Optimal. Leaf size=79 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k x^3+(-k-1) x^2+x}}{x-1}\right )}{d^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k x^3+(-k-1) x^2+x}}{x-1}\right )}{d^{3/4}} \]
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Rubi [F] time = 12.58, 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 {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {align*} \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+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 {-1+2 (-1+k) x+k x^2}{\sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \left (-1+(3+d) x-(3+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 (-1+2 (-1+k) x^4+k x^8\right )}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-1+(3+d) x^4-(3+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 {x^2}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (1-3 \left (1+\frac {d}{3}\right ) x^4+3 \left (1+\frac {d k}{3}\right ) x^8-x^{12}\right )}+\frac {2 (1-k) x^6}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (1-3 \left (1+\frac {d}{3}\right ) x^4+3 \left (1+\frac {d k}{3}\right ) x^8-x^{12}\right )}+\frac {k x^{10}}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-1+3 \left (1+\frac {d}{3}\right ) x^4-3 \left (1+\frac {d k}{3}\right ) 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}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (1-3 \left (1+\frac {d}{3}\right ) x^4+3 \left (1+\frac {d k}{3}\right ) 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 (1-3 \left (1+\frac {d}{3}\right ) x^4+3 \left (1+\frac {d k}{3}\right ) 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 (-1+3 \left (1+\frac {d}{3}\right ) x^4-3 \left (1+\frac {d k}{3}\right ) 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 = 3.32, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 0.26, size = 79, 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}}{-1+x}\right )}{d^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+x}\right )}{d^{3/4}} \end {gather*}
Antiderivative was successfully verified.
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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.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {k x^{2} + 2 \, {\left (k - 1\right )} x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{4}} {\left ({\left (d k + 3\right )} x^{2} - x^{3} - {\left (d + 3\right )} x + 1\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {-1+2 \left (-1+k \right ) x +k \,x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{4}} \left (-1+\left (3+d \right ) x -\left (d k +3\right ) x^{2}+x^{3}\right )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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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 - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{4}} {\left ({\left (d k + 3\right )} x^{2} - x^{3} - {\left (d + 3\right )} x + 1\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {2\,x\,\left (k-1\right )+k\,x^2-1}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/4}\,\left (x^3+\left (-d\,k-3\right )\,x^2+\left (d+3\right )\,x-1\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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