Integrand size = 18, antiderivative size = 113 \[ \int x (d+e x)^2 \left (a+b x^2\right )^p \, dx=\frac {\left (b d^2-a e^2\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {e^2 \left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)}+\frac {2}{3} d e x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right ) \]
[Out]
Time = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1666, 455, 45, 12, 372, 371} \[ \int x (d+e x)^2 \left (a+b x^2\right )^p \, dx=\frac {\left (b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)}+\frac {e^2 \left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}+\frac {2}{3} d e x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right ) \]
[In]
[Out]
Rule 12
Rule 45
Rule 371
Rule 372
Rule 455
Rule 1666
Rubi steps \begin{align*} \text {integral}& = \int 2 d e x^2 \left (a+b x^2\right )^p \, dx+\int x \left (a+b x^2\right )^p \left (d^2+e^2 x^2\right ) \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int (a+b x)^p \left (d^2+e^2 x\right ) \, dx,x,x^2\right )+(2 d e) \int x^2 \left (a+b x^2\right )^p \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {\left (b d^2-a e^2\right ) (a+b x)^p}{b}+\frac {e^2 (a+b x)^{1+p}}{b}\right ) \, dx,x,x^2\right )+\left (2 d e \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int x^2 \left (1+\frac {b x^2}{a}\right )^p \, dx \\ & = \frac {\left (b d^2-a e^2\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {e^2 \left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)}+\frac {2}{3} d e x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right ) \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.63 \[ \int x (d+e x)^2 \left (a+b x^2\right )^p \, dx=\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (3 b^2 x^2 \left (1+\frac {b x^2}{a}\right )^p \left (d^2 (2+p)+e^2 (1+p) x^2\right )-3 a^2 e^2 \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )+3 a b \left (e^2 p x^2 \left (1+\frac {b x^2}{a}\right )^p+d^2 (2+p) \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )\right )+4 b^2 d e \left (2+3 p+p^2\right ) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )\right )}{6 b^2 (1+p) (2+p)} \]
[In]
[Out]
\[\int x \left (e x +d \right )^{2} \left (b \,x^{2}+a \right )^{p}d x\]
[In]
[Out]
\[ \int x (d+e x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p} x \,d x } \]
[In]
[Out]
Time = 4.84 (sec) , antiderivative size = 408, normalized size of antiderivative = 3.61 \[ \int x (d+e x)^2 \left (a+b x^2\right )^p \, dx=\frac {2 a^{p} d e x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + d^{2} \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} - \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases}\right ) \]
[In]
[Out]
\[ \int x (d+e x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p} x \,d x } \]
[In]
[Out]
\[ \int x (d+e x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p} x \,d x } \]
[In]
[Out]
Timed out. \[ \int x (d+e x)^2 \left (a+b x^2\right )^p \, dx=\int x\,{\left (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^2 \,d x \]
[In]
[Out]