Integrand size = 22, antiderivative size = 146 \[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx=\frac {\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (2+p+q)}-\frac {(b c (1+p)+a d (1+q)) \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^q \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (1+p,-q,2+p,-\frac {d \left (a+b x^2\right )}{b c-a d}\right )}{2 b^2 d (1+p) (2+p+q)} \]
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Time = 0.09 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {457, 81, 72, 71} \[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx=\frac {\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^{q+1}}{2 b d (p+q+2)}-\frac {\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^q (a d (q+1)+b c (p+1)) \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (p+1,-q,p+2,-\frac {d \left (b x^2+a\right )}{b c-a d}\right )}{2 b^2 d (p+1) (p+q+2)} \]
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Rule 71
Rule 72
Rule 81
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x (a+b x)^p (c+d x)^q \, dx,x,x^2\right ) \\ & = \frac {\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (2+p+q)}-\frac {(b c (1+p)+a d (1+q)) \text {Subst}\left (\int (a+b x)^p (c+d x)^q \, dx,x,x^2\right )}{2 b d (2+p+q)} \\ & = \frac {\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (2+p+q)}-\frac {\left ((b c (1+p)+a d (1+q)) \left (c+d x^2\right )^q \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q}\right ) \text {Subst}\left (\int (a+b x)^p \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^q \, dx,x,x^2\right )}{2 b d (2+p+q)} \\ & = \frac {\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (2+p+q)}-\frac {(b c (1+p)+a d (1+q)) \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^q \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \, _2F_1\left (1+p,-q;2+p;-\frac {d \left (a+b x^2\right )}{b c-a d}\right )}{2 b^2 d (1+p) (2+p+q)} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.81 \[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx=\frac {\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^q \left (b \left (c+d x^2\right )-\frac {(b c (1+p)+a d (1+q)) \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (1+p,-q,2+p,\frac {d \left (a+b x^2\right )}{-b c+a d}\right )}{1+p}\right )}{2 b^2 d (2+p+q)} \]
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\[\int x^{3} \left (b \,x^{2}+a \right )^{p} \left (d \,x^{2}+c \right )^{q}d x\]
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\[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx=\int { {\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q} x^{3} \,d x } \]
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Timed out. \[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx=\text {Timed out} \]
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\[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx=\int { {\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q} x^{3} \,d x } \]
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\[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx=\int { {\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q} x^{3} \,d x } \]
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Timed out. \[ \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx=\int x^3\,{\left (b\,x^2+a\right )}^p\,{\left (d\,x^2+c\right )}^q \,d x \]
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