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

Optimal. Leaf size=190 \[ -\frac {\log \left (d^{2/3} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}+\left (\sqrt [3]{d} x+\sqrt [3]{d}\right ) \sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}+x^2+2 x+1\right )}{2 \sqrt [3]{d}}+\frac {\log \left (-\sqrt [3]{d} \sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}+x+1\right )}{\sqrt [3]{d}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x+\sqrt {3}}{2 \sqrt [3]{d} \sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}+x+1}\right )}{\sqrt [3]{d}} \]

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Rubi [F]  time = 8.09, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-3-2 \left (1+k^2\right ) x+\left (1+k^2\right ) x^2+4 k^2 x^3+k^2 x^4}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

((5*d + k^(-2))*x*(1 - x^2)^(2/3)*(1 - k^2*x^2)^(2/3)*AppellF1[1/2, 2/3, 2/3, 3/2, x^2, k^2*x^2])/(d^2*((1 - x
^2)*(1 - k^2*x^2))^(2/3)) - (3^(3/4)*Sqrt[2 + Sqrt[3]]*Sqrt[(-1 - k^2 + 2*k^2*x^2)^2]*((-1 + k^2)^(2/3) + 2^(2
/3)*k^(2/3)*((1 - x^2)*(1 - k^2*x^2))^(1/3))*Sqrt[((-1 + k^2)^(4/3) - 2^(2/3)*k^(2/3)*(-1 + k^2)^(2/3)*((1 - x
^2)*(1 - k^2*x^2))^(1/3) + 2*2^(1/3)*k^(4/3)*((1 - x^2)*(1 - k^2*x^2))^(2/3))/((1 + Sqrt[3])*(-1 + k^2)^(2/3)
+ 2^(2/3)*k^(2/3)*((1 - x^2)*(1 - k^2*x^2))^(1/3))^2]*EllipticF[ArcSin[((1 - Sqrt[3])*(-1 + k^2)^(2/3) + 2^(2/
3)*k^(2/3)*((1 - x^2)*(1 - k^2*x^2))^(1/3))/((1 + Sqrt[3])*(-1 + k^2)^(2/3) + 2^(2/3)*k^(2/3)*((1 - x^2)*(1 -
k^2*x^2))^(1/3))], -7 - 4*Sqrt[3]])/(2^(2/3)*d*k^(2/3)*(1 + k^2 - 2*k^2*x^2)*Sqrt[(-1 - k^2*(1 - 2*x^2))^2]*Sq
rt[((-1 + k^2)^(2/3)*((-1 + k^2)^(2/3) + 2^(2/3)*k^(2/3)*((1 - x^2)*(1 - k^2*x^2))^(1/3)))/((1 + Sqrt[3])*(-1
+ k^2)^(2/3) + 2^(2/3)*k^(2/3)*((1 - x^2)*(1 - k^2*x^2))^(1/3))^2]) - ((1 - d + 5*d*k^2 - 8*d^2*k^2)*(1 - x^2)
^(2/3)*(1 - k^2*x^2)^(2/3)*Defer[Int][1/((1 - x^2)^(2/3)*(1 - k^2*x^2)^(2/3)*(1 - d + (2 + d)*x + (1 + d*k^2)*
x^2 - d*k^2*x^3)), x])/(d^2*k^2*((1 - x^2)*(1 - k^2*x^2))^(2/3)) - ((2 + d + 11*d*k^2 + 2*d^2*k^2*(1 - k^2))*(
1 - x^2)^(2/3)*(1 - k^2*x^2)^(2/3)*Defer[Int][x/((1 - x^2)^(2/3)*(1 - k^2*x^2)^(2/3)*(1 - d + (2 + d)*x + (1 +
d*k^2)*x^2 - d*k^2*x^3)), x])/(d^2*k^2*((1 - x^2)*(1 - k^2*x^2))^(2/3)) - ((1 + 8*d*k^2 + 2*d^2*(k^2 + 3*k^4)
)*(1 - x^2)^(2/3)*(1 - k^2*x^2)^(2/3)*Defer[Int][x^2/((1 - x^2)^(2/3)*(1 - k^2*x^2)^(2/3)*(1 - d + (2 + d)*x +
(1 + d*k^2)*x^2 - d*k^2*x^3)), x])/(d^2*k^2*((1 - x^2)*(1 - k^2*x^2))^(2/3))

