Optimal. Leaf size=221 \[ -\frac {\left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac {(d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^2}-\frac {2 b (d e-c f) \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d^2}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f}+\frac {a b f x}{d}-\frac {b^2 (d e-c f) \text {Li}_2\left (-\frac {c+d x+1}{-c-d x+1}\right )}{d^2}+\frac {b^2 f \log \left (1-(c+d x)^2\right )}{2 d^2}+\frac {b^2 f (c+d x) \tanh ^{-1}(c+d x)}{d^2} \]
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Rubi [A] time = 0.46, antiderivative size = 220, 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 = {6111, 5928, 5910, 260, 6048, 5948, 5984, 5918, 2402, 2315} \[ -\frac {b^2 (d e-c f) \text {PolyLog}\left (2,-\frac {c+d x+1}{-c-d x+1}\right )}{d^2}+\frac {\left (-\frac {\left (c^2+1\right ) f}{d}+2 c e-\frac {d e^2}{f}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac {(d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^2}-\frac {2 b (d e-c f) \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d^2}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f}+\frac {a b f x}{d}+\frac {b^2 f \log \left (1-(c+d x)^2\right )}{2 d^2}+\frac {b^2 f (c+d x) \tanh ^{-1}(c+d x)}{d^2} \]
Antiderivative was successfully verified.
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Rule 260
Rule 2315
Rule 2402
Rule 5910
Rule 5918
Rule 5928
Rule 5948
Rule 5984
Rule 6048
Rule 6111
Rubi steps
\begin {align*} \int (e+f x) \left (a+b \tanh ^{-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 \tanh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f}-\frac {b \operatorname {Subst}\left (\int \left (-\frac {f^2 \left (a+b \tanh ^{-1}(x)\right )}{d^2}+\frac {\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2+2 f (d e-c f) x\right ) \left (a+b \tanh ^{-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 \tanh ^{-1}(c+d x)\right )^2}{2 f}-\frac {b \operatorname {Subst}\left (\int \frac {\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2+2 f (d e-c f) x\right ) \left (a+b \tanh ^{-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 \tanh ^{-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 \tanh ^{-1}(c+d x)\right )^2}{2 f}-\frac {b \operatorname {Subst}\left (\int \left (\frac {d^2 e^2 \left (1+\frac {f \left (-2 c d e+f+c^2 f\right )}{d^2 e^2}\right ) \left (a+b \tanh ^{-1}(x)\right )}{1-x^2}-\frac {2 f (-d e+c f) x \left (a+b \tanh ^{-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 \tanh ^{-1}(x) \, dx,x,c+d x\right )}{d^2}\\ &=\frac {a b f x}{d}+\frac {b^2 f (c+d x) \tanh ^{-1}(c+d x)}{d^2}+\frac {(e+f x)^2 \left (a+b \tanh ^{-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 \tanh ^{-1}(x)\right )}{1-x^2} \, dx,x,c+d x\right )}{d^2}-\frac {\left (b \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )\right ) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-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) \tanh ^{-1}(c+d x)}{d^2}+\frac {(d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^2}-\frac {\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac {(e+f x)^2 \left (a+b \tanh ^{-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 \tanh ^{-1}(x)}{1-x} \, dx,x,c+d x\right )}{d^2}\\ &=\frac {a b f x}{d}+\frac {b^2 f (c+d x) \tanh ^{-1}(c+d x)}{d^2}+\frac {(d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^2}-\frac {\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f}-\frac {2 b (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-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-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) \tanh ^{-1}(c+d x)}{d^2}+\frac {(d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^2}-\frac {\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f}-\frac {2 b (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-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 (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c-d x}\right )}{d^2}\\ &=\frac {a b f x}{d}+\frac {b^2 f (c+d x) \tanh ^{-1}(c+d x)}{d^2}+\frac {(d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^2}-\frac {\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f}-\frac {2 b (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d^2}+\frac {b^2 f \log \left (1-(c+d x)^2\right )}{2 d^2}-\frac {b^2 (d e-c f) \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{d^2}\\ \end {align*}
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Mathematica [A] time = 0.54, size = 271, normalized size = 1.23 \[ \frac {-a^2 c^2 f+2 a^2 c d e+2 a^2 d^2 e x+a^2 d^2 f x^2+2 b \tanh ^{-1}(c+d x) \left (-((c+d x) (a c f-a d (2 e+f x)-b f))-2 b (d e-c f) \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )\right )-4 a b d e \log \left (\frac {1}{\sqrt {1-(c+d x)^2}}\right )+a b f \log (-c-d x+1)-a b f \log (c+d x+1)+4 a b c f \log \left (\frac {1}{\sqrt {1-(c+d x)^2}}\right )+2 a b c f+2 a b d f x+2 b^2 (d e-c f) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )+b^2 (c+d x-1) \tanh ^{-1}(c+d x)^2 (-c f+2 d e+d f x+f)-2 b^2 f \log \left (\frac {1}{\sqrt {1-(c+d x)^2}}\right )}{2 d^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{2} f x + a^{2} e + {\left (b^{2} f x + b^{2} e\right )} \operatorname {artanh}\left (d x + c\right )^{2} + 2 \, {\left (a b f x + a b e\right )} \operatorname {artanh}\left (d x + c\right ), x\right ) \]
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 \[ \int {\left (f x + e\right )} {\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.06, size = 857, normalized size = 3.88 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.62, size = 415, normalized size = 1.88 \[ \frac {1}{2} \, a^{2} f x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {2 \, x}{d^{2}} - \frac {{\left (c^{2} + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{3}} + \frac {{\left (c^{2} - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{3}}\right )}\right )} a b f + a^{2} e x + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {artanh}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} a b e}{d} + \frac {{\left (d e - c f\right )} {\left (\log \left (d x + c + 1\right ) \log \left (-\frac {1}{2} \, d x - \frac {1}{2} \, c + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, d x + \frac {1}{2} \, c + \frac {1}{2}\right )\right )} b^{2}}{d^{2}} + \frac {{\left (c f + f\right )} b^{2} \log \left (d x + c + 1\right )}{2 \, d^{2}} - \frac {{\left (c f - f\right )} b^{2} \log \left (d x + c - 1\right )}{2 \, d^{2}} + \frac {4 \, b^{2} d f x \log \left (d x + c + 1\right ) + {\left (b^{2} d^{2} f x^{2} + 2 \, b^{2} d^{2} e x - {\left (c^{2} f - 2 \, {\left (d e - f\right )} c - 2 \, d e + f\right )} b^{2}\right )} \log \left (d x + c + 1\right )^{2} + {\left (b^{2} d^{2} f x^{2} + 2 \, b^{2} d^{2} e x - {\left (c^{2} f - 2 \, {\left (d e + f\right )} c + 2 \, d e + f\right )} b^{2}\right )} \log \left (-d x - c + 1\right )^{2} - 2 \, {\left (2 \, b^{2} d f x + {\left (b^{2} d^{2} f x^{2} + 2 \, b^{2} d^{2} e x - {\left (c^{2} f - 2 \, {\left (d e - f\right )} c - 2 \, d e + f\right )} b^{2}\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )}{8 \, d^{2}} \]
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 \[ \int \left (e+f\,x\right )\,{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^2 \,d x \]
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 \[ \int \left (a + b \operatorname {atanh}{\left (c + d x \right )}\right )^{2} \left (e + f x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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