3.19.24 \(\int \frac {1+k^{3/2} x^3}{\sqrt {(1-x^2) (1-k^2 x^2)} (-1+k^{3/2} x^3)} \, dx\) [1824]

3.19.24.1 Optimal result
3.19.24.2 Mathematica [C] (warning: unable to verify)
3.19.24.3 Rubi [C] (verified)
3.19.24.4 Maple [A] (verified)
3.19.24.5 Fricas [B] (verification not implemented)
3.19.24.6 Sympy [F]
3.19.24.7 Maxima [F]
3.19.24.8 Giac [F(-2)]
3.19.24.9 Mupad [F(-1)]

3.19.24.1 Optimal result

Integrand size = 47, antiderivative size = 124 \[ \int \frac {1+k^{3/2} x^3}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^{3/2} x^3\right )} \, dx=-\frac {2 \arctan \left (\frac {(-1+k) x}{1-2 \sqrt {k} x+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{3 (-1+k)}-\frac {4 \arctan \left (\frac {\sqrt {1+k+k^2} x}{1+\sqrt {k} x+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{3 \sqrt {1+k+k^2}} \]

output
-2*arctan((-1+k)*x/(1-2*k^(1/2)*x+k*x^2+(1+(-k^2-1)*x^2+k^2*x^4)^(1/2)))/( 
-3+3*k)-4/3*arctan((k^2+k+1)^(1/2)*x/(1+k^(1/2)*x+k*x^2+(1+(-k^2-1)*x^2+k^ 
2*x^4)^(1/2)))/(k^2+k+1)^(1/2)
 
3.19.24.2 Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 23.96 (sec) , antiderivative size = 558, normalized size of antiderivative = 4.50 \[ \int \frac {1+k^{3/2} x^3}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^{3/2} x^3\right )} \, dx=\frac {(1+i) \sqrt {-1+x^2} \sqrt {-1+k^2 x^2} \left (-\frac {(1-i) \arctan \left (\frac {\sqrt {-1+k^2 x^2}}{\sqrt {k} \sqrt {-1+x^2}}\right )}{-1+k}+\frac {\left (-1-i \sqrt {3}+k-i \sqrt {3} k\right ) \arctan \left (\frac {(1+i) \sqrt {1+k+k^2} \sqrt {-1+k^2 x^2}}{\sqrt {k} \sqrt {-\sqrt {2+2 i \sqrt {3}}-4 i k+\left (-i+\sqrt {3}\right ) k^2} \sqrt {-1+x^2}}\right )}{\sqrt {1+k+k^2} \sqrt {-\sqrt {2+2 i \sqrt {3}}-4 i k+\left (-i+\sqrt {3}\right ) k^2}}+\frac {\left (1-i \sqrt {3}+\left (-1-i \sqrt {3}\right ) k\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {1+k+k^2} \sqrt {-1+k^2 x^2}}{\sqrt {k} \sqrt {-\sqrt {2-2 i \sqrt {3}}+4 i k+\left (i+\sqrt {3}\right ) k^2} \sqrt {-1+x^2}}\right )}{\sqrt {1+k+k^2} \sqrt {-\sqrt {2-2 i \sqrt {3}}+4 i k+\left (i+\sqrt {3}\right ) k^2}}\right )+3 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \operatorname {EllipticF}\left (\arcsin (x),k^2\right )-2 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \operatorname {EllipticPi}\left (k,\arcsin (x),k^2\right )-2 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \operatorname {EllipticPi}\left (-\sqrt [3]{-1} k,\arcsin (x),k^2\right )-2 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \operatorname {EllipticPi}\left (\frac {1}{2} i \left (i+\sqrt {3}\right ) k,\arcsin (x),k^2\right )}{3 \sqrt {\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}} \]

input
Integrate[(1 + k^(3/2)*x^3)/(Sqrt[(1 - x^2)*(1 - k^2*x^2)]*(-1 + k^(3/2)*x 
^3)),x]
 
