Optimal. Leaf size=230 \[ \frac {d (b c+a d) x^{1+m}}{2 a c (b c-a d)^2 \left (c+d x^2\right )}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {b^2 (a d (5-m)-b (c-c m)) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^3 (1+m)}-\frac {d^2 (a d (1-m)-b c (5-m)) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{2 c^2 (b c-a d)^3 (1+m)} \]
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Rubi [A]
time = 0.27, antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {483, 593, 598,
371} \begin {gather*} -\frac {b^2 x^{m+1} (a d (5-m)-b (c-c m)) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{2 a^2 (m+1) (b c-a d)^3}-\frac {d^2 x^{m+1} (a d (1-m)-b c (5-m)) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )}{2 c^2 (m+1) (b c-a d)^3}+\frac {d x^{m+1} (a d+b c)}{2 a c \left (c+d x^2\right ) (b c-a d)^2}+\frac {b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)} \end {gather*}
Antiderivative was successfully verified.
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Rule 371
Rule 483
Rule 593
Rule 598
Rubi steps
\begin {align*} \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx &=\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {\int \frac {x^m \left (2 a d-b c (1-m)-b d (3-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{2 a (b c-a d)}\\ &=\frac {d (b c+a d) x^{1+m}}{2 a c (b c-a d)^2 \left (c+d x^2\right )}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {\int \frac {x^m \left (2 \left (4 a b c d-b^2 c^2 (1-m)-a^2 d^2 (1-m)\right )-2 b d (b c+a d) (1-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{4 a c (b c-a d)^2}\\ &=\frac {d (b c+a d) x^{1+m}}{2 a c (b c-a d)^2 \left (c+d x^2\right )}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {\int \left (\frac {2 b^2 c (-b c (1-m)+a d (5-m)) x^m}{(b c-a d) \left (a+b x^2\right )}+\frac {2 a d^2 (a d (1-m)-b c (5-m)) x^m}{(b c-a d) \left (c+d x^2\right )}\right ) \, dx}{4 a c (b c-a d)^2}\\ &=\frac {d (b c+a d) x^{1+m}}{2 a c (b c-a d)^2 \left (c+d x^2\right )}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {\left (d^2 (a d (1-m)-b c (5-m))\right ) \int \frac {x^m}{c+d x^2} \, dx}{2 c (b c-a d)^3}+\frac {\left (b^2 (b c (1-m)-a d (5-m))\right ) \int \frac {x^m}{a+b x^2} \, dx}{2 a (b c-a d)^3}\\ &=\frac {d (b c+a d) x^{1+m}}{2 a c (b c-a d)^2 \left (c+d x^2\right )}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {b^2 (b c (1-m)-a d (5-m)) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^3 (1+m)}-\frac {d^2 (a d (1-m)-b c (5-m)) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{2 c^2 (b c-a d)^3 (1+m)}\\ \end {align*}
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Mathematica [A]
time = 0.41, size = 173, normalized size = 0.75 \begin {gather*} \frac {x^{1+m} \left (2 a b^2 c^2 d \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )-2 a^2 b c d^2 \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )-(b c-a d) \left (b^2 c^2 \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+a^2 d^2 \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )\right )\right )}{a^2 c^2 (-b c+a d)^3 (1+m)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{m}}{\left (b \,x^{2}+a \right )^{2} \left (d \,x^{2}+c \right )^{2}}\, dx\]
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*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
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*} \text {could not integrate} \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 = -0.00 \begin {gather*} \int \frac {x^m}{{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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