3.30 \(\int \frac{(2+3 x^2) (5+x^4)^{3/2}}{x^2} \, dx\)

Optimal. Leaf size=199 \[ \frac{6 \sqrt [4]{5} \left (14+5 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right ),\frac{1}{2}\right )}{7 \sqrt{x^4+5}}-\frac{\left (14-3 x^2\right ) \left (x^4+5\right )^{3/2}}{7 x}+\frac{6}{35} x \left (14 x^2+25\right ) \sqrt{x^4+5}+\frac{24 x \sqrt{x^4+5}}{x^2+\sqrt{5}}-\frac{24 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}} \]

[Out]

(24*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) + (6*x*(25 + 14*x^2)*Sqrt[5 + x^4])/35 - ((14 - 3*x^2)*(5 + x^4)^(3/2))/(
7*x) - (24*5^(1/4)*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticE[2*ArcTan[x/5^(1/4)], 1/2])/Sqrt
[5 + x^4] + (6*5^(1/4)*(14 + 5*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x
/5^(1/4)], 1/2])/(7*Sqrt[5 + x^4])

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Rubi [A]  time = 0.0883871, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1272, 1177, 1198, 220, 1196} \[ -\frac{\left (14-3 x^2\right ) \left (x^4+5\right )^{3/2}}{7 x}+\frac{6}{35} x \left (14 x^2+25\right ) \sqrt{x^4+5}+\frac{24 x \sqrt{x^4+5}}{x^2+\sqrt{5}}+\frac{6 \sqrt [4]{5} \left (14+5 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{7 \sqrt{x^4+5}}-\frac{24 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}} \]

Antiderivative was successfully verified.

[In]

Int[((2 + 3*x^2)*(5 + x^4)^(3/2))/x^2,x]

[Out]

(24*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) + (6*x*(25 + 14*x^2)*Sqrt[5 + x^4])/35 - ((14 - 3*x^2)*(5 + x^4)^(3/2))/(
7*x) - (24*5^(1/4)*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticE[2*ArcTan[x/5^(1/4)], 1/2])/Sqrt
[5 + x^4] + (6*5^(1/4)*(14 + 5*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x
/5^(1/4)], 1/2])/(7*Sqrt[5 + x^4])

Rule 1272

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((f*x)^(m + 1)*(a
+ c*x^4)^p*(d*(m + 4*p + 3) + e*(m + 1)*x^2))/(f*(m + 1)*(m + 4*p + 3)), x] + Dist[(4*p)/(f^2*(m + 1)*(m + 4*p
+ 3)), Int[(f*x)^(m + 2)*(a + c*x^4)^(p - 1)*(a*e*(m + 1) - c*d*(m + 4*p + 3)*x^2), x], x] /; FreeQ[{a, c, d,
e, f}, x] && GtQ[p, 0] && LtQ[m, -1] && m + 4*p + 3 != 0 && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1177

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(d*(4*p + 3) + e*(4*p + 1)*x^2)*(a
+ c*x^4)^p)/((4*p + 1)*(4*p + 3)), x] + Dist[(2*p)/((4*p + 1)*(4*p + 3)), Int[Simp[2*a*d*(4*p + 3) + (2*a*e*(4
*p + 1))*x^2, x]*(a + c*x^4)^(p - 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] &
& FractionQ[p] && IntegerQ[2*p]

Rule 1198

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/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 (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^2} \, dx &=-\frac{\left (14-3 x^2\right ) \left (5+x^4\right )^{3/2}}{7 x}-\frac{6}{7} \int \left (-15-14 x^2\right ) \sqrt{5+x^4} \, dx\\ &=\frac{6}{35} x \left (25+14 x^2\right ) \sqrt{5+x^4}-\frac{\left (14-3 x^2\right ) \left (5+x^4\right )^{3/2}}{7 x}-\frac{2}{35} \int \frac{-750-420 x^2}{\sqrt{5+x^4}} \, dx\\ &=\frac{6}{35} x \left (25+14 x^2\right ) \sqrt{5+x^4}-\frac{\left (14-3 x^2\right ) \left (5+x^4\right )^{3/2}}{7 x}-\left (24 \sqrt{5}\right ) \int \frac{1-\frac{x^2}{\sqrt{5}}}{\sqrt{5+x^4}} \, dx+\frac{1}{7} \left (12 \left (25+14 \sqrt{5}\right )\right ) \int \frac{1}{\sqrt{5+x^4}} \, dx\\ &=\frac{24 x \sqrt{5+x^4}}{\sqrt{5}+x^2}+\frac{6}{35} x \left (25+14 x^2\right ) \sqrt{5+x^4}-\frac{\left (14-3 x^2\right ) \left (5+x^4\right )^{3/2}}{7 x}-\frac{24 \sqrt [4]{5} \left (\sqrt{5}+x^2\right ) \sqrt{\frac{5+x^4}{\left (\sqrt{5}+x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{5+x^4}}+\frac{6 \sqrt [4]{5} \left (14+5 \sqrt{5}\right ) \left (\sqrt{5}+x^2\right ) \sqrt{\frac{5+x^4}{\left (\sqrt{5}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{7 \sqrt{5+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0279748, size = 53, normalized size = 0.27 \[ 15 \sqrt{5} x \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};-\frac{x^4}{5}\right )-\frac{10 \sqrt{5} \, _2F_1\left (-\frac{3}{2},-\frac{1}{4};\frac{3}{4};-\frac{x^4}{5}\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[((2 + 3*x^2)*(5 + x^4)^(3/2))/x^2,x]

