3.13.48 \(\int \frac {(-3+k^2) x+2 k^2 x^3}{\sqrt [4]{(1-x^2) (1-k^2 x^2)} (-1+d+(3-d k^2) x^2-3 x^4+x^6)} \, dx\) [1248]

3.13.48.1 Optimal result

Integrand size = 64, antiderivative size = 90 \[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{-1+x^2}\right )}{d^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{-1+x^2}\right )}{d^{3/4}} \]

output

arctan(d^(1/4)*(1+(-k^2-1)*x^2+k^2*x^4)^(1/4)/(x^2-1))/d^(3/4)-arctanh(d^( 
1/4)*(1+(-k^2-1)*x^2+k^2*x^4)^(1/4)/(x^2-1))/d^(3/4)
 

3.13.48.2 Mathematica [A] (verified)

Time = 20.01 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.20 \[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=\frac {\sqrt [4]{-1+x^2} \sqrt [4]{-1+k^2 x^2} \left (\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{-1+k^2 x^2}}{\left (-1+x^2\right )^{3/4}}\right )-\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{-1+k^2 x^2}}{\left (-1+x^2\right )^{3/4}}\right )\right )}{d^{3/4} \sqrt [4]{\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}} \]

input

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

output

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

3.13.48.3 Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {2 k^2 x^3+\left (k^2-3\right ) x}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (x^2 \left (3-d k^2\right )+d+x^6-3 x^4-1\right )} \, dx\)

\(\Big \downarrow \) 2027

\(\displaystyle \int \frac {x \left (2 k^2 x^2+k^2-3\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (x^2 \left (3-d k^2\right )+d+x^6-3 x^4-1\right )}dx\)

\(\Big \downarrow \) 2048

\(\displaystyle \int \frac {x \left (2 k^2 x^2+k^2-3\right )}{\sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1} \left (x^2 \left (3-d k^2\right )+d+x^6-3 x^4-1\right )}dx\)

\(\Big \downarrow \) 7266

\(\displaystyle \frac {1}{2} \int \frac {-2 x^2 k^2-k^2+3}{\sqrt [4]{k^2 x^4-\left (k^2+1\right ) x^2+1} \left (-x^6+3 x^4-\left (3-d k^2\right ) x^2-d+1\right )}dx^2\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{2} \int \left (\frac {2 k^2 x^2}{\sqrt [4]{k^2 x^4-\left (k^2+1\right ) x^2+1} \left (x^6-3 x^4+\left (3-d k^2\right ) x^2+d-1\right )}+\frac {3 \left (1-\frac {k^2}{3}\right )}{\sqrt [4]{k^2 x^4-\left (k^2+1\right ) x^2+1} \left (-x^6+3 x^4-\left (3-d k^2\right ) x^2-d+1\right )}\right )dx^2\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (2 k^2 \int \frac {x^2}{\sqrt [4]{k^2 x^4-\left (k^2+1\right ) x^2+1} \left (x^6-3 x^4+\left (3-d k^2\right ) x^2+d-1\right )}dx^2+\left (3-k^2\right ) \int \frac {1}{\sqrt [4]{k^2 x^4-\left (k^2+1\right ) x^2+1} \left (-x^6+3 x^4-\left (3-d k^2\right ) x^2-d+1\right )}dx^2\right )\)

input

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

output

$Aborted
 

3.13.48.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2027

Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ 
(p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & 
& PosQ[s - r] &&  !(EqQ[p, 1] && EqQ[u, 1])
 

rule 2048

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))*((c_) + (d_.)*(x_)^(n_.)))^(p_) 
, x_Symbol] :> Int[u*(a*c*e + (b*c + a*d)*e*x^n + b*d*e*x^(2*n))^p, x] /; F 
reeQ[{a, b, c, d, e, n, p}, x]
 

rule 7266

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

rule 7293

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

3.13.48.4 Maple [F]

\[\int \frac {\left (k^{2}-3\right ) x +2 k^{2} x^{3}}{{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )}^{\frac {1}{4}} \left (-1+d +\left (-d \,k^{2}+3\right ) x^{2}-3 x^{4}+x^{6}\right )}d x\]

input

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

output

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

3.13.48.5 Fricas [F(-1)]

Timed out. \[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.13.48.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.13.48.7 Maxima [F]

\[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=\int { \frac {2 \, k^{2} x^{3} + {\left (k^{2} - 3\right )} x}{{\left (x^{6} - 3 \, x^{4} - {\left (d k^{2} - 3\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{4}}} \,d x } \]

input

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

output

integrate((2*k^2*x^3 + (k^2 - 3)*x)/((x^6 - 3*x^4 - (d*k^2 - 3)*x^2 + d - 
1)*((k^2*x^2 - 1)*(x^2 - 1))^(1/4)), x)
 

3.13.48.8 Giac [F]

\[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=\int { \frac {2 \, k^{2} x^{3} + {\left (k^{2} - 3\right )} x}{{\left (x^{6} - 3 \, x^{4} - {\left (d k^{2} - 3\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{4}}} \,d x } \]

input

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

output

integrate((2*k^2*x^3 + (k^2 - 3)*x)/((x^6 - 3*x^4 - (d*k^2 - 3)*x^2 + d - 
1)*((k^2*x^2 - 1)*(x^2 - 1))^(1/4)), x)
 

3.13.48.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=-\int \frac {x\,\left (k^2-3\right )+2\,k^2\,x^3}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{1/4}\,\left (-x^6+3\,x^4+\left (d\,k^2-3\right )\,x^2-d+1\right )} \,d x \]

input

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

output

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