Optimal. Leaf size=222 \[ \frac {i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d^2}-\frac {(-c f+d e+f) (d e-(c+1) f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac {2 b (d e-c f) \log \left (\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d^2}+\frac {(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 f}-\frac {a b f x}{d}+\frac {i b^2 (d e-c f) \text {Li}_2\left (1-\frac {2}{i (c+d x)+1}\right )}{d^2}+\frac {b^2 f \log \left ((c+d x)^2+1\right )}{2 d^2}-\frac {b^2 f (c+d x) \tan ^{-1}(c+d x)}{d^2} \]
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Rubi [A] time = 0.37, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {5047, 4864, 4846, 260, 4984, 4884, 4920, 4854, 2402, 2315} \[ \frac {i b^2 (d e-c f) \text {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{d^2}+\frac {i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d^2}-\frac {(-c f+d e+f) (d e-(c+1) f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac {2 b (d e-c f) \log \left (\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d^2}+\frac {(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 f}-\frac {a b f x}{d}+\frac {b^2 f \log \left ((c+d x)^2+1\right )}{2 d^2}-\frac {b^2 f (c+d x) \tan ^{-1}(c+d x)}{d^2} \]
Antiderivative was successfully verified.
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Rule 260
Rule 2315
Rule 2402
Rule 4846
Rule 4854
Rule 4864
Rule 4884
Rule 4920
Rule 4984
Rule 5047
Rubi steps
\begin {align*} \int (e+f x) \left (a+b \tan ^{-1}(c+d x)\right )^2 \, 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 )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 f}-\frac {b \operatorname {Subst}\left (\int \left (\frac {f^2 \left (a+b \tan ^{-1}(x)\right )}{d^2}+\frac {((d e-f-c f) (d e+f-c f)+2 f (d e-c f) x) \left (a+b \tan ^{-1}(x)\right )}{d^2 \left (1+x^2\right )}\right ) \, dx,x,c+d x\right )}{f}\\ &=\frac {(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{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) \left (a+b \tan ^{-1}(x)\right )}{1+x^2} \, dx,x,c+d x\right )}{d^2 f}-\frac {(b f) \operatorname {Subst}\left (\int \left (a+b \tan ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d^2}\\ &=-\frac {a b f x}{d}+\frac {(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 f}-\frac {b \operatorname {Subst}\left (\int \left (\frac {(d e+f-c f) (d e-(1+c) f) \left (a+b \tan ^{-1}(x)\right )}{1+x^2}-\frac {2 f (-d e+c f) x \left (a+b \tan ^{-1}(x)\right )}{1+x^2}\right ) \, dx,x,c+d x\right )}{d^2 f}-\frac {\left (b^2 f\right ) \operatorname {Subst}\left (\int \tan ^{-1}(x) \, dx,x,c+d x\right )}{d^2}\\ &=-\frac {a b f x}{d}-\frac {b^2 f (c+d x) \tan ^{-1}(c+d x)}{d^2}+\frac {(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 f}+\frac {\left (b^2 f\right ) \operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,c+d x\right )}{d^2}-\frac {(2 b (d e-c f)) \operatorname {Subst}\left (\int \frac {x \left (a+b \tan ^{-1}(x)\right )}{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 {a+b \tan ^{-1}(x)}{1+x^2} \, dx,x,c+d x\right )}{d^2 f}\\ &=-\frac {a b f x}{d}-\frac {b^2 f (c+d x) \tan ^{-1}(c+d x)}{d^2}+\frac {i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d^2}-\frac {(d e+f-c f) (d e-(1+c) f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac {(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 f}+\frac {b^2 f \log \left (1+(c+d x)^2\right )}{2 d^2}+\frac {(2 b (d e-c f)) \operatorname {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{i-x} \, dx,x,c+d x\right )}{d^2}\\ &=-\frac {a b f x}{d}-\frac {b^2 f (c+d x) \tan ^{-1}(c+d x)}{d^2}+\frac {i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d^2}-\frac {(d e+f-c f) (d e-(1+c) f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac {(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 f}+\frac {2 b (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (\frac {2}{1+i (c+d x)}\right )}{d^2}+\frac {b^2 f \log \left (1+(c+d x)^2\right )}{2 d^2}-\frac {\left (2 b^2 (d e-c f)\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d^2}\\ &=-\frac {a b f x}{d}-\frac {b^2 f (c+d x) \tan ^{-1}(c+d x)}{d^2}+\frac {i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d^2}-\frac {(d e+f-c f) (d e-(1+c) f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac {(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 f}+\frac {2 b (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (\frac {2}{1+i (c+d x)}\right )}{d^2}+\frac {b^2 f \log \left (1+(c+d x)^2\right )}{2 d^2}+\frac {\left (2 i b^2 (d e-c f)\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i (c+d x)}\right )}{d^2}\\ &=-\frac {a b f x}{d}-\frac {b^2 f (c+d x) \tan ^{-1}(c+d x)}{d^2}+\frac {i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d^2}-\frac {(d e+f-c f) (d e-(1+c) f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac {(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 f}+\frac {2 b (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (\frac {2}{1+i (c+d x)}\right )}{d^2}+\frac {b^2 f \log \left (1+(c+d x)^2\right )}{2 d^2}+\frac {i b^2 (d e-c f) \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{d^2}\\ \end {align*}
