Optimal. Leaf size=292 \[ -\frac {(e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right ) (a d (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))-b c (a B (m+1)+A b (3-m)) (a d (m+1)+b (c-c m)))}{8 a^3 b^3 e (m+1)}+\frac {d (e x)^{m+1} (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))}{8 a^2 b^3 e (m+1)}+\frac {(e x)^{m+1} (b c-a d) \left (c (a B (m+1)+A b (3-m))-d x^2 (A b (m+1)-a B (m+5))\right )}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac {\left (c+d x^2\right )^2 (e x)^{m+1} (A b-a B)}{4 a b e \left (a+b x^2\right )^2} \]
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Rubi [A] time = 0.41, antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {577, 459, 364} \[ -\frac {(e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right ) (a d (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))-b c (a B (m+1)+A b (3-m)) (a d (m+1)+b (c-c m)))}{8 a^3 b^3 e (m+1)}+\frac {(e x)^{m+1} (b c-a d) \left (c (a B (m+1)+A b (3-m))-d x^2 (A b (m+1)-a B (m+5))\right )}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac {d (e x)^{m+1} (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))}{8 a^2 b^3 e (m+1)}+\frac {\left (c+d x^2\right )^2 (e x)^{m+1} (A b-a B)}{4 a b e \left (a+b x^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 364
Rule 459
Rule 577
Rubi steps
\begin {align*} \int \frac {(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^2}{\left (a+b x^2\right )^3} \, dx &=\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}-\frac {\int \frac {(e x)^m \left (c+d x^2\right ) \left (-c (A b (3-m)+a B (1+m))+d (A b (1+m)-a B (5+m)) x^2\right )}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}+\frac {(b c-a d) (e x)^{1+m} \left (c (A b (3-m)+a B (1+m))-d (A b (1+m)-a B (5+m)) x^2\right )}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac {\int \frac {(e x)^m \left (c (A b (3-m)+a B (1+m)) (b c (1-m)+a d (1+m))+d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m)) x^2\right )}{a+b x^2} \, dx}{8 a^2 b^2}\\ &=\frac {d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m)) (e x)^{1+m}}{8 a^2 b^3 e (1+m)}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}+\frac {(b c-a d) (e x)^{1+m} \left (c (A b (3-m)+a B (1+m))-d (A b (1+m)-a B (5+m)) x^2\right )}{8 a^2 b^2 e \left (a+b x^2\right )}-\frac {\left (\frac {a d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m))}{b}-c (A b (3-m)+a B (1+m)) (a d (1+m)+b (c-c m))\right ) \int \frac {(e x)^m}{a+b x^2} \, dx}{8 a^2 b^2}\\ &=\frac {d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m)) (e x)^{1+m}}{8 a^2 b^3 e (1+m)}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}+\frac {(b c-a d) (e x)^{1+m} \left (c (A b (3-m)+a B (1+m))-d (A b (1+m)-a B (5+m)) x^2\right )}{8 a^2 b^2 e \left (a+b x^2\right )}-\frac {\left (\frac {a d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m))}{b}-c (A b (3-m)+a B (1+m)) (a d (1+m)+b (c-c m))\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{8 a^3 b^2 e (1+m)}\\ \end {align*}
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Mathematica [A] time = 0.21, size = 165, normalized size = 0.57 \[ \frac {x (e x)^m \left (\frac {(A b-a B) (b c-a d)^2 \, _2F_1\left (3,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a^3}+\frac {(b c-a d) \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right ) (-3 a B d+2 A b d+b B c)}{a^2}+\frac {d \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right ) (-3 a B d+A b d+2 b B c)}{a}+B d^2\right )}{b^3 (m+1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B d^{2} x^{6} + {\left (2 \, B c d + A d^{2}\right )} x^{4} + A c^{2} + {\left (B c^{2} + 2 \, A c d\right )} x^{2}\right )} \left (e x\right )^{m}}{b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}}, x\right ) \]
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 \[ \int \frac {{\left (B x^{2} + A\right )} {\left (d x^{2} + c\right )}^{2} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \,x^{2}+A \right ) \left (d \,x^{2}+c \right )^{2} \left (e x \right )^{m}}{\left (b \,x^{2}+a \right )^{3}}\, dx \]
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 \[ \int \frac {{\left (B x^{2} + A\right )} {\left (d x^{2} + c\right )}^{2} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{3}}\,{d x} \]
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 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,{\left (d\,x^2+c\right )}^2}{{\left (b\,x^2+a\right )}^3} \,d x \]
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 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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