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

Optimal. Leaf size=71 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{k x^3+(-k-1) x^2+x}}\right )}{d^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{k x^3+(-k-1) x^2+x}}\right )}{d^{3/4}} \]

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Rubi [F]  time = 18.11, 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 {x^2 \left (3-2 (1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(4*(k^2 - 2*d*(1 + k))*(1 - x)^(3/4)*x*(1 - k*x)^(3/4)*AppellF1[1/4, 3/4, 3/4, 5/4, x, k*x])/(d^2*((1 - x)*x*(
1 - k*x))^(3/4)) + (4*k*(1 - x)^(3/4)*x^2*(1 - k*x)^(3/4)*AppellF1[5/4, 3/4, 3/4, 9/4, x, k*x])/(5*d*((1 - x)*
x*(1 - k*x))^(3/4)) - (4*(k^2 - 2*d*(1 + k))*(1 - x)^(3/4)*x^(3/4)*(1 - k*x)^(3/4)*Defer[Subst][Defer[Int][1/(
(1 - x^4)^(3/4)*(1 - k*x^4)^(3/4)*(1 - (1 + k)*x^4 + k*x^8 - d*x^12)), x], x, x^(1/4)])/(d^2*((1 - x)*x*(1 - k
*x))^(3/4)) + (4*(k^2*(1 + k) - d*(2 + 5*k + 2*k^2))*(1 - x)^(3/4)*x^(3/4)*(1 - k*x)^(3/4)*Defer[Subst][Defer[
Int][x^4/((1 - x^4)^(3/4)*(1 - k*x^4)^(3/4)*(1 - (1 + k)*x^4 + k*x^8 - d*x^12)), x], x, x^(1/4)])/(d^2*((1 - x
)*x*(1 - k*x))^(3/4)) - (4*(3*d^2 + k^3 - 3*d*k*(1 + k))*(1 - x)^(3/4)*x^(3/4)*(1 - k*x)^(3/4)*Defer[Subst][De
fer[Int][x^8/((1 - x^4)^(3/4)*(1 - k*x^4)^(3/4)*(1 - (1 + k)*x^4 + k*x^8 - d*x^12)), x], x, x^(1/4)])/(d^2*((1
- x)*x*(1 - k*x))^(3/4))

