3.849 \(\int \frac {1}{x^6 (a+b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=126 \[ -\frac {77 b^{5/2} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 a^{7/2} \sqrt [4]{a+b x^2}}-\frac {77 b^2}{60 a^3 x \sqrt [4]{a+b x^2}}+\frac {11 b}{30 a^2 x^3 \sqrt [4]{a+b x^2}}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^2}} \]

[Out]

-1/5/a/x^5/(b*x^2+a)^(1/4)+11/30*b/a^2/x^3/(b*x^2+a)^(1/4)-77/60*b^2/a^3/x/(b*x^2+a)^(1/4)-77/20*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^(7/2)/(b*x^2+a)^(1/4)

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Rubi [A]  time = 0.04, 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 = {286, 197, 196} \[ -\frac {77 b^2}{60 a^3 x \sqrt [4]{a+b x^2}}-\frac {77 b^{5/2} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 a^{7/2} \sqrt [4]{a+b x^2}}+\frac {11 b}{30 a^2 x^3 \sqrt [4]{a+b x^2}}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(a + b*x^2)^(5/4)),x]

[Out]

-1/(5*a*x^5*(a + b*x^2)^(1/4)) + (11*b)/(30*a^2*x^3*(a + b*x^2)^(1/4)) - (77*b^2)/(60*a^3*x*(a + b*x^2)^(1/4))
- (77*b^(5/2)*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(20*a^(7/2)*(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 197

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 286

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^6 \left (a+b x^2\right )^{5/4}} \, dx &=-\frac {1}{5 a x^5 \sqrt [4]{a+b x^2}}-\frac {(11 b) \int \frac {1}{x^4 \left (a+b x^2\right )^{5/4}} \, dx}{10 a}\\ &=-\frac {1}{5 a x^5 \sqrt [4]{a+b x^2}}+\frac {11 b}{30 a^2 x^3 \sqrt [4]{a+b x^2}}+\frac {\left (77 b^2\right ) \int \frac {1}{x^2 \left (a+b x^2\right )^{5/4}} \, dx}{60 a^2}\\ &=-\frac {1}{5 a x^5 \sqrt [4]{a+b x^2}}+\frac {11 b}{30 a^2 x^3 \sqrt [4]{a+b x^2}}-\frac {77 b^2}{60 a^3 x \sqrt [4]{a+b x^2}}-\frac {\left (77 b^3\right ) \int \frac {1}{\left (a+b x^2\right )^{5/4}} \, dx}{40 a^3}\\ &=-\frac {1}{5 a x^5 \sqrt [4]{a+b x^2}}+\frac {11 b}{30 a^2 x^3 \sqrt [4]{a+b x^2}}-\frac {77 b^2}{60 a^3 x \sqrt [4]{a+b x^2}}-\frac {\left (77 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^4 \sqrt [4]{a+b x^2}}\\ &=-\frac {1}{5 a x^5 \sqrt [4]{a+b x^2}}+\frac {11 b}{30 a^2 x^3 \sqrt [4]{a+b x^2}}-\frac {77 b^2}{60 a^3 x \sqrt [4]{a+b x^2}}-\frac {77 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^{7/2} \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 54, normalized size = 0.43 \[ -\frac {\sqrt [4]{\frac {b x^2}{a}+1} \, _2F_1\left (-\frac {5}{2},\frac {5}{4};-\frac {3}{2};-\frac {b x^2}{a}\right )}{5 a x^5 \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(a + b*x^2)^(5/4)),x]

[Out]

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

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{b^{2} x^{10} + 2 \, a b x^{8} + a^{2} x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/4)/(b^2*x^10 + 2*a*b*x^8 + a^2*x^6), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(5/4)*x^6), x)

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {5}{4}} x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(b*x^2+a)^(5/4),x)

[Out]

int(1/x^6/(b*x^2+a)^(5/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(5/4)*x^6), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^6\,{\left (b\,x^2+a\right )}^{5/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a + b*x^2)^(5/4)),x)

[Out]

int(1/(x^6*(a + b*x^2)^(5/4)), x)

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sympy [C]  time = 1.40, size = 32, normalized size = 0.25 \[ - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {5}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac {5}{4}} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(b*x**2+a)**(5/4),x)

[Out]

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

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