Integrand size = 18, antiderivative size = 206 \[ \int x^2 (a+b x)^n (c+d x)^p \, dx=-\frac {(b c (2+n)+a d (2+p)) (a+b x)^{1+n} (c+d x)^{1+p}}{b^2 d^2 (2+n+p) (3+n+p)}+\frac {x (a+b x)^{1+n} (c+d x)^{1+p}}{b d (3+n+p)}-\frac {\left (b^2 c^2 \left (2+3 n+n^2\right )+2 a b c d (1+n) (1+p)+a^2 d^2 \left (2+3 p+p^2\right )\right ) (a+b x)^{1+n} (c+d x)^{1+p} \operatorname {Hypergeometric2F1}\left (1,2+n+p,2+p,\frac {b (c+d x)}{b c-a d}\right )}{b^2 d^2 (b c-a d) (1+p) (2+n+p) (3+n+p)} \]
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Time = 0.12 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {92, 81, 72, 71} \[ \int x^2 (a+b x)^n (c+d x)^p \, dx=\frac {(a+b x)^{n+1} (c+d x)^p \left (a^2 d^2 \left (p^2+3 p+2\right )+2 a b c d (n+1) (p+1)+b^2 c^2 \left (n^2+3 n+2\right )\right ) \left (\frac {b (c+d x)}{b c-a d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (n+1,-p,n+2,-\frac {d (a+b x)}{b c-a d}\right )}{b^3 d^2 (n+1) (n+p+2) (n+p+3)}-\frac {(a+b x)^{n+1} (c+d x)^{p+1} (a d (p+2)+b c (n+2))}{b^2 d^2 (n+p+2) (n+p+3)}+\frac {x (a+b x)^{n+1} (c+d x)^{p+1}}{b d (n+p+3)} \]
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Rule 71
Rule 72
Rule 81
Rule 92
Rubi steps \begin{align*} \text {integral}& = \frac {x (a+b x)^{1+n} (c+d x)^{1+p}}{b d (3+n+p)}+\frac {\int (a+b x)^n (c+d x)^p (-a c-(b c (2+n)+a d (2+p)) x) \, dx}{b d (3+n+p)} \\ & = -\frac {(b c (2+n)+a d (2+p)) (a+b x)^{1+n} (c+d x)^{1+p}}{b^2 d^2 (2+n+p) (3+n+p)}+\frac {x (a+b x)^{1+n} (c+d x)^{1+p}}{b d (3+n+p)}+\frac {\left (b^2 c^2 \left (2+3 n+n^2\right )+2 a b c d (1+n) (1+p)+a^2 d^2 \left (2+3 p+p^2\right )\right ) \int (a+b x)^n (c+d x)^p \, dx}{b^2 d^2 (2+n+p) (3+n+p)} \\ & = -\frac {(b c (2+n)+a d (2+p)) (a+b x)^{1+n} (c+d x)^{1+p}}{b^2 d^2 (2+n+p) (3+n+p)}+\frac {x (a+b x)^{1+n} (c+d x)^{1+p}}{b d (3+n+p)}+\frac {\left (\left (b^2 c^2 \left (2+3 n+n^2\right )+2 a b c d (1+n) (1+p)+a^2 d^2 \left (2+3 p+p^2\right )\right ) (c+d x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-p}\right ) \int (a+b x)^n \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^p \, dx}{b^2 d^2 (2+n+p) (3+n+p)} \\ & = -\frac {(b c (2+n)+a d (2+p)) (a+b x)^{1+n} (c+d x)^{1+p}}{b^2 d^2 (2+n+p) (3+n+p)}+\frac {x (a+b x)^{1+n} (c+d x)^{1+p}}{b d (3+n+p)}+\frac {\left (b^2 c^2 \left (2+3 n+n^2\right )+2 a b c d (1+n) (1+p)+a^2 d^2 \left (2+3 p+p^2\right )\right ) (a+b x)^{1+n} (c+d x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-p} \, _2F_1\left (1+n,-p;2+n;-\frac {d (a+b x)}{b c-a d}\right )}{b^3 d^2 (1+n) (2+n+p) (3+n+p)} \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.86 \[ \int x^2 (a+b x)^n (c+d x)^p \, dx=\frac {(a+b x)^{1+n} (c+d x)^p \left (-\frac {(b c (2+n)+a d (2+p)) (c+d x)}{b d (2+n+p)}+x (c+d x)+\frac {\left (b^2 c^2 \left (2+3 n+n^2\right )+2 a b c d (1+n) (1+p)+a^2 d^2 \left (2+3 p+p^2\right )\right ) \left (\frac {b (c+d x)}{b c-a d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1+n,-p,2+n,\frac {d (a+b x)}{-b c+a d}\right )}{b^2 d (1+n) (2+n+p)}\right )}{b d (3+n+p)} \]
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\[\int x^{2} \left (b x +a \right )^{n} \left (d x +c \right )^{p}d x\]
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\[ \int x^2 (a+b x)^n (c+d x)^p \, dx=\int { {\left (b x + a\right )}^{n} {\left (d x + c\right )}^{p} x^{2} \,d x } \]
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Exception generated. \[ \int x^2 (a+b x)^n (c+d x)^p \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int x^2 (a+b x)^n (c+d x)^p \, dx=\int { {\left (b x + a\right )}^{n} {\left (d x + c\right )}^{p} x^{2} \,d x } \]
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\[ \int x^2 (a+b x)^n (c+d x)^p \, dx=\int { {\left (b x + a\right )}^{n} {\left (d x + c\right )}^{p} x^{2} \,d x } \]
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Timed out. \[ \int x^2 (a+b x)^n (c+d x)^p \, dx=\int x^2\,{\left (a+b\,x\right )}^n\,{\left (c+d\,x\right )}^p \,d x \]
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