output
((1 + I)*Sqrt[-1 + x^2]*Sqrt[-1 + k^2*x^2]*(((-1 + I)*ArcTan[Sqrt[-1 + k^2 
*x^2]/(Sqrt[k]*Sqrt[-1 + x^2])])/(-1 + k) + ((-1 - I*Sqrt[3] + k - I*Sqrt[ 
3]*k)*ArcTan[((1 + I)*Sqrt[1 + k + k^2]*Sqrt[-1 + k^2*x^2])/(Sqrt[k]*Sqrt[ 
-Sqrt[2 + (2*I)*Sqrt[3]] - (4*I)*k + (-I + Sqrt[3])*k^2]*Sqrt[-1 + x^2])]) 
/(Sqrt[1 + k + k^2]*Sqrt[-Sqrt[2 + (2*I)*Sqrt[3]] - (4*I)*k + (-I + Sqrt[3 
])*k^2]) + ((1 - I*Sqrt[3] + (-1 - I*Sqrt[3])*k)*ArcTanh[((1 + I)*Sqrt[1 + 
 k + k^2]*Sqrt[-1 + k^2*x^2])/(Sqrt[k]*Sqrt[-Sqrt[2 - (2*I)*Sqrt[3]] + (4* 
I)*k + (I + Sqrt[3])*k^2]*Sqrt[-1 + x^2])])/(Sqrt[1 + k + k^2]*Sqrt[-Sqrt[ 
2 - (2*I)*Sqrt[3]] + (4*I)*k + (I + Sqrt[3])*k^2])) + 3*Sqrt[1 - x^2]*Sqrt 
[1 - k^2*x^2]*EllipticF[ArcSin[x], k^2] - 2*Sqrt[1 - x^2]*Sqrt[1 - k^2*x^2 
]*EllipticPi[k, ArcSin[x], k^2] - 2*Sqrt[1 - x^2]*Sqrt[1 - k^2*x^2]*Ellipt 
icPi[-((-1)^(1/3)*k), ArcSin[x], k^2] - 2*Sqrt[1 - x^2]*Sqrt[1 - k^2*x^2]* 
EllipticPi[(I/2)*(I + Sqrt[3])*k, ArcSin[x], k^2])/(3*Sqrt[(-1 + x^2)*(-1 
+ k^2*x^2)])
 
3.19.24.3 Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 2.52 (sec) , antiderivative size = 709, normalized size of antiderivative = 5.72, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {2048, 7276, 2009}

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.

\(\displaystyle \int \frac {k^{3/2} x^3+1}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (k^{3/2} x^3-1\right )} \, dx\)

\(\Big \downarrow \) 2048

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

\(\Big \downarrow \) 7276

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\arctan \left (\frac {(1-k) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{3 (1-k)}-\frac {2 \arctan \left (\frac {\sqrt {k^2+k+1} x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{3 \sqrt {k^2+k+1}}-\frac {\arctan \left (\frac {(1-k) \left (k x^2+1\right )}{2 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{3 (1-k)}-\frac {\left (1-\sqrt [3]{-1}\right ) \left (k x^2+1\right ) \sqrt {\frac {k^2 x^4-\left (k^2+1\right ) x^2+1}{\left (k x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(k+1)^2}{4 k}\right )}{3 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}-\frac {\left (k x^2+1\right ) \sqrt {\frac {k^2 x^4-\left (k^2+1\right ) x^2+1}{\left (k x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(k+1)^2}{4 k}\right )}{3 \left (1-\sqrt [3]{-1}\right ) \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}+\frac {\left (k x^2+1\right ) \sqrt {\frac {k^2 x^4-\left (k^2+1\right ) x^2+1}{\left (k x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(k+1)^2}{4 k}\right )}{3 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}+\frac {(-1)^{2/3} \sqrt {2} \text {arctanh}\left (\frac {-\left (\sqrt [3]{-1} \left (k^2+1\right )+2 k\right ) k x^2+k^2+2 \sqrt [3]{-1} k+1}{\sqrt {2} \sqrt {k} \sqrt {\left (1+i \sqrt {3}\right ) \left (k^2+k+1\right )} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{3 \sqrt {\left (1+i \sqrt {3}\right ) \left (k^2+k+1\right )}}-\frac {\sqrt [3]{-1} \sqrt {2} \text {arctanh}\left (\frac {-\left (2 k-(-1)^{2/3} \left (k^2+1\right )\right ) k x^2+k^2-2 (-1)^{2/3} k+1}{\sqrt {2} \sqrt {k} \sqrt {\left (1-i \sqrt {3}\right ) \left (k^2+k+1\right )} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{3 \sqrt {\left (1-i \sqrt {3}\right ) \left (k^2+k+1\right )}}\)

