Integrand size = 47, antiderivative size = 125 \[ \int \frac {-1+k^{3/2} x^3}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^{3/2} x^3\right )} \, dx=-\frac {4 \arctan \left (\frac {\sqrt {1+k+k^2} x}{1-\sqrt {k} x+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{3 \sqrt {1+k+k^2}}-\frac {2 \arctan \left (\frac {(-1+k) x}{1+2 \sqrt {k} x+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{3 (-1+k)} \]
-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)-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)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 16.25 (sec) , antiderivative size = 558, normalized size of antiderivative = 4.46 \[ \int \frac {-1+k^{3/2} x^3}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^{3/2} x^3\right )} \, dx=\frac {(-1-i) \sqrt {-1+x^2} \sqrt {-1+k^2 x^2} \left (-\frac {(1-i) \arctan \left (\frac {\sqrt {-1+k^2 x^2}}{\sqrt {k} \sqrt {-1+x^2}}\right )}{-1+k}+\frac {\left (-1-i \sqrt {3}+k-i \sqrt {3} k\right ) \arctan \left (\frac {(1+i) \sqrt {1+k+k^2} \sqrt {-1+k^2 x^2}}{\sqrt {k} \sqrt {-\sqrt {2+2 i \sqrt {3}}-4 i k+\left (-i+\sqrt {3}\right ) k^2} \sqrt {-1+x^2}}\right )}{\sqrt {1+k+k^2} \sqrt {-\sqrt {2+2 i \sqrt {3}}-4 i k+\left (-i+\sqrt {3}\right ) k^2}}+\frac {\left (1-i \sqrt {3}+\left (-1-i \sqrt {3}\right ) k\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {1+k+k^2} \sqrt {-1+k^2 x^2}}{\sqrt {k} \sqrt {-\sqrt {2-2 i \sqrt {3}}+4 i k+\left (i+\sqrt {3}\right ) k^2} \sqrt {-1+x^2}}\right )}{\sqrt {1+k+k^2} \sqrt {-\sqrt {2-2 i \sqrt {3}}+4 i k+\left (i+\sqrt {3}\right ) k^2}}\right )+3 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \operatorname {EllipticF}\left (\arcsin (x),k^2\right )-2 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \operatorname {EllipticPi}\left (k,\arcsin (x),k^2\right )-2 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \operatorname {EllipticPi}\left (-\sqrt [3]{-1} k,\arcsin (x),k^2\right )-2 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \operatorname {EllipticPi}\left (\frac {1}{2} i \left (i+\sqrt {3}\right ) k,\arcsin (x),k^2\right )}{3 \sqrt {\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}} \]
((-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]*Sqr t[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]*Ellip ticPi[-((-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)])
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 2.54 (sec) , antiderivative size = 709, normalized size of antiderivative = 5.67, 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 |
| \(\displaystyle -\frac {\arctan \left (\frac {(1-k) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{3 (1-k)}-\frac {2 \arctan \left (\frac {\sqrt {k^2+k+1} x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{3 \sqrt {k^2+k+1}}+\frac {\arctan \left (\frac {(1-k) \left (k x^2+1\right )}{2 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{3 (1-k)}-\frac {\left (1-\sqrt [3]{-1}\right ) \left (k x^2+1\right ) \sqrt {\frac {k^2 x^4-\left (k^2+1\right ) x^2+1}{\left (k x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(k+1)^2}{4 k}\right )}{3 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}-\frac {\left (k x^2+1\right ) \sqrt {\frac {k^2 x^4-\left (k^2+1\right ) x^2+1}{\left (k x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(k+1)^2}{4 k}\right )}{3 \left (1-\sqrt [3]{-1}\right ) \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}+\frac {\left (k x^2+1\right ) \sqrt {\frac {k^2 x^4-\left (k^2+1\right ) x^2+1}{\left (k x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(k+1)^2}{4 k}\right )}{3 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}-\frac {(-1)^{2/3} \sqrt {2} \text {arctanh}\left (\frac {-\left (\sqrt [3]{-1} \left (k^2+1\right )+2 k\right ) k x^2+k^2+2 \sqrt [3]{-1} k+1}{\sqrt {2} \sqrt {k} \sqrt {\left (1+i \sqrt {3}\right ) \left (k^2+k+1\right )} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{3 \sqrt {\left (1+i \sqrt {3}\right ) \left (k^2+k+1\right )}}+\frac {\sqrt [3]{-1} \sqrt {2} \text {arctanh}\left (\frac {-\left (2 k-(-1)^{2/3} \left (k^2+1\right )\right ) k x^2+k^2-2 (-1)^{2/3} k+1}{\sqrt {2} \sqrt {k} \sqrt {\left (1-i \sqrt {3}\right ) \left (k^2+k+1\right )} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{3 \sqrt {\left (1-i \sqrt {3}\right ) \left (k^2+k+1\right )}}\) |
