3.27.90 \(\int \frac {(2-k^2) x-2 x^3+k^2 x^5}{((1-x^2) (1-k^2 x^2))^{2/3} (-1+d+(-2 d+k^2) x^2+d x^4)} \, dx\)

Optimal. Leaf size=243 \[ -\frac {\log \left (\sqrt [3]{d} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}+k^2 x^2-1\right )}{2 d^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}}{\sqrt [3]{d} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}-2 k^2 x^2+2}\right )}{2 d^{2/3}}+\frac {\log \left (d^{2/3} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{4/3}+\left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3} \left (\sqrt [3]{d}-\sqrt [3]{d} k^2 x^2\right )+k^4 x^4-2 k^2 x^2+1\right )}{4 d^{2/3}} \]

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Rubi [C]  time = 2.44, antiderivative size = 251, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 7, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {1594, 6715, 6719, 1586, 6728, 137, 136} \begin {gather*} \frac {3 \left (1-x^2\right )^2 \left (\frac {1-k^2 x^2}{1-k^2}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 d \left (1-x^2\right )}{k^2-\sqrt {k^4-4 d k^2+4 d}}\right )}{8 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {3 \left (1-x^2\right )^2 \left (\frac {1-k^2 x^2}{1-k^2}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 d \left (1-x^2\right )}{k^2+\sqrt {k^4-4 d k^2+4 d}}\right )}{8 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[((2 - k^2)*x - 2*x^3 + k^2*x^5)/(((1 - x^2)*(1 - k^2*x^2))^(2/3)*(-1 + d + (-2*d + k^2)*x^2 + d*x^4)),x]

[Out]

(3*(1 - x^2)^2*((1 - k^2*x^2)/(1 - k^2))^(2/3)*AppellF1[4/3, 2/3, 1, 7/3, -((k^2*(1 - x^2))/(1 - k^2)), (2*d*(
1 - x^2))/(k^2 - Sqrt[4*d - 4*d*k^2 + k^4])])/(8*((1 - x^2)*(1 - k^2*x^2))^(2/3)) + (3*(1 - x^2)^2*((1 - k^2*x
^2)/(1 - k^2))^(2/3)*AppellF1[4/3, 2/3, 1, 7/3, -((k^2*(1 - x^2))/(1 - k^2)), (2*d*(1 - x^2))/(k^2 + Sqrt[4*d
- 4*d*k^2 + k^4])])/(8*((1 - x^2)*(1 - k^2*x^2))^(2/3))

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 137

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), 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]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x^2+d x^4\right )} \, dx &=\int \frac {x \left (2-k^2-2 x^2+k^2 x^4\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x^2+d x^4\right )} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {2-k^2-2 x+k^2 x^2}{\left ((1-x) \left (1-k^2 x\right )\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x+d x^2\right )} \, dx,x,x^2\right )\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {2-k^2-2 x+k^2 x^2}{(1-x)^{2/3} \left (1-k^2 x\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x+d x^2\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{1-x} \left (2-k^2-k^2 x\right )}{\left (1-k^2 x\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x+d x^2\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \left (\frac {\left (-k^2+\sqrt {4 d-4 d k^2+k^4}\right ) \sqrt [3]{1-x}}{\left (-2 d+k^2-\sqrt {4 d-4 d k^2+k^4}+2 d x\right ) \left (1-k^2 x\right )^{2/3}}+\frac {\left (-k^2-\sqrt {4 d-4 d k^2+k^4}\right ) \sqrt [3]{1-x}}{\left (-2 d+k^2+\sqrt {4 d-4 d k^2+k^4}+2 d x\right ) \left (1-k^2 x\right )^{2/3}}\right ) \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (-k^2-\sqrt {4 d-4 d k^2+k^4}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{1-x}}{\left (-2 d+k^2+\sqrt {4 d-4 d k^2+k^4}+2 d x\right ) \left (1-k^2 x\right )^{2/3}} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (-k^2+\sqrt {4 d-4 d k^2+k^4}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{1-x}}{\left (-2 d+k^2-\sqrt {4 d-4 d k^2+k^4}+2 d x\right ) \left (1-k^2 x\right )^{2/3}} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (-k^2-\sqrt {4 d-4 d k^2+k^4}\right ) \left (1-x^2\right )^{2/3} \left (\frac {-1+k^2 x^2}{-1+k^2}\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{1-x}}{\left (-2 d+k^2+\sqrt {4 d-4 d k^2+k^4}+2 d x\right ) \left (-\frac {1}{-1+k^2}+\frac {k^2 x}{-1+k^2}\right )^{2/3}} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (-k^2+\sqrt {4 d-4 d k^2+k^4}\right ) \left (1-x^2\right )^{2/3} \left (\frac {-1+k^2 x^2}{-1+k^2}\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{1-x}}{\left (-2 d+k^2-\sqrt {4 d-4 d k^2+k^4}+2 d x\right ) \left (-\frac {1}{-1+k^2}+\frac {k^2 x}{-1+k^2}\right )^{2/3}} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {3 \left (1-x^2\right )^2 \left (\frac {1-k^2 x^2}{1-k^2}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 d \left (1-x^2\right )}{k^2-\sqrt {4 d-4 d k^2+k^4}}\right )}{8 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {3 \left (1-x^2\right )^2 \left (\frac {1-k^2 x^2}{1-k^2}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 d \left (1-x^2\right )}{k^2+\sqrt {4 d-4 d k^2+k^4}}\right )}{8 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ \end {align*}

