3.9.47 \(\int \frac {1}{x^2 (a+b x^2)^{5/4}} \, dx\) [847]

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

[Out]

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

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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 = {292, 203, 202} \begin {gather*} -\frac {3 \sqrt {b} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{a^{3/2} \sqrt [4]{a+b x^2}}-\frac {1}{a x \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^2)^(5/4)),x]

[Out]

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

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Dist[(1 + b*(x^2/a))^(1/4)/(a*(a + b*x^2)^(1/4)), Int[1/(1 + b*
(x^2/a))^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a] && PosQ[b/a]

Rule 292

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[(c*x)^(m + 1)/(a*c*(m + 1)*(a + b*x^2)^(1
/4)), x] - Dist[b*((2*m + 1)/(2*a*c^2*(m + 1))), Int[(c*x)^(m + 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c
}, x] && PosQ[b/a] && IntegerQ[2*m] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (a+b x^2\right )^{5/4}} \, dx &=-\frac {1}{a x \sqrt [4]{a+b x^2}}-\frac {(3 b) \int \frac {1}{\left (a+b x^2\right )^{5/4}} \, dx}{2 a}\\ &=-\frac {1}{a x \sqrt [4]{a+b x^2}}-\frac {\left (3 b \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{2 a^2 \sqrt [4]{a+b x^2}}\\ &=-\frac {1}{a x \sqrt [4]{a+b x^2}}-\frac {3 \sqrt {b} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{a^{3/2} \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 7.41, size = 52, normalized size = 0.68 \begin {gather*} -\frac {\sqrt [4]{1+\frac {b x^2}{a}} \, _2F_1\left (-\frac {1}{2},\frac {5}{4};\frac {1}{2};-\frac {b x^2}{a}\right )}{a x \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x^2)^(5/4)),x]

[Out]

-(((1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[-1/2, 5/4, 1/2, -((b*x^2)/a)])/(a*x*(a + b*x^2)^(1/4)))

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{2} \left (b \,x^{2}+a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b*x^2+a)^(5/4),x)

[Out]

int(1/x^2/(b*x^2+a)^(5/4),x)

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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^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(5/4)*x^2), x)

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Fricas [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^2+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/4)/(b^2*x^6 + 2*a*b*x^4 + a^2*x^2), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.57, size = 27, normalized size = 0.36 \begin {gather*} - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {5}{4}} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b*x**2+a)**(5/4),x)

[Out]

-hyper((-1/2, 5/4), (1/2,), b*x**2*exp_polar(I*pi)/a)/(a**(5/4)*x)

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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^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(5/4)*x^2), x)

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Mupad [B]
time = 5.16, size = 40, normalized size = 0.53 \begin {gather*} -\frac {2\,{\left (\frac {a}{b\,x^2}+1\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{4},\frac {7}{4};\ \frac {11}{4};\ -\frac {a}{b\,x^2}\right )}{7\,x\,{\left (b\,x^2+a\right )}^{5/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x^2)^(5/4)),x)

[Out]

-(2*(a/(b*x^2) + 1)^(5/4)*hypergeom([5/4, 7/4], 11/4, -a/(b*x^2)))/(7*x*(a + b*x^2)^(5/4))

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