3.10.81 \(\int \frac {3-9 x^4+2 x^6}{x (1+x^2)^2 (-1+2 x^2) \sqrt {\frac {1-2 x^2}{1+2 x^2}} (1+2 x^2)} \, dx\)

Optimal. Leaf size=74 \[ 3 \tanh ^{-1}\left (\sqrt {\frac {1-2 x^2}{2 x^2+1}}\right )-\frac {2 \sqrt {\frac {1-2 x^2}{2 x^2+1}} \left (2 x^4-x^2-1\right )}{3 \left (2 x^4+x^2-1\right )} \]

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Rubi [A]  time = 3.70, antiderivative size = 129, normalized size of antiderivative = 1.74, number of steps used = 17, number of rules used = 12, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {6688, 6719, 1586, 1606, 37, 96, 92, 206, 103, 21, 93, 203} \begin {gather*} -\frac {4}{3 \left (x^2+1\right ) \sqrt {\frac {1-2 x^2}{2 x^2+1}}}+\frac {2}{3 \sqrt {\frac {1-2 x^2}{2 x^2+1}}}+\frac {3 \sqrt {1-2 x^2} \tanh ^{-1}\left (\sqrt {1-2 x^2} \sqrt {2 x^2+1}\right )}{2 \sqrt {\frac {1-2 x^2}{2 x^2+1}} \sqrt {2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 - 9*x^4 + 2*x^6)/(x*(1 + x^2)^2*(-1 + 2*x^2)*Sqrt[(1 - 2*x^2)/(1 + 2*x^2)]*(1 + 2*x^2)),x]

[Out]

2/(3*Sqrt[(1 - 2*x^2)/(1 + 2*x^2)]) - 4/(3*(1 + x^2)*Sqrt[(1 - 2*x^2)/(1 + 2*x^2)]) + (3*Sqrt[1 - 2*x^2]*ArcTa
nh[Sqrt[1 - 2*x^2]*Sqrt[1 + 2*x^2]])/(2*Sqrt[(1 - 2*x^2)/(1 + 2*x^2)]*Sqrt[1 + 2*x^2])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1606

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.)*((g_.) + (h_.)*(x_)
)^(q_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[
{a, b, c, d, e, f, g, h, m, n, p, q}, x] && PolyQ[Px, x] && IntegersQ[m, n]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps

