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

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

[Out]

20/21*x*(x^4+5)^(1/2)+2/3*x^3*(x^4+5)^(1/2)+1/21*x^5*(7*x^2+6)*(x^4+5)^(1/2)-10*x*(x^4+5)^(1/2)/(x^2+5^(1/2))+
10*5^(1/4)*(cos(2*arctan(1/5*x*5^(3/4)))^2)^(1/2)/cos(2*arctan(1/5*x*5^(3/4)))*EllipticE(sin(2*arctan(1/5*x*5^
(3/4))),1/2*2^(1/2))*(x^2+5^(1/2))*((x^4+5)/(x^2+5^(1/2))^2)^(1/2)/(x^4+5)^(1/2)-5/21*5^(1/4)*(cos(2*arctan(1/
5*x*5^(3/4)))^2)^(1/2)/cos(2*arctan(1/5*x*5^(3/4)))*EllipticF(sin(2*arctan(1/5*x*5^(3/4))),1/2*2^(1/2))*(x^2+5
^(1/2))*(21+2*5^(1/2))*((x^4+5)/(x^2+5^(1/2))^2)^(1/2)/(x^4+5)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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 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])

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*}

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Mathematica [C]  time = 0.03, size = 82, normalized size = 0.39 \[ \frac {1}{21} x \left (-30 \sqrt {5} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {x^4}{5}\right )-35 \sqrt {5} x^2 \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {x^4}{5}\right )+6 \left (x^4+5\right )^{3/2}+7 \left (x^4+5\right )^{3/2} x^2\right ) \]

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, -1/5*x^4] - 35*Sq
rt[5]*x^2*Hypergeometric2F1[-1/2, 3/4, 7/4, -1/5*x^4]))/21

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (3 \, x^{6} + 2 \, x^{4}\right )} \sqrt {x^{4} + 5}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x^{4} + 5} {\left (3 \, x^{2} + 2\right )} x^{4}\,{d x} \]

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)

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maple [C]  time = 0.09, size = 192, normalized size = 0.92 \[ \frac {\sqrt {x^{4}+5}\, x^{7}}{3}+\frac {2 \sqrt {x^{4}+5}\, x^{5}}{7}+\frac {2 \sqrt {x^{4}+5}\, x^{3}}{3}+\frac {20 \sqrt {x^{4}+5}\, x}{21}-\frac {4 \sqrt {5}\, \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )}{21 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {2 i \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \left (-\EllipticE \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )+\EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )\right )}{\sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x^{4} + 5} {\left (3 \, x^{2} + 2\right )} x^{4}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^4\,\sqrt {x^4+5}\,\left (3\,x^2+2\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 2.33, size = 78, normalized size = 0.38 \[ \frac {3 \sqrt {5} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} + \frac {\sqrt {5} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} \]

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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