3.802 \(\int \frac{(a+c x^4)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=252 \[ \frac{6 \sqrt [4]{a} c^{5/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{5 \sqrt{a+c x^4}}+\frac{12 c^{3/2} x \sqrt{a+c x^4}}{5 \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{12 \sqrt [4]{a} c^{5/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+c x^4}}-\frac{6 c \sqrt{a+c x^4}}{5 x}-\frac{\left (a+c x^4\right )^{3/2}}{5 x^5} \]

[Out]

(-6*c*Sqrt[a + c*x^4])/(5*x) + (12*c^(3/2)*x*Sqrt[a + c*x^4])/(5*(Sqrt[a] + Sqrt[c]*x^2)) - (a + c*x^4)^(3/2)/
(5*x^5) - (12*a^(1/4)*c^(5/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*
ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(5*Sqrt[a + c*x^4]) + (6*a^(1/4)*c^(5/4)*(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])/(5*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.0855572, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {277, 305, 220, 1196} \[ \frac{12 c^{3/2} x \sqrt{a+c x^4}}{5 \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{6 \sqrt [4]{a} c^{5/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+c x^4}}-\frac{12 \sqrt [4]{a} c^{5/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+c x^4}}-\frac{6 c \sqrt{a+c x^4}}{5 x}-\frac{\left (a+c x^4\right )^{3/2}}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^(3/2)/x^6,x]

[Out]

(-6*c*Sqrt[a + c*x^4])/(5*x) + (12*c^(3/2)*x*Sqrt[a + c*x^4])/(5*(Sqrt[a] + Sqrt[c]*x^2)) - (a + c*x^4)^(3/2)/
(5*x^5) - (12*a^(1/4)*c^(5/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*
ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(5*Sqrt[a + c*x^4]) + (6*a^(1/4)*c^(5/4)*(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])/(5*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 305

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

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 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{\left (a+c x^4\right )^{3/2}}{x^6} \, dx &=-\frac{\left (a+c x^4\right )^{3/2}}{5 x^5}+\frac{1}{5} (6 c) \int \frac{\sqrt{a+c x^4}}{x^2} \, dx\\ &=-\frac{6 c \sqrt{a+c x^4}}{5 x}-\frac{\left (a+c x^4\right )^{3/2}}{5 x^5}+\frac{1}{5} \left (12 c^2\right ) \int \frac{x^2}{\sqrt{a+c x^4}} \, dx\\ &=-\frac{6 c \sqrt{a+c x^4}}{5 x}-\frac{\left (a+c x^4\right )^{3/2}}{5 x^5}+\frac{1}{5} \left (12 \sqrt{a} c^{3/2}\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx-\frac{1}{5} \left (12 \sqrt{a} c^{3/2}\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+c x^4}} \, dx\\ &=-\frac{6 c \sqrt{a+c x^4}}{5 x}+\frac{12 c^{3/2} x \sqrt{a+c x^4}}{5 \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\left (a+c x^4\right )^{3/2}}{5 x^5}-\frac{12 \sqrt [4]{a} c^{5/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+c x^4}}+\frac{6 \sqrt [4]{a} c^{5/4} \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 )}{5 \sqrt{a+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.009079, size = 52, normalized size = 0.21 \[ -\frac{a \sqrt{a+c x^4} \, _2F_1\left (-\frac{3}{2},-\frac{5}{4};-\frac{1}{4};-\frac{c x^4}{a}\right )}{5 x^5 \sqrt{\frac{c x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^4)^(3/2)/x^6,x]

[Out]

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

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Maple [C]  time = 0.012, size = 128, normalized size = 0.5 \begin{align*} -{\frac{a}{5\,{x}^{5}}\sqrt{c{x}^{4}+a}}-{\frac{7\,c}{5\,x}\sqrt{c{x}^{4}+a}}+{{\frac{12\,i}{5}}{c}^{{\frac{3}{2}}}\sqrt{a}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}} \left ({\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) \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^6,x)

[Out]

-1/5*a*(c*x^4+a)^(1/2)/x^5-7/5*c*(c*x^4+a)^(1/2)/x+12/5*I*c^(3/2)*a^(1/2)/(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
)-EllipticE(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^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x^6,x, algorithm="maxima")

[Out]

integrate((c*x^4 + a)^(3/2)/x^6, 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^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x^6,x, algorithm="fricas")

[Out]

integral((c*x^4 + a)^(3/2)/x^6, x)

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Sympy [C]  time = 1.47777, size = 46, normalized size = 0.18 \begin{align*} \frac{a^{\frac{3}{2}} \Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, - \frac{5}{4} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+a)**(3/2)/x**6,x)

[Out]

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

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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^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x^6,x, algorithm="giac")

[Out]

integrate((c*x^4 + a)^(3/2)/x^6, x)