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

Optimal. Leaf size=75 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k x^3+(-k-1) x^2+x}}{x}\right )}{d^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k x^3+(-k-1) x^2+x}}{x}\right )}{d^{3/4}} \]

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Rubi [F]  time = 19.46, 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 (1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(8*(1 + k)*(1 - x)^(1/4)*x^(1/4)*(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][x^6/((1 - x^4)^(1/4)*(1 - k*x^4)^(1/4
)*(d - d*(1 + k)*x^4 + d*k*x^8 - x^12)), x], x, x^(1/4)])/((1 - x)*x*(1 - k*x))^(1/4) + (12*(1 - x)^(1/4)*x^(1
/4)*(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][x^2/((1 - x^4)^(1/4)*(1 - k*x^4)^(1/4)*(-d + d*(1 + k)*x^4 - d*k*x
^8 + x^12)), x], x, x^(1/4)])/((1 - x)*x*(1 - k*x))^(1/4) + (4*k*(1 - x)^(1/4)*x^(1/4)*(1 - k*x)^(1/4)*Defer[S
ubst][Defer[Int][x^10/((1 - x^4)^(1/4)*(1 - k*x^4)^(1/4)*(-d + d*(1 + k)*x^4 - d*k*x^8 + x^12)), x], x, x^(1/4
)])/((1 - x)*x*(1 - k*x))^(1/4)

Rubi steps

\begin {align*} \int \frac {3-2 (1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx &=\frac {\left (\sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \int \frac {3-2 (1+k) x+k x^2}{\sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx}{\sqrt [4]{(1-x) x (1-k x)}}\\ &=\frac {\left (4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (3-2 (1+k) x^4+k x^8\right )}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-d+d (1+k) x^4-d k x^8+x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\\ &=\frac {\left (4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \left (\frac {2 (1+k) x^6}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (d-d (1+k) x^4+d k x^8-x^{12}\right )}+\frac {3 x^2}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-d+d (1+k) x^4-d k x^8+x^{12}\right )}+\frac {k x^{10}}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-d+d (1+k) x^4-d k x^8+x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\\ &=\frac {\left (12 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-d+d (1+k) x^4-d k x^8+x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (4 k \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^{10}}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-d+d (1+k) x^4-d k x^8+x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (8 (1+k) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (d-d (1+k) x^4+d k x^8-x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\\ \end {align*}

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Mathematica [F]  time = 3.46, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3-2 (1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.32, size = 75, normalized size = 1.00 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{x}\right )}{d^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{x}\right )}{d^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

Timed out

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giac [B]  time = 0.68, size = 288, normalized size = 3.84 \begin {gather*} -\frac {\sqrt {2} \left (-d^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {1}{d}\right )^{\frac {1}{4}} + 2 \, {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {1}{d}\right )^{\frac {1}{4}}}\right )}{d^{3}} - \frac {\sqrt {2} \left (-d^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {1}{d}\right )^{\frac {1}{4}} - 2 \, {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {1}{d}\right )^{\frac {1}{4}}}\right )}{d^{3}} + \frac {\sqrt {2} \left (-d^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}} \left (-\frac {1}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}} + \sqrt {-\frac {1}{d}}\right )}{2 \, d^{3}} - \frac {\sqrt {2} \left (-d^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}} \left (-\frac {1}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}} + \sqrt {-\frac {1}{d}}\right )}{2 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {3-2 \left (1+k \right ) x +k \,x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{4}} \left (-d +d \left (1+k \right ) x -d k \,x^{2}+x^{3}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {k\,x^2-2\,x\,\left (k+1\right )+3}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/4}\,\left (-x^3+d\,k\,x^2-d\,\left (k+1\right )\,x+d\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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