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

Optimal. Leaf size=231 \[ -\frac {\log \left (-\sqrt [3]{d} k x-\sqrt [3]{d}+\sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}\right )}{d^{2/3}}+\frac {\log \left (d^{2/3} k^2 x^2+2 d^{2/3} k x+d^{2/3}+\sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1} \left (\sqrt [3]{d} k x+\sqrt [3]{d}\right )+\left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}\right )}{2 d^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} k x+\sqrt {3} \sqrt [3]{d}}{\sqrt [3]{d} k x+\sqrt [3]{d}+2 \sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}\right )}{d^{2/3}} \]

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Rubi [F]  time = 9.49, 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 k+2 \left (1+k^2\right ) x-k \left (1+k^2\right ) x^2-4 k^2 x^3-k^3 x^4}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-((k*(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])/((1 - x^2)*
(1 - k^2*x^2))^(2/3)) + (3^(3/4)*Sqrt[2 + Sqrt[3]]*k^(4/3)*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)*(1 + k^2 - 2*k^2*x^2)*Sqrt[(-1 - k^2*(1 - 2*x^2))^2]*Sqrt[((-
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]) + (k*(8 - d^2*k^2 - d*(5 - 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 - (1 + 2*d)*k*x - (1 + d*k^2)*x^
2 + k*x^3)), x])/((1 - x^2)*(1 - k^2*x^2))^(2/3) + ((2 - (2 + 11*d)*k^2 - d*(1 + 2*d)*k^4)*(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 - (1 + 2*d)*k*x - (1 + d*k^2)*x^2 +
k*x^3)), x])/((1 - x^2)*(1 - k^2*x^2))^(2/3) - (k*(6 + (2 + 8*d)*k^2 + d^2*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 - (1 + 2*d)*k*x - (1 + d*k^2)*x^2 + k*x^3)),
x])/((1 - x^2)*(1 - k^2*x^2))^(2/3)

Rubi steps

\begin {align*} \int \frac {3 k+2 \left (1+k^2\right ) x-k \left (1+k^2\right ) x^2-4 k^2 x^3-k^3 x^4}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k 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 k+2 \left (1+k^2\right ) x-k \left (1+k^2\right ) x^2-4 k^2 x^3-k^3 x^4}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k 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 {k \left (5+d k^2\right )}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}}-\frac {k^2 x}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}}+\frac {k \left (8-d^2 k^2-d \left (5-k^2\right )\right )+\left (2-(2+11 d) k^2-d (1+2 d) k^4\right ) x-k \left (6+(2+8 d) k^2+d^2 k^4\right ) x^2}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k 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 {k \left (8-d^2 k^2-d \left (5-k^2\right )\right )+\left (2-(2+11 d) k^2-d (1+2 d) k^4\right ) x-k \left (6+(2+8 d) k^2+d^2 k^4\right ) x^2}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}-\frac {\left (k^2 \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}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}-\frac {\left (k \left (5+d 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}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=-\frac {k \left (5+d 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 )}{\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 {k \left (8-d^2 k^2-d \left (5-k^2\right )\right )}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )}+\frac {\left (2-(2+11 d) k^2-d (1+2 d) k^4\right ) x}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )}+\frac {k \left (-6-(2+8 d) k^2-d^2 k^4\right ) x^2}{\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )}\right ) \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}-\frac {\left (k^2 \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 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=-\frac {k \left (5+d 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 )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}-\frac {1}{2} k^2 \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 )-\frac {\left (k \left (6+(2+8 d) k^2+d^2 k^4\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-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (2-(2+11 d) k^2-d (1+2 d) k^4\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-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (k \left (8-d^2 k^2-d \left (5-k^2\right )\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-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=-\frac {k \left (5+d 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 )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}-\frac {\left (k \left (6+(2+8 d) k^2+d^2 k^4\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-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (2-(2+11 d) k^2-d (1+2 d) k^4\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-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (k \left (8-d^2 k^2-d \left (5-k^2\right )\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-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}-\frac {\left (3 k^2 \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 \left (-1-k^2+2 k^2 x^2\right )}\\ &=-\frac {k \left (5+d 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 )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {3^{3/4} \sqrt {2+\sqrt {3}} k^{4/3} \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} \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 (k \left (6+(2+8 d) k^2+d^2 k^4\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-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (2-(2+11 d) k^2-d (1+2 d) k^4\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-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (k \left (8-d^2 k^2-d \left (5-k^2\right )\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-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\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 = 0.64, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 k+2 \left (1+k^2\right ) x-k \left (1+k^2\right ) x^2-4 k^2 x^3-k^3 x^4}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 8.11, size = 231, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d}+\sqrt {3} \sqrt [3]{d} k x}{\sqrt [3]{d}+\sqrt [3]{d} k x+2 \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{d^{2/3}}-\frac {\log \left (-\sqrt [3]{d}-\sqrt [3]{d} k x+\sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}\right )}{d^{2/3}}+\frac {\log \left (d^{2/3}+2 d^{2/3} k x+d^{2/3} k^2 x^2+\left (\sqrt [3]{d}+\sqrt [3]{d} k x\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 d^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

[Out]

Timed out

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