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Mathematica [A] time = 0.45, size = 264, normalized size = 1.19 \[ \frac {-a^2 c^2 f+2 a^2 c d e+2 a^2 d^2 e x+a^2 d^2 f x^2-2 b \tan ^{-1}(c+d x) \left (a \left (c^2 f-2 c d e-2 d^2 e x-f \left (d^2 x^2+1\right )\right )-2 b (d e-c f) \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )+b f (c+d x)\right )+4 a b d e \log \left (\frac {1}{\sqrt {(c+d x)^2+1}}\right )-4 a b c f \log \left (\frac {1}{\sqrt {(c+d x)^2+1}}\right )-2 a b c f-2 a b d f x-2 i b^2 (d e-c f) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c+d x)}\right )+b^2 (c+d x-i) \tan ^{-1}(c+d x)^2 (-c f+2 d e+d f x+i f)-2 b^2 f \log \left (\frac {1}{\sqrt {(c+d x)^2+1}}\right )}{2 d^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{2} f x + a^{2} e + {\left (b^{2} f x + b^{2} e\right )} \arctan \left (d x + c\right )^{2} + 2 \, {\left (a b f x + a b e\right )} \arctan \left (d x + c\right ), x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.14, size = 748, normalized size = 3.37 \[ -\frac {i b^{2} \ln \left (d x +c -i\right )^{2} c f}{4 d^{2}}+\frac {i b^{2} \ln \left (d x +c +i\right )^{2} c f}{4 d^{2}}-\frac {i b^{2} \ln \left (1+\left (d x +c \right )^{2}\right ) \ln \left (d x +c -i\right ) e}{2 d}-\frac {i b^{2} \ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right ) e}{2 d}+\frac {i b^{2} \ln \left (1+\left (d x +c \right )^{2}\right ) \ln \left (d x +c +i\right ) e}{2 d}+\frac {i b^{2} \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right ) \ln \left (d x +c -i\right ) e}{2 d}-\frac {i b^{2} \dilog \left (-\frac {i \left (d x +c +i\right )}{2}\right ) c f}{2 d^{2}}+\frac {i b^{2} \dilog \left (\frac {i \left (d x +c -i\right )}{2}\right ) c f}{2 d^{2}}-\frac {i b^{2} \ln \left (1+\left (d x +c \right )^{2}\right ) \ln \left (d x +c +i\right ) c f}{2 d^{2}}+\frac {i b^{2} \ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right ) c f}{2 d^{2}}+\frac {i b^{2} \ln \left (1+\left (d x +c \right )^{2}\right ) \ln \left (d x +c -i\right ) c f}{2 d^{2}}-\frac {i b^{2} \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right ) \ln \left (d x +c -i\right ) c f}{2 d^{2}}+\frac {i b^{2} \ln \left (d x +c -i\right )^{2} e}{4 d}-\frac {i b^{2} \dilog \left (\frac {i \left (d x +c -i\right )}{2}\right ) e}{2 d}+\frac {i b^{2} \dilog \left (-\frac {i \left (d x +c +i\right )}{2}\right ) e}{2 d}-\frac {i b^{2} \ln \left (d x +c +i\right )^{2} e}{4 d}-\frac {a b f x}{d}+\frac {b^{2} f \ln \left (1+\left (d x +c \right )^{2}\right )}{2 d^{2}}+\frac {b^{2} \arctan \left (d x +c \right )^{2} f}{2 d^{2}}+\frac {b^{2} \arctan \left (d x +c \right )^{2} f \,x^{2}}{2}+\arctan \left (d x +c \right )^{2} x \,b^{2} e +\frac {a^{2} c e}{d}-\frac {a^{2} f \,c^{2}}{2 d^{2}}+\frac {a b \ln \left (1+\left (d x +c \right )^{2}\right ) c f}{d^{2}}+\frac {2 \arctan \left (d x +c \right ) a b c e}{d}+\frac {b^{2} \ln \left (1+\left (d x +c \right )^{2}\right ) \arctan \left (d x +c \right ) c f}{d^{2}}-\frac {a b f \arctan \left (d x +c \right ) c^{2}}{d^{2}}+\frac {a b f \arctan \left (d x +c \right )}{d^{2}}+\frac {\arctan \left (d x +c \right )^{2} b^{2} c e}{d}-\frac {b^{2} \arctan \left (d x +c \right )^{2} f \,c^{2}}{2 d^{2}}-\frac {b^{2} \arctan \left (d x +c \right ) f x}{d}-\frac {b^{2} \arctan \left (d x +c \right ) f c}{d^{2}}+2 \arctan \left (d x +c \right ) x a b e +a b f \arctan \left (d x +c \right ) x^{2}-\frac {b^{2} \ln \left (1+\left (d x +c \right )^{2}\right ) \arctan \left (d x +c \right ) e}{d}-\frac {a b \ln \left (1+\left (d x +c \right )^{2}\right ) e}{d}-\frac {a b c f}{d^{2}}+\frac {a^{2} x^{2} f}{2}+a^{2} x e \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (e+f\,x\right )\,{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {atan}{\left (c + d x \right )}\right )^{2} \left (e + f x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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