3.821 \(\int \frac {1}{x^6 \sqrt [4]{a+b x^2}} \, dx\)

Optimal. Leaf size=148 \[ -\frac {7 b^{5/2} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 a^{5/2} \sqrt [4]{a+b x^2}}+\frac {7 b^3 x}{20 a^3 \sqrt [4]{a+b x^2}}-\frac {7 b^2 \left (a+b x^2\right )^{3/4}}{20 a^3 x}+\frac {7 b \left (a+b x^2\right )^{3/4}}{30 a^2 x^3}-\frac {\left (a+b x^2\right )^{3/4}}{5 a x^5} \]

[Out]

7/20*b^3*x/a^3/(b*x^2+a)^(1/4)-1/5*(b*x^2+a)^(3/4)/a/x^5+7/30*b*(b*x^2+a)^(3/4)/a^2/x^3-7/20*b^2*(b*x^2+a)^(3/
4)/a^3/x-7/20*b^(5/2)*(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))/a^(5/2)/(b*x^2+a)^(1/4)

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Rubi [A]  time = 0.05, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {325, 229, 227, 196} \[ \frac {7 b^3 x}{20 a^3 \sqrt [4]{a+b x^2}}-\frac {7 b^2 \left (a+b x^2\right )^{3/4}}{20 a^3 x}-\frac {7 b^{5/2} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 a^{5/2} \sqrt [4]{a+b x^2}}+\frac {7 b \left (a+b x^2\right )^{3/4}}{30 a^2 x^3}-\frac {\left (a+b x^2\right )^{3/4}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(a + b*x^2)^(1/4)),x]

[Out]

(7*b^3*x)/(20*a^3*(a + b*x^2)^(1/4)) - (a + b*x^2)^(3/4)/(5*a*x^5) + (7*b*(a + b*x^2)^(3/4))/(30*a^2*x^3) - (7
*b^2*(a + b*x^2)^(3/4))/(20*a^3*x) - (7*b^(5/2)*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2,
2])/(20*a^(5/2)*(a + b*x^2)^(1/4))

Rule 196

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

Rule 227

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

Rule 229

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + (b*x^2
)/a)^(1/4), 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^6 \sqrt [4]{a+b x^2}} \, dx &=-\frac {\left (a+b x^2\right )^{3/4}}{5 a x^5}-\frac {(7 b) \int \frac {1}{x^4 \sqrt [4]{a+b x^2}} \, dx}{10 a}\\ &=-\frac {\left (a+b x^2\right )^{3/4}}{5 a x^5}+\frac {7 b \left (a+b x^2\right )^{3/4}}{30 a^2 x^3}+\frac {\left (7 b^2\right ) \int \frac {1}{x^2 \sqrt [4]{a+b x^2}} \, dx}{20 a^2}\\ &=-\frac {\left (a+b x^2\right )^{3/4}}{5 a x^5}+\frac {7 b \left (a+b x^2\right )^{3/4}}{30 a^2 x^3}-\frac {7 b^2 \left (a+b x^2\right )^{3/4}}{20 a^3 x}+\frac {\left (7 b^3\right ) \int \frac {1}{\sqrt [4]{a+b x^2}} \, dx}{40 a^3}\\ &=-\frac {\left (a+b x^2\right )^{3/4}}{5 a x^5}+\frac {7 b \left (a+b x^2\right )^{3/4}}{30 a^2 x^3}-\frac {7 b^2 \left (a+b x^2\right )^{3/4}}{20 a^3 x}+\frac {\left (7 b^3 \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx}{40 a^3 \sqrt [4]{a+b x^2}}\\ &=\frac {7 b^3 x}{20 a^3 \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{5 a x^5}+\frac {7 b \left (a+b x^2\right )^{3/4}}{30 a^2 x^3}-\frac {7 b^2 \left (a+b x^2\right )^{3/4}}{20 a^3 x}-\frac {\left (7 b^3 \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{40 a^3 \sqrt [4]{a+b x^2}}\\ &=\frac {7 b^3 x}{20 a^3 \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{5 a x^5}+\frac {7 b \left (a+b x^2\right )^{3/4}}{30 a^2 x^3}-\frac {7 b^2 \left (a+b x^2\right )^{3/4}}{20 a^3 x}-\frac {7 b^{5/2} \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 )}{20 a^{5/2} \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 51, normalized size = 0.34 \[ -\frac {\sqrt [4]{\frac {b x^2}{a}+1} \, _2F_1\left (-\frac {5}{2},\frac {1}{4};-\frac {3}{2};-\frac {b x^2}{a}\right )}{5 x^5 \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.12, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{b x^{8} + a x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(3/4)/(b*x^8 + a*x^6), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a + b*x^2)^(1/4)),x)

[Out]

int(1/(x^6*(a + b*x^2)^(1/4)), x)

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sympy [C]  time = 1.11, size = 32, normalized size = 0.22 \[ - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {1}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 \sqrt [4]{a} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-hyper((-5/2, 1/4), (-3/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(1/4)*x**5)

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