[Out]

(-10*Sqrt[5]*Hypergeometric2F1[-3/2, -1/4, 3/4, -x^4/5])/x + 15*Sqrt[5]*x*Hypergeometric2F1[-3/2, 1/4, 5/4, -x
^4/5]

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Maple [C]  time = 0.015, size = 192, normalized size = 1. \begin{align*}{\frac{3\,{x}^{5}}{7}\sqrt{{x}^{4}+5}}+{\frac{45\,x}{7}\sqrt{{x}^{4}+5}}+{\frac{12\,\sqrt{5}}{7\,\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}}{\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ){\frac{1}{\sqrt{{x}^{4}+5}}}}-10\,{\frac{\sqrt{{x}^{4}+5}}{x}}+{\frac{2\,{x}^{3}}{5}\sqrt{{x}^{4}+5}}+{\frac{{\frac{24\,i}{5}}}{\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}} \left ({\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) -{\it EllipticE} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) \right ){\frac{1}{\sqrt{{x}^{4}+5}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2+2)*(x^4+5)^(3/2)/x^2,x)

[Out]

3/7*x^5*(x^4+5)^(1/2)+45/7*x*(x^4+5)^(1/2)+12/7*5^(1/2)/(I*5^(1/2))^(1/2)*(25-5*I*5^(1/2)*x^2)^(1/2)*(25+5*I*5
^(1/2)*x^2)^(1/2)/(x^4+5)^(1/2)*EllipticF(1/5*x*5^(1/2)*(I*5^(1/2))^(1/2),I)-10*(x^4+5)^(1/2)/x+2/5*x^3*(x^4+5
)^(1/2)+24/5*I/(I*5^(1/2))^(1/2)*(25-5*I*5^(1/2)*x^2)^(1/2)*(25+5*I*5^(1/2)*x^2)^(1/2)/(x^4+5)^(1/2)*(Elliptic
F(1/5*x*5^(1/2)*(I*5^(1/2))^(1/2),I)-EllipticE(1/5*x*5^(1/2)*(I*5^(1/2))^(1/2),I))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x^{4} + 5\right )}^{\frac{3}{2}}{\left (3 \, x^{2} + 2\right )}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)*(x^4+5)^(3/2)/x^2,x, algorithm="maxima")

[Out]

integrate((x^4 + 5)^(3/2)*(3*x^2 + 2)/x^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (3 \, x^{6} + 2 \, x^{4} + 15 \, x^{2} + 10\right )} \sqrt{x^{4} + 5}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)*(x^4+5)^(3/2)/x^2,x, algorithm="fricas")

[Out]

integral((3*x^6 + 2*x^4 + 15*x^2 + 10)*sqrt(x^4 + 5)/x^2, x)

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Sympy [C]  time = 3.37316, size = 160, normalized size = 0.8 \begin{align*} \frac{3 \sqrt{5} x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac{9}{4}\right )} + \frac{\sqrt{5} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{2 \Gamma \left (\frac{7}{4}\right )} + \frac{15 \sqrt{5} x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{5 \sqrt{5} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{2 x \Gamma \left (\frac{3}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2+2)*(x**4+5)**(3/2)/x**2,x)

[Out]

3*sqrt(5)*x**5*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), x**4*exp_polar(I*pi)/5)/(4*gamma(9/4)) + sqrt(5)*x**3*gam
ma(3/4)*hyper((-1/2, 3/4), (7/4,), x**4*exp_polar(I*pi)/5)/(2*gamma(7/4)) + 15*sqrt(5)*x*gamma(1/4)*hyper((-1/
2, 1/4), (5/4,), x**4*exp_polar(I*pi)/5)/(4*gamma(5/4)) + 5*sqrt(5)*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), x*
*4*exp_polar(I*pi)/5)/(2*x*gamma(3/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x^{4} + 5\right )}^{\frac{3}{2}}{\left (3 \, x^{2} + 2\right )}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)*(x^4+5)^(3/2)/x^2,x, algorithm="giac")

[Out]

integrate((x^4 + 5)^(3/2)*(3*x^2 + 2)/x^2, x)