Rubi steps

\begin {align*} \int \frac {x^2 \left (3-2 (1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx &=\frac {\left ((1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \int \frac {x^{5/4} \left (3-2 (1+k) x+k x^2\right )}{(1-x)^{3/4} (1-k x)^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^8 \left (3-2 (1+k) x^4+k x^8\right )}{\left (1-x^4\right )^{3/4} \left (1-k x^4\right )^{3/4} \left (-1+(1+k) x^4-k x^8+d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \left (\frac {k^2-2 d (1+k)}{d^2 \left (1-x^4\right )^{3/4} \left (1-k x^4\right )^{3/4}}+\frac {k x^4}{d \left (1-x^4\right )^{3/4} \left (1-k x^4\right )^{3/4}}-\frac {-k^2+2 d (1+k)-\left (2 d+5 d k-k^2+2 d k^2-k^3\right ) x^4-\left (3 d^2+k^3-3 d k (1+k)\right ) x^8}{d^2 \left (1-x^4\right )^{3/4} \left (1-k x^4\right )^{3/4} \left (-1+(1+k) x^4-k x^8+d x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}\\ &=-\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {-k^2+2 d (1+k)-\left (2 d+5 d k-k^2+2 d k^2-k^3\right ) x^4-\left (3 d^2+k^3-3 d k (1+k)\right ) x^8}{\left (1-x^4\right )^{3/4} \left (1-k x^4\right )^{3/4} \left (-1+(1+k) x^4-k x^8+d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{d^2 ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 k (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (1-x^4\right )^{3/4} \left (1-k x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{d ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 \left (k^2-2 d (1+k)\right ) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^4\right )^{3/4} \left (1-k x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{d^2 ((1-x) x (1-k x))^{3/4}}\\ &=\frac {4 \left (k^2-2 d (1+k)\right ) (1-x)^{3/4} x (1-k x)^{3/4} F_1\left (\frac {1}{4};\frac {3}{4},\frac {3}{4};\frac {5}{4};x,k x\right )}{d^2 ((1-x) x (1-k x))^{3/4}}+\frac {4 k (1-x)^{3/4} x^2 (1-k x)^{3/4} F_1\left (\frac {5}{4};\frac {3}{4},\frac {3}{4};\frac {9}{4};x,k x\right )}{5 d ((1-x) x (1-k x))^{3/4}}-\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \left (\frac {2 d (-1-k) \left (1-\frac {k^2}{2 d+2 d k}\right )}{\left (1-x^4\right )^{3/4} \left (1-k x^4\right )^{3/4} \left (1-(1+k) x^4+k x^8-d x^{12}\right )}+\frac {\left (2 d+5 d k-k^2+2 d k^2-k^3\right ) x^4}{\left (1-x^4\right )^{3/4} \left (1-k x^4\right )^{3/4} \left (1-(1+k) x^4+k x^8-d x^{12}\right )}+\frac {\left (3 d^2+k^3-3 d k (1+k)\right ) x^8}{\left (1-x^4\right )^{3/4} \left (1-k x^4\right )^{3/4} \left (1-(1+k) x^4+k x^8-d x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{d^2 ((1-x) x (1-k x))^{3/4}}\\ &=\frac {4 \left (k^2-2 d (1+k)\right ) (1-x)^{3/4} x (1-k x)^{3/4} F_1\left (\frac {1}{4};\frac {3}{4},\frac {3}{4};\frac {5}{4};x,k x\right )}{d^2 ((1-x) x (1-k x))^{3/4}}+\frac {4 k (1-x)^{3/4} x^2 (1-k x)^{3/4} F_1\left (\frac {5}{4};\frac {3}{4},\frac {3}{4};\frac {9}{4};x,k x\right )}{5 d ((1-x) x (1-k x))^{3/4}}-\frac {\left (4 \left (k^2-2 d (1+k)\right ) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^4\right )^{3/4} \left (1-k x^4\right )^{3/4} \left (1-(1+k) x^4+k x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{d^2 ((1-x) x (1-k x))^{3/4}}-\frac {\left (4 \left (3 d^2+k^3-3 d k (1+k)\right ) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^8}{\left (1-x^4\right )^{3/4} \left (1-k x^4\right )^{3/4} \left (1-(1+k) x^4+k x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{d^2 ((1-x) x (1-k x))^{3/4}}-\frac {\left (4 \left (-k^2 (1+k)+d \left (2+5 k+2 k^2\right )\right ) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (1-x^4\right )^{3/4} \left (1-k x^4\right )^{3/4} \left (1-(1+k) x^4+k x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{d^2 ((1-x) x (1-k x))^{3/4}}\\ \end {align*}

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Mathematica [F]  time = 1.14, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^2 \left (3-2 (1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 3.33, size = 71, normalized size = 1.00 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{x+(-1-k) x^2+k x^3}}\right )}{d^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{x+(-1-k) x^2+k x^3}}\right )}{d^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

[Out]

Timed out

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giac [B]  time = 1.42, size = 264, normalized size = 3.72 \begin {gather*} -\frac {\sqrt {2} \left (-d\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-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 (-d\right )^{\frac {1}{4}}}\right )}{d} - \frac {\sqrt {2} \left (-d\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-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 (-d\right )^{\frac {1}{4}}}\right )}{d} - \frac {\sqrt {2} \left (-d\right )^{\frac {1}{4}} \log \left (\sqrt {2} \left (-d\right )^{\frac {1}{4}} {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-d} + \sqrt {\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}}\right )}{2 \, d} + \frac {\sqrt {2} \left (-d\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \left (-d\right )^{\frac {1}{4}} {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-d} + \sqrt {\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}}\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \left (3-2 \left (1+k \right ) x +k \,x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {3}{4}} \left (-1+\left (1+k \right ) x -k \,x^{2}+d \,x^{3}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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