Optimal. Leaf size=326 \[ -\frac {3 b^2 (d e-c f) \text {Li}_2\left (1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d^2}-\frac {3 b^2 f \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d^2}-\frac {\left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d^2 f}+\frac {(d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{d^2}-\frac {3 b (d e-c f) \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^2}+\frac {3 b f \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {3 b f (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 f}+\frac {3 b^3 (d e-c f) \text {Li}_3\left (1-\frac {2}{-c-d x+1}\right )}{2 d^2}-\frac {3 b^3 f \text {Li}_2\left (-\frac {c+d x+1}{-c-d x+1}\right )}{2 d^2} \]
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Rubi [A] time = 0.73, antiderivative size = 325, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {6111, 5928, 5910, 5984, 5918, 2402, 2315, 6048, 5948, 6058, 6610} \[ -\frac {3 b^2 (d e-c f) \text {PolyLog}\left (2,1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d^2}+\frac {3 b^3 (d e-c f) \text {PolyLog}\left (3,1-\frac {2}{-c-d x+1}\right )}{2 d^2}-\frac {3 b^3 f \text {PolyLog}\left (2,-\frac {c+d x+1}{-c-d x+1}\right )}{2 d^2}-\frac {3 b^2 f \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d^2}+\frac {\left (-\frac {\left (c^2+1\right ) f}{d}+2 c e-\frac {d e^2}{f}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}+\frac {(d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{d^2}-\frac {3 b (d e-c f) \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^2}+\frac {3 b f \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {3 b f (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 f} \]
Antiderivative was successfully verified.
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Rule 2315
Rule 2402
Rule 5910
Rule 5918
Rule 5928
Rule 5948
Rule 5984
Rule 6048
Rule 6058
Rule 6111
Rule 6610
Rubi steps
\begin {align*} \int (e+f x) \left (a+b \tanh ^{-1}(c+d x)\right )^3 \, 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 )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 f}-\frac {(3 b) \operatorname {Subst}\left (\int \left (-\frac {f^2 \left (a+b \tanh ^{-1}(x)\right )^2}{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 )^2}{d^2 \left (1-x^2\right )}\right ) \, dx,x,c+d x\right )}{2 f}\\ &=\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 f}-\frac {(3 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 )^2}{1-x^2} \, dx,x,c+d x\right )}{2 d^2 f}+\frac {(3 b f) \operatorname {Subst}\left (\int \left (a+b \tanh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{2 d^2}\\ &=\frac {3 b f (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 f}-\frac {(3 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 )^2}{1-x^2}-\frac {2 f (-d e+c f) x \left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2}\right ) \, dx,x,c+d x\right )}{2 d^2 f}-\frac {\left (3 b^2 f\right ) \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 {3 b f \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {3 b f (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 f}-\frac {\left (3 b^2 f\right ) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{1-x} \, dx,x,c+d x\right )}{d^2}-\frac {(3 b (d e-c f)) \operatorname {Subst}\left (\int \frac {x \left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{d^2}-\frac {\left (3 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 {\left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{2 d^2 f}\\ &=\frac {3 b f \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {3 b f (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {(d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{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 )^3}{2 d^2 f}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 f}-\frac {3 b^2 f \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d^2}+\frac {\left (3 b^3 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 {(3 b (d e-c f)) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{1-x} \, dx,x,c+d x\right )}{d^2}\\ &=\frac {3 b f \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {3 b f (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {(d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{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 )^3}{2 d^2 f}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 f}-\frac {3 b^2 f \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d^2}-\frac {3 b (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1-c-d x}\right )}{d^2}-\frac {\left (3 b^3 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 {\left (6 b^2 (d e-c f)\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right ) \log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d^2}\\ &=\frac {3 b f \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {3 b f (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {(d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{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 )^3}{2 d^2 f}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 f}-\frac {3 b^2 f \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d^2}-\frac {3 b (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1-c-d x}\right )}{d^2}-\frac {3 b^3 f \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{2 d^2}-\frac {3 b^2 (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{d^2}+\frac {\left (3 b^3 (d