3.11.49 \(\int \frac {x^2}{(-1+x^4) \sqrt [4]{x^2+x^4}} \, dx\)

Optimal. Leaf size=79 \[ \frac {\left (x^4+x^2\right )^{3/4}}{x \left (x^2+1\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+x^2}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+x^2}}\right )}{2 \sqrt [4]{2}} \]

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Rubi [C]  time = 0.17, antiderivative size = 45, normalized size of antiderivative = 0.57, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2056, 1254, 466, 510} \begin {gather*} -\frac {2 x^3 \, _2F_1\left (1,\frac {5}{4};\frac {9}{4};\frac {2 x^2}{x^2+1}\right )}{5 \left (x^2+1\right ) \sqrt [4]{x^4+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/((-1 + x^4)*(x^2 + x^4)^(1/4)),x]

[Out]

(-2*x^3*Hypergeometric2F1[1, 5/4, 9/4, (2*x^2)/(1 + x^2)])/(5*(1 + x^2)*(x^2 + x^4)^(1/4))

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1254

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[(f*x)^m*(d +
e*x^2)^(q + p)*(a/d + (c*x^2)/e)^p, x] /; FreeQ[{a, c, d, e, f, q, m, q}, x] && EqQ[c*d^2 + a*e^2, 0] && Integ
erQ[p]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx &=\frac {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \int \frac {x^{3/2}}{\sqrt [4]{1+x^2} \left (-1+x^4\right )} \, dx}{\sqrt [4]{x^2+x^4}}\\ &=\frac {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \int \frac {x^{3/2}}{\left (-1+x^2\right ) \left (1+x^2\right )^{5/4}} \, dx}{\sqrt [4]{x^2+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (-1+x^4\right ) \left (1+x^4\right )^{5/4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}}\\ &=-\frac {2 x^3 \, _2F_1\left (1,\frac {5}{4};\frac {9}{4};\frac {2 x^2}{1+x^2}\right )}{5 \left (1+x^2\right ) \sqrt [4]{x^2+x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 68, normalized size = 0.86 \begin {gather*} \frac {x \left (2^{3/4} \sqrt [4]{\frac {1}{x^2}+1} \left (\tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{x^2}+1}}{\sqrt [4]{2}}\right )-\tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{x^2}+1}}{\sqrt [4]{2}}\right )\right )+4\right )}{4 \sqrt [4]{x^4+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/((-1 + x^4)*(x^2 + x^4)^(1/4)),x]

[Out]

(x*(4 + 2^(3/4)*(1 + x^(-2))^(1/4)*(ArcTan[(1 + x^(-2))^(1/4)/2^(1/4)] - ArcTanh[(1 + x^(-2))^(1/4)/2^(1/4)]))
)/(4*(x^2 + x^4)^(1/4))

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IntegrateAlgebraic [A]  time = 0.25, size = 79, normalized size = 1.00 \begin {gather*} \frac {\left (x^2+x^4\right )^{3/4}}{x \left (1+x^2\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2/((-1 + x^4)*(x^2 + x^4)^(1/4)),x]

[Out]

(x^2 + x^4)^(3/4)/(x*(1 + x^2)) - ArcTan[(2^(1/4)*x)/(x^2 + x^4)^(1/4)]/(2*2^(1/4)) - ArcTanh[(2^(1/4)*x)/(x^2
+ x^4)^(1/4)]/(2*2^(1/4))

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fricas [B]  time = 1.80, size = 258, normalized size = 3.27 \begin {gather*} \frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{3} + x\right )} \arctan \left (\frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (2 \cdot 2^{\frac {3}{4}} \sqrt {x^{4} + x^{2}} x + 2^{\frac {1}{4}} {\left (3 \, x^{3} + x\right )}\right )} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{2 \, {\left (x^{3} - x\right )}}\right ) - 2^{\frac {3}{4}} {\left (x^{3} + x\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{3} + x\right )} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) + 2^{\frac {3}{4}} {\left (x^{3} + x\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{3} + x\right )} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) + 16 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{16 \, {\left (x^{3} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*2^(3/4)*(x^3 + x)*arctan(1/2*(4*2^(3/4)*(x^4 + x^2)^(1/4)*x^2 + 2^(3/4)*(2*2^(3/4)*sqrt(x^4 + x^2)*x +
2^(1/4)*(3*x^3 + x)) + 4*2^(1/4)*(x^4 + x^2)^(3/4))/(x^3 - x)) - 2^(3/4)*(x^3 + x)*log((4*sqrt(2)*(x^4 + x^2)
^(1/4)*x^2 + 2^(3/4)*(3*x^3 + x) + 4*2^(1/4)*sqrt(x^4 + x^2)*x + 4*(x^4 + x^2)^(3/4))/(x^3 - x)) + 2^(3/4)*(x^
3 + x)*log((4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 - 2^(3/4)*(3*x^3 + x) - 4*2^(1/4)*sqrt(x^4 + x^2)*x + 4*(x^4 + x^2
)^(3/4))/(x^3 - x)) + 16*(x^4 + x^2)^(3/4))/(x^3 + x)