\begin {align*} \int \frac {3-9 x^4+2 x^6}{x \left (1+x^2\right )^2 \left (-1+2 x^2\right ) \sqrt {\frac {1-2 x^2}{1+2 x^2}} \left (1+2 x^2\right )} \, dx &=\int \frac {-3+9 x^4-2 x^6}{x \left (1+x^2\right )^2 \sqrt {\frac {1-2 x^2}{1+2 x^2}} \left (1-4 x^4\right )} \, dx\\ &=\frac {\sqrt {1-2 x^2} \int \frac {\sqrt {1+2 x^2} \left (-3+9 x^4-2 x^6\right )}{x \sqrt {1-2 x^2} \left (1+x^2\right )^2 \left (1-4 x^4\right )} \, dx}{\sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}\\ &=\frac {\sqrt {1-2 x^2} \int \frac {-3+9 x^4-2 x^6}{x \left (1-2 x^2\right )^{3/2} \left (1+x^2\right )^2 \sqrt {1+2 x^2}} \, dx}{\sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}\\ &=\frac {\sqrt {1-2 x^2} \operatorname {Subst}\left (\int \frac {-3+9 x^2-2 x^3}{(1-2 x)^{3/2} x (1+x)^2 \sqrt {1+2 x}} \, dx,x,x^2\right )}{2 \sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}\\ &=\frac {\sqrt {1-2 x^2} \operatorname {Subst}\left (\int \left (-\frac {2}{(1-2 x)^{3/2} \sqrt {1+2 x}}-\frac {3}{(1-2 x)^{3/2} x \sqrt {1+2 x}}-\frac {8}{(1-2 x)^{3/2} (1+x)^2 \sqrt {1+2 x}}+\frac {16}{(1-2 x)^{3/2} (1+x) \sqrt {1+2 x}}\right ) \, dx,x,x^2\right )}{2 \sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}\\ &=-\frac {\sqrt {1-2 x^2} \operatorname {Subst}\left (\int \frac {1}{(1-2 x)^{3/2} \sqrt {1+2 x}} \, dx,x,x^2\right )}{\sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}-\frac {\left (3 \sqrt {1-2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{(1-2 x)^{3/2} x \sqrt {1+2 x}} \, dx,x,x^2\right )}{2 \sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}-\frac {\left (4 \sqrt {1-2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{(1-2 x)^{3/2} (1+x)^2 \sqrt {1+2 x}} \, dx,x,x^2\right )}{\sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}+\frac {\left (8 \sqrt {1-2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{(1-2 x)^{3/2} (1+x) \sqrt {1+2 x}} \, dx,x,x^2\right )}{\sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}\\ &=\frac {2}{3 \sqrt {\frac {1-2 x^2}{1+2 x^2}}}-\frac {4}{3 \left (1+x^2\right ) \sqrt {\frac {1-2 x^2}{1+2 x^2}}}-\frac {\left (4 \sqrt {1-2 x^2}\right ) \operatorname {Subst}\left (\int \frac {2-4 x}{(1-2 x)^{3/2} (1+x) \sqrt {1+2 x}} \, dx,x,x^2\right )}{3 \sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}-\frac {\left (3 \sqrt {1-2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-2 x} x \sqrt {1+2 x}} \, dx,x,x^2\right )}{2 \sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}+\frac {\left (8 \sqrt {1-2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-2 x} (1+x) \sqrt {1+2 x}} \, dx,x,x^2\right )}{3 \sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}\\ &=\frac {2}{3 \sqrt {\frac {1-2 x^2}{1+2 x^2}}}-\frac {4}{3 \left (1+x^2\right ) \sqrt {\frac {1-2 x^2}{1+2 x^2}}}-\frac {\left (8 \sqrt {1-2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-2 x} (1+x) \sqrt {1+2 x}} \, dx,x,x^2\right )}{3 \sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}+\frac {\left (3 \sqrt {1-2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{2-2 x^2} \, dx,x,\sqrt {1-2 x^2} \sqrt {1+2 x^2}\right )}{\sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}+\frac {\left (16 \sqrt {1-2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+3 x^2} \, dx,x,\frac {\sqrt {1+2 x^2}}{\sqrt {1-2 x^2}}\right )}{3 \sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}\\ &=\frac {2}{3 \sqrt {\frac {1-2 x^2}{1+2 x^2}}}-\frac {4}{3 \left (1+x^2\right ) \sqrt {\frac {1-2 x^2}{1+2 x^2}}}+\frac {16 \sqrt {1-2 x^2} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt {1+2 x^2}}{\sqrt {1-2 x^2}}\right )}{3 \sqrt {3} \sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}+\frac {3 \sqrt {1-2 x^2} \tanh ^{-1}\left (\sqrt {1-2 x^2} \sqrt {1+2 x^2}\right )}{2 \sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}-\frac {\left (16 \sqrt {1-2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+3 x^2} \, dx,x,\frac {\sqrt {1+2 x^2}}{\sqrt {1-2 x^2}}\right )}{3 \sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}\\ &=\frac {2}{3 \sqrt {\frac {1-2 x^2}{1+2 x^2}}}-\frac {4}{3 \left (1+x^2\right ) \sqrt {\frac {1-2 x^2}{1+2 x^2}}}+\frac {3 \sqrt {1-2 x^2} \tanh ^{-1}\left (\sqrt {1-2 x^2} \sqrt {1+2 x^2}\right )}{2 \sqrt {\frac {1-2 x^2}{1+2 x^2}} \sqrt {1+2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 80, normalized size = 1.08 \begin {gather*} \frac {8 x^4-4 x^2+9 \sqrt {4-\frac {1}{x^4}} \left (x^2+1\right ) x^2 \sin ^{-1}\left (\frac {1}{2 x^2}\right )-4}{6 \sqrt {\frac {1-2 x^2}{2 x^2+1}} \left (2 x^4+3 x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - 9*x^4 + 2*x^6)/(x*(1 + x^2)^2*(-1 + 2*x^2)*Sqrt[(1 - 2*x^2)/(1 + 2*x^2)]*(1 + 2*x^2)),x]

[Out]

(-4 - 4*x^2 + 8*x^4 + 9*Sqrt[4 - x^(-4)]*x^2*(1 + x^2)*ArcSin[1/(2*x^2)])/(6*Sqrt[(1 - 2*x^2)/(1 + 2*x^2)]*(1
+ 3*x^2 + 2*x^4))

