3.1247 \(\int \frac {1}{x^7 (a-b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=108 \[ \frac {5 b^{3/2} \left (1-\frac {b x^4}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{12 a^{3/2} \left (a-b x^4\right )^{3/4}}-\frac {5 b \sqrt [4]{a-b x^4}}{12 a^2 x^2}-\frac {\sqrt [4]{a-b x^4}}{6 a x^6} \]

[Out]

-1/6*(-b*x^4+a)^(1/4)/a/x^6-5/12*b*(-b*x^4+a)^(1/4)/a^2/x^2+5/12*b^(3/2)*(1-b*x^4/a)^(3/4)*(cos(1/2*arcsin(x^2
*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arcsin(x^2*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arcsin(x^2*b^(1/2)/a^(1/2))
),2^(1/2))/a^(3/2)/(-b*x^4+a)^(3/4)

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Rubi [A]  time = 0.07, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {275, 325, 233, 232} \[ \frac {5 b^{3/2} \left (1-\frac {b x^4}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{12 a^{3/2} \left (a-b x^4\right )^{3/4}}-\frac {5 b \sqrt [4]{a-b x^4}}{12 a^2 x^2}-\frac {\sqrt [4]{a-b x^4}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(a - b*x^4)^(3/4)),x]

[Out]

-(a - b*x^4)^(1/4)/(6*a*x^6) - (5*b*(a - b*x^4)^(1/4))/(12*a^2*x^2) + (5*b^(3/2)*(1 - (b*x^4)/a)^(3/4)*Ellipti
cF[ArcSin[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(12*a^(3/2)*(a - b*x^4)^(3/4))

Rule 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(3/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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^7 \left (a-b x^4\right )^{3/4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^4 \left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt [4]{a-b x^4}}{6 a x^6}+\frac {(5 b) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right )}{12 a}\\ &=-\frac {\sqrt [4]{a-b x^4}}{6 a x^6}-\frac {5 b \sqrt [4]{a-b x^4}}{12 a^2 x^2}+\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right )}{24 a^2}\\ &=-\frac {\sqrt [4]{a-b x^4}}{6 a x^6}-\frac {5 b \sqrt [4]{a-b x^4}}{12 a^2 x^2}+\frac {\left (5 b^2 \left (1-\frac {b x^4}{a}\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-\frac {b x^2}{a}\right )^{3/4}} \, dx,x,x^2\right )}{24 a^2 \left (a-b x^4\right )^{3/4}}\\ &=-\frac {\sqrt [4]{a-b x^4}}{6 a x^6}-\frac {5 b \sqrt [4]{a-b x^4}}{12 a^2 x^2}+\frac {5 b^{3/2} \left (1-\frac {b x^4}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{12 a^{3/2} \left (a-b x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.48 \[ -\frac {\left (1-\frac {b x^4}{a}\right )^{3/4} \, _2F_1\left (-\frac {3}{2},\frac {3}{4};-\frac {1}{2};\frac {b x^4}{a}\right )}{6 x^6 \left (a-b x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(a - b*x^4)^(3/4)),x]

[Out]

-1/6*((1 - (b*x^4)/a)^(3/4)*Hypergeometric2F1[-3/2, 3/4, -1/2, (b*x^4)/a])/(x^6*(a - b*x^4)^(3/4))

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{b x^{11} - a x^{7}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(-b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

integral(-(-b*x^4 + a)^(1/4)/(b*x^11 - a*x^7), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {3}{4}} x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(-b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((-b*x^4 + a)^(3/4)*x^7), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{4}+a \right )^{\frac {3}{4}} x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(-b*x^4+a)^(3/4),x)

[Out]

int(1/x^7/(-b*x^4+a)^(3/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {3}{4}} x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(-b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((-b*x^4 + a)^(3/4)*x^7), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^7\,{\left (a-b\,x^4\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(a - b*x^4)^(3/4)),x)

[Out]

int(1/(x^7*(a - b*x^4)^(3/4)), x)

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sympy [C]  time = 2.77, size = 34, normalized size = 0.31 \[ - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{6 a^{\frac {3}{4}} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(-b*x**4+a)**(3/4),x)

[Out]

-hyper((-3/2, 3/4), (-1/2,), b*x**4*exp_polar(2*I*pi)/a)/(6*a**(3/4)*x**6)

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