input
Int[(1 + k^(3/2)*x^3)/(Sqrt[(1 - x^2)*(1 - k^2*x^2)]*(-1 + k^(3/2)*x^3)),x 
]
 
output
-1/3*ArcTan[((1 - k)*x)/Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4]]/(1 - k) - (2*Ar 
cTan[(Sqrt[1 + k + k^2]*x)/Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4]])/(3*Sqrt[1 + 
 k + k^2]) - ArcTan[((1 - k)*(1 + k*x^2))/(2*Sqrt[k]*Sqrt[1 - (1 + k^2)*x^ 
2 + k^2*x^4])]/(3*(1 - k)) + ((-1)^(2/3)*Sqrt[2]*ArcTanh[(1 + 2*(-1)^(1/3) 
*k + k^2 - k*(2*k + (-1)^(1/3)*(1 + k^2))*x^2)/(Sqrt[2]*Sqrt[k]*Sqrt[(1 + 
I*Sqrt[3])*(1 + k + k^2)]*Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4])])/(3*Sqrt[(1 
+ I*Sqrt[3])*(1 + k + k^2)]) - ((-1)^(1/3)*Sqrt[2]*ArcTanh[(1 - 2*(-1)^(2/ 
3)*k + k^2 - k*(2*k - (-1)^(2/3)*(1 + k^2))*x^2)/(Sqrt[2]*Sqrt[k]*Sqrt[(1 
- I*Sqrt[3])*(1 + k + k^2)]*Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4])])/(3*Sqrt[( 
1 - I*Sqrt[3])*(1 + k + k^2)]) + ((1 + k*x^2)*Sqrt[(1 - (1 + k^2)*x^2 + k^ 
2*x^4)/(1 + k*x^2)^2]*EllipticF[2*ArcTan[Sqrt[k]*x], (1 + k)^2/(4*k)])/(3* 
Sqrt[k]*Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4]) - ((1 + k*x^2)*Sqrt[(1 - (1 + k 
^2)*x^2 + k^2*x^4)/(1 + k*x^2)^2]*EllipticF[2*ArcTan[Sqrt[k]*x], (1 + k)^2 
/(4*k)])/(3*(1 - (-1)^(1/3))*Sqrt[k]*Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4]) - 
((1 - (-1)^(1/3))*(1 + k*x^2)*Sqrt[(1 - (1 + k^2)*x^2 + k^2*x^4)/(1 + k*x^ 
2)^2]*EllipticF[2*ArcTan[Sqrt[k]*x], (1 + k)^2/(4*k)])/(3*Sqrt[k]*Sqrt[1 - 
 (1 + k^2)*x^2 + k^2*x^4])
 

3.19.24.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 7276
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 
3.19.24.4 Maple [A] (verified)

Time = 1.17 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.76

method result size
pseudoelliptic \(-\frac {2 \left (\sqrt {-\left (-1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-k^{2}-k -1}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-k^{\frac {3}{2}} x^{2}-\sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1+\sqrt {k}\, x +k \,x^{2}}\right )+\frac {\sqrt {-k^{2}-k -1}\, \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+2 k^{\frac {3}{2}} x^{2}+2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1-2 \sqrt {k}\, x +k \,x^{2}}\right )}{2}+\ln \left (2\right ) \left (\sqrt {-\left (-1+k \right )^{2}}+\frac {\sqrt {-k^{2}-k -1}}{2}\right )\right )}{3 \sqrt {-k^{2}-k -1}\, \sqrt {-\left (-1+k \right )^{2}}}\) \(218\)
elliptic \(\text {Expression too large to display}\) \(1045\)
default \(\text {Expression too large to display}\) \(1398\)

input
int((1+k^(3/2)*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(-1+k^(3/2)*x^3),x,metho 
d=_RETURNVERBOSE)
 
output
-2/3*((-(-1+k)^2)^(1/2)*ln(((-k^2-k-1)^(1/2)*((x^2-1)*(k^2*x^2-1))^(1/2)-k 
^(3/2)*x^2-k^(1/2)+(-k^2-2*k-1)*x)/(1+k^(1/2)*x+k*x^2))+1/2*(-k^2-k-1)^(1/ 
2)*ln(((-(-1+k)^2)^(1/2)*((x^2-1)*(k^2*x^2-1))^(1/2)+2*k^(3/2)*x^2+2*k^(1/ 
2)+(-k^2-2*k-1)*x)/(1-2*k^(1/2)*x+k*x^2))+ln(2)*((-(-1+k)^2)^(1/2)+1/2*(-k 
^2-k-1)^(1/2)))/(-k^2-k-1)^(1/2)/(-(-1+k)^2)^(1/2)
 
3.19.24.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (106) = 212\).

