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

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

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 64, normalized size of antiderivative = 1.00, 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 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]

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]]

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*}

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Mathematica [A]  time = 0.24, 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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fricas [B]  time = 0.80, size = 232, normalized size = 3.62 \[ \left [\frac {3 \, {\left (a^{2} x^{8} + 2 \, a b x^{4} + b^{2}\right )} \sqrt {a} \log \left (-2 \, a x^{4} - 2 \, \sqrt {a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}} - b\right ) - 2 \, {\left (4 \, a^{2} x^{8} + 3 \, a b x^{4}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{12 \, {\left (a^{5} x^{8} + 2 \, a^{4} b x^{4} + a^{3} b^{2}\right )}}, -\frac {3 \, {\left (a^{2} x^{8} + 2 \, a b x^{4} + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{a x^{4} + b}\right ) + {\left (4 \, a^{2} x^{8} + 3 \, a b x^{4}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{6 \, {\left (a^{5} x^{8} + 2 \, a^{4} b x^{4} + a^{3} b^{2}\right )}}\right ] \]

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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giac [A]  time = 0.26, size = 63, normalized size = 0.98 \[ -\frac {{\left (\frac {4 \, x^{4}}{a} + \frac {3 \, b}{a^{2}}\right )} x^{2}}{6 \, {\left (a x^{4} + b\right )}^{\frac {3}{2}}} - \frac {\log \left ({\left | -\sqrt {a} x^{2} + \sqrt {a x^{4} + b} \right |}\right )}{2 \, a^{\frac {5}{2}}} + \frac {\log \left ({\left | b \right |}\right )}{4 \, a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.03, size = 221, normalized size = 3.45 \[ -\frac {\left (a \,x^{4}+b \right )^{\frac {5}{2}} \left (-3 a^{5} x^{8} \ln \left (\sqrt {a}\, x^{2}+\sqrt {a \,x^{4}+b}\right )+4 \sqrt {-\frac {\left (-a \,x^{2}+\sqrt {-a b}\right ) \left (a \,x^{2}+\sqrt {-a b}\right )}{a}}\, a^{\frac {9}{2}} x^{6}-6 a^{4} b \,x^{4} \ln \left (\sqrt {a}\, x^{2}+\sqrt {a \,x^{4}+b}\right )+3 \sqrt {-\frac {\left (-a \,x^{2}+\sqrt {-a b}\right ) \left (a \,x^{2}+\sqrt {-a b}\right )}{a}}\, a^{\frac {7}{2}} b \,x^{2}-3 a^{3} b^{2} \ln \left (\sqrt {a}\, x^{2}+\sqrt {a \,x^{4}+b}\right )\right )}{6 \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} \left (-a \,x^{2}+\sqrt {-a b}\right )^{2} \left (a \,x^{2}+\sqrt {-a b}\right )^{2} a^{\frac {7}{2}} x^{10}} \]

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)*(-(-a*x^2+(-a*b)^(1/2))*(a*x^2+(-a*b)^(1/2))/a)^(1/2)*x^6-3*ln(a^(1/2)*x^2+(a*
x^4+b)^(1/2))*x^8*a^5+3*a^(7/2)*b*(-(-a*x^2+(-a*b)^(1/2))*(a*x^2+(-a*b)^(1/2))/a)^(1/2)*x^2-6*a^4*b*ln(a^(1/2)
*x^2+(a*x^4+b)^(1/2))*x^4-3*a^3*b^2*ln(a^(1/2)*x^2+(a*x^4+b)^(1/2)))/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 [A]  time = 1.92, size = 62, normalized size = 0.97 \[ -\frac {\log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{4}}} + \sqrt {a}}\right )}{4 \, a^{\frac {5}{2}}} - \frac {4 \, a + \frac {3 \, b}{x^{4}}}{6 \, {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.38, size = 48, normalized size = 0.75 \[ \frac {\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right )}{2\,a^{5/2}}-\frac {\frac {a+\frac {b}{x^4}}{a^2}+\frac {1}{3\,a}}{2\,{\left (a+\frac {b}{x^4}\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b/x^4)^(5/2)),x)

