Optimal. Leaf size=177 \[ -\frac{i b d (e+f x)^{m+2} \text{Hypergeometric2F1}\left (1,m+2,m+3,\frac{d (e+f x)}{-c f+d e+i f}\right )}{2 f (m+1) (m+2) (d e+(-c+i) f)}+\frac{i b d (e+f x)^{m+2} \text{Hypergeometric2F1}\left (1,m+2,m+3,\frac{d (e+f x)}{d e-(c+i) f}\right )}{2 f (m+1) (m+2) (d e-(c+i) f)}+\frac{(e+f x)^{m+1} \left (a+b \tan ^{-1}(c+d x)\right )}{f (m+1)} \]
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Rubi [A] time = 0.246324, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {5047, 4862, 712, 68} \[ -\frac{i b d (e+f x)^{m+2} \text{Hypergeometric2F1}\left (1,m+2,m+3,\frac{d (e+f x)}{-c f+d e+i f}\right )}{2 f (m+1) (m+2) (d e+(-c+i) f)}+\frac{i b d (e+f x)^{m+2} \text{Hypergeometric2F1}\left (1,m+2,m+3,\frac{d (e+f x)}{d e-(c+i) f}\right )}{2 f (m+1) (m+2) (d e-(c+i) f)}+\frac{(e+f x)^{m+1} \left (a+b \tan ^{-1}(c+d x)\right )}{f (m+1)} \]
Antiderivative was successfully verified.
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Rule 5047
Rule 4862
Rule 712
Rule 68
Rubi steps
\begin{align*} \int (e+f x)^m \left (a+b \tan ^{-1}(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \left (\frac{d e-c f}{d}+\frac{f x}{d}\right )^m \left (a+b \tan ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{(e+f x)^{1+m} \left (a+b \tan ^{-1}(c+d x)\right )}{f (1+m)}-\frac{b \operatorname{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 \tan ^{-1}(c+d x)\right )}{f (1+m)}-\frac{b \operatorname{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 \tan ^{-1}(c+d x)\right )}{f (1+m)}-\frac{(i b) \operatorname{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) \operatorname{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 \tan ^{-1}(c+d x)\right )}{f (1+m)}-\frac{i b d (e+f x)^{2+m} \, _2F_1\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} \, _2F_1\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*}
Mathematica [A] time = 0.287765, size = 162, normalized size = 0.92 \[ \frac{(e+f x)^{m+1} \left (2 \left (a+b \tan ^{-1}(c+d x)\right )+\frac{b d (e+f x) \left ((d e-(c+i) f) \text{Hypergeometric2F1}\left (1,m+2,m+3,\frac{d (e+f x)}{d e-(c-i) f}\right )+(-d e+(c-i) f) \text{Hypergeometric2F1}\left (1,m+2,m+3,\frac{d (e+f x)}{d e-(c+i) f}\right )\right )}{(m+2) (-i c f+i d e+f) (d e-(c-i) f)}\right )}{2 f (m+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.284, size = 0, normalized size = 0. \begin{align*} \int \left ( fx+e \right ) ^{m} \left ( a+b\arctan \left ( dx+c \right ) \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \arctan \left (d x + c\right ) + a\right )}{\left (f x + e\right )}^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arctan \left (d x + c\right ) + a\right )}{\left (f x + e\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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