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IntegrateAlgebraic [A]  time = 0.15, size = 74, normalized size = 1.00 \begin {gather*} -\frac {2 \sqrt {\frac {1-2 x^2}{1+2 x^2}} \left (-1-x^2+2 x^4\right )}{3 \left (-1+x^2+2 x^4\right )}+3 \tanh ^{-1}\left (\sqrt {\frac {1-2 x^2}{1+2 x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(3 - 9*x^4 + 2*x^6)/(x*(1 + x^2)^2*(-1 + 2*x^2)*Sqrt[(1 - 2*x^2)/(1 + 2*x^2)]*(1 + 2*x^2)),
x]

[Out]

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

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fricas [A]  time = 0.59, size = 107, normalized size = 1.45 \begin {gather*} -\frac {8 \, x^{4} + 4 \, x^{2} + 9 \, {\left (2 \, x^{4} + x^{2} - 1\right )} \log \left (\frac {{\left (2 \, x^{2} + 1\right )} \sqrt {-\frac {2 \, x^{2} - 1}{2 \, x^{2} + 1}} - 1}{x^{2}}\right ) + 4 \, {\left (2 \, x^{4} - x^{2} - 1\right )} \sqrt {-\frac {2 \, x^{2} - 1}{2 \, x^{2} + 1}} - 4}{6 \, {\left (2 \, x^{4} + x^{2} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^6-9*x^4+3)/x/(x^2+1)^2/(2*x^2-1)/((-2*x^2+1)/(2*x^2+1))^(1/2)/(2*x^2+1),x, algorithm="fricas")

[Out]

-1/6*(8*x^4 + 4*x^2 + 9*(2*x^4 + x^2 - 1)*log(((2*x^2 + 1)*sqrt(-(2*x^2 - 1)/(2*x^2 + 1)) - 1)/x^2) + 4*(2*x^4
- x^2 - 1)*sqrt(-(2*x^2 - 1)/(2*x^2 + 1)) - 4)/(2*x^4 + x^2 - 1)

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giac [A]  time = 0.55, size = 44, normalized size = 0.59 \begin {gather*} -\frac {32}{3 \, {\left (\frac {{\left (\sqrt {-4 \, x^{4} + 1} - 1\right )}^{3}}{x^{6}} + 8\right )}} - \frac {3}{2} \, \log \left (-\frac {\sqrt {-4 \, x^{4} + 1} - 1}{2 \, x^{2}}\right ) + \frac {2}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^6-9*x^4+3)/x/(x^2+1)^2/(2*x^2-1)/((-2*x^2+1)/(2*x^2+1))^(1/2)/(2*x^2+1),x, algorithm="giac")

[Out]

-32/3/((sqrt(-4*x^4 + 1) - 1)^3/x^6 + 8) - 3/2*log(-1/2*(sqrt(-4*x^4 + 1) - 1)/x^2) + 2/3

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maple [A]  time = 0.30, size = 102, normalized size = 1.38

method	result	size
trager	\(-\frac {2 \left (x^{2}-1\right ) \left (2 x^{2}+1\right ) \sqrt {-\frac {2 x^{2}-1}{2 x^{2}+1}}}{3 \left (2 x^{4}+x^{2}-1\right )}+\frac {3 \ln \left (\frac {2 \sqrt {-\frac {2 x^{2}-1}{2 x^{2}+1}}\, x^{2}+\sqrt {-\frac {2 x^{2}-1}{2 x^{2}+1}}+1}{x^{2}}\right )}{2}\)	\(102\)
default	\(-\frac {64 \sqrt {-4 x^{4}+1}\, x^{6}+16 \left (-4 x^{4}+1\right )^{\frac {3}{2}} x^{2}-32 \sqrt {-4 x^{4}+1}\, x^{4}+162 \arctanh \left (\frac {1}{\sqrt {-4 x^{4}+1}}\right ) x^{4}-8 \left (-4 x^{4}+1\right )^{\frac {3}{2}}-52 x^{2} \sqrt {-4 x^{4}+1}+81 \arctanh \left (\frac {1}{\sqrt {-4 x^{4}+1}}\right ) x^{2}+44 \sqrt {-4 x^{4}+1}-81 \arctanh \left (\frac {1}{\sqrt {-4 x^{4}+1}}\right )}{54 \left (x^{2}+1\right ) \sqrt {-\left (2 x^{2}+1\right ) \left (2 x^{2}-1\right )}\, \sqrt {-\frac {2 x^{2}-1}{2 x^{2}+1}}}\)	\(169\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^6-9*x^4+3)/x/(x^2+1)^2/(2*x^2-1)/((-2*x^2+1)/(2*x^2+1))^(1/2)/(2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

