Optimal. Leaf size=85 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \left (k x^3+(-k-1) x^2+x\right )^{3/4}}{(x-1) x}\right )}{d^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \left (k x^3+(-k-1) x^2+x\right )^{3/4}}{(x-1) x}\right )}{d^{3/4}} \]
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Rubi [F] time = 15.03, 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 {\left (-1-2 (-1+k) x+k x^2\right ) \left (1-2 k x+k^2 x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {\left (-1-2 (-1+k) x+k x^2\right ) \left (1-2 k x+k^2 x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx &=\int \frac {(-1+k x)^2 \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx\\ &=\frac {\left ((1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \int \frac {(-1+k x)^2 \left (-1-2 (-1+k) x+k x^2\right )}{(1-x)^{3/4} x^{3/4} (1-k x)^{3/4} \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {\left ((1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \int \frac {(1-k x)^{5/4} \left (-1-2 (-1+k) x+k x^2\right )}{(1-x)^{3/4} x^{3/4} \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 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 {\left (1-k x^4\right )^{5/4} \left (-1-2 (-1+k) x^4+k x^8\right )}{\left (1-x^4\right )^{3/4} \left (-d+(1+3 d k) x^4-\left (1+3 d k^2\right ) x^8+d k^3 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 \frac {\left (1-k x^4\right )^{5/4} \left (-1-2 (-1+k) x^4+k x^8\right )}{\left (1-x^4\right )^{3/4} \left (x^4-x^8+d \left (-1+k x^4\right )^3\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 {\left (1-k x^4\right )^{5/4}}{\left (1-x^4\right )^{3/4} \left (d-(1+3 d k) x^4+\left (1+3 d k^2\right ) x^8-d k^3 x^{12}\right )}+\frac {2 (-1+k) x^4 \left (1-k x^4\right )^{5/4}}{\left (1-x^4\right )^{3/4} \left (d-(1+3 d k) x^4+\left (1+3 d k^2\right ) x^8-d k^3 x^{12}\right )}+\frac {k x^8 \left (1-k x^4\right )^{5/4}}{\left (1-x^4\right )^{3/4} \left (-d+(1+3 d k) x^4-\left (1+3 d k^2\right ) x^8+d k^3 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 {\left (1-k x^4\right )^{5/4}}{\left (1-x^4\right )^{3/4} \left (d-(1+3 d k) x^4+\left (1+3 d k^2\right ) x^8-d k^3 x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}-\frac {\left (8 (1-k) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (1-k x^4\right )^{5/4}}{\left (1-x^4\right )^{3/4} \left (d-(1+3 d k) x^4+\left (1+3 d k^2\right ) x^8-d k^3 x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{((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^8 \left (1-k x^4\right )^{5/4}}{\left (1-x^4\right )^{3/4} \left (-d+(1+3 d k) x^4-\left (1+3 d k^2\right ) x^8+d k^3 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 \frac {\left (1-k x^4\right )^{5/4}}{\left (1-x^4\right )^{3/4} \left (-x^4+x^8-d \left (-1+k x^4\right )^3\right )} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}-\frac {\left (8 (1-k) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (1-k x^4\right )^{5/4}}{\left (1-x^4\right )^{3/4} \left (-x^4+x^8-d \left (-1+k x^4\right )^3\right )} \, dx,x,\sqrt [4]{x}\right )}{((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^8 \left (1-k x^4\right )^{5/4}}{\left (1-x^4\right )^{3/4} \left (x^4-x^8+d \left (-1+k x^4\right )^3\right )} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}\\ \end {align*}
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Mathematica [F] time = 1.08, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-1-2 (-1+k) x+k x^2\right ) \left (1-2 k x+k^2 x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 3.16, size = 85, normalized size = 1.00 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \left (x+(-1-k) x^2+k x^3\right )^{3/4}}{(-1+x) x}\right )}{d^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \left (x+(-1-k) x^2+k x^3\right )^{3/4}}{(-1+x) 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 {{\left (k^{2} x^{2} - 2 \, k x + 1\right )} {\left (k x^{2} - 2 \, {\left (k - 1\right )} x - 1\right )}}{{\left (d k^{3} x^{3} - {\left (3 \, d k^{2} + 1\right )} x^{2} + {\left (3 \, d k + 1\right )} x - d\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.
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maple [F] time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (-1-2 \left (-1+k \right ) x +k \,x^{2}\right ) \left (k^{2} x^{2}-2 k x +1\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {3}{4}} \left (-d +\left (3 d k +1\right ) x -\left (3 d \,k^{2}+1\right ) x^{2}+d \,k^{3} x^{3}\right )}\, dx\]
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 {{\left (k^{2} x^{2} - 2 \, k x + 1\right )} {\left (k x^{2} - 2 \, {\left (k - 1\right )} x - 1\right )}}{{\left (d k^{3} x^{3} - {\left (3 \, d k^{2} + 1\right )} x^{2} + {\left (3 \, d k + 1\right )} x - d\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.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (2\,x\,\left (k-1\right )-k\,x^2+1\right )\,\left (k^2\,x^2-2\,k\,x+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{3/4}\,\left (d+x^2\,\left (3\,d\,k^2+1\right )-x\,\left (3\,d\,k+1\right )-d\,k^3\,x^3\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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