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Mathematica [F]  time = 1.88, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x^2+d x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((2 - k^2)*x - 2*x^3 + k^2*x^5)/(((1 - x^2)*(1 - k^2*x^2))^(2/3)*(-1 + d + (-2*d + k^2)*x^2 + d*x^4)
),x]

[Out]

Integrate[((2 - k^2)*x - 2*x^3 + k^2*x^5)/(((1 - x^2)*(1 - k^2*x^2))^(2/3)*(-1 + d + (-2*d + k^2)*x^2 + d*x^4)
), x]

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IntegrateAlgebraic [A]  time = 5.58, size = 243, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}{2-2 k^2 x^2+\sqrt [3]{d} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}\right )}{2 d^{2/3}}-\frac {\log \left (-1+k^2 x^2+\sqrt [3]{d} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 d^{2/3}}+\frac {\log \left (1-2 k^2 x^2+k^4 x^4+\left (\sqrt [3]{d}-\sqrt [3]{d} k^2 x^2\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}+d^{2/3} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{4/3}\right )}{4 d^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((2 - k^2)*x - 2*x^3 + k^2*x^5)/(((1 - x^2)*(1 - k^2*x^2))^(2/3)*(-1 + d + (-2*d + k^2)*x^2
+ d*x^4)),x]

[Out]

-1/2*(Sqrt[3]*ArcTan[(Sqrt[3]*d^(1/3)*(1 + (-1 - k^2)*x^2 + k^2*x^4)^(2/3))/(2 - 2*k^2*x^2 + d^(1/3)*(1 + (-1
- k^2)*x^2 + k^2*x^4)^(2/3))])/d^(2/3) - Log[-1 + k^2*x^2 + d^(1/3)*(1 + (-1 - k^2)*x^2 + k^2*x^4)^(2/3)]/(2*d
^(2/3)) + Log[1 - 2*k^2*x^2 + k^4*x^4 + (d^(1/3) - d^(1/3)*k^2*x^2)*(1 + (-1 - k^2)*x^2 + k^2*x^4)^(2/3) + d^(
2/3)*(1 + (-1 - k^2)*x^2 + k^2*x^4)^(4/3)]/(4*d^(2/3))

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fricas [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(((-k^2+2)*x-2*x^3+k^2*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(-1+d+(k^2-2*d)*x^2+d*x^4),x, algorithm="fr
icas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {k^{2} x^{5} - 2 \, x^{3} - {\left (k^{2} - 2\right )} x}{{\left (d x^{4} + {\left (k^{2} - 2 \, d\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {2}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-k^2+2)*x-2*x^3+k^2*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(-1+d+(k^2-2*d)*x^2+d*x^4),x, algorithm="gi
ac")

[Out]

integrate((k^2*x^5 - 2*x^3 - (k^2 - 2)*x)/((d*x^4 + (k^2 - 2*d)*x^2 + d - 1)*((k^2*x^2 - 1)*(x^2 - 1))^(2/3)),
x)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (-k^{2}+2\right ) x -2 x^{3}+k^{2} x^{5}}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {2}{3}} \left (-1+d +\left (k^{2}-2 d \right ) x^{2}+d \,x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-k^2+2)*x-2*x^3+k^2*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(-1+d+(k^2-2*d)*x^2+d*x^4),x)

[Out]

int(((-k^2+2)*x-2*x^3+k^2*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(-1+d+(k^2-2*d)*x^2+d*x^4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {k^{2} x^{5} - 2 \, x^{3} - {\left (k^{2} - 2\right )} x}{{\left (d x^{4} + {\left (k^{2} - 2 \, d\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {2}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-k^2+2)*x-2*x^3+k^2*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(-1+d+(k^2-2*d)*x^2+d*x^4),x, algorithm="ma
xima")

[Out]

integrate((k^2*x^5 - 2*x^3 - (k^2 - 2)*x)/((d*x^4 + (k^2 - 2*d)*x^2 + d - 1)*((k^2*x^2 - 1)*(x^2 - 1))^(2/3)),
x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*(k^2 - 2) - k^2*x^5 + 2*x^3)/(((x^2 - 1)*(k^2*x^2 - 1))^(2/3)*(d - x^2*(2*d - k^2) + d*x^4 - 1)),x)

[Out]

-int((x*(k^2 - 2) - k^2*x^5 + 2*x^3)/(((x^2 - 1)*(k^2*x^2 - 1))^(2/3)*(d - x^2*(2*d - k^2) + d*x^4 - 1)), x)

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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(((-k**2+2)*x-2*x**3+k**2*x**5)/((-x**2+1)*(-k**2*x**2+1))**(2/3)/(-1+d+(k**2-2*d)*x**2+d*x**4),x)

[Out]

Timed out

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