Integrand size = 23, antiderivative size = 118 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {1}{8} \left (3 a^2+30 a b+35 b^2\right ) x-\frac {(a+b) (a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}-\frac {\left (a^2+10 a b+13 b^2\right ) \tanh (c+d x)}{4 d}+\frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {b^2 \tanh ^3(c+d x)}{3 d} \]
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Time = 0.10 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3744, 474, 466, 1167, 212} \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {\left (a^2+10 a b+13 b^2\right ) \tanh (c+d x)}{4 d}+\frac {1}{8} x \left (3 a^2+30 a b+35 b^2\right )+\frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {(a+b) (a+9 b) \sinh (c+d x) \cosh (c+d x)}{8 d}-\frac {b^2 \tanh ^3(c+d x)}{3 d} \]
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Rule 212
Rule 466
Rule 474
Rule 1167
Rule 3744
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4 \left (a+b x^2\right )^2}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = \frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {\text {Subst}\left (\int \frac {x^4 \left (a^2+10 a b+5 b^2+4 b^2 x^2\right )}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d} \\ & = -\frac {(a+b) (a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {\text {Subst}\left (\int \frac {-((a+b) (a+9 b))-2 (a+b) (a+9 b) x^2-8 b^2 x^4}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d} \\ & = -\frac {(a+b) (a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {\text {Subst}\left (\int \left (2 \left (a^2+10 a b+13 b^2\right )+8 b^2 x^2+\frac {-3 a^2-30 a b-35 b^2}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{8 d} \\ & = -\frac {(a+b) (a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}-\frac {\left (a^2+10 a b+13 b^2\right ) \tanh (c+d x)}{4 d}+\frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {b^2 \tanh ^3(c+d x)}{3 d}+\frac {\left (3 a^2+30 a b+35 b^2\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d} \\ & = \frac {1}{8} \left (3 a^2+30 a b+35 b^2\right ) x-\frac {(a+b) (a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}-\frac {\left (a^2+10 a b+13 b^2\right ) \tanh (c+d x)}{4 d}+\frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {b^2 \tanh ^3(c+d x)}{3 d} \\ \end{align*}
Time = 2.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.80 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {12 \left (3 a^2+30 a b+35 b^2\right ) (c+d x)-24 \left (a^2+4 a b+3 b^2\right ) \sinh (2 (c+d x))+3 (a+b)^2 \sinh (4 (c+d x))+32 b \left (-6 a-10 b+b \text {sech}^2(c+d x)\right ) \tanh (c+d x)}{96 d} \]
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Time = 3.17 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.41
method | result | size |
derivativedivides | \(\frac {a^{2} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a b \left (\frac {\sinh \left (d x +c \right )^{5}}{4 \cosh \left (d x +c \right )}-\frac {5 \sinh \left (d x +c \right )^{3}}{8 \cosh \left (d x +c \right )}+\frac {15 d x}{8}+\frac {15 c}{8}-\frac {15 \tanh \left (d x +c \right )}{8}\right )+b^{2} \left (\frac {\sinh \left (d x +c \right )^{7}}{4 \cosh \left (d x +c \right )^{3}}-\frac {7 \sinh \left (d x +c \right )^{5}}{8 \cosh \left (d x +c \right )^{3}}+\frac {35 d x}{8}+\frac {35 c}{8}-\frac {35 \tanh \left (d x +c \right )}{8}-\frac {35 \tanh \left (d x +c \right )^{3}}{24}\right )}{d}\) | \(166\) |
default | \(\frac {a^{2} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a b \left (\frac {\sinh \left (d x +c \right )^{5}}{4 \cosh \left (d x +c \right )}-\frac {5 \sinh \left (d x +c \right )^{3}}{8 \cosh \left (d x +c \right )}+\frac {15 d x}{8}+\frac {15 c}{8}-\frac {15 \tanh \left (d x +c \right )}{8}\right )+b^{2} \left (\frac {\sinh \left (d x +c \right )^{7}}{4 \cosh \left (d x +c \right )^{3}}-\frac {7 \sinh \left (d x +c \right )^{5}}{8 \cosh \left (d x +c \right )^{3}}+\frac {35 d x}{8}+\frac {35 c}{8}-\frac {35 \tanh \left (d x +c \right )}{8}-\frac {35 \tanh \left (d x +c \right )^{3}}{24}\right )}{d}\) | \(166\) |
risch | \(\frac {3 a^{2} x}{8}+\frac {15 a b x}{4}+\frac {35 b^{2} x}{8}+\frac {{\mathrm e}^{4 d x +4 c} a^{2}}{64 d}+\frac {{\mathrm e}^{4 d x +4 c} a b}{32 d}+\frac {{\mathrm e}^{4 d x +4 c} b^{2}}{64 d}-\frac {{\mathrm e}^{2 d x +2 c} a^{2}}{8 d}-\frac {{\mathrm e}^{2 d x +2 c} a b}{2 d}-\frac {3 \,{\mathrm e}^{2 d x +2 c} b^{2}}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} a^{2}}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} a b}{2 d}+\frac {3 \,{\mathrm e}^{-2 d x -2 c} b^{2}}{8 d}-\frac {{\mathrm e}^{-4 d x -4 c} a^{2}}{64 d}-\frac {{\mathrm e}^{-4 d x -4 c} a b}{32 d}-\frac {{\mathrm e}^{-4 d x -4 c} b^{2}}{64 d}+\frac {4 b \left (3 a \,{\mathrm e}^{4 d x +4 c}+6 b \,{\mathrm e}^{4 d x +4 c}+6 \,{\mathrm e}^{2 d x +2 c} a +9 b \,{\mathrm e}^{2 d x +2 c}+3 a +5 b \right )}{3 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3}}\) | \(293\) |
