Optimal. Leaf size=304 \[ \frac {d (A b c-2 a B c+a A d) (e x)^{1+m}}{2 a c (b c-a d)^2 e \left (c+d x^2\right )}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {b (A b (b c (1-m)-a d (5-m))+a B (a d (3-m)+b c (1+m))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^3 e (1+m)}-\frac {d (b c (B c (3-m)-A d (5-m))+a d (A d (1-m)+B c (1+m))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{2 c^2 (b c-a d)^3 e (1+m)} \]
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Rubi [A]
time = 0.54, antiderivative size = 304, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {593, 598, 371}
\begin {gather*} \frac {b (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right ) (A b (b c (1-m)-a d (5-m))+a B (a d (3-m)+b c (m+1)))}{2 a^2 e (m+1) (b c-a d)^3}-\frac {d (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) (a d (A d (1-m)+B c (m+1))+b c (B c (3-m)-A d (5-m)))}{2 c^2 e (m+1) (b c-a d)^3}+\frac {d (e x)^{m+1} (a A d-2 a B c+A b c)}{2 a c e \left (c+d x^2\right ) (b c-a d)^2}+\frac {(e x)^{m+1} (A b-a B)}{2 a e \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 593
Rule 598
Rubi steps
\begin {align*} \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx &=\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {\int \frac {(e x)^m \left (2 a A d-A b c (1-m)-a B c (1+m)-(A b-a 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 (A b c-2 a B c+a A d) (e x)^{1+m}}{2 a c (b c-a d)^2 e \left (c+d x^2\right )}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {\int \frac {(e x)^m \left (2 \left (A \left (4 a b c d-b^2 c^2 (1-m)-a^2 d^2 (1-m)\right )-a B c (b c+a d) (1+m)\right )-2 b d (A b c-2 a B c+a 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 (A b c-2 a B c+a A d) (e x)^{1+m}}{2 a c (b c-a d)^2 e \left (c+d x^2\right )}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {\int \left (\frac {2 b c (-A b (b c (1-m)-a d (5-m))-a B (a d (3-m)+b c (1+m))) (e x)^m}{(b c-a d) \left (a+b x^2\right )}+\frac {2 a d (b c (B c (3-m)-A d (5-m))+a d (A d (1-m)+B c (1+m))) (e x)^m}{(b c-a d) \left (c+d x^2\right )}\right ) \, dx}{4 a c (b c-a d)^2}\\ &=\frac {d (A b c-2 a B c+a A d) (e x)^{1+m}}{2 a c (b c-a d)^2 e \left (c+d x^2\right )}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {(b (A b (b c (1-m)-a d (5-m))+a B (a d (3-m)+b c (1+m)))) \int \frac {(e x)^m}{a+b x^2} \, dx}{2 a (b c-a d)^3}-\frac {(d (b c (B c (3-m)-A d (5-m))+a d (A d (1-m)+B c (1+m)))) \int \frac {(e x)^m}{c+d x^2} \, dx}{2 c (b c-a d)^3}\\ &=\frac {d (A b c-2 a B c+a A d) (e x)^{1+m}}{2 a c (b c-a d)^2 e \left (c+d x^2\right )}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {b (A b (b c (1-m)-a d (5-m))+a B (a d (3-m)+b c (1+m))) (e 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 e (1+m)}-\frac {d (b c (B c (3-m)-A d (5-m))+a d (A d (1-m)+B c (1+m))) (e 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 e (1+m)}\\ \end {align*}
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Mathematica [A]
time = 1.17, size = 207, normalized size = 0.68 \begin {gather*} \frac {x (e x)^m \left (-a b c^2 (b B c-2 A b d+a B d) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+a^2 c d (b B c-2 A b d+a B d) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )-(b c-a d) \left (b (A b-a B) c^2 \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+a^2 d (-B c+A d) \, _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 \[\int \frac {\left (e x \right )^{m} \left (B \,x^{2}+A \right )}{\left (b \,x^{2}+a \right )^{2} \left (d \,x^{2}+c \right )^{2}}\, dx\]
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 {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^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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