Integrand size = 18, antiderivative size = 77 \[ \int \frac {(a+b x)^n \left (c+d x^2\right )}{x} \, dx=-\frac {a d (a+b x)^{1+n}}{b^2 (1+n)}+\frac {d (a+b x)^{2+n}}{b^2 (2+n)}-\frac {c (a+b x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b x}{a}\right )}{a (1+n)} \]
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Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {966, 81, 67} \[ \int \frac {(a+b x)^n \left (c+d x^2\right )}{x} \, dx=-\frac {a d (a+b x)^{n+1}}{b^2 (n+1)}+\frac {d (a+b x)^{n+2}}{b^2 (n+2)}-\frac {c (a+b x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b x}{a}+1\right )}{a (n+1)} \]
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Rule 67
Rule 81
Rule 966
Rubi steps \begin{align*} \text {integral}& = \frac {d (a+b x)^{2+n}}{b^2 (2+n)}+\frac {\int \frac {(a+b x)^n \left (b^2 c (2+n)-a b d (2+n) x\right )}{x} \, dx}{b^2 (2+n)} \\ & = -\frac {a d (a+b x)^{1+n}}{b^2 (1+n)}+\frac {d (a+b x)^{2+n}}{b^2 (2+n)}+c \int \frac {(a+b x)^n}{x} \, dx \\ & = -\frac {a d (a+b x)^{1+n}}{b^2 (1+n)}+\frac {d (a+b x)^{2+n}}{b^2 (2+n)}-\frac {c (a+b x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {b x}{a}\right )}{a (1+n)} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.83 \[ \int \frac {(a+b x)^n \left (c+d x^2\right )}{x} \, dx=-\frac {(a+b x)^{1+n} \left (a d (a-b (1+n) x)+b^2 c (2+n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b x}{a}\right )\right )}{a b^2 (1+n) (2+n)} \]
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\[\int \frac {\left (b x +a \right )^{n} \left (d \,x^{2}+c \right )}{x}d x\]
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\[ \int \frac {(a+b x)^n \left (c+d x^2\right )}{x} \, dx=\int { \frac {{\left (d x^{2} + c\right )} {\left (b x + a\right )}^{n}}{x} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (61) = 122\).
Time = 2.29 (sec) , antiderivative size = 279, normalized size of antiderivative = 3.62 \[ \int \frac {(a+b x)^n \left (c+d x^2\right )}{x} \, dx=d \left (\begin {cases} \frac {a^{n} x^{2}}{2} & \text {for}\: b = 0 \\\frac {a \log {\left (\frac {a}{b} + x \right )}}{a b^{2} + b^{3} x} + \frac {a}{a b^{2} + b^{3} x} + \frac {b x \log {\left (\frac {a}{b} + x \right )}}{a b^{2} + b^{3} x} & \text {for}\: n = -2 \\- \frac {a \log {\left (\frac {a}{b} + x \right )}}{b^{2}} + \frac {x}{b} & \text {for}\: n = -1 \\- \frac {a^{2} \left (a + b x\right )^{n}}{b^{2} n^{2} + 3 b^{2} n + 2 b^{2}} + \frac {a b n x \left (a + b x\right )^{n}}{b^{2} n^{2} + 3 b^{2} n + 2 b^{2}} + \frac {b^{2} n x^{2} \left (a + b x\right )^{n}}{b^{2} n^{2} + 3 b^{2} n + 2 b^{2}} + \frac {b^{2} x^{2} \left (a + b x\right )^{n}}{b^{2} n^{2} + 3 b^{2} n + 2 b^{2}} & \text {otherwise} \end {cases}\right ) - \frac {b^{n + 1} c n \left (\frac {a}{b} + x\right )^{n + 1} \Phi \left (1 + \frac {b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{a \Gamma \left (n + 2\right )} - \frac {b^{n + 1} c \left (\frac {a}{b} + x\right )^{n + 1} \Phi \left (1 + \frac {b x}{a}, 1, n + 1\right ) \Gamma \left (n + 1\right )}{a \Gamma \left (n + 2\right )} \]
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\[ \int \frac {(a+b x)^n \left (c+d x^2\right )}{x} \, dx=\int { \frac {{\left (d x^{2} + c\right )} {\left (b x + a\right )}^{n}}{x} \,d x } \]
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\[ \int \frac {(a+b x)^n \left (c+d x^2\right )}{x} \, dx=\int { \frac {{\left (d x^{2} + c\right )} {\left (b x + a\right )}^{n}}{x} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^n \left (c+d x^2\right )}{x} \, dx=\int \frac {\left (d\,x^2+c\right )\,{\left (a+b\,x\right )}^n}{x} \,d x \]
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