Optimal. Leaf size=97 \[ \frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{2 f}-\frac{b (d e-c f) \log \left ((c+d x)^2+1\right )}{2 d^2}-\frac{b (-c f+d e+f) (d e-(c+1) f) \tan ^{-1}(c+d x)}{2 d^2 f}-\frac{b f x}{2 d} \]
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Rubi [A] time = 0.112068, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {5047, 4862, 702, 635, 203, 260} \[ \frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{2 f}-\frac{b (d e-c f) \log \left ((c+d x)^2+1\right )}{2 d^2}-\frac{b (-c f+d e+f) (d e-(c+1) f) \tan ^{-1}(c+d x)}{2 d^2 f}-\frac{b f x}{2 d} \]
Antiderivative was successfully verified.
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Rule 5047
Rule 4862
Rule 702
Rule 635
Rule 203
Rule 260
Rubi steps
\begin{align*} \int (e+f x) \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 ) \left (a+b \tan ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{2 f}-\frac{b \operatorname{Subst}\left (\int \frac{\left (\frac{d e-c f}{d}+\frac{f x}{d}\right )^2}{1+x^2} \, dx,x,c+d x\right )}{2 f}\\ &=\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{2 f}-\frac{b \operatorname{Subst}\left (\int \left (\frac{f^2}{d^2}+\frac{(d e-f-c f) (d e+f-c f)+2 f (d e-c f) x}{d^2 \left (1+x^2\right )}\right ) \, dx,x,c+d x\right )}{2 f}\\ &=-\frac{b f x}{2 d}+\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{2 f}-\frac{b \operatorname{Subst}\left (\int \frac{(d e-f-c f) (d e+f-c f)+2 f (d e-c f) x}{1+x^2} \, dx,x,c+d x\right )}{2 d^2 f}\\ &=-\frac{b f x}{2 d}+\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{2 f}-\frac{(b (d e-c f)) \operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,c+d x\right )}{d^2}-\frac{(b (d e+f-c f) (d e-(1+c) f)) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,c+d x\right )}{2 d^2 f}\\ &=-\frac{b f x}{2 d}-\frac{b (d e+f-c f) (d e-(1+c) f) \tan ^{-1}(c+d x)}{2 d^2 f}+\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{2 f}-\frac{b (d e-c f) \log \left (1+(c+d x)^2\right )}{2 d^2}\\ \end{align*}
Mathematica [C] time = 0.0548483, size = 163, normalized size = 1.68 \[ a e x+\frac{1}{2} a f x^2-\frac{b e \left (\log \left (c^2+2 c d x+d^2 x^2+1\right )-2 c \tan ^{-1}(c+d x)\right )}{2 d}+\frac{b f \left (\frac{1}{2} d \left (\frac{c+d x}{d}-\frac{c}{d}\right )^2 \tan ^{-1}(c+d x)-\frac{1}{2} d \left (-\frac{i (-c+i)^2 \log (-c-d x+i)}{2 d^2}+\frac{i (c+i)^2 \log (c+d x+i)}{2 d^2}+\frac{x}{d}\right )\right )}{d}+b e x \tan ^{-1}(c+d x) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.039, size = 146, normalized size = 1.5 \begin{align*}{\frac{a{x}^{2}f}{2}}-{\frac{a{c}^{2}f}{2\,{d}^{2}}}+aex+{\frac{ace}{d}}+{\frac{bf\arctan \left ( dx+c \right ){x}^{2}}{2}}-{\frac{\arctan \left ( dx+c \right ) b{c}^{2}f}{2\,{d}^{2}}}+\arctan \left ( dx+c \right ) xbe+{\frac{b\arctan \left ( dx+c \right ) ce}{d}}-{\frac{bfx}{2\,d}}-{\frac{bcf}{2\,{d}^{2}}}+{\frac{b\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) cf}{2\,{d}^{2}}}-{\frac{b\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) e}{2\,d}}+{\frac{\arctan \left ( dx+c \right ) bf}{2\,{d}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.48311, size = 157, normalized size = 1.62 \begin{align*} \frac{1}{2} \, a f x^{2} + \frac{1}{2} \,{\left (x^{2} \arctan \left (d x + c\right ) - d{\left (\frac{x}{d^{2}} + \frac{{\left (c^{2} - 1\right )} \arctan \left (\frac{d^{2} x + c d}{d}\right )}{d^{3}} - \frac{c \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{3}}\right )}\right )} b f + a e x + \frac{{\left (2 \,{\left (d x + c\right )} \arctan \left (d x + c\right ) - \log \left ({\left (d x + c\right )}^{2} + 1\right )\right )} b e}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.42201, size = 232, normalized size = 2.39 \begin{align*} \frac{a d^{2} f x^{2} +{\left (2 \, a d^{2} e - b d f\right )} x +{\left (b d^{2} f x^{2} + 2 \, b d^{2} e x + 2 \, b c d e -{\left (b c^{2} - b\right )} f\right )} \arctan \left (d x + c\right ) -{\left (b d e - b c f\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{2 \, d^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 2.4684, size = 177, normalized size = 1.82 \begin{align*} \begin{cases} a e x + \frac{a f x^{2}}{2} - \frac{b c^{2} f \operatorname{atan}{\left (c + d x \right )}}{2 d^{2}} + \frac{b c e \operatorname{atan}{\left (c + d x \right )}}{d} + \frac{b c f \log{\left (\frac{c^{2}}{d^{2}} + \frac{2 c x}{d} + x^{2} + \frac{1}{d^{2}} \right )}}{2 d^{2}} + b e x \operatorname{atan}{\left (c + d x \right )} + \frac{b f x^{2} \operatorname{atan}{\left (c + d x \right )}}{2} - \frac{b e \log{\left (\frac{c^{2}}{d^{2}} + \frac{2 c x}{d} + x^{2} + \frac{1}{d^{2}} \right )}}{2 d} - \frac{b f x}{2 d} + \frac{b f \operatorname{atan}{\left (c + d x \right )}}{2 d^{2}} & \text{for}\: d \neq 0 \\\left (a + b \operatorname{atan}{\left (c \right )}\right ) \left (e x + \frac{f x^{2}}{2}\right ) & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.20523, size = 269, normalized size = 2.77 \begin{align*} \frac{2 \, b d^{2} f x^{2} \arctan \left (d x + c\right ) + 2 \, a d^{2} f x^{2} + 4 \, b d^{2} x \arctan \left (d x + c\right ) e - \pi b c^{2} f \mathrm{sgn}\left (d x + c\right ) - 4 \, \pi b c d e \mathrm{sgn}\left (d x + c\right ) + \pi b c^{2} f + 2 \, b c^{2} f \arctan \left (\frac{1}{d x + c}\right ) + 4 \, a d^{2} x e + 4 \, b c d \arctan \left (d x + c\right ) e - 2 \, b d f x + 2 \, b c f \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) - 2 \, b d e \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) + \pi b f \mathrm{sgn}\left (d x + c\right ) - \pi b f - 2 \, b f \arctan \left (\frac{1}{d x + c}\right )}{4 \, d^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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