Time = 0.55 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.72 \[ \int \frac {1+k^{3/2} x^3}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^{3/2} x^3\right )} \, dx=\frac {2 \, \sqrt {k^{2} + k + 1} {\left (k - 1\right )} \arctan \left (-\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1} \sqrt {k^{2} + k + 1} {\left ({\left (k^{2} + 2 \, k + 1\right )} x - {\left (k x^{2} + 1\right )} \sqrt {k}\right )}}{k^{3} x^{4} - {\left (k^{4} + 4 \, k^{3} + 4 \, k^{2} + 4 \, k + 1\right )} x^{2} + k}\right ) - {\left (k^{2} + k + 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1} {\left ({\left (k^{3} + k^{2} - k - 1\right )} x + 2 \, {\left ({\left (k^{2} - k\right )} x^{2} + k - 1\right )} \sqrt {k}\right )}}{4 \, k^{3} x^{4} - {\left (k^{4} + 4 \, k^{3} - 2 \, k^{2} + 4 \, k + 1\right )} x^{2} + 4 \, k}\right )}{3 \, {\left (k^{3} - 1\right )}} \]

input
integrate((1+k^(3/2)*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(-1+k^(3/2)*x^3),x 
, algorithm="fricas")
 
output
1/3*(2*sqrt(k^2 + k + 1)*(k - 1)*arctan(-sqrt(k^2*x^4 - (k^2 + 1)*x^2 + 1) 
*sqrt(k^2 + k + 1)*((k^2 + 2*k + 1)*x - (k*x^2 + 1)*sqrt(k))/(k^3*x^4 - (k 
^4 + 4*k^3 + 4*k^2 + 4*k + 1)*x^2 + k)) - (k^2 + k + 1)*arctan(sqrt(k^2*x^ 
4 - (k^2 + 1)*x^2 + 1)*((k^3 + k^2 - k - 1)*x + 2*((k^2 - k)*x^2 + k - 1)* 
sqrt(k))/(4*k^3*x^4 - (k^4 + 4*k^3 - 2*k^2 + 4*k + 1)*x^2 + 4*k)))/(k^3 - 
1)
 
3.19.24.6 Sympy [F]

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

input
integrate((1+k**(3/2)*x**3)/((-x**2+1)*(-k**2*x**2+1))**(1/2)/(-1+k**(3/2) 
*x**3),x)
 
output
Integral((sqrt(k)*x + 1)*(-sqrt(k)*x + k*x**2 + 1)/(sqrt((x - 1)*(x + 1)*( 
k*x - 1)*(k*x + 1))*(sqrt(k)*x - 1)*(sqrt(k)*x + k*x**2 + 1)), x)
 
3.19.24.7 Maxima [F]

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

input
integrate((1+k^(3/2)*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(-1+k^(3/2)*x^3),x 
, algorithm="maxima")
 
output
integrate((k^(3/2)*x^3 + 1)/((k^(3/2)*x^3 - 1)*sqrt((k^2*x^2 - 1)*(x^2 - 1 
))), x)
 
3.19.24.8 Giac [F(-2)]

Exception generated. \[ \int \frac {1+k^{3/2} x^3}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^{3/2} x^3\right )} \, dx=\text {Exception raised: TypeError} \]

input
integrate((1+k^(3/2)*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(-1+k^(3/2)*x^3),x 
, algorithm="giac")
 
output
Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 
3.19.24.9 Mupad [F(-1)]

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

input
int((k^(3/2)*x^3 + 1)/((k^(3/2)*x^3 - 1)*((x^2 - 1)*(k^2*x^2 - 1))^(1/2)), 
x)
 
output
int((k^(3/2)*x^3 + 1)/((k^(3/2)*x^3 - 1)*((x^2 - 1)*(k^2*x^2 - 1))^(1/2)), 
 x)