Optimal. Leaf size=296 \[ -\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (-9 a^2 b c d^2 (2 p+7)+15 a^3 d^3+3 a b^2 c^2 d \left (4 p^2+24 p+35\right )-b^3 c^3 \left (8 p^3+60 p^2+142 p+105\right )\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{b^3 (2 p+3) (2 p+5) (2 p+7)}+\frac{d x \left (a+b x^2\right )^{p+1} \left (15 a^2 d^2-8 a b c d (p+6)+b^2 c^2 \left (4 p^2+28 p+57\right )\right )}{b^3 (2 p+3) (2 p+5) (2 p+7)}-\frac{d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1} (5 a d-b c (2 p+11))}{b^2 (2 p+5) (2 p+7)}+\frac{d x \left (c+d x^2\right )^2 \left (a+b x^2\right )^{p+1}}{b (2 p+7)} \]
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Rubi [A] time = 0.27632, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {416, 528, 388, 246, 245} \[ -\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (-9 a^2 b c d^2 (2 p+7)+15 a^3 d^3+3 a b^2 c^2 d \left (4 p^2+24 p+35\right )-b^3 c^3 \left (8 p^3+60 p^2+142 p+105\right )\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{b^3 (2 p+3) (2 p+5) (2 p+7)}+\frac{d x \left (a+b x^2\right )^{p+1} \left (15 a^2 d^2-8 a b c d (p+6)+b^2 c^2 \left (4 p^2+28 p+57\right )\right )}{b^3 (2 p+3) (2 p+5) (2 p+7)}-\frac{d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1} (5 a d-b c (2 p+11))}{b^2 (2 p+5) (2 p+7)}+\frac{d x \left (c+d x^2\right )^2 \left (a+b x^2\right )^{p+1}}{b (2 p+7)} \]
Antiderivative was successfully verified.
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Rule 416
Rule 528
Rule 388
Rule 246
Rule 245
Rubi steps
\begin{align*} \int \left (a+b x^2\right )^p \left (c+d x^2\right )^3 \, dx &=\frac{d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^2}{b (7+2 p)}+\frac{\int \left (a+b x^2\right )^p \left (c+d x^2\right ) \left (-c (a d-b c (7+2 p))-d (5 a d-b c (11+2 p)) x^2\right ) \, dx}{b (7+2 p)}\\ &=-\frac{d (5 a d-b c (11+2 p)) x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )}{b^2 (5+2 p) (7+2 p)}+\frac{d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^2}{b (7+2 p)}+\frac{\int \left (a+b x^2\right )^p \left (c \left (5 a^2 d^2-4 a b c d (4+p)+b^2 c^2 \left (35+24 p+4 p^2\right )\right )+d \left (15 a^2 d^2-8 a b c d (6+p)+b^2 c^2 \left (57+28 p+4 p^2\right )\right ) x^2\right ) \, dx}{b^2 (5+2 p) (7+2 p)}\\ &=\frac{d \left (15 a^2 d^2-8 a b c d (6+p)+b^2 c^2 \left (57+28 p+4 p^2\right )\right ) x \left (a+b x^2\right )^{1+p}}{b^3 (3+2 p) (5+2 p) (7+2 p)}-\frac{d (5 a d-b c (11+2 p)) x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )}{b^2 (5+2 p) (7+2 p)}+\frac{d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^2}{b (7+2 p)}-\frac{\left (15 a^3 d^3-9 a^2 b c d^2 (7+2 p)+3 a b^2 c^2 d \left (35+24 p+4 p^2\right )-b^3 c^3 \left (105+142 p+60 p^2+8 p^3\right )\right ) \int \left (a+b x^2\right )^p \, dx}{b^3 (3+2 p) (5+2 p) (7+2 p)}\\ &=\frac{d \left (15 a^2 d^2-8 a b c d (6+p)+b^2 c^2 \left (57+28 p+4 p^2\right )\right ) x \left (a+b x^2\right )^{1+p}}{b^3 (3+2 p) (5+2 p) (7+2 p)}-\frac{d (5 a d-b c (11+2 p)) x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )}{b^2 (5+2 p) (7+2 p)}+\frac{d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^2}{b (7+2 p)}-\frac{\left (\left (15 a^3 d^3-9 a^2 b c d^2 (7+2 p)+3 a b^2 c^2 d \left (35+24 p+4 p^2\right )-b^3 c^3 \left (105+142 p+60 p^2+8 p^3\right )\right ) \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p}\right ) \int \left (1+\frac{b x^2}{a}\right )^p \, dx}{b^3 (3+2 p) (5+2 p) (7+2 p)}\\ &=\frac{d \left (15 a^2 d^2-8 a b c d (6+p)+b^2 c^2 \left (57+28 p+4 p^2\right )\right ) x \left (a+b x^2\right )^{1+p}}{b^3 (3+2 p) (5+2 p) (7+2 p)}-\frac{d (5 a d-b c (11+2 p)) x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )}{b^2 (5+2 p) (7+2 p)}+\frac{d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^2}{b (7+2 p)}-\frac{\left (15 a^3 d^3-9 a^2 b c d^2 (7+2 p)+3 a b^2 c^2 d \left (35+24 p+4 p^2\right )-b^3 c^3 \left (105+142 p+60 p^2+8 p^3\right )\right ) x \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{b^3 (3+2 p) (5+2 p) (7+2 p)}\\ \end{align*}
Mathematica [A] time = 5.06293, size = 136, normalized size = 0.46 \[ \frac{1}{35} x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (d x^2 \left (35 c^2 \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )+d x^2 \left (21 c \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )+5 d x^2 \, _2F_1\left (\frac{7}{2},-p;\frac{9}{2};-\frac{b x^2}{a}\right )\right )\right )+35 c^3 \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 0.053, size = 0, normalized size = 0. \begin{align*} \int \left ( b{x}^{2}+a \right ) ^{p} \left ( d{x}^{2}+c \right ) ^{3}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x^{2} + c\right )}^{3}{\left (b x^{2} + a\right )}^{p}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (d^{3} x^{6} + 3 \, c d^{2} x^{4} + 3 \, c^{2} d x^{2} + c^{3}\right )}{\left (b x^{2} + a\right )}^{p}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [C] time = 52.5896, size = 121, normalized size = 0.41 \begin{align*} a^{p} c^{3} x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )} + a^{p} c^{2} d x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, - p \\ \frac{5}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )} + \frac{3 a^{p} c d^{2} x^{5}{{}_{2}F_{1}\left (\begin{matrix} \frac{5}{2}, - p \\ \frac{7}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{5} + \frac{a^{p} d^{3} x^{7}{{}_{2}F_{1}\left (\begin{matrix} \frac{7}{2}, - p \\ \frac{9}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{7} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x^{2} + c\right )}^{3}{\left (b x^{2} + a\right )}^{p}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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