Optimal. Leaf size=143 \[ -\frac {3 b^2 \text {Li}_2\left (\frac {2}{c+d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}+\frac {3 b \log \left (2-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^2}-\frac {3 b^3 \text {Li}_3\left (\frac {2}{c+d x+1}-1\right )}{2 d e^2} \]
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Rubi [A] time = 0.30, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {6107, 12, 5916, 5988, 5932, 5948, 6056, 6610} \[ -\frac {3 b^2 \text {PolyLog}\left (2,\frac {2}{c+d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^2}-\frac {3 b^3 \text {PolyLog}\left (3,\frac {2}{c+d x+1}-1\right )}{2 d e^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}+\frac {3 b \log \left (2-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 5916
Rule 5932
Rule 5948
Rule 5988
Rule 6056
Rule 6107
Rule 6610
Rubi steps
\begin {align*} \int \frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{(c e+d e x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^3}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^3}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{x \left (1-x^2\right )} \, dx,x,c+d x\right )}{d e^2}\\ &=\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{x (1+x)} \, dx,x,c+d x\right )}{d e^2}\\ &=\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^2}-\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right ) \log \left (2-\frac {2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^2}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c+d x}\right )}{d e^2}+\frac {\left (3 b^3\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 e^2}\\ &=\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^2}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c+d x}\right )}{d e^2}-\frac {3 b^3 \text {Li}_3\left (-1+\frac {2}{1+c+d x}\right )}{2 d e^2}\\ \end {align*}
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Mathematica [C] time = 0.61, size = 248, normalized size = 1.73 \[ \frac {-\frac {2 a^3}{c+d x}-3 a^2 b \log \left (-c^2-2 c d x-d^2 x^2+1\right )+6 a^2 b \log (c+d x)-\frac {6 a^2 b \tanh ^{-1}(c+d x)}{c+d x}+6 a b^2 \left (\tanh ^{-1}(c+d x) \left (\left (1-\frac {1}{c+d x}\right ) \tanh ^{-1}(c+d x)+2 \log \left (1-e^{-2 \tanh ^{-1}(c+d x)}\right )\right )-\text {Li}_2\left (e^{-2 \tanh ^{-1}(c+d x)}\right )\right )+2 b^3 \left (3 \tanh ^{-1}(c+d x) \text {Li}_2\left (e^{2 \tanh ^{-1}(c+d x)}\right )-\frac {3}{2} \text {Li}_3\left (e^{2 \tanh ^{-1}(c+d x)}\right )-\frac {\tanh ^{-1}(c+d x)^3}{c+d x}-\tanh ^{-1}(c+d x)^3+3 \tanh ^{-1}(c+d x)^2 \log \left (1-e^{2 \tanh ^{-1}(c+d x)}\right )+\frac {i \pi ^3}{8}\right )}{2 d e^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {artanh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname {artanh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname {artanh}\left (d x + c\right ) + a^{3}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.51, size = 2001, normalized size = 13.99 \[ \text {result too large to display} \]
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 \[ -\frac {3}{2} \, {\left (d {\left (\frac {\log \left (d x + c + 1\right )}{d^{2} e^{2}} - \frac {2 \, \log \left (d x + c\right )}{d^{2} e^{2}} + \frac {\log \left (d x + c - 1\right )}{d^{2} e^{2}}\right )} + \frac {2 \, \operatorname {artanh}\left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}}\right )} a^{2} b - \frac {a^{3}}{d^{2} e^{2} x + c d e^{2}} - \frac {{\left (b^{3} d x + b^{3} {\left (c - 1\right )}\right )} \log \left (-d x - c + 1\right )^{3} + 3 \, {\left (2 \, a b^{2} + {\left (b^{3} d x + b^{3} {\left (c + 1\right )}\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )^{2}}{8 \, {\left (d^{2} e^{2} x + c d e^{2}\right )}} - \int -\frac {{\left (b^{3} d x + b^{3} {\left (c - 1\right )}\right )} \log \left (d x + c + 1\right )^{3} + 6 \, {\left (a b^{2} d x + a b^{2} {\left (c - 1\right )}\right )} \log \left (d x + c + 1\right )^{2} + 3 \, {\left (4 \, a b^{2} d x + 4 \, a b^{2} c - {\left (b^{3} d x + b^{3} {\left (c - 1\right )}\right )} \log \left (d x + c + 1\right )^{2} + 2 \, {\left (b^{3} d^{2} x^{2} + {\left (c^{2} + c\right )} b^{3} - 2 \, a b^{2} {\left (c - 1\right )} + {\left ({\left (2 \, c d + d\right )} b^{3} - 2 \, a b^{2} d\right )} x\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )}{8 \, {\left (d^{3} e^{2} x^{3} + c^{3} e^{2} - c^{2} e^{2} + {\left (3 \, c d^{2} e^{2} - d^{2} e^{2}\right )} x^{2} + {\left (3 \, c^{2} d e^{2} - 2 \, c d e^{2}\right )} x\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \]
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 \[ \frac {\int \frac {a^{3}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {b^{3} \operatorname {atanh}^{3}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a b^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a^{2} b \operatorname {atanh}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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