Integrand size = 20, antiderivative size = 159 \[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^2} \, dx=\frac {e^3 \left (a+b x^2\right )^{1+p}}{2 b (1+p)}-\frac {d^3 \left (a+b x^2\right )^{1+p}}{a x}+\frac {d \left (3 a e^2+b d^2 (1+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{a}-\frac {3 d^2 e \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a (1+p)} \]
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Time = 0.13 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1821, 1666, 457, 81, 67, 12, 252, 251} \[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^2} \, dx=-\frac {d^3 \left (a+b x^2\right )^{p+1}}{a x}+\frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a e^2+b d^2 (2 p+1)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{a}-\frac {3 d^2 e \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{2 a (p+1)}+\frac {e^3 \left (a+b x^2\right )^{p+1}}{2 b (p+1)} \]
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Rule 12
Rule 67
Rule 81
Rule 251
Rule 252
Rule 457
Rule 1666
Rule 1821
Rubi steps \begin{align*} \text {integral}& = -\frac {d^3 \left (a+b x^2\right )^{1+p}}{a x}-\frac {\int \frac {\left (a+b x^2\right )^p \left (-3 a d^2 e-d \left (3 a e^2+b d^2 (1+2 p)\right ) x-a e^3 x^2\right )}{x} \, dx}{a} \\ & = -\frac {d^3 \left (a+b x^2\right )^{1+p}}{a x}+\frac {\int d \left (3 a e^2+b d^2 (1+2 p)\right ) \left (a+b x^2\right )^p \, dx}{a}-\frac {\int \frac {\left (a+b x^2\right )^p \left (-3 a d^2 e-a e^3 x^2\right )}{x} \, dx}{a} \\ & = -\frac {d^3 \left (a+b x^2\right )^{1+p}}{a x}-\frac {\text {Subst}\left (\int \frac {(a+b x)^p \left (-3 a d^2 e-a e^3 x\right )}{x} \, dx,x,x^2\right )}{2 a}+\frac {\left (d \left (3 a e^2+b d^2 (1+2 p)\right )\right ) \int \left (a+b x^2\right )^p \, dx}{a} \\ & = \frac {e^3 \left (a+b x^2\right )^{1+p}}{2 b (1+p)}-\frac {d^3 \left (a+b x^2\right )^{1+p}}{a x}+\frac {1}{2} \left (3 d^2 e\right ) \text {Subst}\left (\int \frac {(a+b x)^p}{x} \, dx,x,x^2\right )+\frac {\left (d \left (3 a e^2+b d^2 (1+2 p)\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}{a} \\ & = \frac {e^3 \left (a+b x^2\right )^{1+p}}{2 b (1+p)}-\frac {d^3 \left (a+b x^2\right )^{1+p}}{a x}+\frac {d \left (3 a e^2+b d^2 (1+2 p)\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 )}{a}-\frac {3 d^2 e \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1+\frac {b x^2}{a}\right )}{2 a (1+p)} \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.97 \[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^2} \, dx=\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (-2 a b d^3 (1+p) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )+e x \left (6 a b d e (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+\left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^p \left (a e^2-3 b d^2 \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )\right )\right )\right )}{2 a b (1+p) x} \]
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\[\int \frac {\left (e x +d \right )^{3} \left (b \,x^{2}+a \right )^{p}}{x^{2}}d x\]
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\[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p}}{x^{2}} \,d x } \]
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Time = 5.47 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.90 \[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^2} \, dx=- \frac {a^{p} d^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} + 3 a^{p} d e^{2} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} - \frac {3 b^{p} d^{2} e x^{2 p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} + e^{3} \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) \]
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\[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p}}{x^{2}} \,d x } \]
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\[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p}}{x^{2}} \,d x } \]
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Timed out. \[ \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^2} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^3}{x^2} \,d x \]
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