Integrand size = 21, antiderivative size = 169 \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\frac {\text {c1} \left (a+2 b x+c x^2\right )^{1-n}}{2 c (1-n)}-\frac {2^{-n} (\text {b1} c-b \text {c1}) \left (-\frac {b-\sqrt {b^2-a c}+c x}{\sqrt {b^2-a c}}\right )^{-1+n} \left (a+2 b x+c x^2\right )^{1-n} \operatorname {Hypergeometric2F1}\left (1-n,n,2-n,\frac {b+\sqrt {b^2-a c}+c x}{2 \sqrt {b^2-a c}}\right )}{c \sqrt {b^2-a c} (1-n)} \]
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Time = 0.06 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {654, 638} \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\frac {\text {c1} \left (a+2 b x+c x^2\right )^{1-n}}{2 c (1-n)}-\frac {2^{-n} (\text {b1} c-b \text {c1}) \left (-\frac {-\sqrt {b^2-a c}+b+c x}{\sqrt {b^2-a c}}\right )^{n-1} \left (a+2 b x+c x^2\right )^{1-n} \operatorname {Hypergeometric2F1}\left (1-n,n,2-n,\frac {b+c x+\sqrt {b^2-a c}}{2 \sqrt {b^2-a c}}\right )}{c (1-n) \sqrt {b^2-a c}} \]
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Rule 638
Rule 654
Rubi steps \begin{align*} \text {integral}& = \frac {\text {c1} \left (a+2 b x+c x^2\right )^{1-n}}{2 c (1-n)}+\frac {(2 \text {b1} c-2 b \text {c1}) \int \left (a+2 b x+c x^2\right )^{-n} \, dx}{2 c} \\ & = \frac {\text {c1} \left (a+2 b x+c x^2\right )^{1-n}}{2 c (1-n)}-\frac {2^{-n} (\text {b1} c-b \text {c1}) \left (-\frac {b-\sqrt {b^2-a c}+c x}{\sqrt {b^2-a c}}\right )^{-1+n} \left (a+2 b x+c x^2\right )^{1-n} \operatorname {Hypergeometric2F1}\left (1-n,n,2-n,\frac {b+\sqrt {b^2-a c}+c x}{2 \sqrt {b^2-a c}}\right )}{c \sqrt {b^2-a c} (1-n)} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.79 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.56 \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\frac {1}{2} (a+x (2 b+c x))^{-n} \left (\text {c1} x^2 \left (\frac {b-\sqrt {b^2-a c}+c x}{b-\sqrt {b^2-a c}}\right )^n \left (\frac {b+\sqrt {b^2-a c}+c x}{b+\sqrt {b^2-a c}}\right )^n \operatorname {AppellF1}\left (2,n,n,3,-\frac {c x}{b+\sqrt {b^2-a c}},\frac {c x}{-b+\sqrt {b^2-a c}}\right )-\frac {2^{1-n} \text {b1} \left (b-\sqrt {b^2-a c}+c x\right ) \left (\frac {b+\sqrt {b^2-a c}+c x}{\sqrt {b^2-a c}}\right )^n \operatorname {Hypergeometric2F1}\left (1-n,n,2-n,\frac {-b+\sqrt {b^2-a c}-c x}{2 \sqrt {b^2-a c}}\right )}{c (-1+n)}\right ) \]
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\[\int \left (\operatorname {c1} x +\operatorname {b1} \right ) \left (c \,x^{2}+2 b x +a \right )^{-n}d x\]
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\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\int { \frac {c_{1} x + b_{1}}{{\left (c x^{2} + 2 \, b x + a\right )}^{n}} \,d x } \]
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Timed out. \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\text {Timed out} \]
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\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\int { \frac {c_{1} x + b_{1}}{{\left (c x^{2} + 2 \, b x + a\right )}^{n}} \,d x } \]
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\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\int { \frac {c_{1} x + b_{1}}{{\left (c x^{2} + 2 \, b x + a\right )}^{n}} \,d x } \]
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Timed out. \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx=\int \frac {b_{1}+c_{1}\,x}{{\left (c\,x^2+2\,b\,x+a\right )}^n} \,d x \]
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