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Leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (108) = 216\).
Time = 0.27 (sec) , antiderivative size = 394, normalized size of antiderivative = 3.34 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{7} + 3 \, {\left (21 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} - 5 \, a^{2} - 26 \, a b - 21 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} + 8 \, {\left (3 \, {\left (3 \, a^{2} + 30 \, a b + 35 \, b^{2}\right )} d x + 48 \, a b + 80 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 24 \, {\left (3 \, {\left (3 \, a^{2} + 30 \, a b + 35 \, b^{2}\right )} d x + 48 \, a b + 80 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (105 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{4} - 30 \, {\left (5 \, a^{2} + 26 \, a b + 21 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} - 63 \, a^{2} - 654 \, a b - 847 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 24 \, {\left (3 \, {\left (3 \, a^{2} + 30 \, a b + 35 \, b^{2}\right )} d x + 48 \, a b + 80 \, b^{2}\right )} \cosh \left (d x + c\right ) + 3 \, {\left (7 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{6} - 5 \, {\left (5 \, a^{2} + 26 \, a b + 21 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} - {\left (63 \, a^{2} + 654 \, a b + 847 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} - 15 \, a^{2} - 190 \, a b - 175 \, b^{2}\right )} \sinh \left (d x + c\right )}{192 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \]
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\[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \sinh ^{4}{\left (c + d x \right )}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (108) = 216\).
Time = 0.20 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.50 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {1}{64} \, a^{2} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {1}{192} \, b^{2} {\left (\frac {840 \, {\left (d x + c\right )}}{d} + \frac {3 \, {\left (24 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}\right )}}{d} - \frac {63 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1487 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2517 \, e^{\left (-6 \, d x - 6 \, c\right )} + 1608 \, e^{\left (-8 \, d x - 8 \, c\right )} - 3}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )}\right )}}\right )} + \frac {1}{32} \, a b {\left (\frac {120 \, {\left (d x + c\right )}}{d} + \frac {16 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} - \frac {15 \, e^{\left (-2 \, d x - 2 \, c\right )} + 144 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (108) = 216\).
Time = 0.40 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.48 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {3 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 24 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 96 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 72 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, {\left (3 \, a^{2} + 30 \, a b + 35 \, b^{2}\right )} {\left (d x + c\right )} - 3 \, {\left (18 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 210 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 32 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 24 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + \frac {256 \, {\left (3 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 9 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a b + 5 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{192 \, d} \]
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Time = 1.98 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.48 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {\frac {4\,\left (b^2+a\,b\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,b^2+a\,b\right )}{3\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+x\,\left (\frac {3\,a^2}{8}+\frac {15\,a\,b}{4}+\frac {35\,b^2}{8}\right )+\frac {\frac {4\,\left (2\,b^2+a\,b\right )}{3\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+a\,b\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,b^2+a\,b\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+\frac {4\,\left (2\,b^2+a\,b\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a^2+4\,a\,b+3\,b^2\right )}{8\,d}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2+4\,a\,b+3\,b^2\right )}{8\,d}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}\,{\left (a+b\right )}^2}{64\,d}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}\,{\left (a+b\right )}^2}{64\,d} \]
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