3.15 \(\int x^4 (2+3 x^2) \sqrt{5+x^4} \, dx\)

Optimal. Leaf size=208 \[ -\frac{5 \sqrt [4]{5} \left (21+2 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right ),\frac{1}{2}\right )}{21 \sqrt{x^4+5}}+\frac{1}{21} \left (7 x^2+6\right ) \sqrt{x^4+5} x^5+\frac{2}{3} \sqrt{x^4+5} x^3-\frac{10 \sqrt{x^4+5} x}{x^2+\sqrt{5}}+\frac{20}{21} \sqrt{x^4+5} x+\frac{10 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}} \]

[Out]

(20*x*Sqrt[5 + x^4])/21 + (2*x^3*Sqrt[5 + x^4])/3 - (10*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) + (x^5*(6 + 7*x^2)*Sq
rt[5 + x^4])/21 + (10*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] - (5*5^(1/4)*(21 + 2*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF
[2*ArcTan[x/5^(1/4)], 1/2])/(21*Sqrt[5 + x^4])

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Rubi [A]  time = 0.124185, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1274, 1280, 1198, 220, 1196} \[ \frac{1}{21} \left (7 x^2+6\right ) \sqrt{x^4+5} x^5+\frac{2}{3} \sqrt{x^4+5} x^3-\frac{10 \sqrt{x^4+5} x}{x^2+\sqrt{5}}+\frac{20}{21} \sqrt{x^4+5} x-\frac{5 \sqrt [4]{5} \left (21+2 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{21 \sqrt{x^4+5}}+\frac{10 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(2 + 3*x^2)*Sqrt[5 + x^4],x]

[Out]

(20*x*Sqrt[5 + x^4])/21 + (2*x^3*Sqrt[5 + x^4])/3 - (10*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) + (x^5*(6 + 7*x^2)*Sq
rt[5 + x^4])/21 + (10*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] - (5*5^(1/4)*(21 + 2*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF
[2*ArcTan[x/5^(1/4)], 1/2])/(21*Sqrt[5 + x^4])

Rule 1274

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*(c*d*(m + 4*p + 3) + c*e*(4*p + m + 1)*x^2))/(c*f*(4*p + m + 1)*(m + 4*p + 3)), x] + Dist[(4*a*p)/(
(4*p + m + 1)*(m + 4*p + 3)), Int[(f*x)^m*(a + c*x^4)^(p - 1)*Simp[d*(m + 4*p + 3) + e*(4*p + m + 1)*x^2, x],
x], x] /; FreeQ[{a, c, d, e, f, m}, x] && GtQ[p, 0] && NeQ[4*p + m + 1, 0] && NeQ[m + 4*p + 3, 0] && IntegerQ[
2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1280

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

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 x^4 \left (2+3 x^2\right ) \sqrt{5+x^4} \, dx &=\frac{1}{21} x^5 \left (6+7 x^2\right ) \sqrt{5+x^4}+\frac{10}{63} \int \frac{x^4 \left (18+21 x^2\right )}{\sqrt{5+x^4}} \, dx\\ &=\frac{2}{3} x^3 \sqrt{5+x^4}+\frac{1}{21} x^5 \left (6+7 x^2\right ) \sqrt{5+x^4}-\frac{2}{63} \int \frac{x^2 \left (315-90 x^2\right )}{\sqrt{5+x^4}} \, dx\\ &=\frac{20}{21} x \sqrt{5+x^4}+\frac{2}{3} x^3 \sqrt{5+x^4}+\frac{1}{21} x^5 \left (6+7 x^2\right ) \sqrt{5+x^4}+\frac{2}{189} \int \frac{-450-945 x^2}{\sqrt{5+x^4}} \, dx\\ &=\frac{20}{21} x \sqrt{5+x^4}+\frac{2}{3} x^3 \sqrt{5+x^4}+\frac{1}{21} x^5 \left (6+7 x^2\right ) \sqrt{5+x^4}+\left (10 \sqrt{5}\right ) \int \frac{1-\frac{x^2}{\sqrt{5}}}{\sqrt{5+x^4}} \, dx-\frac{1}{21} \left (10 \left (10+21 \sqrt{5}\right )\right ) \int \frac{1}{\sqrt{5+x^4}} \, dx\\ &=\frac{20}{21} x \sqrt{5+x^4}+\frac{2}{3} x^3 \sqrt{5+x^4}-\frac{10 x \sqrt{5+x^4}}{\sqrt{5}+x^2}+\frac{1}{21} x^5 \left (6+7 x^2\right ) \sqrt{5+x^4}+\frac{10 \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{5 \sqrt [4]{5} \left (21+2 \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 )}{21 \sqrt{5+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0387653, size = 82, normalized size = 0.39 \[ \frac{1}{21} x \left (-35 \sqrt{5} x^2 \, _2F_1\left (-\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{x^4}{5}\right )-30 \sqrt{5} \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{x^4}{5}\right )+7 \left (x^4+5\right )^{3/2} x^2+6 \left (x^4+5\right )^{3/2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(6*(5 + x^4)^(3/2) + 7*x^2*(5 + x^4)^(3/2) - 30*Sqrt[5]*Hypergeometric2F1[-1/2, 1/4, 5/4, -x^4/5] - 35*Sqrt
[5]*x^2*Hypergeometric2F1[-1/2, 3/4, 7/4, -x^4/5]))/21

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Maple [C]  time = 0.083, size = 192, normalized size = 0.9 \begin{align*}{\frac{{x}^{7}}{3}\sqrt{{x}^{4}+5}}+{\frac{2\,{x}^{3}}{3}\sqrt{{x}^{4}+5}}-{\frac{2\,i}{\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}}}}+{\frac{2\,{x}^{5}}{7}\sqrt{{x}^{4}+5}}+{\frac{20\,x}{21}\sqrt{{x}^{4}+5}}-{\frac{4\,\sqrt{5}}{21\,\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}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(3*x^2+2)*(x^4+5)^(1/2),x)

[Out]

1/3*x^7*(x^4+5)^(1/2)+2/3*x^3*(x^4+5)^(1/2)-2*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)*(EllipticF(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))+2/7*x^5*(x^4+5)^(1/2)+20/21*x*(x^4+5)^(1/2)-4/21*5^(1/2)/(I*5^(1/2))^(1/2)*(25-5*I*5^(1/2)*x^2)^(1/2)*(2
5+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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(3*x^2+2)*(x^4+5)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(3*x^2+2)*(x^4+5)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 2.02494, size = 78, normalized size = 0.38 \begin{align*} \frac{3 \sqrt{5} x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac{11}{4}\right )} + \frac{\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 )}}{2 \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(3*x**2+2)*(x**4+5)**(1/2),x)

[Out]

3*sqrt(5)*x**7*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), x**4*exp_polar(I*pi)/5)/(4*gamma(11/4)) + sqrt(5)*x**5*g
amma(5/4)*hyper((-1/2, 5/4), (9/4,), x**4*exp_polar(I*pi)/5)/(2*gamma(9/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(3*x^2+2)*(x^4+5)^(1/2),x, algorithm="giac")

[Out]

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