Optimal. Leaf size=171 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) (a f (5 c f+d e)+b e (c f+d e))}{16 e^{7/2} f^{5/2}}+\frac {x (a f (5 c f+d e)+b e (c f+d e))}{16 e^3 f^2 \left (e+f x^2\right )}-\frac {x (3 b e (c f+d e)-a f (5 c f+d e))}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac {x \left (a+b x^2\right ) (d e-c f)}{6 e f \left (e+f x^2\right )^3} \]
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Rubi [A] time = 0.16, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {526, 385, 199, 205} \begin {gather*} \frac {x (a f (5 c f+d e)+b e (c f+d e))}{16 e^3 f^2 \left (e+f x^2\right )}-\frac {x (3 b e (c f+d e)-a f (5 c f+d e))}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) (a f (5 c f+d e)+b e (c f+d e))}{16 e^{7/2} f^{5/2}}-\frac {x \left (a+b x^2\right ) (d e-c f)}{6 e f \left (e+f x^2\right )^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 199
Rule 205
Rule 385
Rule 526
Rubi steps
\begin {align*} \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx &=-\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {\int \frac {-a (d e+5 c f)-3 b (d e+c f) x^2}{\left (e+f x^2\right )^3} \, dx}{6 e f}\\ &=-\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \int \frac {1}{\left (e+f x^2\right )^2} \, dx}{8 e^2 f^2}\\ &=-\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \int \frac {1}{e+f x^2} \, dx}{16 e^3 f^2}\\ &=-\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{5/2}}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 171, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) (a f (5 c f+d e)+b e (c f+d e))}{16 e^{7/2} f^{5/2}}+\frac {x (a f (5 c f+d e)+b e (c f+d e))}{16 e^3 f^2 \left (e+f x^2\right )}+\frac {x (a f (5 c f+d e)+b e (c f-7 d e))}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {x (b e-a f) (d e-c f)}{6 e f^2 \left (e+f x^2\right )^3} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [B] time = 1.47, size = 642, normalized size = 3.75 \begin {gather*} \left [\frac {6 \, {\left (b d e^{3} f^{3} + 5 \, a c e f^{5} + {\left (b c + a d\right )} e^{2} f^{4}\right )} x^{5} - 16 \, {\left (b d e^{4} f^{2} - 5 \, a c e^{2} f^{4} - {\left (b c + a d\right )} e^{3} f^{3}\right )} x^{3} - 3 \, {\left (b d e^{5} + 5 \, a c e^{3} f^{2} + {\left (b d e^{2} f^{3} + 5 \, a c f^{5} + {\left (b c + a d\right )} e f^{4}\right )} x^{6} + {\left (b c + a d\right )} e^{4} f + 3 \, {\left (b d e^{3} f^{2} + 5 \, a c e f^{4} + {\left (b c + a d\right )} e^{2} f^{3}\right )} x^{4} + 3 \, {\left (b d e^{4} f + 5 \, a c e^{2} f^{3} + {\left (b c + a d\right )} e^{3} f^{2}\right )} x^{2}\right )} \sqrt {-e f} \log \left (\frac {f x^{2} - 2 \, \sqrt {-e f} x - e}{f x^{2} + e}\right ) - 6 \, {\left (b d e^{5} f - 11 \, a c e^{3} f^{3} + {\left (b c + a d\right )} e^{4} f^{2}\right )} x}{96 \, {\left (e^{4} f^{6} x^{6} + 3 \, e^{5} f^{5} x^{4} + 3 \, e^{6} f^{4} x^{2} + e^{7} f^{3}\right )}}, \frac {3 \, {\left (b d e^{3} f^{3} + 5 \, a c e f^{5} + {\left (b c + a d\right )} e^{2} f^{4}\right )} x^{5} - 8 \, {\left (b d e^{4} f^{2} - 5 \, a c e^{2} f^{4} - {\left (b c + a d\right )} e^{3} f^{3}\right )} x^{3} + 3 \, {\left (b d e^{5} + 5 \, a c e^{3} f^{2} + {\left (b d e^{2} f^{3} + 5 \, a c f^{5} + {\left (b c + a d\right )} e f^{4}\right )} x^{6} + {\left (b c + a d\right )} e^{4} f + 3 \, {\left (b d e^{3} f^{2} + 5 \, a c e f^{4} + {\left (b c + a d\right )} e^{2} f^{3}\right )} x^{4} + 3 \, {\left (b d e^{4} f + 5 \, a c e^{2} f^{3} + {\left (b c + a d\right )} e^{3} f^{2}\right )} x^{2}\right )} \sqrt {e f} \arctan \left (\frac {\sqrt {e f} x}{e}\right ) - 3 \, {\left (b d e^{5} f - 11 \, a c e^{3} f^{3} + {\left (b c + a d\right )} e^{4} f^{2}\right )} x}{48 \, {\left (e^{4} f^{6} x^{6} + 3 \, e^{5} f^{5} x^{4} + 3 \, e^{6} f^{4} x^{2} + e^{7} f^{3}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.34, size = 184, normalized size = 1.08 \begin {gather*} \frac {{\left (5 \, a c f^{2} + b c f e + a d f e + b d e^{2}\right )} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {7}{2}\right )}}{16 \, f^{\frac {5}{2}}} + \frac {{\left (15 \, a c f^{4} x^{5} + 3 \, b c f^{3} x^{5} e + 3 \, a d f^{3} x^{5} e + 3 \, b d f^{2} x^{5} e^{2} + 40 \, a c f^{3} x^{3} e + 8 \, b c f^{2} x^{3} e^{2} + 8 \, a d f^{2} x^{3} e^{2} - 8 \, b d f x^{3} e^{3} + 33 \, a c f^{2} x e^{2} - 3 \, b c f x e^{3} - 3 \, a d f x e^{3} - 3 \, b d x e^{4}\right )} e^{\left (-3\right )}}{48 \, {\left (f x^{2} + e\right )}^{3} f^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 210, normalized size = 1.23 \begin {gather*} \frac {5 a c \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \sqrt {e f}\, e^{3}}+\frac {a d \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \sqrt {e f}\, e^{2} f}+\frac {b c \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \sqrt {e f}\, e^{2} f}+\frac {b d \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \sqrt {e f}\, e \,f^{2}}+\frac {\frac {\left (5 a c \,f^{2}+a d e f +b c e f +b d \,e^{2}\right ) x^{5}}{16 e^{3}}+\frac {\left (5 a c \,f^{2}+a d e f +b c e f -b d \,e^{2}\right ) x^{3}}{6 e^{2} f}+\frac {\left (11 a c \,f^{2}-a d e f -b c e f -b d \,e^{2}\right ) x}{16 e \,f^{2}}}{\left (f \,x^{2}+e \right )^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.57, size = 194, normalized size = 1.13 \begin {gather*} \frac {3 \, {\left (b d e^{2} f^{2} + 5 \, a c f^{4} + {\left (b c + a d\right )} e f^{3}\right )} x^{5} - 8 \, {\left (b d e^{3} f - 5 \, a c e f^{3} - {\left (b c + a d\right )} e^{2} f^{2}\right )} x^{3} - 3 \, {\left (b d e^{4} - 11 \, a c e^{2} f^{2} + {\left (b c + a d\right )} e^{3} f\right )} x}{48 \, {\left (e^{3} f^{5} x^{6} + 3 \, e^{4} f^{4} x^{4} + 3 \, e^{5} f^{3} x^{2} + e^{6} f^{2}\right )}} + \frac {{\left (b d e^{2} + 5 \, a c f^{2} + {\left (b c + a d\right )} e f\right )} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \, \sqrt {e f} e^{3} f^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.18, size = 176, normalized size = 1.03 \begin {gather*} \frac {\frac {x^5\,\left (5\,a\,c\,f^2+b\,d\,e^2+a\,d\,e\,f+b\,c\,e\,f\right )}{16\,e^3}-\frac {x\,\left (b\,d\,e^2-11\,a\,c\,f^2+a\,d\,e\,f+b\,c\,e\,f\right )}{16\,e\,f^2}+\frac {x^3\,\left (5\,a\,c\,f^2-b\,d\,e^2+a\,d\,e\,f+b\,c\,e\,f\right )}{6\,e^2\,f}}{e^3+3\,e^2\,f\,x^2+3\,e\,f^2\,x^4+f^3\,x^6}+\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x}{\sqrt {e}}\right )\,\left (5\,a\,c\,f^2+b\,d\,e^2+a\,d\,e\,f+b\,c\,e\,f\right )}{16\,e^{7/2}\,f^{5/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 6.66, size = 313, normalized size = 1.83 \begin {gather*} - \frac {\sqrt {- \frac {1}{e^{7} f^{5}}} \left (5 a c f^{2} + a d e f + b c e f + b d e^{2}\right ) \log {\left (- e^{4} f^{2} \sqrt {- \frac {1}{e^{7} f^{5}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{e^{7} f^{5}}} \left (5 a c f^{2} + a d e f + b c e f + b d e^{2}\right ) \log {\left (e^{4} f^{2} \sqrt {- \frac {1}{e^{7} f^{5}}} + x \right )}}{32} + \frac {x^{5} \left (15 a c f^{4} + 3 a d e f^{3} + 3 b c e f^{3} + 3 b d e^{2} f^{2}\right ) + x^{3} \left (40 a c e f^{3} + 8 a d e^{2} f^{2} + 8 b c e^{2} f^{2} - 8 b d e^{3} f\right ) + x \left (33 a c e^{2} f^{2} - 3 a d e^{3} f - 3 b c e^{3} f - 3 b d e^{4}\right )}{48 e^{6} f^{2} + 144 e^{5} f^{3} x^{2} + 144 e^{4} f^{4} x^{4} + 48 e^{3} f^{5} x^{6}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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