3.866 \(\int \frac {1}{x^8 (a+b x^4)^{3/2}} \, dx\)

Optimal. Leaf size=153 \[ \frac {15 b^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{28 a^{13/4} \sqrt {a+b x^4}}+\frac {15 b \sqrt {a+b x^4}}{14 a^3 x^3}-\frac {9 \sqrt {a+b x^4}}{14 a^2 x^7}+\frac {1}{2 a x^7 \sqrt {a+b x^4}} \]

[Out]

1/2/a/x^7/(b*x^4+a)^(1/2)-9/14*(b*x^4+a)^(1/2)/a^2/x^7+15/14*b*(b*x^4+a)^(1/2)/a^3/x^3+15/28*b^(7/4)*(cos(2*ar
ctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),
1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/a^(13/4)/(b*x^4+a)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {290, 325, 220} \[ \frac {15 b^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{28 a^{13/4} \sqrt {a+b x^4}}+\frac {15 b \sqrt {a+b x^4}}{14 a^3 x^3}-\frac {9 \sqrt {a+b x^4}}{14 a^2 x^7}+\frac {1}{2 a x^7 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^8*(a + b*x^4)^(3/2)),x]

[Out]

1/(2*a*x^7*Sqrt[a + b*x^4]) - (9*Sqrt[a + b*x^4])/(14*a^2*x^7) + (15*b*Sqrt[a + b*x^4])/(14*a^3*x^3) + (15*b^(
7/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4
)], 1/2])/(28*a^(13/4)*Sqrt[a + b*x^4])

Rule 220

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

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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^8 \left (a+b x^4\right )^{3/2}} \, dx &=\frac {1}{2 a x^7 \sqrt {a+b x^4}}+\frac {9 \int \frac {1}{x^8 \sqrt {a+b x^4}} \, dx}{2 a}\\ &=\frac {1}{2 a x^7 \sqrt {a+b x^4}}-\frac {9 \sqrt {a+b x^4}}{14 a^2 x^7}-\frac {(45 b) \int \frac {1}{x^4 \sqrt {a+b x^4}} \, dx}{14 a^2}\\ &=\frac {1}{2 a x^7 \sqrt {a+b x^4}}-\frac {9 \sqrt {a+b x^4}}{14 a^2 x^7}+\frac {15 b \sqrt {a+b x^4}}{14 a^3 x^3}+\frac {\left (15 b^2\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{14 a^3}\\ &=\frac {1}{2 a x^7 \sqrt {a+b x^4}}-\frac {9 \sqrt {a+b x^4}}{14 a^2 x^7}+\frac {15 b \sqrt {a+b x^4}}{14 a^3 x^3}+\frac {15 b^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{28 a^{13/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 54, normalized size = 0.35 \[ -\frac {\sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (-\frac {7}{4},\frac {3}{2};-\frac {3}{4};-\frac {b x^4}{a}\right )}{7 a x^7 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^8*(a + b*x^4)^(3/2)),x]

[Out]

-1/7*(Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[-7/4, 3/2, -3/4, -((b*x^4)/a)])/(a*x^7*Sqrt[a + b*x^4])

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{4} + a}}{b^{2} x^{16} + 2 \, a b x^{12} + a^{2} x^{8}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^4 + a)/(b^2*x^16 + 2*a*b*x^12 + a^2*x^8), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(3/2)*x^8), x)

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maple [C]  time = 0.02, size = 135, normalized size = 0.88 \[ \frac {b^{2} x}{2 \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}\, a^{3}}+\frac {15 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b^{2} \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{14 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, a^{3}}+\frac {4 \sqrt {b \,x^{4}+a}\, b}{7 a^{3} x^{3}}-\frac {\sqrt {b \,x^{4}+a}}{7 a^{2} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^8/(b*x^4+a)^(3/2),x)

[Out]

-1/7*(b*x^4+a)^(1/2)/a^2/x^7+4/7*b*(b*x^4+a)^(1/2)/a^3/x^3+1/2*b^2/a^3*x/((x^4+a/b)*b)^(1/2)+15/14/a^3*b^2/(I/
a^(1/2)*b^(1/2))^(1/2)*(-I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)/(b*x^4+a)^(1/2)*Ellipt
icF((I/a^(1/2)*b^(1/2))^(1/2)*x,I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(b*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(3/2)*x^8), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^8\,{\left (b\,x^4+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^8*(a + b*x^4)^(3/2)),x)

[Out]

int(1/(x^8*(a + b*x^4)^(3/2)), x)

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sympy [C]  time = 1.56, size = 44, normalized size = 0.29 \[ \frac {\Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, \frac {3}{2} \\ - \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} x^{7} \Gamma \left (- \frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**8/(b*x**4+a)**(3/2),x)

[Out]

gamma(-7/4)*hyper((-7/4, 3/2), (-3/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*x**7*gamma(-3/4))

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