2)*b*x**8 +
36*a**(15/2)*b**2*x**4 + 12*a**(13/2)*b**3) - 14*a**6*b*x**8*sqrt(1 + b/(a*x**4))/(12*a**(19/2)*x**12 + 36*a*
*(17/2)*b*x**8 + 36*a**(15/2)*b**2*x**4 + 12*a**(13/2)*b**3) - 9*a**6*b*x**8*log(b/(a*x**4))/(12*a**(19/2)*x**
12 + 36*a**(17/2)*b*x**8 + 36*a**(15/2)*b**2*x**4 + 12*a**(13/2)*b**3) + 18*a**6*b*x**8*log(sqrt(1 + b/(a*x**4
)) + 1)/(12*a**(19/2)*x**12 + 36*a**(17/2)*b*x**8 + 36*a**(15/2)*b**2*x**4 + 12*a**(13/2)*b**3) - 6*a**5*b**2*
x**4*sqrt(1 + b/(a*x**4))/(12*a**(19/2)*x**12 + 36*a**(17/2)*b*x**8 + 36*a**(15/2)*b**2*x**4 + 12*a**(13/2)*b*
*3) - 9*a**5*b**2*x**4*log(b/(a*x**4))/(12*a**(19/2)*x**12 + 36*a**(17/2)*b*x**8 + 36*a**(15/2)*b**2*x**4 + 12
*a**(13/2)*b**3) + 18*a**5*b**2*x**4*log(sqrt(1 + b/(a*x**4)) + 1)/(12*a**(19/2)*x**12 + 36*a**(17/2)*b*x**8 +
36*a**(15/2)*b**2*x**4 + 12*a**(13/2)*b**3) - 3*a**4*b**3*log(b/(a*x**4))/(12*a**(19/2)*x**12 + 36*a**(17/2)*
b*x**8 + 36*a**(15/2)*b**2*x**4 + 12*a**(13/2)*b**3) + 6*a**4*b**3*log(sqrt(1 + b/(a*x**4)) + 1)/(12*a**(19/2)
*x**12 + 36*a**(17/2)*b*x**8 + 36*a**(15/2)*b**2*x**4 + 12*a**(13/2)*b**3)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a + \frac{b}{x^{4}}\right )}^{\frac{5}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((a + b/x^4)^(5/2)*x), x)