-2/3*(x^2-1)*(2*x^2+1)/(2*x^4+x^2-1)*(-(2*x^2-1)/(2*x^2+1))^(1/2)+3/2*ln((2*(-(2*x^2-1)/(2*x^2+1))^(1/2)*x^2+(
-(2*x^2-1)/(2*x^2+1))^(1/2)+1)/x^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{6} - 9 \, x^{4} + 3}{{\left (2 \, x^{2} + 1\right )} {\left (2 \, x^{2} - 1\right )} {\left (x^{2} + 1\right )}^{2} x \sqrt {-\frac {2 \, x^{2} - 1}{2 \, x^{2} + 1}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^6-9*x^4+3)/x/(x^2+1)^2/(2*x^2-1)/((-2*x^2+1)/(2*x^2+1))^(1/2)/(2*x^2+1),x, algorithm="maxima")

[Out]

integrate((2*x^6 - 9*x^4 + 3)/((2*x^2 + 1)*(2*x^2 - 1)*(x^2 + 1)^2*x*sqrt(-(2*x^2 - 1)/(2*x^2 + 1))), x)

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mupad [B]  time = 1.42, size = 270, normalized size = 3.65 \begin {gather*} 3\,\mathrm {atanh}\left (\sqrt {-\frac {2\,x^2-1}{2\,x^2+1}}\right )-\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}\,\sqrt {-\frac {2\,x^2-1}{2\,x^2+1}}}{3}\right )+\frac {2\,\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}\,\sqrt {1-2\,x^2}\,\sqrt {\frac {1}{2\,x^2+1}}}{3}\right )}{2\,x^2+2}-\frac {\left (x^2+\frac {1}{2}\right )\,\left (\frac {x^2}{3}-\frac {1}{3}\right )\,\sqrt {-\frac {2\,x^2-1}{2\,x^2+1}}}{2\,x^4+x^2-1}+\frac {3\,x^2}{\sqrt {1-2\,x^2}\,\left (2\,x^2+2\right )\,\sqrt {\frac {1}{2\,x^2+1}}}+\frac {2\,\sqrt {3}\,x^2\,\mathrm {atan}\left (\frac {\sqrt {3}\,\sqrt {1-2\,x^2}\,\sqrt {\frac {1}{2\,x^2+1}}}{3}\right )}{2\,x^2+2}-\frac {2{}\mathrm {i}}{\sqrt {-\frac {2\,x^2-1}{2\,x^2+1}}\,3{}\mathrm {i}+{\left (-\frac {2\,x^2-1}{2\,x^2+1}\right )}^{3/2}\,1{}\mathrm {i}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^6 - 9*x^4 + 3)/(x*(x^2 + 1)^2*(2*x^2 - 1)*(2*x^2 + 1)*(-(2*x^2 - 1)/(2*x^2 + 1))^(1/2)),x)

[Out]

3*atanh((-(2*x^2 - 1)/(2*x^2 + 1))^(1/2)) - 2i/((-(2*x^2 - 1)/(2*x^2 + 1))^(1/2)*3i + (-(2*x^2 - 1)/(2*x^2 + 1
))^(3/2)*1i) - 3^(1/2)*atan((3^(1/2)*(-(2*x^2 - 1)/(2*x^2 + 1))^(1/2))/3) + (2*3^(1/2)*atan((3^(1/2)*(1 - 2*x^
2)^(1/2)*(1/(2*x^2 + 1))^(1/2))/3))/(2*x^2 + 2) - ((x^2 + 1/2)*(x^2/3 - 1/3)*(-(2*x^2 - 1)/(2*x^2 + 1))^(1/2))
/(x^2 + 2*x^4 - 1) + (3*x^2)/((1 - 2*x^2)^(1/2)*(2*x^2 + 2)*(1/(2*x^2 + 1))^(1/2)) + (2*3^(1/2)*x^2*atan((3^(1
/2)*(1 - 2*x^2)^(1/2)*(1/(2*x^2 + 1))^(1/2))/3))/(2*x^2 + 2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**6-9*x**4+3)/x/(x**2+1)**2/(2*x**2-1)/((-2*x**2+1)/(2*x**2+1))**(1/2)/(2*x**2+1),x)

[Out]

Timed out

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