3.1.50 \(\int \frac {x^4 (2+3 x^2)}{(5+x^4)^{3/2}} \, dx\) [50]

Optimal. Leaf size=196 \[ -\frac {x^3 \left (15-2 x^2\right )}{10 \sqrt {5+x^4}}-\frac {1}{5} x \sqrt {5+x^4}+\frac {9 x \sqrt {5+x^4}}{2 \left (\sqrt {5}+x^2\right )}-\frac {9 \sqrt [4]{5} \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{2 \sqrt {5+x^4}}+\frac {\left (2+9 \sqrt {5}\right ) \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{5} \sqrt {5+x^4}} \]

[Out]

-1/10*x^3*(-2*x^2+15)/(x^4+5)^(1/2)-1/5*x*(x^4+5)^(1/2)+9/2*x*(x^4+5)^(1/2)/(x^2+5^(1/2))-9/2*5^(1/4)*(cos(2*a
rctan(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))*(x^2+5^(1/2))*(2+9*5^(1/2))*((x^4
+5)/(x^2+5^(1/2))^2)^(1/2)*5^(3/4)/(x^4+5)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1290, 1294, 1212, 226, 1210} \begin {gather*} \frac {\left (2+9 \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 \sqrt [4]{5} \sqrt {x^4+5}}-\frac {9 \sqrt [4]{5} \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 )}{2 \sqrt {x^4+5}}-\frac {1}{5} \sqrt {x^4+5} x+\frac {9 \sqrt {x^4+5} x}{2 \left (x^2+\sqrt {5}\right )}-\frac {\left (15-2 x^2\right ) x^3}{10 \sqrt {x^4+5}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/10*(x^3*(15 - 2*x^2))/Sqrt[5 + x^4] - (x*Sqrt[5 + x^4])/5 + (9*x*Sqrt[5 + x^4])/(2*(Sqrt[5] + x^2)) - (9*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])/(2*Sqrt[5 + x^4])
+ ((2 + 9*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^(1/4)*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])

Rule 1294

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 \frac {x^4 \left (2+3 x^2\right )}{\left (5+x^4\right )^{3/2}} \, dx &=-\frac {x^3 \left (15-2 x^2\right )}{10 \sqrt {5+x^4}}+\frac {1}{10} \int \frac {x^2 \left (45-6 x^2\right )}{\sqrt {5+x^4}} \, dx\\ &=-\frac {x^3 \left (15-2 x^2\right )}{10 \sqrt {5+x^4}}-\frac {1}{5} x \sqrt {5+x^4}-\frac {1}{30} \int \frac {-30-135 x^2}{\sqrt {5+x^4}} \, dx\\ &=-\frac {x^3 \left (15-2 x^2\right )}{10 \sqrt {5+x^4}}-\frac {1}{5} x \sqrt {5+x^4}-\frac {1}{2} \left (9 \sqrt {5}\right ) \int \frac {1-\frac {x^2}{\sqrt {5}}}{\sqrt {5+x^4}} \, dx-\frac {1}{2} \left (-2-9 \sqrt {5}\right ) \int \frac {1}{\sqrt {5+x^4}} \, dx\\ &=-\frac {x^3 \left (15-2 x^2\right )}{10 \sqrt {5+x^4}}-\frac {1}{5} x \sqrt {5+x^4}+\frac {9 x \sqrt {5+x^4}}{2 \left (\sqrt {5}+x^2\right )}-\frac {9 \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 )}{2 \sqrt {5+x^4}}+\frac {\left (2+9 \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 \sqrt [4]{5} \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 = 70, normalized size = 0.36 \begin {gather*} \frac {x \left (-1+3 x^2\right )}{\sqrt {5+x^4}}+\frac {x \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {x^4}{5}\right )}{\sqrt {5}}-\frac {3 x^3 \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {x^4}{5}\right )}{\sqrt {5}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-1 + 3*x^2))/Sqrt[5 + x^4] + (x*Hypergeometric2F1[1/4, 1/2, 5/4, -1/5*x^4])/Sqrt[5] - (3*x^3*Hypergeometri
c2F1[3/4, 3/2, 7/4, -1/5*x^4])/Sqrt[5]

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Maple [C] Result contains complex when optimal does not.
time = 0.14, size = 168, normalized size = 0.86

method	result	size
meijerg	\(\frac {3 \sqrt {5}\, x^{7} \hypergeom \left (\left [\frac {3}{2}, \frac {7}{4}\right ], \left [\frac {11}{4}\right ], -\frac {x^{4}}{5}\right )}{175}+\frac {2 \sqrt {5}\, x^{5} \hypergeom \left (\left [\frac {5}{4}, \frac {3}{2}\right ], \left [\frac {9}{4}\right ], -\frac {x^{4}}{5}\right )}{125}\)	\(40\)
risch	\(-\frac {x \left (3 x^{2}+2\right )}{2 \sqrt {x^{4}+5}}+\frac {9 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{10 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {\sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(163\)
elliptic	\(-\frac {2 \left (\frac {3}{4} x^{3}+\frac {1}{2} x \right )}{\sqrt {x^{4}+5}}+\frac {9 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{10 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {\sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(164\)
default	\(-\frac {3 x^{3}}{2 \sqrt {x^{4}+5}}+\frac {9 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{10 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {x}{\sqrt {x^{4}+5}}+\frac {\sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(168\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*x^3/(x^4+5)^(1/2)+9/10*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))-x/(x^4+5)^(1
/2)+1/25*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)*Ellipti
cF(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^4*(3*x^2+2)/(x^4+5)^(3/2),x, algorithm="maxima")

[Out]

integrate((3*x^2 + 2)*x^4/(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^4*(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 = 2.42, size = 75, normalized size = 0.38 \begin {gather*} \frac {3 \sqrt {5} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{100 \Gamma \left (\frac {11}{4}\right )} + \frac {\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 )}}{50 \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(5)*x**7*gamma(7/4)*hyper((3/2, 7/4), (11/4,), x**4*exp_polar(I*pi)/5)/(100*gamma(11/4)) + sqrt(5)*x**5*
gamma(5/4)*hyper((5/4, 3/2), (9/4,), x**4*exp_polar(I*pi)/5)/(50*gamma(9/4))

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

[Out]

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

[Out]

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

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