e-c f)\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (1-\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d^2}\\ &=\frac {3 b f \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {3 b f (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^2}+\frac {(d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{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 )^3}{2 d^2 f}+\frac {(e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 f}-\frac {3 b^2 f \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d^2}-\frac {3 b (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1-c-d x}\right )}{d^2}-\frac {3 b^3 f \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{2 d^2}-\frac {3 b^2 (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{d^2}+\frac {3 b^3 (d e-c f) \text {Li}_3\left (1-\frac {2}{1-c-d x}\right )}{2 d^2}\\ \end {align*}
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Mathematica [A] time = 1.15, size = 566, normalized size = 1.74 \[ \frac {2 a^3 f (c+d x)^2+2 a^2 (c+d x) (-2 a c f+2 a d e+3 b f)+3 a^2 b (-2 c f+2 d e+f) \log (-c-d x+1)+3 a^2 b (2 d e-(2 c+1) f) \log (c+d x+1)-6 a^2 b (c+d x) \tanh ^{-1}(c+d x) (c f-d (2 e+f x))+12 a b^2 d e \left (\text {Li}_2\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )+\tanh ^{-1}(c+d x) \left ((c+d x-1) \tanh ^{-1}(c+d x)-2 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )\right )\right )-12 a b^2 c f \left (\text {Li}_2\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )+\tanh ^{-1}(c+d x) \left ((c+d x-1) \tanh ^{-1}(c+d x)-2 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )\right )\right )+12 a b^2 f \left (-\log \left (\frac {1}{\sqrt {1-(c+d x)^2}}\right )-\frac {1}{2} \left (1-(c+d x)^2\right ) \tanh ^{-1}(c+d x)^2+(c+d x) \tanh ^{-1}(c+d x)\right )+2 b^3 f \left (\tanh ^{-1}(c+d x) \left (\left (c^2+2 c d x+d^2 x^2-1\right ) \tanh ^{-1}(c+d x)^2+3 (c+d x-1) \tanh ^{-1}(c+d x)-6 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )\right )+3 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )\right )+4 b^3 d e \left (3 \tanh ^{-1}(c+d x) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )+\frac {3}{2} \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )+\tanh ^{-1}(c+d x)^2 \left ((c+d x-1) \tanh ^{-1}(c+d x)-3 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )\right )\right )-4 b^3 c f \left (3 \tanh ^{-1}(c+d x) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )+\frac {3}{2} \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )+\tanh ^{-1}(c+d x)^2 \left ((c+d x-1) \tanh ^{-1}(c+d x)-3 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )\right )\right )}{4 d^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{3} f x + a^{3} e + {\left (b^{3} f x + b^{3} e\right )} \operatorname {artanh}\left (d x + c\right )^{3} + 3 \, {\left (a b^{2} f x + a b^{2} e\right )} \operatorname {artanh}\left (d x + c\right )^{2} + 3 \, {\left (a^{2} b f x + a^{2} b e\right )} \operatorname {artanh}\left (d x + c\right ), x\right ) \]
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 \[ \int {\left (f x + e\right )} {\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 1.68, size = 20255, normalized size = 62.13 \[ \text {output too large to display} \]
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 \[ \frac {1}{2} \, a^{3} f x^{2} + \frac {3}{4} \, {\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^{2} b f + a^{3} e x + \frac {3 \, {\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^{2} b e}{2 \, d} - \frac {{\left (b^{3} d^{2} f x^{2} + 2 \, b^{3} d^{2} e x - {\left (c^{2} f - 2 \, {\left (d e + f\right )} c + 2 \, d e + f\right )} b^{3}\right )} \log \left (-d x - c + 1\right )^{3} - 3 \, {\left (2 \, a b^{2} d^{2} f x^{2} + 2 \, {\left (2 \, a b^{2} d^{2} e + b^{3} d f\right )} x + {\left (b^{3} d^{2} f x^{2} + 2 \, b^{3} d^{2} e x - {\left (c^{2} f - 2 \, {\left (d e - f\right )} c - 2 \, d e + f\right )} b^{3}\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )^{2}}{16 \, d^{2}} - \int -\frac {{\left (b^{3} d^{2} f x^{2} + {\left (d^{2} e + c d f - d f\right )} b^{3} x + {\left (c d e - d e\right )} b^{3}\right )} \log \left (d x + c + 1\right )^{3} + 6 \, {\left (a b^{2} d^{2} f x^{2} + {\left (d^{2} e + c d f - d f\right )} a b^{2} x + {\left (c d e - d e\right )} a b^{2}\right )} \log \left (d x + c + 1\right )^{2} - 3 \, {\left (2 \, a b^{2} d^{2} f x^{2} + {\left (b^{3} d^{2} f x^{2} + {\left (d^{2} e + c d f - d f\right )} b^{3} x + {\left (c d e - d e\right )} b^{3}\right )} \log \left (d x + c + 1\right )^{2} + 2 \, {\left (2 \, a b^{2} d^{2} e + b^{3} d f\right )} x + {\left (4 \, {\left (c d e - d e\right )} a b^{2} - {\left (c^{2} f - 2 \, {\left (d e - f\right )} c - 2 \, d e + f\right )} b^{3} + {\left (4 \, a b^{2} d^{2} f + b^{3} d^{2} f\right )} x^{2} + 2 \, {\left (b^{3} d^{2} e + 2 \, {\left (d^{2} e + c d f - d f\right )} a b^{2}\right )} x\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )}{8 \, {\left (d^{2} x + c d - d\right )}}\,{d x} \]
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 \[ \int \left (e+f\,x\right )\,{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3 \,d x \]
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 \[ \int \left (a + b \operatorname {atanh}{\left (c + d x \right )}\right )^{3} \left (e + f x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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