Optimal. Leaf size=498 \[ \frac {e x^3 \left (-3 b e \left (a e (4 p+5)+c^2 \left (8 p^2+26 p+21\right )\right )+3 b^2 c^2 \left (16 p^2+48 p+35\right )+c^2 e^2 \left (4 p^2+16 p+15\right )\right ) \left (\frac {2 b x^2}{c-\sqrt {c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac {2 b x^2}{\sqrt {c^2-4 a b}+c}+1\right )^{-p} F_1\left (\frac {3}{2};-p,-p;\frac {5}{2};-\frac {2 b x^2}{c-\sqrt {c^2-4 a b}},-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}}\right )}{3 b^2 (4 p+5) (4 p+7)}+\frac {c x \left (-3 a b e^2 (4 p+7)+a e^3 (2 p+5)+b^2 c^2 \left (16 p^2+48 p+35\right )\right ) \left (\frac {2 b x^2}{c-\sqrt {c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac {2 b x^2}{\sqrt {c^2-4 a b}+c}+1\right )^{-p} F_1\left (\frac {1}{2};-p,-p;\frac {3}{2};-\frac {2 b x^2}{c-\sqrt {c^2-4 a b}},-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}}\right )}{b^2 (4 p+5) (4 p+7)}+\frac {c e^2 x (12 b p+21 b-2 e p-5 e) \left (a+b x^4+c x^2\right )^{p+1}}{b^2 (4 p+5) (4 p+7)}+\frac {e^3 x^3 \left (a+b x^4+c x^2\right )^{p+1}}{b (4 p+7)} \]
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Rubi [A] time = 0.81, antiderivative size = 498, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1206, 1679, 1203, 1105, 429, 1141, 510} \[ \frac {c x \left (-3 a b e^2 (4 p+7)+a e^3 (2 p+5)+b^2 c^2 \left (16 p^2+48 p+35\right )\right ) \left (\frac {2 b x^2}{c-\sqrt {c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac {2 b x^2}{\sqrt {c^2-4 a b}+c}+1\right )^{-p} F_1\left (\frac {1}{2};-p,-p;\frac {3}{2};-\frac {2 b x^2}{c-\sqrt {c^2-4 a b}},-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}}\right )}{b^2 (4 p+5) (4 p+7)}+\frac {e x^3 \left (-3 b e \left (a e (4 p+5)+c^2 \left (8 p^2+26 p+21\right )\right )+3 b^2 c^2 \left (16 p^2+48 p+35\right )+c^2 e^2 \left (4 p^2+16 p+15\right )\right ) \left (\frac {2 b x^2}{c-\sqrt {c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac {2 b x^2}{\sqrt {c^2-4 a b}+c}+1\right )^{-p} F_1\left (\frac {3}{2};-p,-p;\frac {5}{2};-\frac {2 b x^2}{c-\sqrt {c^2-4 a b}},-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}}\right )}{3 b^2 (4 p+5) (4 p+7)}+\frac {c e^2 x (12 b p+21 b-2 e p-5 e) \left (a+b x^4+c x^2\right )^{p+1}}{b^2 (4 p+5) (4 p+7)}+\frac {e^3 x^3 \left (a+b x^4+c x^2\right )^{p+1}}{b (4 p+7)} \]
Antiderivative was successfully verified.
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Rule 429
Rule 510
Rule 1105
Rule 1141
Rule 1203
Rule 1206
Rule 1679
Rubi steps
\begin {align*} \int \left (c+e x^2\right )^3 \left (a+c x^2+b x^4\right )^p \, dx &=\frac {e^3 x^3 \left (a+c x^2+b x^4\right )^{1+p}}{b (7+4 p)}+\frac {\int \left (a+c x^2+b x^4\right )^p \left (b c^3 (7+4 p)-3 e \left (a e^2-b c^2 (7+4 p)\right ) x^2+c e^2 (21 b-5 e+12 b p-2 e p) x^4\right ) \, dx}{b (7+4 p)}\\ &=\frac {c e^2 (21 b-5 e+12 b p-2 e p) x \left (a+c x^2+b x^4\right )^{1+p}}{b^2 (5+4 p) (7+4 p)}+\frac {e^3 x^3 \left (a+c x^2+b x^4\right )^{1+p}}{b (7+4 p)}+\frac {\int \left (c \left (a e^3 (5+2 p)-3 a b e^2 (7+4 p)+b^2 c^2 \left (35+48 p+16 p^2\right )\right )+e \left (c^2 e^2 \left (15+16 p+4 p^2\right )+3 b^2 c^2 \left (35+48 p+16 p^2\right )-3 b e \left (a e (5+4 p)+c^2 \left (21+26 p+8 p^2\right )\right )\right ) x^2\right ) \left (a+c x^2+b x^4\right )^p \, dx}{b^2 (5+4 p) (7+4 p)}\\ &=\frac {c e^2 (21 b-5 e+12 b p-2 e p) x \left (a+c x^2+b x^4\right )^{1+p}}{b^2 (5+4 p) (7+4 p)}+\frac {e^3 x^3 \left (a+c x^2+b x^4\right )^{1+p}}{b (7+4 p)}+\frac {\int \left (c \left (a e^3 (5+2 p)-3 a b e^2 (7+4 p)+b^2 c^2 \left (35+48 p+16 p^2\right )\right ) \left (a+c x^2+b x^4\right )^p+e \left (c^2 e^2 \left (15+16 p+4 p^2\right )+3 b^2 c^2 \left (35+48 p+16 p^2\right )-3 b e \left (a e (5+4 p)+c^2 \left (21+26 p+8 p^2\right )\right )\right ) x^2 \left (a+c x^2+b x^4\right )^p\right ) \, dx}{b^2 (5+4 p) (7+4 p)}\\ &=\frac {c e^2 (21 b-5 e+12 b p-2 e p) x \left (a+c x^2+b x^4\right )^{1+p}}{b^2 (5+4 p) (7+4 p)}+\frac {e^3 x^3 \left (a+c x^2+b x^4\right )^{1+p}}{b (7+4 p)}+\frac {\left (c \left (a e^3 (5+2 p)-3 a b e^2 (7+4 p)+b^2 c^2 \left (35+48 p+16 p^2\right )\right )\right ) \int \left (a+c x^2+b x^4\right )^p \, dx}{b^2 (5+4 p) (7+4 p)}+\frac {\left (e \left (c^2 e^2 \left (15+16 p+4 p^2\right )+3 b^2 c^2 \left (35+48 p+16 p^2\right )-3 b e \left (a e (5+4 p)+c^2 \left (21+26 p+8 p^2\right )\right )\right )\right ) \int x^2 \left (a+c x^2+b x^4\right )^p \, dx}{b^2 (5+4 p) (7+4 p)}\\ &=\frac {c e^2 (21 b-5 e+12 b p-2 e p) x \left (a+c x^2+b x^4\right )^{1+p}}{b^2 (5+4 p) (7+4 p)}+\frac {e^3 x^3 \left (a+c x^2+b x^4\right )^{1+p}}{b (7+4 p)}+\frac {\left (c \left (a e^3 (5+2 p)-3 a b e^2 (7+4 p)+b^2 c^2 \left (35+48 p+16 p^2\right )\right ) \left (1+\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (1+\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p\right ) \int \left (1+\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^p \left (1+\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^p \, dx}{b^2 (5+4 p) (7+4 p)}+\frac {\left (e \left (c^2 e^2 \left (15+16 p+4 p^2\right )+3 b^2 c^2 \left (35+48 p+16 p^2\right )-3 b e \left (a e (5+4 p)+c^2 \left (21+26 p+8 p^2\right )\right )\right ) \left (1+\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (1+\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p\right ) \int x^2 \left (1+\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^p \left (1+\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^p \, dx}{b^2 (5+4 p) (7+4 p)}\\ &=\frac {c e^2 (21 b-5 e+12 b p-2 e p) x \left (a+c x^2+b x^4\right )^{1+p}}{b^2 (5+4 p) (7+4 p)}+\frac {e^3 x^3 \left (a+c x^2+b x^4\right )^{1+p}}{b (7+4 p)}+\frac {c \left (a e^3 (5+2 p)-3 a b e^2 (7+4 p)+b^2 c^2 \left (35+48 p+16 p^2\right )\right ) x \left (1+\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (1+\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p F_1\left (\frac {1}{2};-p,-p;\frac {3}{2};-\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )}{b^2 (5+4 p) (7+4 p)}+\frac {e \left (c^2 e^2 \left (15+16 p+4 p^2\right )+3 b^2 c^2 \left (35+48 p+16 p^2\right )-3 b e \left (a e (5+4 p)+c^2 \left (21+26 p+8 p^2\right )\right )\right ) x^3 \left (1+\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (1+\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p F_1\left (\frac {3}{2};-p,-p;\frac {5}{2};-\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )}{3 b^2 (5+4 p) (7+4 p)}\\ \end {align*}
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Mathematica [A] time = 0.51, size = 373, normalized size = 0.75 \[ \frac {1}{35} x \left (\frac {-\sqrt {c^2-4 a b}+2 b x^2+c}{c-\sqrt {c^2-4 a b}}\right )^{-p} \left (\frac {\sqrt {c^2-4 a b}+2 b x^2+c}{\sqrt {c^2-4 a b}+c}\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (e x^2 \left (e x^2 \left (5 e x^2 F_1\left (\frac {7}{2};-p,-p;\frac {9}{2};-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}},\frac {2 b x^2}{\sqrt {c^2-4 a b}-c}\right )+21 c F_1\left (\frac {5}{2};-p,-p;\frac {7}{2};-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}},\frac {2 b x^2}{\sqrt {c^2-4 a b}-c}\right )\right )+35 c^2 F_1\left (\frac {3}{2};-p,-p;\frac {5}{2};-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}},\frac {2 b x^2}{\sqrt {c^2-4 a b}-c}\right )\right )+35 c^3 F_1\left (\frac {1}{2};-p,-p;\frac {3}{2};-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}},\frac {2 b x^2}{\sqrt {c^2-4 a b}-c}\right )\right ) \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{3} x^{6} + 3 \, c e^{2} x^{4} + 3 \, c^{2} e x^{2} + c^{3}\right )} {\left (b x^{4} + c x^{2} + a\right )}^{p}, x\right ) \]
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 \[ \int {\left (e x^{2} + c\right )}^{3} {\left (b x^{4} + c x^{2} + a\right )}^{p}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.09, size = 0, normalized size = 0.00 \[ \int \left (e \,x^{2}+c \right )^{3} \left (b \,x^{4}+c \,x^{2}+a \right )^{p}\, dx \]
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 \[ \int {\left (e x^{2} + c\right )}^{3} {\left (b x^{4} + c x^{2} + a\right )}^{p}\,{d x} \]
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 \[ \int {\left (e\,x^2+c\right )}^3\,{\left (b\,x^4+c\,x^2+a\right )}^p \,d x \]
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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