[Out]

atanh((a + b/x^4)^(1/2)/a^(1/2))/(2*a^(5/2)) - ((a + b/x^4)/a^2 + 1/(3*a))/(2*(a + b/x^4)^(3/2))

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sympy [B]  time = 3.69, size = 743, normalized size = 11.61 \[ - \frac {8 a^{7} x^{12} \sqrt {1 + \frac {b}{a x^{4}}}}{12 a^{\frac {19}{2}} x^{12} + 36 a^{\frac {17}{2}} b x^{8} + 36 a^{\frac {15}{2}} b^{2} x^{4} + 12 a^{\frac {13}{2}} b^{3}} - \frac {3 a^{7} x^{12} \log {\left (\frac {b}{a x^{4}} \right )}}{12 a^{\frac {19}{2}} x^{12} + 36 a^{\frac {17}{2}} b x^{8} + 36 a^{\frac {15}{2}} b^{2} x^{4} + 12 a^{\frac {13}{2}} b^{3}} + \frac {6 a^{7} x^{12} \log {\left (\sqrt {1 + \frac {b}{a x^{4}}} + 1 \right )}}{12 a^{\frac {19}{2}} x^{12} + 36 a^{\frac {17}{2}} b x^{8} + 36 a^{\frac {15}{2}} b^{2} x^{4} + 12 a^{\frac {13}{2}} b^{3}} - \frac {14 a^{6} b x^{8} \sqrt {1 + \frac {b}{a x^{4}}}}{12 a^{\frac {19}{2}} x^{12} + 36 a^{\frac {17}{2}} b x^{8} + 36 a^{\frac {15}{2}} b^{2} x^{4} + 12 a^{\frac {13}{2}} b^{3}} - \frac {9 a^{6} b x^{8} \log {\left (\frac {b}{a x^{4}} \right )}}{12 a^{\frac {19}{2}} x^{12} + 36 a^{\frac {17}{2}} b x^{8} + 36 a^{\frac {15}{2}} b^{2} x^{4} + 12 a^{\frac {13}{2}} b^{3}} + \frac {18 a^{6} b x^{8} \log {\left (\sqrt {1 + \frac {b}{a x^{4}}} + 1 \right )}}{12 a^{\frac {19}{2}} x^{12} + 36 a^{\frac {17}{2}} b x^{8} + 36 a^{\frac {15}{2}} b^{2} x^{4} + 12 a^{\frac {13}{2}} b^{3}} - \frac {6 a^{5} b^{2} x^{4} \sqrt {1 + \frac {b}{a x^{4}}}}{12 a^{\frac {19}{2}} x^{12} + 36 a^{\frac {17}{2}} b x^{8} + 36 a^{\frac {15}{2}} b^{2} x^{4} + 12 a^{\frac {13}{2}} b^{3}} - \frac {9 a^{5} b^{2} x^{4} \log {\left (\frac {b}{a x^{4}} \right )}}{12 a^{\frac {19}{2}} x^{12} + 36 a^{\frac {17}{2}} b x^{8} + 36 a^{\frac {15}{2}} b^{2} x^{4} + 12 a^{\frac {13}{2}} b^{3}} + \frac {18 a^{5} b^{2} x^{4} \log {\left (\sqrt {1 + \frac {b}{a x^{4}}} + 1 \right )}}{12 a^{\frac {19}{2}} x^{12} + 36 a^{\frac {17}{2}} b x^{8} + 36 a^{\frac {15}{2}} b^{2} x^{4} + 12 a^{\frac {13}{2}} b^{3}} - \frac {3 a^{4} b^{3} \log {\left (\frac {b}{a x^{4}} \right )}}{12 a^{\frac {19}{2}} x^{12} + 36 a^{\frac {17}{2}} b x^{8} + 36 a^{\frac {15}{2}} b^{2} x^{4} + 12 a^{\frac {13}{2}} b^{3}} + \frac {6 a^{4} b^{3} \log {\left (\sqrt {1 + \frac {b}{a x^{4}}} + 1 \right )}}{12 a^{\frac {19}{2}} x^{12} + 36 a^{\frac {17}{2}} b x^{8} + 36 a^{\frac {15}{2}} b^{2} x^{4} + 12 a^{\frac {13}{2}} b^{3}} \]

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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