Optimal. Leaf size=292 \[ \frac {(e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) (a d (a d (1-m)+b c (m+1)) (A d (3-m)+B c (m+1))-b c (a d (m+1)-b c (m+3)) (A d (m+1)-B c (m+5)))}{8 c^3 d^3 e (m+1)}+\frac {b (e x)^{m+1} (a d (m+1)-b c (m+3)) (A d (m+1)-B c (m+5))}{8 c^2 d^3 e (m+1)}-\frac {(e x)^{m+1} (b c-a d) \left (a (A d (3-m)+B c (m+1))-b x^2 (A d (m+1)-B c (m+5))\right )}{8 c^2 d^2 e \left (c+d x^2\right )}-\frac {\left (a+b x^2\right )^2 (e x)^{m+1} (B c-A d)}{4 c d e \left (c+d x^2\right )^2} \]
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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} \[ \frac {(e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) (a d (a d (1-m)+b c (m+1)) (A d (3-m)+B c (m+1))-b c (a d (m+1)-b c (m+3)) (A d (m+1)-B c (m+5)))}{8 c^3 d^3 e (m+1)}-\frac {(e x)^{m+1} (b c-a d) \left (a (A d (3-m)+B c (m+1))-b x^2 (A d (m+1)-B c (m+5))\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac {b (e x)^{m+1} (a d (m+1)-b c (m+3)) (A d (m+1)-B c (m+5))}{8 c^2 d^3 e (m+1)}-\frac {\left (a+b x^2\right )^2 (e x)^{m+1} (B c-A d)}{4 c d e \left (c+d x^2\right )^2} \]
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 )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx &=-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac {\int \frac {(e x)^m \left (a+b x^2\right ) \left (-a (A d (3-m)+B c (1+m))+b (A d (1+m)-B c (5+m)) x^2\right )}{\left (c+d x^2\right )^2} \, dx}{4 c d}\\ &=-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac {(b c-a d) (e x)^{1+m} \left (a (A d (3-m)+B c (1+m))-b (A d (1+m)-B c (5+m)) x^2\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac {\int \frac {(e x)^m \left (a (a d (1-m)+b c (1+m)) (A d (3-m)+B c (1+m))+b (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m)) x^2\right )}{c+d x^2} \, dx}{8 c^2 d^2}\\ &=\frac {b (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m)) (e x)^{1+m}}{8 c^2 d^3 e (1+m)}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac {(b c-a d) (e x)^{1+m} \left (a (A d (3-m)+B c (1+m))-b (A d (1+m)-B c (5+m)) x^2\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac {\left (a (a d (1-m)+b c (1+m)) (A d (3-m)+B c (1+m))-\frac {b c (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m))}{d}\right ) \int \frac {(e x)^m}{c+d x^2} \, dx}{8 c^2 d^2}\\ &=\frac {b (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m)) (e x)^{1+m}}{8 c^2 d^3 e (1+m)}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{4 c d e \left (c+d x^2\right )^2}-\frac {(b c-a d) (e x)^{1+m} \left (a (A d (3-m)+B c (1+m))-b (A d (1+m)-B c (5+m)) x^2\right )}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac {\left (a (a d (1-m)+b c (1+m)) (A d (3-m)+B c (1+m))-\frac {b c (a d (1+m)-b c (3+m)) (A d (1+m)-B c (5+m))}{d}\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{8 c^3 d^2 e (1+m)}\\ \end {align*}
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Mathematica [A] time = 0.20, size = 169, normalized size = 0.58 \[ \frac {x (e x)^m \left (-\frac {(b c-a d)^2 (B c-A d) \, _2F_1\left (3,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )}{c^3}+\frac {(b c-a d) \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) (-a B d-2 A b d+3 b B c)}{c^2}-\frac {b \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) (-2 a B d-A b d+3 b B c)}{c}+b^2 B\right )}{d^3 (m+1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B b^{2} x^{6} + {\left (2 \, B a b + A b^{2}\right )} x^{4} + A a^{2} + {\left (B a^{2} + 2 \, A a b\right )} x^{2}\right )} \left (e x\right )^{m}}{d^{3} x^{6} + 3 \, c d^{2} x^{4} + 3 \, c^{2} d x^{2} + c^{3}}, x\right ) \]
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 \[ \int \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}}\,{d x} \]
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 \[ \int \frac {\left (b \,x^{2}+a \right )^{2} \left (B \,x^{2}+A \right ) \left (e x \right )^{m}}{\left (d \,x^{2}+c \right )^{3}}\, dx \]
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 \[ \int \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}}\,{d x} \]
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 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,{\left (b\,x^2+a\right )}^2}{{\left (d\,x^2+c\right )}^3} \,d x \]
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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