3.833 \(\int \frac {1}{x^2 (a+b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=76 \[ -\frac {\sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\sqrt {a} \left (a+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a+b x^2}}{a x} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {325, 233, 231} \[ -\frac {\sqrt [4]{a+b x^2}}{a x}-\frac {\sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^2)^(3/4)),x]

[Out]

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

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + (b*x^2
)/a)^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[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]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 49, normalized size = 0.64 \[ -\frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {1}{2};-\frac {b x^2}{a}\right )}{x \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{b x^{4} + a x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*x^2), x)

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maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b*x^2+a)^(3/4),x)

[Out]

int(1/x^2/(b*x^2+a)^(3/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*x^2), x)

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mupad [B]  time = 5.07, size = 40, normalized size = 0.53 \[ -\frac {2\,{\left (\frac {a}{b\,x^2}+1\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{4},\frac {5}{4};\ \frac {9}{4};\ -\frac {a}{b\,x^2}\right )}{5\,x\,{\left (b\,x^2+a\right )}^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x^2)^(3/4)),x)

[Out]

-(2*(a/(b*x^2) + 1)^(3/4)*hypergeom([3/4, 5/4], 9/4, -a/(b*x^2)))/(5*x*(a + b*x^2)^(3/4))

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sympy [C]  time = 0.96, size = 27, normalized size = 0.36 \[ - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {3}{4}} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b*x**2+a)**(3/4),x)

[Out]

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

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