3.1.51 \(\int \frac {x^2 (2+3 x^2)}{(5+x^4)^{3/2}} \, dx\) [51]

Optimal. Leaf size=177 \[ -\frac {x \left (15-2 x^2\right )}{10 \sqrt {5+x^4}}-\frac {x \sqrt {5+x^4}}{5 \left (\sqrt {5}+x^2\right )}+\frac {\left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{5^{3/4} \sqrt {5+x^4}}-\frac {\left (2-3 \sqrt {5}\right ) \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{4\ 5^{3/4} \sqrt {5+x^4}} \]

[Out]

-1/10*x*(-2*x^2+15)/(x^4+5)^(1/2)-1/5*x*(x^4+5)^(1/2)/(x^2+5^(1/2))+1/5*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)-1/20*(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))*(2-3*5^(1/2))*(x^2+5^(1/2))*((x^4+5)/(x^2+5^(1/2))^2)^(
1/2)*5^(1/4)/(x^4+5)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1290, 1212, 226, 1210} \begin {gather*} -\frac {\left (2-3 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \text {ArcTan}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{4\ 5^{3/4} \sqrt {x^4+5}}+\frac {\left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \text {ArcTan}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{5^{3/4} \sqrt {x^4+5}}-\frac {\sqrt {x^4+5} x}{5 \left (x^2+\sqrt {5}\right )}-\frac {\left (15-2 x^2\right ) x}{10 \sqrt {x^4+5}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1212

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 1290

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

Rubi steps

\begin {align*} \int \frac {x^2 \left (2+3 x^2\right )}{\left (5+x^4\right )^{3/2}} \, dx &=-\frac {x \left (15-2 x^2\right )}{10 \sqrt {5+x^4}}+\frac {1}{10} \int \frac {15-2 x^2}{\sqrt {5+x^4}} \, dx\\ &=-\frac {x \left (15-2 x^2\right )}{10 \sqrt {5+x^4}}+\frac {\int \frac {1-\frac {x^2}{\sqrt {5}}}{\sqrt {5+x^4}} \, dx}{\sqrt {5}}+\frac {1}{10} \left (15-2 \sqrt {5}\right ) \int \frac {1}{\sqrt {5+x^4}} \, dx\\ &=-\frac {x \left (15-2 x^2\right )}{10 \sqrt {5+x^4}}-\frac {x \sqrt {5+x^4}}{5 \left (\sqrt {5}+x^2\right )}+\frac {\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 )}{5^{3/4} \sqrt {5+x^4}}-\frac {\left (2-3 \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 )}{4\ 5^{3/4} \sqrt {5+x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.03, size = 68, normalized size = 0.38 \begin {gather*} \frac {1}{150} x \left (-\frac {225}{\sqrt {5+x^4}}+45 \sqrt {5} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {x^4}{5}\right )+4 \sqrt {5} x^2 \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {x^4}{5}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-225/Sqrt[5 + x^4] + 45*Sqrt[5]*Hypergeometric2F1[1/4, 1/2, 5/4, -1/5*x^4] + 4*Sqrt[5]*x^2*Hypergeometric2
F1[3/4, 3/2, 7/4, -1/5*x^4]))/150

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Maple [C] Result contains complex when optimal does not.
time = 0.14, size = 168, normalized size = 0.95

method	result	size
meijerg	\(\frac {3 \sqrt {5}\, x^{5} \hypergeom \left (\left [\frac {5}{4}, \frac {3}{2}\right ], \left [\frac {9}{4}\right ], -\frac {x^{4}}{5}\right )}{125}+\frac {2 \sqrt {5}\, x^{3} \hypergeom \left (\left [\frac {3}{4}, \frac {3}{2}\right ], \left [\frac {7}{4}\right ], -\frac {x^{4}}{5}\right )}{75}\)	\(40\)
risch	\(\frac {x \left (2 x^{2}-15\right )}{10 \sqrt {x^{4}+5}}-\frac {i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {3 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{50 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(163\)
elliptic	\(-\frac {2 \left (-\frac {1}{10} x^{3}+\frac {3}{4} x \right )}{\sqrt {x^{4}+5}}-\frac {i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {3 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{50 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(164\)
default	\(-\frac {3 x}{2 \sqrt {x^{4}+5}}+\frac {3 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{50 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {x^{3}}{5 \sqrt {x^{4}+5}}-\frac {i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(168\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*x/(x^4+5)^(1/2)+3/50*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)+1/5*x^3/(x^4+5)^(1/2)-1/25*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)-Elli
pticE(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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((3*x^2 + 2)*x^2/(x^4 + 5)^(3/2), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [C] Result contains complex when optimal does not.
time = 2.24, size = 75, normalized size = 0.42 \begin {gather*} \frac {3 \sqrt {5} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{100 \Gamma \left (\frac {9}{4}\right )} + \frac {\sqrt {5} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{50 \Gamma \left (\frac {7}{4}\right )} \end {gather*}

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**5*gamma(5/4)*hyper((5/4, 3/2), (9/4,), x**4*exp_polar(I*pi)/5)/(100*gamma(9/4)) + sqrt(5)*x**3*ga
mma(3/4)*hyper((3/4, 3/2), (7/4,), x**4*exp_polar(I*pi)/5)/(50*gamma(7/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((3*x^2 + 2)*x^2/(x^4 + 5)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\left (3\,x^2+2\right )}{{\left (x^4+5\right )}^{3/2}} \,d x \end {gather*}

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]

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

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