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

Optimal. Leaf size=219 \[ \frac {300}{77} \sqrt {x^4+5} x+\frac {40 \sqrt {x^4+5} x}{3 \left (x^2+\sqrt {5}\right )}+\frac {10 \sqrt [4]{5} \left (154-45 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{231 \sqrt {x^4+5}}-\frac {40 \sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{3 \sqrt {x^4+5}}+\frac {1}{99} \left (27 x^2+22\right ) \left (x^4+5\right )^{3/2} x^3+\frac {2}{231} \left (135 x^2+154\right ) \sqrt {x^4+5} x^3 \]

[Out]

1/99*x^3*(27*x^2+22)*(x^4+5)^(3/2)+300/77*x*(x^4+5)^(1/2)+2/231*x^3*(135*x^2+154)*(x^4+5)^(1/2)+40/3*x*(x^4+5)
^(1/2)/(x^2+5^(1/2))-40/3*5^(1/4)*(cos(2*arctan(1/5*x*5^(3/4)))^2)^(1/2)/cos(2*arctan(1/5*x*5^(3/4)))*Elliptic
E(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)+10/231
*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))*(154-45*5^(1/2))*(x^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 = 219, 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} \[ \frac {1}{99} \left (27 x^2+22\right ) \left (x^4+5\right )^{3/2} x^3+\frac {2}{231} \left (135 x^2+154\right ) \sqrt {x^4+5} x^3+\frac {40 \sqrt {x^4+5} x}{3 \left (x^2+\sqrt {5}\right )}+\frac {300}{77} \sqrt {x^4+5} x+\frac {10 \sqrt [4]{5} \left (154-45 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{231 \sqrt {x^4+5}}-\frac {40 \sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{3 \sqrt {x^4+5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(300*x*Sqrt[5 + x^4])/77 + (40*x*Sqrt[5 + x^4])/(3*(Sqrt[5] + x^2)) + (2*x^3*(154 + 135*x^2)*Sqrt[5 + x^4])/23
1 + (x^3*(22 + 27*x^2)*(5 + x^4)^(3/2))/99 - (40*5^(1/4)*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*Ell
ipticE[2*ArcTan[x/5^(1/4)], 1/2])/(3*Sqrt[5 + x^4]) + (10*5^(1/4)*(154 - 45*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 +
x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x/5^(1/4)], 1/2])/(231*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^2 \left (2+3 x^2\right ) \left (5+x^4\right )^{3/2} \, dx &=\frac {1}{99} x^3 \left (22+27 x^2\right ) \left (5+x^4\right )^{3/2}+\frac {10}{33} \int x^2 \left (22+27 x^2\right ) \sqrt {5+x^4} \, dx\\ &=\frac {2}{231} x^3 \left (154+135 x^2\right ) \sqrt {5+x^4}+\frac {1}{99} x^3 \left (22+27 x^2\right ) \left (5+x^4\right )^{3/2}+\frac {20}{231} \int \frac {x^2 \left (154+135 x^2\right )}{\sqrt {5+x^4}} \, dx\\ &=\frac {300}{77} x \sqrt {5+x^4}+\frac {2}{231} x^3 \left (154+135 x^2\right ) \sqrt {5+x^4}+\frac {1}{99} x^3 \left (22+27 x^2\right ) \left (5+x^4\right )^{3/2}-\frac {20}{693} \int \frac {675-462 x^2}{\sqrt {5+x^4}} \, dx\\ &=\frac {300}{77} x \sqrt {5+x^4}+\frac {2}{231} x^3 \left (154+135 x^2\right ) \sqrt {5+x^4}+\frac {1}{99} x^3 \left (22+27 x^2\right ) \left (5+x^4\right )^{3/2}-\frac {1}{3} \left (40 \sqrt {5}\right ) \int \frac {1-\frac {x^2}{\sqrt {5}}}{\sqrt {5+x^4}} \, dx-\frac {1}{231} \left (20 \left (225-154 \sqrt {5}\right )\right ) \int \frac {1}{\sqrt {5+x^4}} \, dx\\ &=\frac {300}{77} x \sqrt {5+x^4}+\frac {40 x \sqrt {5+x^4}}{3 \left (\sqrt {5}+x^2\right )}+\frac {2}{231} x^3 \left (154+135 x^2\right ) \sqrt {5+x^4}+\frac {1}{99} x^3 \left (22+27 x^2\right ) \left (5+x^4\right )^{3/2}-\frac {40 \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 )}{3 \sqrt {5+x^4}}+\frac {10 \sqrt [4]{5} \left (154-45 \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 )}{231 \sqrt {5+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 68, normalized size = 0.31 \[ \frac {1}{33} x \left (-225 \sqrt {5} \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {x^4}{5}\right )+110 \sqrt {5} x^2 \, _2F_1\left (-\frac {3}{2},\frac {3}{4};\frac {7}{4};-\frac {x^4}{5}\right )+9 \left (x^4+5\right )^{5/2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(9*(5 + x^4)^(5/2) - 225*Sqrt[5]*Hypergeometric2F1[-3/2, 1/4, 5/4, -1/5*x^4] + 110*Sqrt[5]*x^2*Hypergeometr
ic2F1[-3/2, 3/4, 7/4, -1/5*x^4]))/33

