Optimal. Leaf size=94 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{d} k x+\sqrt [4]{d}}{\sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}\right )}{d^{3/4}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{d} k x+\sqrt [4]{d}}{\sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}\right )}{d^{3/4}} \]
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Rubi [F] time = 35.35, 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 (-2 k+(-1+k) (1+k) x+2 k x^2\right ) \left (1+2 k x+k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {\left (-2 k+(-1+k) (1+k) x+2 k x^2\right ) \left (1+2 k x+k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx &=\int \frac {(1+k x)^2 \left (-2 k+(-1+k) (1+k) x+2 k x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx\\ &=\frac {\left (\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {(1+k x)^2 \left (-2 k+(-1+k) (1+k) x+2 k x^2\right )}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}\\ &=\frac {\left (\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {(1+k x)^2 \left (2 k+(1-k) (1+k) x-2 k x^2\right )}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )} \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}\\ &=\frac {\left (\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \left (-\frac {k \left (5+(1+3 d) k^2-d k^4\right )}{\left (1-d k^2\right )^2 \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}}-\frac {2 k^2 x}{\left (1-d k^2\right ) \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}}+\frac {k \left (7+k^2-3 d^2 k^2 \left (1-k^2\right )-d \left (5+2 k^2+k^4\right )\right )+\left (1-k^4-6 d^2 k^4 \left (1-k^2\right )-d k^2 \left (19+14 k^2-k^4\right )\right ) x-k \left (5+3 (1+8 d) k^2+3 d^2 k^4-3 d^2 k^6\right ) x^2}{\left (-1+d k^2\right )^2 \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )}\right ) \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}\\ &=-\frac {\left (2 k^2 \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {x}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}} \, dx}{\left (1-d k^2\right ) \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}+\frac {\left (\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {k \left (7+k^2-3 d^2 k^2 \left (1-k^2\right )-d \left (5+2 k^2+k^4\right )\right )+\left (1-k^4-6 d^2 k^4 \left (1-k^2\right )-d k^2 \left (19+14 k^2-k^4\right )\right ) x-k \left (5+3 (1+8 d) k^2+3 d^2 k^4-3 d^2 k^6\right ) x^2}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )} \, dx}{\left (-1+d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}-\frac {\left (k \left (5+(1+3 d) k^2-d k^4\right ) \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {1}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}} \, dx}{\left (1-d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}\\ &=-\frac {k \left (5+(1+3 d) k^2-d k^4\right ) x \left (\frac {1-x^2}{1-k^2 x^2}\right )^{3/4} \left (1-k^2 x^2\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\frac {\left (1-k^2\right ) x^2}{1-k^2 x^2}\right )}{\left (1-d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}-\frac {\left (k^2 \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{(1-x)^{3/4} \left (1-k^2 x\right )^{3/4}} \, dx,x,x^2\right )}{\left (1-d k^2\right ) \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}+\frac {\left (\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \left (\frac {k \left (7+k^2-3 d^2 k^2 \left (1-k^2\right )-d \left (5+2 k^2+k^4\right )\right )}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )}+\frac {\left (1-k^4-6 d^2 k^4 \left (1-k^2\right )-d k^2 \left (19+14 k^2-k^4\right )\right ) x}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )}+\frac {k \left (-5-3 (1+8 d) k^2-3 d^2 k^4+3 d^2 k^6\right ) x^2}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )}\right ) \, dx}{\left (-1+d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}\\ &=-\frac {k \left (5+(1+3 d) k^2-d k^4\right ) x \left (\frac {1-x^2}{1-k^2 x^2}\right )^{3/4} \left (1-k^2 x^2\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\frac {\left (1-k^2\right ) x^2}{1-k^2 x^2}\right )}{\left (1-d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}-\frac {k^2 \operatorname {Subst}\left (\int \frac {1}{\left (1+\left (-1-k^2\right ) x+k^2 x^2\right )^{3/4}} \, dx,x,x^2\right )}{1-d k^2}-\frac {\left (k \left (5+3 (1+8 d) k^2+3 d^2 k^4-3 d^2 k^6\right ) \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {x^2}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )} \, dx}{\left (-1+d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}+\frac {\left (\left (1-k^4-6 d^2 k^4 \left (1-k^2\right )-d k^2 \left (19+14 k^2-k^4\right )\right ) \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {x}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )} \, dx}{\left (-1+d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}+\frac {\left (k \left (7+k^2-3 d^2 k^2 \left (1-k^2\right )-d \left (5+2 k^2+k^4\right )\right ) \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {1}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )} \, dx}{\left (-1+d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}\\ &=-\frac {k \left (5+(1+3 d) k^2-d k^4\right ) x \left (\frac {1-x^2}{1-k^2 x^2}\right )^{3/4} \left (1-k^2 x^2\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\frac {\left (1-k^2\right ) x^2}{1-k^2 x^2}\right )}{\left (1-d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}-\frac {\left (k \left (5+3 (1+8 d) k^2+3 d^2 k^4-3 d^2 k^6\right ) \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {x^2}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )} \, dx}{\left (-1+d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}+\frac {\left (\left (1-k^4-6 d^2 k^4 \left (1-k^2\right )-d k^2 \left (19+14 k^2-k^4\right )\right ) \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {x}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )} \, dx}{\left (-1+d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}+\frac {\left (k \left (7+k^2-3 d^2 k^2 \left (1-k^2\right )-d \left (5+2 k^2+k^4\right )\right ) \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {1}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )} \, dx}{\left (-1+d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}-\frac {\left (4 k^2 \sqrt {\left (-1-k^2+2 k^2 x^2\right )^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-4 k^2+\left (-1-k^2\right )^2+4 k^2 x^4}} \, dx,x,\sqrt [4]{\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}\right )}{\left (1-d k^2\right ) \left (-1-k^2+2 k^2 x^2\right )}\\ &=\frac {\sqrt {2} k^{3/2} \sqrt {-1+k^2} \sqrt {\left (-1-k^2+2 k^2 x^2\right )^2} \sqrt {\frac {\left (1+k^2 \left (1-2 x^2\right )\right )^2}{\left (1-k^2\right )^2 \left (1-\frac {2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}{1-k^2}\right )^2}} \left (1-\frac {2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}{1-k^2}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {k} \sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}{\sqrt {-1+k^2}}\right )|\frac {1}{2}\right )}{\left (1-d k^2\right ) \left (1+k^2-2 k^2 x^2\right ) \sqrt {\left (-1-k^2 \left (1-2 x^2\right )\right )^2}}-\frac {k \left (5+(1+3 d) k^2-d k^4\right ) x \left (\frac {1-x^2}{1-k^2 x^2}\right )^{3/4} \left (1-k^2 x^2\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\frac {\left (1-k^2\right ) x^2}{1-k^2 x^2}\right )}{\left (1-d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}-\frac {\left (k \left (5+3 (1+8 d) k^2+3 d^2 k^4-3 d^2 k^6\right ) \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {x^2}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )} \, dx}{\left (-1+d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}+\frac {\left (\left (1-k^4-6 d^2 k^4 \left (1-k^2\right )-d k^2 \left (19+14 k^2-k^4\right )\right ) \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {x}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )} \, dx}{\left (-1+d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}+\frac {\left (k \left (7+k^2-3 d^2 k^2 \left (1-k^2\right )-d \left (5+2 k^2+k^4\right )\right ) \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {1}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right )} \, dx}{\left (-1+d k^2\right )^2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}\\ \end {align*}
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Mathematica [F] time = 1.88, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-2 k+(-1+k) (1+k) x+2 k x^2\right ) \left (1+2 k x+k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 15.76, size = 94, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{d}+\sqrt [4]{d} k x}{\sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{d^{3/4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{d}+\sqrt [4]{d} k x}{\sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\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 ({\left (k + 1\right )} {\left (k - 1\right )} x + 2 \, k x^{2} - 2 \, k\right )}}{{\left ({\left (d k^{2} - 1\right )} k x^{3} + {\left (3 \, d + 1\right )} k x + {\left (3 \, d k^{2} + 1\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\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 = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (-2 k +\left (-1+k \right ) \left (1+k \right ) x +2 k \,x^{2}\right ) \left (k^{2} x^{2}+2 k x +1\right )}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {3}{4}} \left (-1+d +\left (1+3 d \right ) k x +\left (3 d \,k^{2}+1\right ) x^{2}+k \left (d \,k^{2}-1\right ) x^{3}\right )}\, dx\]
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 ({\left (k + 1\right )} {\left (k - 1\right )} x + 2 \, k x^{2} - 2 \, k\right )}}{{\left ({\left (d k^{2} - 1\right )} k x^{3} + {\left (3 \, d + 1\right )} k x + {\left (3 \, d k^{2} + 1\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\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\,k\,x^2+\left (k-1\right )\,\left (k+1\right )\,x-2\,k\right )\,\left (k^2\,x^2+2\,k\,x+1\right )}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{3/4}\,\left (k\,\left (d\,k^2-1\right )\,x^3+\left (3\,d\,k^2+1\right )\,x^2+k\,\left (3\,d+1\right )\,x+d-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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