Optimal. Leaf size=160 \[ -\frac {3 b^2 e \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d}+\frac {3 b e \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac {3 b e (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}-\frac {e \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}-\frac {3 b^3 e \text {Li}_2\left (-\frac {c+d x+1}{-c-d x+1}\right )}{2 d} \]
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Rubi [A] time = 0.26, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {6107, 12, 5916, 5980, 5910, 5984, 5918, 2402, 2315, 5948} \[ -\frac {3 b^3 e \text {PolyLog}\left (2,-\frac {c+d x+1}{-c-d x+1}\right )}{2 d}-\frac {3 b^2 e \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d}+\frac {3 b e \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac {3 b e (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}-\frac {e \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2315
Rule 2402
Rule 5910
Rule 5916
Rule 5918
Rule 5948
Rule 5980
Rule 5984
Rule 6107
Rubi steps
\begin {align*} \int (c e+d e x) \left (a+b \tanh ^{-1}(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int e x \left (a+b \tanh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac {e \operatorname {Subst}\left (\int x \left (a+b \tanh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}-\frac {(3 b e) \operatorname {Subst}\left (\int \frac {x^2 \left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{2 d}\\ &=\frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}+\frac {(3 b e) \operatorname {Subst}\left (\int \left (a+b \tanh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{2 d}-\frac {(3 b e) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{2 d}\\ &=\frac {3 b e (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}-\frac {e \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}+\frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}-\frac {\left (3 b^2 e\right ) \operatorname {Subst}\left (\int \frac {x \left (a+b \tanh ^{-1}(x)\right )}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {3 b e \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac {3 b e (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}-\frac {e \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}+\frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}-\frac {\left (3 b^2 e\right ) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{1-x} \, dx,x,c+d x\right )}{d}\\ &=\frac {3 b e \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac {3 b e (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}-\frac {e \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}+\frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}-\frac {3 b^2 e \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d}+\frac {\left (3 b^3 e\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {3 b e \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac {3 b e (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}-\frac {e \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}+\frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}-\frac {3 b^2 e \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d}-\frac {\left (3 b^3 e\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c-d x}\right )}{d}\\ &=\frac {3 b e \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac {3 b e (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}-\frac {e \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}+\frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{2 d}-\frac {3 b^2 e \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d}-\frac {3 b^3 e \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{2 d}\\ \end {align*}
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Mathematica [A] time = 1.13, size = 213, normalized size = 1.33 \[ \frac {e \left (a \left (-3 a b \left (c^2-1\right ) \log (-c-d x+1)+3 a b \left (c^2-1\right ) \log (c+d x+1)+2 a d x (2 a c+a d x+3 b)-12 b^2 \log \left (\frac {1}{\sqrt {1-(c+d x)^2}}\right )\right )+6 b^2 (c+d x-1) \tanh ^{-1}(c+d x)^2 (a (c+d x+1)+b)+6 b \tanh ^{-1}(c+d x) \left (a (a d x (2 c+d x)+2 b (c+d x))-2 b^2 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )\right )+2 b^3 \left (c^2+2 c d x+d^2 x^2-1\right ) \tanh ^{-1}(c+d x)^3+6 b^3 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )\right )}{4 d} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{3} d e x + a^{3} c e + {\left (b^{3} d e x + b^{3} c e\right )} \operatorname {artanh}\left (d x + c\right )^{3} + 3 \, {\left (a b^{2} d e x + a b^{2} c e\right )} \operatorname {artanh}\left (d x + c\right )^{2} + 3 \, {\left (a^{2} b d e x + a^{2} b c e\right )} \operatorname {artanh}\left (d x + c\right ), x\right ) \]
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 \[ \int {\left (d e x + c e\right )} {\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.61, size = 6834, normalized size = 42.71 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.64, size = 629, normalized size = 3.93 \[ \frac {1}{2} \, a^{3} d e 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 d e + a^{3} c 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 c e}{2 \, d} + \frac {3 \, {\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^{3} e}{2 \, d} + \frac {3 \, {\left (c e + e\right )} a b^{2} \log \left (d x + c + 1\right )}{2 \, d} - \frac {3 \, {\left (c e - e\right )} a b^{2} \log \left (d x + c - 1\right )}{2 \, d} + \frac {24 \, a b^{2} d e x \log \left (d x + c + 1\right ) + {\left (b^{3} d^{2} e x^{2} + 2 \, b^{3} c d e x + {\left (c^{2} e - e\right )} b^{3}\right )} \log \left (d x + c + 1\right )^{3} - {\left (b^{3} d^{2} e x^{2} + 2 \, b^{3} c d e x + {\left (c^{2} e - e\right )} b^{3}\right )} \log \left (-d x - c + 1\right )^{3} + 6 \, {\left (a b^{2} d^{2} e x^{2} + {\left (c^{2} e - e\right )} a b^{2} + {\left (c e + e\right )} b^{3} + {\left (2 \, a b^{2} c d e + b^{3} d e\right )} x\right )} \log \left (d x + c + 1\right )^{2} + 3 \, {\left (2 \, a b^{2} d^{2} e x^{2} + 2 \, {\left (c^{2} e - e\right )} a b^{2} + 2 \, {\left (c e - e\right )} b^{3} + 2 \, {\left (2 \, a b^{2} c d e + b^{3} d e\right )} x + {\left (b^{3} d^{2} e x^{2} + 2 \, b^{3} c d e x + {\left (c^{2} e - e\right )} b^{3}\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )^{2} - 3 \, {\left (8 \, a b^{2} d e x + {\left (b^{3} d^{2} e x^{2} + 2 \, b^{3} c d e x + {\left (c^{2} e - e\right )} b^{3}\right )} \log \left (d x + c + 1\right )^{2} + 4 \, {\left (a b^{2} d^{2} e x^{2} + {\left (c^{2} e - e\right )} a b^{2} + {\left (c e + e\right )} b^{3} + {\left (2 \, a b^{2} c d e + b^{3} d e\right )} x\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (c\,e+d\,e\,x\right )\,{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3 \,d x \]
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 \[ e \left (\int a^{3} c\, dx + \int a^{3} d x\, dx + \int b^{3} c \operatorname {atanh}^{3}{\left (c + d x \right )}\, dx + \int 3 a b^{2} c \operatorname {atanh}^{2}{\left (c + d x \right )}\, dx + \int 3 a^{2} b c \operatorname {atanh}{\left (c + d x \right )}\, dx + \int b^{3} d x \operatorname {atanh}^{3}{\left (c + d x \right )}\, dx + \int 3 a b^{2} d x \operatorname {atanh}^{2}{\left (c + d x \right )}\, dx + \int 3 a^{2} b d x \operatorname {atanh}{\left (c + d x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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