Integrand size = 18, antiderivative size = 177 \[ \int (e+f x)^m \left (a+b \cot ^{-1}(c+d x)\right ) \, dx=\frac {(e+f x)^{1+m} \left (a+b \cot ^{-1}(c+d x)\right )}{f (1+m)}+\frac {i b d (e+f x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {d (e+f x)}{d e+i f-c f}\right )}{2 f (d e+(i-c) f) (1+m) (2+m)}-\frac {i b d (e+f x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {d (e+f x)}{d e-(i+c) f}\right )}{2 f (d e-(i+c) f) (1+m) (2+m)} \]
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Time = 0.21 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5156, 4973, 726, 70} \[ \int (e+f x)^m \left (a+b \cot ^{-1}(c+d x)\right ) \, dx=\frac {(e+f x)^{m+1} \left (a+b \cot ^{-1}(c+d x)\right )}{f (m+1)}+\frac {i b d (e+f x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {d (e+f x)}{d e-c f+i f}\right )}{2 f (m+1) (m+2) (d e+(-c+i) f)}-\frac {i b d (e+f x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {d (e+f x)}{d e-(c+i) f}\right )}{2 f (m+1) (m+2) (d e-(c+i) f)} \]
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Rule 70
Rule 726
Rule 4973
Rule 5156
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^m \left (a+b \cot ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d} \\ & = \frac {(e+f x)^{1+m} \left (a+b \cot ^{-1}(c+d x)\right )}{f (1+m)}+\frac {b \text {Subst}\left (\int \frac {\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{1+x^2} \, dx,x,c+d x\right )}{f (1+m)} \\ & = \frac {(e+f x)^{1+m} \left (a+b \cot ^{-1}(c+d x)\right )}{f (1+m)}+\frac {b \text {Subst}\left (\int \left (\frac {i \left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{2 (i-x)}+\frac {i \left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{2 (i+x)}\right ) \, dx,x,c+d x\right )}{f (1+m)} \\ & = \frac {(e+f x)^{1+m} \left (a+b \cot ^{-1}(c+d x)\right )}{f (1+m)}+\frac {(i b) \text {Subst}\left (\int \frac {\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{i-x} \, dx,x,c+d x\right )}{2 f (1+m)}+\frac {(i b) \text {Subst}\left (\int \frac {\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{i+x} \, dx,x,c+d x\right )}{2 f (1+m)} \\ & = \frac {(e+f x)^{1+m} \left (a+b \cot ^{-1}(c+d x)\right )}{f (1+m)}+\frac {i b d (e+f x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {d (e+f x)}{d e+i f-c f}\right )}{2 f (d e+(i-c) f) (1+m) (2+m)}-\frac {i b d (e+f x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {d (e+f x)}{d e-(i+c) f}\right )}{2 f (d e-(i+c) f) (1+m) (2+m)} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.92 \[ \int (e+f x)^m \left (a+b \cot ^{-1}(c+d x)\right ) \, dx=\frac {(e+f x)^{1+m} \left (2 \left (a+b \cot ^{-1}(c+d x)\right )+\frac {b d (e+f x) \left ((d e-(i+c) f) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {d (e+f x)}{d e-(-i+c) f}\right )+(-d e+(-i+c) f) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {d (e+f x)}{d e-(i+c) f}\right )\right )}{(-i d e+f+i c f) (d e-(i+c) f) (2+m)}\right )}{2 f (1+m)} \]
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\[\int \left (f x +e \right )^{m} \left (a +b \,\operatorname {arccot}\left (d x +c \right )\right )d x\]
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\[ \int (e+f x)^m \left (a+b \cot ^{-1}(c+d x)\right ) \, dx=\int { {\left (b \operatorname {arccot}\left (d x + c\right ) + a\right )} {\left (f x + e\right )}^{m} \,d x } \]
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Timed out. \[ \int (e+f x)^m \left (a+b \cot ^{-1}(c+d x)\right ) \, dx=\text {Timed out} \]
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\[ \int (e+f x)^m \left (a+b \cot ^{-1}(c+d x)\right ) \, dx=\int { {\left (b \operatorname {arccot}\left (d x + c\right ) + a\right )} {\left (f x + e\right )}^{m} \,d x } \]
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\[ \int (e+f x)^m \left (a+b \cot ^{-1}(c+d x)\right ) \, dx=\int { {\left (b \operatorname {arccot}\left (d x + c\right ) + a\right )} {\left (f x + e\right )}^{m} \,d x } \]
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Timed out. \[ \int (e+f x)^m \left (a+b \cot ^{-1}(c+d x)\right ) \, dx=\int {\left (e+f\,x\right )}^m\,\left (a+b\,\mathrm {acot}\left (c+d\,x\right )\right ) \,d x \]
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