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

Optimal. Leaf size=111 \[ 2 \sqrt [4]{d} \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k x^3+(-k-1) x^2+x}}{x}\right )-2 \sqrt [4]{d} \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k x^3+(-k-1) x^2+x}}{x}\right )+\frac {4 \left (k x^3-k x^2-x^2+x\right )^{3/4}}{(x-1) (k x-1)} \]

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Rubi [F]  time = 38.57, 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^3 \left (3-2 (1+k) x+k x^2\right )}{(-1+x) \sqrt [4]{(1-x) x (1-k 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[(x^3*(3 - 2*(1 + k)*x + k*x^2))/((-1 + x)*((1 - x)*x*(1 - k*x))^(1/4)*(-1 + k*x)*(-d + d*(1 + k)*x - d*k*x
^2 + x^3)),x]

[Out]

(4*(3 + d^2*k^3 - 3*d*k*(1 + k))*(1 - x)^(1/4)*x*(1 - k*x)^(1/4)*AppellF1[3/4, 5/4, 5/4, 7/4, x, k*x])/(3*((1
- x)*x*(1 - k*x))^(1/4)) - (4*(2 + 2*k - d*k^2)*(1 - x)^(1/4)*x^2*(1 - k*x)^(1/4)*AppellF1[7/4, 5/4, 5/4, 11/4
, x, k*x])/(7*((1 - x)*x*(1 - k*x))^(1/4)) + (4*k*(1 - x)^(1/4)*x^3*(1 - k*x)^(1/4)*AppellF1[11/4, 5/4, 5/4, 1
5/4, x, k*x])/(11*((1 - x)*x*(1 - k*x))^(1/4)) - (4*d*(3 + d^2*k^3 - 3*d*k*(1 + k))*(1 - x)^(1/4)*x^(1/4)*(1 -
k*x)^(1/4)*Defer[Subst][Defer[Int][x^2/((1 - x^4)^(5/4)*(1 - k*x^4)^(5/4)*(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*d*(5 + (5 - 3*d)*k - 7*d*k^2 - (3 - d)*d*k^3 + d^2*k^4)*
(1 - x)^(1/4)*x^(1/4)*(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][x^6/((1 - x^4)^(5/4)*(1 - k*x^4)^(5/4)*(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*d*(2 + 8*k + (2 - 4*d)*k^2 - 4
*d*k^3 + d^2*k^4)*(1 - x)^(1/4)*x^(1/4)*(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][x^10/((1 - x^4)^(5/4)*(1 - k*x
^4)^(5/4)*(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 {x^3 \left (3-2 (1+k) x+k x^2\right )}{(-1+x) \sqrt [4]{(1-x) x (1-k 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 {x^{11/4} \left (3-2 (1+k) x+k x^2\right )}{\sqrt [4]{1-x} (-1+x) \sqrt [4]{1-k x} (-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 (\sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \int \frac {x^{11/4} \left (3-2 (1+k) x+k x^2\right )}{(1-x)^{5/4} \sqrt [4]{1-k x} (-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 (\sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \int \frac {x^{11/4} \left (3-2 (1+k) x+k x^2\right )}{(1-x)^{5/4} (1-k x)^{5/4} \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^{14} \left (3-2 (1+k) x^4+k x^8\right )}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4} \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 {\left (3+d^2 k^3-3 d k (1+k)\right ) x^2}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4}}-\frac {\left (2+2 k-d k^2\right ) x^6}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4}}+\frac {k x^{10}}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4}}+\frac {x^2 \left (d \left (3+d^2 k^3-3 d k (1+k)\right )-d \left (5+(5-3 d) k-7 d k^2-(3-d) d k^3+d^2 k^4\right ) x^4+d \left (2+8 k+(2-4 d) k^2-4 d k^3+d^2 k^4\right ) x^8\right )}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4} \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 (4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (d \left (3+d^2 k^3-3 d k (1+k)\right )-d \left (5+(5-3 d) k-7 d k^2-(3-d) d k^3+d^2 k^4\right ) x^4+d \left (2+8 k+(2-4 d) k^2-4 d k^3+d^2 k^4\right ) x^8\right )}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4} \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}}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (4 \left (-2-2 k+d k^2\right ) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (4 \left (3+d^2 k^3-3 d k (1+k)\right ) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\\ &=\frac {4 \left (3+d^2 k^3-3 d k (1+k)\right ) \sqrt [4]{1-x} x \sqrt [4]{1-k x} F_1\left (\frac {3}{4};\frac {5}{4},\frac {5}{4};\frac {7}{4};x,k x\right )}{3 \sqrt [4]{(1-x) x (1-k x)}}-\frac {4 \left (2+2 k-d k^2\right ) \sqrt [4]{1-x} x^2 \sqrt [4]{1-k x} F_1\left (\frac {7}{4};\frac {5}{4},\frac {5}{4};\frac {11}{4};x,k x\right )}{7 \sqrt [4]{(1-x) x (1-k x)}}+\frac {4 k \sqrt [4]{1-x} x^3 \sqrt [4]{1-k x} F_1\left (\frac {11}{4};\frac {5}{4},\frac {5}{4};\frac {15}{4};x,k x\right )}{11 \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 {d \left (-3-d^2 k^3+3 d k (1+k)\right ) x^2}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4} \left (d-d (1+k) x^4+d k x^8-x^{12}\right )}+\frac {d \left (5+(5-3 d) k-7 d k^2-(3-d) d k^3+d^2 k^4\right ) x^6}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4} \left (d-d (1+k) x^4+d k x^8-x^{12}\right )}+\frac {d \left (-2-8 k-(2-4 d) k^2+4 d k^3-d^2 k^4\right ) x^{10}}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4} \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 {4 \left (3+d^2 k^3-3 d k (1+k)\right ) \sqrt [4]{1-x} x \sqrt [4]{1-k x} F_1\left (\frac {3}{4};\frac {5}{4},\frac {5}{4};\frac {7}{4};x,k x\right )}{3 \sqrt [4]{(1-x) x (1-k x)}}-\frac {4 \left (2+2 k-d k^2\right ) \sqrt [4]{1-x} x^2 \sqrt [4]{1-k x} F_1\left (\frac {7}{4};\frac {5}{4},\frac {5}{4};\frac {11}{4};x,k x\right )}{7 \sqrt [4]{(1-x) x (1-k x)}}+\frac {4 k \sqrt [4]{1-x} x^3 \sqrt [4]{1-k x} F_1\left (\frac {11}{4};\frac {5}{4},\frac {5}{4};\frac {15}{4};x,k x\right )}{11 \sqrt [4]{(1-x) x (1-k x)}}-\frac {\left (4 d \left (2+8 k+(2-4 d) k^2-4 d k^3+d^2 k^4\right ) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^{10}}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4} \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 d \left (5+(5-3 d) k-7 d k^2-(3-d) d k^3+d^2 k^4\right ) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4} \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 d \left (3+d^2 k^3-3 d k (1+k)\right ) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (1-x^4\right )^{5/4} \left (1-k x^4\right )^{5/4} \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 = 7.89, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3 \left (3-2 (1+k) x+k x^2\right )}{(-1+x) \sqrt [4]{(1-x) x (1-k 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[(x^3*(3 - 2*(1 + k)*x + k*x^2))/((-1 + x)*((1 - x)*x*(1 - k*x))^(1/4)*(-1 + k*x)*(-d + d*(1 + k)*x -
d*k*x^2 + x^3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.94, size = 111, normalized size = 1.00 \begin {gather*} \frac {4 \left (x-x^2-k x^2+k x^3\right )^{3/4}}{(-1+x) (-1+k x)}+2 \sqrt [4]{d} \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{x}\right )-2 \sqrt [4]{d} \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

Timed out

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giac [B]  time = 0.38, size = 312, normalized size = 2.81 \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^{2}} - \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^{2}} + \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^{2}} - \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^{2}} + \frac {4}{{\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \left (3-2 \left (1+k \right ) x +k \,x^{2}\right )}{\left (-1+x \right ) \left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{4}} \left (k x -1\right ) \left (-d +d \left (1+k \right ) x -d k \,x^{2}+x^{3}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

Timed out

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