Integrand size = 14, antiderivative size = 74 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=(a+b)^3 x-\frac {b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^2 (3 a+b) \tanh ^3(c+d x)}{3 d}-\frac {b^3 \tanh ^5(c+d x)}{5 d} \]
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Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3742, 398, 212} \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^2 (3 a+b) \tanh ^3(c+d x)}{3 d}+x (a+b)^3-\frac {b^3 \tanh ^5(c+d x)}{5 d} \]
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Rule 212
Rule 398
Rule 3742
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^3}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (-b \left (3 a^2+3 a b+b^2\right )-b^2 (3 a+b) x^2-b^3 x^4+\frac {(a+b)^3}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d} \\ & = -\frac {b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^2 (3 a+b) \tanh ^3(c+d x)}{3 d}-\frac {b^3 \tanh ^5(c+d x)}{5 d}+\frac {(a+b)^3 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = (a+b)^3 x-\frac {b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^2 (3 a+b) \tanh ^3(c+d x)}{3 d}-\frac {b^3 \tanh ^5(c+d x)}{5 d} \\ \end{align*}
Time = 0.72 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.28 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {\tanh (c+d x) \left (\frac {15 (a+b)^3 \text {arctanh}\left (\sqrt {\tanh ^2(c+d x)}\right )}{\sqrt {\tanh ^2(c+d x)}}-b \left (45 a^2+15 a b \left (3+\tanh ^2(c+d x)\right )+b^2 \left (15+5 \tanh ^2(c+d x)+3 \tanh ^4(c+d x)\right )\right )\right )}{15 d} \]
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Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.51
method | result | size |
parallelrisch | \(-\frac {3 b^{3} \tanh \left (d x +c \right )^{5}+15 a \,b^{2} \tanh \left (d x +c \right )^{3}+5 b^{3} \tanh \left (d x +c \right )^{3}-15 a^{3} d x -45 a^{2} b d x -45 a \,b^{2} d x -15 b^{3} d x +45 a^{2} b \tanh \left (d x +c \right )+45 a \,b^{2} \tanh \left (d x +c \right )+15 b^{3} \tanh \left (d x +c \right )}{15 d}\) | \(112\) |
derivativedivides | \(\frac {-3 a^{2} b \tanh \left (d x +c \right )-3 a \,b^{2} \tanh \left (d x +c \right )-a \,b^{2} \tanh \left (d x +c \right )^{3}-\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )}{2}-\frac {b^{3} \tanh \left (d x +c \right )^{3}}{3}-b^{3} \tanh \left (d x +c \right )-\frac {b^{3} \tanh \left (d x +c \right )^{5}}{5}+\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )}{2}}{d}\) | \(141\) |
default | \(\frac {-3 a^{2} b \tanh \left (d x +c \right )-3 a \,b^{2} \tanh \left (d x +c \right )-a \,b^{2} \tanh \left (d x +c \right )^{3}-\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )}{2}-\frac {b^{3} \tanh \left (d x +c \right )^{3}}{3}-b^{3} \tanh \left (d x +c \right )-\frac {b^{3} \tanh \left (d x +c \right )^{5}}{5}+\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )}{2}}{d}\) | \(141\) |
parts | \(a^{3} x +\frac {b^{3} \left (-\frac {\tanh \left (d x +c \right )^{5}}{5}-\frac {\tanh \left (d x +c \right )^{3}}{3}-\tanh \left (d x +c \right )-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {3 a^{2} b \left (-\tanh \left (d x +c \right )-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {3 a \,b^{2} \left (-\frac {\tanh \left (d x +c \right )^{3}}{3}-\tanh \left (d x +c \right )-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}\) | \(155\) |
risch | \(a^{3} x +3 b \,a^{2} x +3 a \,b^{2} x +b^{3} x +\frac {2 b \left (45 a^{2} {\mathrm e}^{8 d x +8 c}+90 a b \,{\mathrm e}^{8 d x +8 c}+45 b^{2} {\mathrm e}^{8 d x +8 c}+180 a^{2} {\mathrm e}^{6 d x +6 c}+270 a b \,{\mathrm e}^{6 d x +6 c}+90 b^{2} {\mathrm e}^{6 d x +6 c}+270 a^{2} {\mathrm e}^{4 d x +4 c}+330 a b \,{\mathrm e}^{4 d x +4 c}+140 \,{\mathrm e}^{4 d x +4 c} b^{2}+180 a^{2} {\mathrm e}^{2 d x +2 c}+210 a b \,{\mathrm e}^{2 d x +2 c}+70 \,{\mathrm e}^{2 d x +2 c} b^{2}+45 a^{2}+60 a b +23 b^{2}\right )}{15 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\) | \(224\) |
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Leaf count of result is larger than twice the leaf count of optimal. 567 vs. \(2 (70) = 140\).
Time = 0.27 (sec) , antiderivative size = 567, normalized size of antiderivative = 7.66 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {{\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} - 5 \, {\left (27 \, a^{2} b + 24 \, a b^{2} + 5 \, b^{3} + 2 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 18 \, a^{2} b + 12 \, a b^{2} + 10 \, b^{3} + 3 \, {\left (27 \, a^{2} b + 24 \, a b^{2} + 5 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \]
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Time = 0.16 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.70 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\begin {cases} a^{3} x + 3 a^{2} b x - \frac {3 a^{2} b \tanh {\left (c + d x \right )}}{d} + 3 a b^{2} x - \frac {a b^{2} \tanh ^{3}{\left (c + d x \right )}}{d} - \frac {3 a b^{2} \tanh {\left (c + d x \right )}}{d} + b^{3} x - \frac {b^{3} \tanh ^{5}{\left (c + d x \right )}}{5 d} - \frac {b^{3} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{3} \tanh {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tanh ^{2}{\left (c \right )}\right )^{3} & \text {otherwise} \end {cases} \]
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Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (70) = 140\).
Time = 0.20 (sec) , antiderivative size = 239, normalized size of antiderivative = 3.23 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {1}{15} \, b^{3} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} + 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} + 45 \, e^{\left (-8 \, d x - 8 \, c\right )} + 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + a b^{2} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + 3 \, a^{2} b {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + a^{3} x \]
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Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (70) = 140\).
Time = 0.31 (sec) , antiderivative size = 241, normalized size of antiderivative = 3.26 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 90 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 45 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 330 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 140 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 210 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 70 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \]
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Time = 0.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.16 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=x\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^3\,\left (b^3+3\,a\,b^2\right )}{3\,d}-\frac {b^3\,{\mathrm {tanh}\left (c+d\,x\right )}^5}{5\,d}-\frac {b\,\mathrm {tanh}\left (c+d\,x\right )\,\left (3\,a^2+3\,a\,b+b^2\right )}{d} \]
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