3.2059 \(\int \sqrt {a+\frac {b}{x^4}} x \, dx\)

Optimal. Leaf size=49 \[ \frac {1}{2} x^2 \sqrt {a+\frac {b}{x^4}}-\frac {1}{2} \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b}}{x^2 \sqrt {a+\frac {b}{x^4}}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {335, 275, 277, 217, 206} \[ \frac {1}{2} x^2 \sqrt {a+\frac {b}{x^4}}-\frac {1}{2} \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b}}{x^2 \sqrt {a+\frac {b}{x^4}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x^4]*x,x]

[Out]

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

Rule 206

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 66, normalized size = 1.35 \[ \frac {x^2 \sqrt {a+\frac {b}{x^4}} \left (\sqrt {a x^4+b}-\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {a x^4+b}}{\sqrt {b}}\right )\right )}{2 \sqrt {a x^4+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x^4]*x,x]

[Out]

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

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fricas [A]  time = 1.19, size = 112, normalized size = 2.29 \[ \left [\frac {1}{2} \, x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}} + \frac {1}{4} \, \sqrt {b} \log \left (\frac {a x^{4} - 2 \, \sqrt {b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}} + 2 \, b}{x^{4}}\right ), \frac {1}{2} \, x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}} + \frac {1}{2} \, \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{b}\right )\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*x^2*sqrt((a*x^4 + b)/x^4) + 1/4*sqrt(b)*log((a*x^4 - 2*sqrt(b)*x^2*sqrt((a*x^4 + b)/x^4) + 2*b)/x^4), 1/2
*x^2*sqrt((a*x^4 + b)/x^4) + 1/2*sqrt(-b)*arctan(sqrt(-b)*x^2*sqrt((a*x^4 + b)/x^4)/b)]

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giac [A]  time = 0.15, size = 36, normalized size = 0.73 \[ \frac {b \arctan \left (\frac {\sqrt {a x^{4} + b}}{\sqrt {-b}}\right )}{2 \, \sqrt {-b}} + \frac {1}{2} \, \sqrt {a x^{4} + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*arctan(sqrt(a*x^4 + b)/sqrt(-b))/sqrt(-b) + 1/2*sqrt(a*x^4 + b)

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maple [A]  time = 0.01, size = 64, normalized size = 1.31 \[ \frac {\sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, \left (-\sqrt {b}\, \ln \left (\frac {2 b +2 \sqrt {a \,x^{4}+b}\, \sqrt {b}}{x^{2}}\right )+\sqrt {a \,x^{4}+b}\right ) x^{2}}{2 \sqrt {a \,x^{4}+b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.04, size = 60, normalized size = 1.22 \[ \frac {1}{2} \, \sqrt {a + \frac {b}{x^{4}}} x^{2} + \frac {1}{4} \, \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} x^{2} - \sqrt {b}}{\sqrt {a + \frac {b}{x^{4}}} x^{2} + \sqrt {b}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,\sqrt {a+\frac {b}{x^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b/x^4)^(1/2),x)

[Out]

int(x*(a + b/x^4)^(1/2), x)

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sympy [A]  time = 1.79, size = 66, normalized size = 1.35 \[ \frac {\sqrt {a} x^{2}}{2 \sqrt {1 + \frac {b}{a x^{4}}}} - \frac {\sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{2}} \right )}}{2} + \frac {b}{2 \sqrt {a} x^{2} \sqrt {1 + \frac {b}{a x^{4}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*x**2/(2*sqrt(1 + b/(a*x**4))) - sqrt(b)*asinh(sqrt(b)/(sqrt(a)*x**2))/2 + b/(2*sqrt(a)*x**2*sqrt(1 + b
/(a*x**4)))

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