Rubi steps

\begin {align*} \int \frac {-3-2 \left (1+k^2\right ) x+\left (1+k^2\right ) x^2+4 k^2 x^3+k^2 x^4}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \int \frac {-3-2 \left (1+k^2\right ) x+\left (1+k^2\right ) x^2+4 k^2 x^3+k^2 x^4}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx}{\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 ) \int \left (\frac {5 d+\frac {1}{k^2}}{d^2 \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}}+\frac {x}{d \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}}+\frac {1-8 d^2 k^2-d \left (1-5 k^2\right )+\left (2+d+11 d k^2+2 d^2 k^2 \left (1-k^2\right )\right ) x+\left (1+8 d k^2+2 d^2 \left (k^2+3 k^4\right )\right ) x^2}{d^2 k^2 \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )}\right ) \, dx}{\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 ) \int \frac {x}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}} \, dx}{d \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (5 d+\frac {1}{k^2}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \int \frac {1}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}} \, dx}{d^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 ) \int \frac {1-8 d^2 k^2-d \left (1-5 k^2\right )+\left (2+d+11 d k^2+2 d^2 k^2 \left (1-k^2\right )\right ) x+\left (1+8 d k^2+2 d^2 \left (k^2+3 k^4\right )\right ) x^2}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx}{d^2 k^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (5 d+\frac {1}{k^2}\right ) x \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} F_1\left (\frac {1}{2};\frac {2}{3},\frac {2}{3};\frac {3}{2};x^2,k^2 x^2\right )}{d^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 {1}{(1-x)^{2/3} \left (1-k^2 x\right )^{2/3}} \, dx,x,x^2\right )}{2 d \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 ) \int \left (\frac {8 d^2 k^2 \left (1+\frac {-1+d-5 d k^2}{8 d^2 k^2}\right )}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )}+\frac {\left (-2-2 d^2 k^2 \left (1-k^2\right )-d \left (1+11 k^2\right )\right ) x}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )}+\frac {\left (-1-8 d k^2-2 d^2 \left (k^2+3 k^4\right )\right ) x^2}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )}\right ) \, dx}{d^2 k^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (5 d+\frac {1}{k^2}\right ) x \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} F_1\left (\frac {1}{2};\frac {2}{3},\frac {2}{3};\frac {3}{2};x^2,k^2 x^2\right )}{d^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1+\left (-1-k^2\right ) x+k^2 x^2\right )^{2/3}} \, dx,x,x^2\right )}{2 d}+\frac {\left (\left (-1+d-5 d k^2+8 d^2 k^2\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \int \frac {1}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )} \, dx}{d^2 k^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (-2-2 d^2 k^2 \left (1-k^2\right )-d \left (1+11 k^2\right )\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \int \frac {x}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )} \, dx}{d^2 k^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (-1-8 d k^2-2 d^2 \left (k^2+3 k^4\right )\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \int \frac {x^2}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )} \, dx}{d^2 k^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (5 d+\frac {1}{k^2}\right ) x \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} F_1\left (\frac {1}{2};\frac {2}{3},\frac {2}{3};\frac {3}{2};x^2,k^2 x^2\right )}{d^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (-1+d-5 d k^2+8 d^2 k^2\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \int \frac {1}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )} \, dx}{d^2 k^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (-2-2 d^2 k^2 \left (1-k^2\right )-d \left (1+11 k^2\right )\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \int \frac {x}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )} \, dx}{d^2 k^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (-1-8 d k^2-2 d^2 \left (k^2+3 k^4\right )\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \int \frac {x^2}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )} \, dx}{d^2 k^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (3 \sqrt {\left (-1-k^2+2 k^2 x^2\right )^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-4 k^2+\left (-1-k^2\right )^2+4 k^2 x^3}} \, dx,x,\sqrt [3]{\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}\right )}{2 d \left (-1-k^2+2 k^2 x^2\right )}\\ &=\frac {\left (5 d+\frac {1}{k^2}\right ) x \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} F_1\left (\frac {1}{2};\frac {2}{3},\frac {2}{3};\frac {3}{2};x^2,k^2 x^2\right )}{d^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}-\frac {3^{3/4} \sqrt {2+\sqrt {3}} \sqrt {\left (-1-k^2+2 k^2 x^2\right )^2} \left (\left (-1+k^2\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}\right ) \sqrt {\frac {\left (-1+k^2\right )^{4/3}-2^{2/3} k^{2/3} \left (-1+k^2\right )^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}+2 \sqrt [3]{2} k^{4/3} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}{\left (\left (1+\sqrt {3}\right ) \left (-1+k^2\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \left (-1+k^2\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}{\left (1+\sqrt {3}\right ) \left (-1+k^2\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\right )|-7-4 \sqrt {3}\right )}{2^{2/3} d k^{2/3} \left (1+k^2-2 k^2 x^2\right ) \sqrt {\left (-1-k^2 \left (1-2 x^2\right )\right )^2} \sqrt {\frac {\left (-1+k^2\right )^{2/3} \left (\left (-1+k^2\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}\right )}{\left (\left (1+\sqrt {3}\right ) \left (-1+k^2\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}\right )^2}}}+\frac {\left (\left (-1+d-5 d k^2+8 d^2 k^2\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \int \frac {1}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )} \, dx}{d^2 k^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (-2-2 d^2 k^2 \left (1-k^2\right )-d \left (1+11 k^2\right )\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \int \frac {x}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )} \, dx}{d^2 k^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (-1-8 d k^2-2 d^2 \left (k^2+3 k^4\right )\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \int \frac {x^2}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )} \, dx}{d^2 k^2 \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 = 3.59, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-3-2 \left (1+k^2\right ) x+\left (1+k^2\right ) x^2+4 k^2 x^3+k^2 x^4}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 8.10, size = 190, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3}+\sqrt {3} x}{1+x+2 \sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{\sqrt [3]{d}}+\frac {\log \left (1+x-\sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}\right )}{\sqrt [3]{d}}-\frac {\log \left (1+2 x+x^2+\left (\sqrt [3]{d}+\sqrt [3]{d} x\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}+d^{2/3} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 \sqrt [3]{d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[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^{4} + 4 \, k^{2} x^{3} + {\left (k^{2} + 1\right )} x^{2} - 2 \, {\left (k^{2} + 1\right )} x - 3}{{\left (d k^{2} x^{3} - {\left (d k^{2} + 1\right )} x^{2} - {\left (d + 2\right )} x + 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((-3-2*(k^2+1)*x+(k^2+1)*x^2+4*k^2*x^3+k^2*x^4)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(-1+d-(2+d)*x-(d*k^2+1)
*x^2+d*k^2*x^3),x, algorithm="giac")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((k^2*x^4 + 4*k^2*x^3 + (k^2 + 1)*x^2 - 2*(k^2 + 1)*x - 3)/((d*k^2*x^3 - (d*k^2 + 1)*x^2 - (d + 2)*x
+ 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 = -0.01 \begin {gather*} \int -\frac {4\,k^2\,x^3-2\,x\,\left (k^2+1\right )+k^2\,x^4+x^2\,\left (k^2+1\right )-3}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{2/3}\,\left (x^2\,\left (d\,k^2+1\right )-d+x\,\left (d+2\right )-d\,k^2\,x^3+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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