3.1009 \(\int \frac {1}{x^2 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx\)

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

[Out]

-(c*x^4+a)^(1/2)/a/x+x*c^(1/2)*(c*x^4+a)^(1/2)/a/(a^(1/2)+x^2*c^(1/2))-c^(1/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)
))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+
x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(3/4)/(c*x^4+a)^(1/2)+1/2*c^(1/4)*(cos(2*arctan(c^(1/
4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2
))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(3/4)/(c*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 = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4, 325, 305, 220, 1196} \[ \frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} \sqrt {a+c x^4}}-\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} \sqrt {a+c x^4}}+\frac {\sqrt {c} x \sqrt {a+c x^4}}{a \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt {a+c x^4}}{a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4

Int[(u_.)*((a_.) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(a + c*x^(2*n))^p, x] /; Fre
eQ[{a, b, c, n, p}, x] && EqQ[j, 2*n] && EqQ[b, 0]

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 305

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 325

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 1196

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

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Mathematica [C]  time = 0.01, size = 49, normalized size = 0.21 \[ -\frac {\sqrt {\frac {c x^4}{a}+1} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-\frac {c x^4}{a}\right )}{x \sqrt {a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + a}}{c x^{6} + a x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^4 + a)/(c*x^6 + a*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{4} + a} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(c*x^4 + a)*x^2), x)

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maple [C]  time = 0.01, size = 115, normalized size = 0.50 \[ \frac {i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right ) \sqrt {c}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {a}}-\frac {\sqrt {c \,x^{4}+a}}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{4} + a} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(c*x^4 + a)*x^2), x)

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mupad [B]  time = 4.56, size = 40, normalized size = 0.17 \[ -\frac {\sqrt {\frac {a}{c\,x^4}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{4};\ \frac {7}{4};\ -\frac {a}{c\,x^4}\right )}{3\,x\,\sqrt {c\,x^4+a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + c*x^4)^(1/2)),x)

[Out]

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

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sympy [C]  time = 1.11, size = 39, normalized size = 0.17 \[ \frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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