3.9.29 \(\int \frac {1}{x^2 \sqrt {a+b x^4}} \, dx\) [829]

Optimal. Leaf size=232 \[ -\frac {\sqrt {a+b x^4}}{a x}+\frac {\sqrt {b} x \sqrt {a+b x^4}}{a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} \sqrt {a+b x^4}} \]

[Out]

-(b*x^4+a)^(1/2)/a/x+x*b^(1/2)*(b*x^4+a)^(1/2)/a/(a^(1/2)+x^2*b^(1/2))-b^(1/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)
))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+
x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/a^(3/4)/(b*x^4+a)^(1/2)+1/2*b^(1/4)*(cos(2*arctan(b^(1/
4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2
))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/a^(3/4)/(b*x^4+a)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {331, 311, 226, 1210} \begin {gather*} \frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} \sqrt {a+b x^4}}-\frac {\sqrt {a+b x^4}}{a x}+\frac {\sqrt {b} x \sqrt {a+b x^4}}{a \left (\sqrt {a}+\sqrt {b} x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[a + b*x^4]),x]

[Out]

-(Sqrt[a + b*x^4]/(a*x)) + (Sqrt[b]*x*Sqrt[a + b*x^4])/(a*(Sqrt[a] + Sqrt[b]*x^2)) - (b^(1/4)*(Sqrt[a] + Sqrt[
b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(a^(3/4)*Sq
rt[a + b*x^4]) + (b^(1/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcT
an[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(3/4)*Sqrt[a + b*x^4])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 331

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

Rule 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \sqrt {a+b x^4}} \, dx &=-\frac {\sqrt {a+b x^4}}{a x}+\frac {b \int \frac {x^2}{\sqrt {a+b x^4}} \, dx}{a}\\ &=-\frac {\sqrt {a+b x^4}}{a x}+\frac {\sqrt {b} \int \frac {1}{\sqrt {a+b x^4}} \, dx}{\sqrt {a}}-\frac {\sqrt {b} \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{\sqrt {a}}\\ &=-\frac {\sqrt {a+b x^4}}{a x}+\frac {\sqrt {b} x \sqrt {a+b x^4}}{a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 49, normalized size = 0.21 \begin {gather*} -\frac {\sqrt {1+\frac {b x^4}{a}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-\frac {b x^4}{a}\right )}{x \sqrt {a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*Sqrt[a + b*x^4]),x]

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.15, size = 115, normalized size = 0.50

method	result	size
default	\(-\frac {\sqrt {b \,x^{4}+a}}{a x}+\frac {i \sqrt {b}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(115\)
risch	\(-\frac {\sqrt {b \,x^{4}+a}}{a x}+\frac {i \sqrt {b}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(115\)
elliptic	\(-\frac {\sqrt {b \,x^{4}+a}}{a x}+\frac {i \sqrt {b}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x^4+a)^(1/2)/a/x+I*b^(1/2)/a^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*
b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*(EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*b^(1/2))^(
1/2),I))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(b*x^4 + a)*x^2), x)

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Fricas [A]
time = 0.09, size = 71, normalized size = 0.31 \begin {gather*} -\frac {\sqrt {a} x \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - \sqrt {a} x \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + \sqrt {b x^{4} + a}}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(a)*x*(-b/a)^(3/4)*elliptic_e(arcsin(x*(-b/a)^(1/4)), -1) - sqrt(a)*x*(-b/a)^(3/4)*elliptic_f(arcsin(x*(
-b/a)^(1/4)), -1) + sqrt(b*x^4 + a))/(a*x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.40, size = 39, normalized size = 0.17 \begin {gather*} \frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x \Gamma \left (\frac {3}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*x*gamma(3/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(b*x^4 + a)*x^2), x)

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Mupad [B]
time = 1.26, size = 40, normalized size = 0.17 \begin {gather*} -\frac {\sqrt {\frac {a}{b\,x^4}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{4};\ \frac {7}{4};\ -\frac {a}{b\,x^4}\right )}{3\,x\,\sqrt {b\,x^4+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x^4)^(1/2)),x)

[Out]

-((a/(b*x^4) + 1)^(1/2)*hypergeom([1/2, 3/4], 7/4, -a/(b*x^4)))/(3*x*(a + b*x^4)^(1/2))

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