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giac [A]  time = 0.51, size = 63, normalized size = 0.80 \begin {gather*} \frac {1}{4} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{8} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{8} \cdot 2^{\frac {3}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} \right |}\right ) + \frac {1}{{\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^4-1)/(x^4+x^2)^(1/4),x, algorithm="giac")

[Out]

1/4*2^(3/4)*arctan(1/2*2^(3/4)*(1/x^2 + 1)^(1/4)) - 1/8*2^(3/4)*log(2^(1/4) + (1/x^2 + 1)^(1/4)) + 1/8*2^(3/4)
*log(abs(-2^(1/4) + (1/x^2 + 1)^(1/4))) + 1/(1/x^2 + 1)^(1/4)

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maple [C]  time = 6.34, size = 241, normalized size = 3.05

method	result	size
risch	\(\frac {x}{\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {\sqrt {x^{4}+x^{2}}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x +2 \left (x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{3}-4 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}-\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x}{x \left (-1+x \right ) \left (1+x \right )}\right )}{8}-\frac {\RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {\sqrt {x^{4}+x^{2}}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x +2 \left (x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+3 \RootOf \left (\textit {\_Z}^{4}-8\right ) x^{3}+4 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}+\RootOf \left (\textit {\_Z}^{4}-8\right ) x}{x \left (-1+x \right ) \left (1+x \right )}\right )}{8}\)	\(241\)
trager	\(\frac {\left (x^{4}+x^{2}\right )^{\frac {3}{4}}}{x \left (x^{2}+1\right )}-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {\sqrt {x^{4}+x^{2}}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x +2 \left (x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{3}-4 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}-\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x}{x \left (-1+x \right ) \left (1+x \right )}\right )}{8}+\frac {\RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {-\sqrt {x^{4}+x^{2}}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x +2 \left (x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-3 \RootOf \left (\textit {\_Z}^{4}-8\right ) x^{3}+4 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}-\RootOf \left (\textit {\_Z}^{4}-8\right ) x}{x \left (-1+x \right ) \left (1+x \right )}\right )}{8}\)	\(250\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(x^2*(x^2+1))^(1/4)-1/8*RootOf(_Z^2+RootOf(_Z^4-8)^2)*ln(-((x^4+x^2)^(1/2)*RootOf(_Z^2+RootOf(_Z^4-8)^2)*Roo
tOf(_Z^4-8)^2*x+2*(x^4+x^2)^(1/4)*RootOf(_Z^4-8)^2*x^2-3*RootOf(_Z^2+RootOf(_Z^4-8)^2)*x^3-4*(x^4+x^2)^(3/4)-R
ootOf(_Z^2+RootOf(_Z^4-8)^2)*x)/x/(-1+x)/(1+x))-1/8*RootOf(_Z^4-8)*ln(((x^4+x^2)^(1/2)*RootOf(_Z^4-8)^3*x+2*(x
^4+x^2)^(1/4)*RootOf(_Z^4-8)^2*x^2+3*RootOf(_Z^4-8)*x^3+4*(x^4+x^2)^(3/4)+RootOf(_Z^4-8)*x)/x/(-1+x)/(1+x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {2 \, {\left (x^{3} + x\right )} x^{\frac {3}{2}}}{3 \, {\left (x^{4} - 1\right )} {\left (x^{2} + 1\right )}^{\frac {1}{4}}} - \int \frac {8 \, {\left (x^{2} + 1\right )}^{\frac {3}{4}} x^{\frac {3}{2}}}{3 \, {\left (x^{8} - 2 \, x^{4} + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^4-1)/(x^4+x^2)^(1/4),x, algorithm="maxima")

[Out]

-2/3*(x^3 + x)*x^(3/2)/((x^4 - 1)*(x^2 + 1)^(1/4)) - integrate(8/3*(x^2 + 1)^(3/4)*x^(3/2)/(x^8 - 2*x^4 + 1),
x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\sqrt [4]{x^{2} \left (x^{2} + 1\right )} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(x**4-1)/(x**4+x**2)**(1/4),x)

[Out]

Integral(x**2/((x**2*(x**2 + 1))**(1/4)*(x - 1)*(x + 1)*(x**2 + 1)), x)

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