3.9.35 \(\int \frac {1}{x^6 (a+b x^2)^{3/4}} \, dx\) [835]

Optimal. Leaf size=126 \[ -\frac {\sqrt [4]{a+b x^2}}{5 a x^5}+\frac {3 b \sqrt [4]{a+b x^2}}{10 a^2 x^3}-\frac {3 b^2 \sqrt [4]{a+b x^2}}{4 a^3 x}-\frac {3 b^{5/2} \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{4 a^{5/2} \left (a+b x^2\right )^{3/4}} \]

[Out]

-1/5*(b*x^2+a)^(1/4)/a/x^5+3/10*b*(b*x^2+a)^(1/4)/a^2/x^3-3/4*b^2*(b*x^2+a)^(1/4)/a^3/x-3/4*b^(5/2)*(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*arct
an(x*b^(1/2)/a^(1/2))),2^(1/2))/a^(5/2)/(b*x^2+a)^(3/4)

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Rubi [A]
time = 0.03, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {331, 239, 237} \begin {gather*} -\frac {3 b^{5/2} \left (\frac {b x^2}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{4 a^{5/2} \left (a+b x^2\right )^{3/4}}-\frac {3 b^2 \sqrt [4]{a+b x^2}}{4 a^3 x}+\frac {3 b \sqrt [4]{a+b x^2}}{10 a^2 x^3}-\frac {\sqrt [4]{a+b x^2}}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(a + b*x^2)^(3/4)),x]

[Out]

-1/5*(a + b*x^2)^(1/4)/(a*x^5) + (3*b*(a + b*x^2)^(1/4))/(10*a^2*x^3) - (3*b^2*(a + b*x^2)^(1/4))/(4*a^3*x) -
(3*b^(5/2)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(4*a^(5/2)*(a + b*x^2)^(3/4))

Rule 237

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

Rule 239

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 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]

Rubi steps

\begin {align*} \int \frac {1}{x^6 \left (a+b x^2\right )^{3/4}} \, dx &=-\frac {\sqrt [4]{a+b x^2}}{5 a x^5}-\frac {(9 b) \int \frac {1}{x^4 \left (a+b x^2\right )^{3/4}} \, dx}{10 a}\\ &=-\frac {\sqrt [4]{a+b x^2}}{5 a x^5}+\frac {3 b \sqrt [4]{a+b x^2}}{10 a^2 x^3}+\frac {\left (3 b^2\right ) \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4}} \, dx}{4 a^2}\\ &=-\frac {\sqrt [4]{a+b x^2}}{5 a x^5}+\frac {3 b \sqrt [4]{a+b x^2}}{10 a^2 x^3}-\frac {3 b^2 \sqrt [4]{a+b x^2}}{4 a^3 x}-\frac {\left (3 b^3\right ) \int \frac {1}{\left (a+b x^2\right )^{3/4}} \, dx}{8 a^3}\\ &=-\frac {\sqrt [4]{a+b x^2}}{5 a x^5}+\frac {3 b \sqrt [4]{a+b x^2}}{10 a^2 x^3}-\frac {3 b^2 \sqrt [4]{a+b x^2}}{4 a^3 x}-\frac {\left (3 b^3 \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}{8 a^3 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {\sqrt [4]{a+b x^2}}{5 a x^5}+\frac {3 b \sqrt [4]{a+b x^2}}{10 a^2 x^3}-\frac {3 b^2 \sqrt [4]{a+b x^2}}{4 a^3 x}-\frac {3 b^{5/2} \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 )}{4 a^{5/2} \left (a+b x^2\right )^{3/4}}\\ \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 = 51, normalized size = 0.40 \begin {gather*} -\frac {\left (1+\frac {b x^2}{a}\right )^{3/4} \, _2F_1\left (-\frac {5}{2},\frac {3}{4};-\frac {3}{2};-\frac {b x^2}{a}\right )}{5 x^5 \left (a+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(a + b*x^2)^(3/4)),x]

[Out]

-1/5*((1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[-5/2, 3/4, -3/2, -((b*x^2)/a)])/(x^5*(a + b*x^2)^(3/4))

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{6} \left (b \,x^{2}+a \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(b*x^2+a)^(3/4),x)

[Out]

int(1/x^6/(b*x^2+a)^(3/4),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^6/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*x^6), 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^6/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/4)/(b*x^8 + a*x^6), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.70, size = 32, normalized size = 0.25 \begin {gather*} - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {3}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac {3}{4}} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(b*x**2+a)**(3/4),x)

[Out]

-hyper((-5/2, 3/4), (-3/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(3/4)*x**5)

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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^6/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*x^6), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^6\,{\left (b\,x^2+a\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a + b*x^2)^(3/4)),x)

[Out]

int(1/(x^6*(a + b*x^2)^(3/4)), x)

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