3.10.47 \(\int \frac {x^{12}}{(1+x^4)^{3/2}} \, dx\) [947]

Optimal. Leaf size=90 \[ -\frac {x^9}{2 \sqrt {1+x^4}}-\frac {15}{14} x \sqrt {1+x^4}+\frac {9}{14} x^5 \sqrt {1+x^4}+\frac {15 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{28 \sqrt {1+x^4}} \]

[Out]

-1/2*x^9/(x^4+1)^(1/2)-15/14*x*(x^4+1)^(1/2)+9/14*x^5*(x^4+1)^(1/2)+15/28*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/c
os(2*arctan(x))*EllipticF(sin(2*arctan(x)),1/2*2^(1/2))*((x^4+1)/(x^2+1)^2)^(1/2)/(x^4+1)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {294, 327, 226} \begin {gather*} \frac {15 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \text {ArcTan}(x)\left |\frac {1}{2}\right .\right )}{28 \sqrt {x^4+1}}-\frac {15}{14} \sqrt {x^4+1} x-\frac {x^9}{2 \sqrt {x^4+1}}+\frac {9}{14} \sqrt {x^4+1} x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^12/(1 + x^4)^(3/2),x]

[Out]

-1/2*x^9/Sqrt[1 + x^4] - (15*x*Sqrt[1 + x^4])/14 + (9*x^5*Sqrt[1 + x^4])/14 + (15*(1 + x^2)*Sqrt[(1 + x^4)/(1
+ x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(28*Sqrt[1 + 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 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {x^{12}}{\left (1+x^4\right )^{3/2}} \, dx &=-\frac {x^9}{2 \sqrt {1+x^4}}+\frac {9}{2} \int \frac {x^8}{\sqrt {1+x^4}} \, dx\\ &=-\frac {x^9}{2 \sqrt {1+x^4}}+\frac {9}{14} x^5 \sqrt {1+x^4}-\frac {45}{14} \int \frac {x^4}{\sqrt {1+x^4}} \, dx\\ &=-\frac {x^9}{2 \sqrt {1+x^4}}-\frac {15}{14} x \sqrt {1+x^4}+\frac {9}{14} x^5 \sqrt {1+x^4}+\frac {15}{14} \int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=-\frac {x^9}{2 \sqrt {1+x^4}}-\frac {15}{14} x \sqrt {1+x^4}+\frac {9}{14} x^5 \sqrt {1+x^4}+\frac {15 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{28 \sqrt {1+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 = 4.19, size = 52, normalized size = 0.58 \begin {gather*} \frac {x \left (-15-6 x^4+2 x^8+15 \sqrt {1+x^4} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-x^4\right )\right )}{14 \sqrt {1+x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^12/(1 + x^4)^(3/2),x]

[Out]

(x*(-15 - 6*x^4 + 2*x^8 + 15*Sqrt[1 + x^4]*Hypergeometric2F1[1/4, 1/2, 5/4, -x^4]))/(14*Sqrt[1 + x^4])

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Maple [C] Result contains complex when optimal does not.
time = 0.15, size = 94, normalized size = 1.04

method	result	size
meijerg	\(\frac {x^{13} \hypergeom \left (\left [\frac {3}{2}, \frac {13}{4}\right ], \left [\frac {17}{4}\right ], -x^{4}\right )}{13}\)	\(17\)
risch	\(\frac {x \left (2 x^{8}-6 x^{4}-15\right )}{14 \sqrt {x^{4}+1}}+\frac {15 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{14 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(84\)
default	\(-\frac {x}{2 \sqrt {x^{4}+1}}+\frac {x^{5} \sqrt {x^{4}+1}}{7}-\frac {4 x \sqrt {x^{4}+1}}{7}+\frac {15 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{14 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(94\)
elliptic	\(-\frac {x}{2 \sqrt {x^{4}+1}}+\frac {x^{5} \sqrt {x^{4}+1}}{7}-\frac {4 x \sqrt {x^{4}+1}}{7}+\frac {15 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{14 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12/(x^4+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*x/(x^4+1)^(1/2)+1/7*x^5*(x^4+1)^(1/2)-4/7*x*(x^4+1)^(1/2)+15/14/(1/2*2^(1/2)+1/2*I*2^(1/2))*(1-I*x^2)^(1/
2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*EllipticF(x*(1/2*2^(1/2)+1/2*I*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^12/(x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^12/(x^4 + 1)^(3/2), x)

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Fricas [C] Result contains complex when optimal does not.
time = 0.08, size = 55, normalized size = 0.61 \begin {gather*} -\frac {15 \, \sqrt {i} {\left (-i \, x^{4} - i\right )} F(\arcsin \left (\frac {\sqrt {i}}{x}\right )\,|\,-1) - {\left (2 \, x^{9} - 6 \, x^{5} - 15 \, x\right )} \sqrt {x^{4} + 1}}{14 \, {\left (x^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12/(x^4+1)^(3/2),x, algorithm="fricas")

[Out]

-1/14*(15*sqrt(I)*(-I*x^4 - I)*elliptic_f(arcsin(sqrt(I)/x), -1) - (2*x^9 - 6*x^5 - 15*x)*sqrt(x^4 + 1))/(x^4
+ 1)

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Sympy [C] Result contains complex when optimal does not.
time = 0.59, size = 29, normalized size = 0.32 \begin {gather*} \frac {x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {17}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**12/(x**4+1)**(3/2),x)

[Out]

x**13*gamma(13/4)*hyper((3/2, 13/4), (17/4,), x**4*exp_polar(I*pi))/(4*gamma(17/4))

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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^12/(x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(x^12/(x^4 + 1)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{12}}{{\left (x^4+1\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12/(x^4 + 1)^(3/2),x)

[Out]

int(x^12/(x^4 + 1)^(3/2), x)

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