3.799 \(\int \frac{(a+c x^4)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=126 \[ \frac{2 c^{7/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{7 \sqrt [4]{a} \sqrt{a+c x^4}}-\frac{2 c \sqrt{a+c x^4}}{7 x^3}-\frac{\left (a+c x^4\right )^{3/2}}{7 x^7} \]

[Out]

(-2*c*Sqrt[a + c*x^4])/(7*x^3) - (a + c*x^4)^(3/2)/(7*x^7) + (2*c^(7/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^
4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(7*a^(1/4)*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.0322656, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {277, 220} \[ \frac{2 c^{7/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{7 \sqrt [4]{a} \sqrt{a+c x^4}}-\frac{2 c \sqrt{a+c x^4}}{7 x^3}-\frac{\left (a+c x^4\right )^{3/2}}{7 x^7} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^(3/2)/x^8,x]

[Out]

(-2*c*Sqrt[a + c*x^4])/(7*x^3) - (a + c*x^4)^(3/2)/(7*x^7) + (2*c^(7/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^
4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(7*a^(1/4)*Sqrt[a + c*x^4])

Rule 277

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

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]

Rubi steps

\begin{align*} \int \frac{\left (a+c x^4\right )^{3/2}}{x^8} \, dx &=-\frac{\left (a+c x^4\right )^{3/2}}{7 x^7}+\frac{1}{7} (6 c) \int \frac{\sqrt{a+c x^4}}{x^4} \, dx\\ &=-\frac{2 c \sqrt{a+c x^4}}{7 x^3}-\frac{\left (a+c x^4\right )^{3/2}}{7 x^7}+\frac{1}{7} \left (4 c^2\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx\\ &=-\frac{2 c \sqrt{a+c x^4}}{7 x^3}-\frac{\left (a+c x^4\right )^{3/2}}{7 x^7}+\frac{2 c^{7/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{7 \sqrt [4]{a} \sqrt{a+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0108799, size = 52, normalized size = 0.41 \[ -\frac{a \sqrt{a+c x^4} \, _2F_1\left (-\frac{7}{4},-\frac{3}{2};-\frac{3}{4};-\frac{c x^4}{a}\right )}{7 x^7 \sqrt{\frac{c x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^4)^(3/2)/x^8,x]

[Out]

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

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Maple [C]  time = 0.014, size = 105, normalized size = 0.8 \begin{align*} -{\frac{a}{7\,{x}^{7}}\sqrt{c{x}^{4}+a}}-{\frac{3\,c}{7\,{x}^{3}}\sqrt{c{x}^{4}+a}}+{\frac{4\,{c}^{2}}{7}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*a*(c*x^4+a)^(1/2)/x^7-3/7*c*(c*x^4+a)^(1/2)/x^3+4/7*c^2/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^
2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + a\right )}^{\frac{3}{2}}}{x^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^4 + a)^(3/2)/x^8, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c x^{4} + a\right )}^{\frac{3}{2}}}{x^{8}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^4 + a)^(3/2)/x^8, x)

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Sympy [C]  time = 1.73654, size = 46, normalized size = 0.37 \begin{align*} \frac{a^{\frac{3}{2}} \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{c x^{4} e^{i \pi }}{a}} \right )}}{4 x^{7} \Gamma \left (- \frac{3}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + a\right )}^{\frac{3}{2}}}{x^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^4 + a)^(3/2)/x^8, x)