-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])
3.19.37.3.1 Defintions of rubi rules used
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]
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]
Time = 1.13 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.74
| method | result | size |
| pseudoelliptic | \(-\frac {2 \sqrt {-\left (-1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-k^{2}-k -1}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+k^{\frac {3}{2}} x^{2}+\sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1-\sqrt {k}\, x +k \,x^{2}}\right )+\sqrt {-k^{2}-k -1}\, \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-2 k^{\frac {3}{2}} x^{2}-2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1+2 \sqrt {k}\, x +k \,x^{2}}\right )+2 \ln \left (2\right ) \left (\sqrt {-\left (-1+k \right )^{2}}+\frac {\sqrt {-k^{2}-k -1}}{2}\right )}{3 \sqrt {-k^{2}-k -1}\, \sqrt {-\left (-1+k \right )^{2}}}\) | \(217\) |
| elliptic | \(\text {Expression too large to display}\) | \(1050\) |
| default | \(\text {Expression too large to display}\) | \(1400\) |
-1/3*(2*(-(-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))+(-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))+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)
Time = 0.54 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.70 \[ \int \frac {-1+k^{3/2} x^3}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^{3/2} x^3\right )} \, dx=-\frac {2 \, \sqrt {k^{2} + k + 1} {\left (k - 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1} \sqrt {k^{2} + k + 1} {\left ({\left (k^{2} + 2 \, k + 1\right )} x + {\left (k x^{2} + 1\right )} \sqrt {k}\right )}}{k^{3} x^{4} - {\left (k^{4} + 4 \, k^{3} + 4 \, k^{2} + 4 \, k + 1\right )} x^{2} + k}\right ) - {\left (k^{2} + k + 1\right )} \arctan \left (-\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1} {\left ({\left (k^{3} + k^{2} - k - 1\right )} x - 2 \, {\left ({\left (k^{2} - k\right )} x^{2} + k - 1\right )} \sqrt {k}\right )}}{4 \, k^{3} x^{4} - {\left (k^{4} + 4 \, k^{3} - 2 \, k^{2} + 4 \, k + 1\right )} x^{2} + 4 \, k}\right )}{3 \, {\left (k^{3} - 1\right )}} \]
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")
-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)
\[ \int \frac {-1+k^{3/2} x^3}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^{3/2} x^3\right )} \, dx=\int \frac {\left (\sqrt {k} x - 1\right ) \left (\sqrt {k} x + k x^{2} + 1\right )}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (\sqrt {k} x + 1\right ) \left (- \sqrt {k} x + k x^{2} + 1\right )}\, dx \]
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)
\[ \int \frac {-1+k^{3/2} x^3}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^{3/2} x^3\right )} \, dx=\int { \frac {k^{\frac {3}{2}} x^{3} - 1}{{\left (k^{\frac {3}{2}} x^{3} + 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}} \,d x } \]
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")
Exception generated. \[ \int \frac {-1+k^{3/2} x^3}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^{3/2} x^3\right )} \, dx=\text {Exception raised: TypeError} \]
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")
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
Timed out. \[ \int \frac {-1+k^{3/2} x^3}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^{3/2} x^3\right )} \, dx=\int \frac {k^{3/2}\,x^3-1}{\left (k^{3/2}\,x^3+1\right )\,\sqrt {\left (x^2-1\right )\,\left (k^2\,x^2-1\right )}} \,d x \]