Optimal. Leaf size=108 \[ \frac {(-b-2 c k) \tan ^{-1}\left (\frac {(k-1) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}\right )}{4 (k-1) k}+\frac {(b-2 c k) \tan ^{-1}\left (\frac {(k+1) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2+1}\right )}{2 k (k+1)} \]
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Rubi [C] time = 1.62, antiderivative size = 192, normalized size of antiderivative = 1.78, number of steps used = 8, number of rules used = 4, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {6719, 6725, 419, 537} \begin {gather*} \frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} (b-2 c k) \Pi \left (-k;\sin ^{-1}(x)|k^2\right )}{2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} (b+2 c k) \Pi \left (k;\sin ^{-1}(x)|k^2\right )}{2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {c \sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \end {gather*}
Antiderivative was successfully verified.
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Rule 419
Rule 537
Rule 6719
Rule 6725
Rubi steps
\begin {align*} \int \frac {c+b x^2+c k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx &=\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {c+b x^2+c k^2 x^4}{\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (-1+k^2 x^4\right )} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \left (\frac {c}{\sqrt {1-x^2} \sqrt {1-k^2 x^2}}+\frac {2 c+b x^2}{\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (-1+k^2 x^4\right )}\right ) \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {2 c+b x^2}{\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (-1+k^2 x^4\right )} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (c \sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1}{\sqrt {1-x^2} \sqrt {1-k^2 x^2}} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {c \sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \left (-\frac {b+2 c k}{2 k \sqrt {1-x^2} \left (1-k x^2\right ) \sqrt {1-k^2 x^2}}+\frac {b-2 c k}{2 k \sqrt {1-x^2} \left (1+k x^2\right ) \sqrt {1-k^2 x^2}}\right ) \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {c \sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left ((b-2 c k) \sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1}{\sqrt {1-x^2} \left (1+k x^2\right ) \sqrt {1-k^2 x^2}} \, dx}{2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left ((b+2 c k) \sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1}{\sqrt {1-x^2} \left (1-k x^2\right ) \sqrt {1-k^2 x^2}} \, dx}{2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {c \sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {(b-2 c k) \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (-k;\sin ^{-1}(x)|k^2\right )}{2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {(b+2 c k) \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (k;\sin ^{-1}(x)|k^2\right )}{2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ \end {align*}
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Mathematica [C] time = 0.51, size = 93, normalized size = 0.86 \begin {gather*} \frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left ((b-2 c k) \Pi \left (-k;\sin ^{-1}(x)|k^2\right )-(b+2 c k) \Pi \left (k;\sin ^{-1}(x)|k^2\right )+2 c k F\left (\sin ^{-1}(x)|k^2\right )\right )}{2 k \sqrt {\left (x^2-1\right ) \left (k^2 x^2-1\right )}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 3.27, size = 108, normalized size = 1.00 \begin {gather*} \frac {(-b-2 c k) \tan ^{-1}\left (\frac {(-1+k) x}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{4 (-1+k) k}+\frac {(b-2 c k) \tan ^{-1}\left (\frac {(1+k) x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{2 k (1+k)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.81, size = 107, normalized size = 0.99 \begin {gather*} \frac {{\left (2 \, c k^{2} - {\left (b + 2 \, c\right )} k + b\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k + 1\right )} x}\right ) + {\left (2 \, c k^{2} + {\left (b + 2 \, c\right )} k + b\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k - 1\right )} x}\right )}{4 \, {\left (k^{3} - k\right )}} \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 {c k^{2} x^{4} + b x^{2} + c}{{\left (k^{2} x^{4} - 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.14, size = 174, normalized size = 1.61
method | result | size |
default | \(\frac {c \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticF \left (x , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}+\frac {\left (-2 c k +b \right ) \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , -k , k\right )}{2 k \sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}-\frac {\left (2 c k +b \right ) \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , k , k\right )}{2 k \sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}\) | \(174\) |
elliptic | \(\frac {\left (\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (-1+k \right )}\right ) c}{-2+2 k}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (-1+k \right )}\right ) b}{4 k \left (-1+k \right )}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (1+k \right )}\right ) c}{2+2 k}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (1+k \right )}\right ) b}{4 k \left (1+k \right )}\right ) \sqrt {2}}{2}\) | \(177\) |
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 {c k^{2} x^{4} + b x^{2} + c}{{\left (k^{2} x^{4} - 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 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 {c\,k^2\,x^4+b\,x^2+c}{\left (k^2\,x^4-1\right )\,\sqrt {\left (x^2-1\right )\,\left (k^2\,x^2-1\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {b x^{2} + c k^{2} x^{4} + c}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (k x^{2} - 1\right ) \left (k x^{2} + 1\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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