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (3 \, x^{8} + 2 \, x^{6} + 15 \, x^{4} + 10 \, x^{2}\right )} \sqrt {x^{4} + 5}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((3*x^8 + 2*x^6 + 15*x^4 + 10*x^2)*sqrt(x^4 + 5), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.01, size = 204, normalized size = 0.93 \[ \frac {3 \sqrt {x^{4}+5}\, x^{9}}{11}+\frac {2 \sqrt {x^{4}+5}\, x^{7}}{9}+\frac {195 \sqrt {x^{4}+5}\, x^{5}}{77}+\frac {22 \sqrt {x^{4}+5}\, x^{3}}{9}+\frac {300 \sqrt {x^{4}+5}\, x}{77}-\frac {60 \sqrt {5}\, \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )}{77 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {8 i \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \left (-\EllipticE \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )+\EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )\right )}{3 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/11*(x^4+5)^(1/2)*x^9+195/77*(x^4+5)^(1/2)*x^5+300/77*(x^4+5)^(1/2)*x-60/77*5^(1/2)/(I*5^(1/2))^(1/2)*(-5*I*5
^(1/2)*x^2+25)^(1/2)*(5*I*5^(1/2)*x^2+25)^(1/2)/(x^4+5)^(1/2)*EllipticF(1/5*5^(1/2)*(I*5^(1/2))^(1/2)*x,I)+2/9
*(x^4+5)^(1/2)*x^7+22/9*(x^4+5)^(1/2)*x^3+8/3*I/(I*5^(1/2))^(1/2)*(-5*I*5^(1/2)*x^2+25)^(1/2)*(5*I*5^(1/2)*x^2
+25)^(1/2)/(x^4+5)^(1/2)*(EllipticF(1/5*5^(1/2)*(I*5^(1/2))^(1/2)*x,I)-EllipticE(1/5*5^(1/2)*(I*5^(1/2))^(1/2)
*x,I))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 3.65, size = 160, normalized size = 0.73 \[ \frac {3 \sqrt {5} x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac {13}{4}\right )} + \frac {\sqrt {5} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{2 \Gamma \left (\frac {11}{4}\right )} + \frac {15 \sqrt {5} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {5 \sqrt {5} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{2 \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(5)*x**9*gamma(9/4)*hyper((-1/2, 9/4), (13/4,), x**4*exp_polar(I*pi)/5)/(4*gamma(13/4)) + sqrt(5)*x**7*g
amma(7/4)*hyper((-1/2, 7/4), (11/4,), x**4*exp_polar(I*pi)/5)/(2*gamma(11/4)) + 15*sqrt(5)*x**5*gamma(5/4)*hyp
er((-1/2, 5/4), (9/4,), x**4*exp_polar(I*pi)/5)/(4*gamma(9/4)) + 5*sqrt(5)*x**3*gamma(3/4)*hyper((-1/2, 3/4),
(7/4,), x**4*exp_polar(I*pi)/5)/(2*gamma(7/4))

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