Optimal. Leaf size=433 \[ -\frac {b \left (2 a^2 d^2 (1+m) (A d (3-m)+B c (1+m))+3 a b c d (3+m) (A d (1+m)-B c (5+m))-b^2 c^2 (5+m) (A d (3+m)-B c (7+m))\right ) (e x)^{1+m}}{8 c^2 d^4 e (1+m)}-\frac {b^2 (a d (3+m) (A d (3-m)+B c (1+m))+b c (5+m) (A d (3+m)-B c (7+m))) (e x)^{3+m}}{8 c^2 d^3 e^3 (3+m)}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^3}{4 c d e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (A d (3+m)-B c (7+m))) (e x)^{1+m} \left (a+b x^2\right )^2}{8 c^2 d^2 e \left (c+d x^2\right )}-\frac {(b c-a d) \left (a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+b^2 c^2 (5+m) (A d (3+m)-B c (7+m))+2 a b c d \left (A d \left (3-2 m-m^2\right )+B c \left (5+6 m+m^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{8 c^3 d^4 e (1+m)} \]
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Rubi [A]
time = 0.76, antiderivative size = 433, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {591, 584, 371}
\begin {gather*} -\frac {b (e x)^{m+1} \left (2 a^2 d^2 (m+1) (A d (3-m)+B c (m+1))+3 a b c d (m+3) (A d (m+1)-B c (m+5))-b^2 c^2 (m+5) (A d (m+3)-B c (m+7))\right )}{8 c^2 d^4 e (m+1)}-\frac {(e x)^{m+1} (b c-a d) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) \left (a^2 d^2 (1-m) (A d (3-m)+B c (m+1))+2 a b c d \left (A d \left (-m^2-2 m+3\right )+B c \left (m^2+6 m+5\right )\right )+b^2 c^2 (m+5) (A d (m+3)-B c (m+7))\right )}{8 c^3 d^4 e (m+1)}-\frac {b^2 (e x)^{m+3} (a d (m+3) (A d (3-m)+B c (m+1))+b c (m+5) (A d (m+3)-B c (m+7)))}{8 c^2 d^3 e^3 (m+3)}+\frac {\left (a+b x^2\right )^2 (e x)^{m+1} (a d (A d (3-m)+B c (m+1))+b c (A d (m+3)-B c (m+7)))}{8 c^2 d^2 e \left (c+d x^2\right )}-\frac {\left (a+b x^2\right )^3 (e x)^{m+1} (B c-A d)}{4 c d e \left (c+d x^2\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 371
Rule 584
Rule 591
Rubi steps
\begin {align*} \int \frac {(e x)^m \left (a+b x^2\right )^3 \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 )^3}{4 c d e \left (c+d x^2\right )^2}-\frac {\int \frac {(e x)^m \left (a+b x^2\right )^2 \left (-a (A d (3-m)+B c (1+m))+b (A d (3+m)-B c (7+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 )^3}{4 c d e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (A d (3+m)-B c (7+m))) (e x)^{1+m} \left (a+b x^2\right )^2}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac {\int \frac {(e x)^m \left (a+b x^2\right ) \left (a (a d (1-m) (A d (3-m)+B c (1+m))-b c (1+m) (A d (3+m)-B c (7+m)))-b (a d (3+m) (A d (3-m)+B c (1+m))+b c (5+m) (A d (3+m)-B c (7+m))) x^2\right )}{c+d x^2} \, dx}{8 c^2 d^2}\\ &=-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^3}{4 c d e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (A d (3+m)-B c (7+m))) (e x)^{1+m} \left (a+b x^2\right )^2}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac {\int \left (-\frac {b \left (2 a^2 d^2 (1+m) (A d (3-m)+B c (1+m))+3 a b c d (3+m) (A d (1+m)-B c (5+m))-b^2 c^2 (5+m) (A d (3+m)-B c (7+m))\right ) (e x)^m}{d^2}-\frac {b^2 (a d (3+m) (A d (3-m)+B c (1+m))+b c (5+m) (A d (3+m)-B c (7+m))) (e x)^{2+m}}{d e^2}+\frac {\left (35 b^3 B c^4-15 A b^3 c^3 d-45 a b^2 B c^3 d+9 a A b^2 c^2 d^2+9 a^2 b B c^2 d^2+3 a^2 A b c d^3+a^3 B c d^3+3 a^3 A d^4+12 b^3 B c^4 m-8 A b^3 c^3 d m-24 a b^2 B c^3 d m+12 a A b^2 c^2 d^2 m+12 a^2 b B c^2 d^2 m-4 a^3 A d^4 m+b^3 B c^4 m^2-A b^3 c^3 d m^2-3 a b^2 B c^3 d m^2+3 a A b^2 c^2 d^2 m^2+3 a^2 b B c^2 d^2 m^2-3 a^2 A b c d^3 m^2-a^3 B c d^3 m^2+a^3 A d^4 m^2\right ) (e x)^m}{d^2 \left (c+d x^2\right )}\right ) \, dx}{8 c^2 d^2}\\ &=-\frac {b \left (2 a^2 d^2 (1+m) (A d (3-m)+B c (1+m))+3 a b c d (3+m) (A d (1+m)-B c (5+m))-b^2 c^2 (5+m) (A d (3+m)-B c (7+m))\right ) (e x)^{1+m}}{8 c^2 d^4 e (1+m)}-\frac {b^2 (a d (3+m) (A d (3-m)+B c (1+m))+b c (5+m) (A d (3+m)-B c (7+m))) (e x)^{3+m}}{8 c^2 d^3 e^3 (3+m)}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^3}{4 c d e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (A d (3+m)-B c (7+m))) (e x)^{1+m} \left (a+b x^2\right )^2}{8 c^2 d^2 e \left (c+d x^2\right )}-\frac {\left ((b c-a d) \left (a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+b^2 c^2 (5+m) (A d (3+m)-B c (7+m))+2 a b c d \left (A d \left (3-2 m-m^2\right )+B c \left (5+6 m+m^2\right )\right )\right )\right ) \int \frac {(e x)^m}{c+d x^2} \, dx}{8 c^2 d^4}\\ &=-\frac {b \left (2 a^2 d^2 (1+m) (A d (3-m)+B c (1+m))+3 a b c d (3+m) (A d (1+m)-B c (5+m))-b^2 c^2 (5+m) (A d (3+m)-B c (7+m))\right ) (e x)^{1+m}}{8 c^2 d^4 e (1+m)}-\frac {b^2 (a d (3+m) (A d (3-m)+B c (1+m))+b c (5+m) (A d (3+m)-B c (7+m))) (e x)^{3+m}}{8 c^2 d^3 e^3 (3+m)}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^3}{4 c d e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (A d (3+m)-B c (7+m))) (e x)^{1+m} \left (a+b x^2\right )^2}{8 c^2 d^2 e \left (c+d x^2\right )}-\frac {(b c-a d) \left (a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+b^2 c^2 (5+m) (A d (3+m)-B c (7+m))+2 a b c d \left (A d \left (3-2 m-m^2\right )+B c \left (5+6 m+m^2\right )\right )\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^4 e (1+m)}\\ \end {align*}
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Mathematica [A]
time = 1.05, size = 581, normalized size = 1.34 \begin {gather*} \frac {x (e x)^m \left (-b^2 c^3 (3 b B c-A b d-3 a B d) (3+m)+b^3 B c^3 d (1+m) x^2+6 b^3 B c^4 (3+m) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )-3 A b^3 c^3 d (3+m) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )-9 a b^2 B c^3 d (3+m) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+3 a A b^2 c^2 d^2 (3+m) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+3 a^2 b B c^2 d^2 (3+m) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )-4 b^3 B c^4 (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+3 A b^3 c^3 d (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+9 a b^2 B c^3 d (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )-6 a A b^2 c^2 d^2 (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )-6 a^2 b B c^2 d^2 (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+3 a^2 A b c d^3 (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+a^3 B c d^3 (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+(b c-a d)^3 (B c-A d) (3+m) \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )\right )}{c^3 d^4 (1+m) (3+m)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right )^{3} \left (B \,x^{2}+A \right )}{\left (d \,x^{2}+c \right )^{3}}\, dx\]
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 \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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e x\right )^{m} \left (A + B x^{2}\right ) \left (a + b x^{2}\right )^{3}}{\left (c + d x^{2}\right )^{3}}\, dx \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 )}^3}{{\left (d\,x^2+c\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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