Integrand size = 21, antiderivative size = 48 \[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {a \tanh (c+d x)}{d}-\frac {(a-b) \tanh ^3(c+d x)}{3 d}-\frac {b \tanh ^5(c+d x)}{5 d} \]
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Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3756, 380} \[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {(a-b) \tanh ^3(c+d x)}{3 d}+\frac {a \tanh (c+d x)}{d}-\frac {b \tanh ^5(c+d x)}{5 d} \]
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Rule 380
Rule 3756
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (1-x^2\right ) \left (a+b x^2\right ) \, dx,x,\tanh (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (a-(a-b) x^2-b x^4\right ) \, dx,x,\tanh (c+d x)\right )}{d} \\ & = \frac {a \tanh (c+d x)}{d}-\frac {(a-b) \tanh ^3(c+d x)}{3 d}-\frac {b \tanh ^5(c+d x)}{5 d} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.79 \[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {a \tanh (c+d x)}{d}+\frac {2 b \tanh (c+d x)}{15 d}+\frac {b \text {sech}^2(c+d x) \tanh (c+d x)}{15 d}-\frac {b \text {sech}^4(c+d x) \tanh (c+d x)}{5 d}-\frac {a \tanh ^3(c+d x)}{3 d} \]
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Time = 8.92 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(-\frac {\frac {\tanh \left (d x +c \right )^{5} b}{5}+\frac {\left (a -b \right ) \tanh \left (d x +c \right )^{3}}{3}-a \tanh \left (d x +c \right )}{d}\) | \(42\) |
default | \(-\frac {\frac {\tanh \left (d x +c \right )^{5} b}{5}+\frac {\left (a -b \right ) \tanh \left (d x +c \right )^{3}}{3}-a \tanh \left (d x +c \right )}{d}\) | \(42\) |
risch | \(-\frac {4 \left (15 a \,{\mathrm e}^{6 d x +6 c}+15 b \,{\mathrm e}^{6 d x +6 c}+35 a \,{\mathrm e}^{4 d x +4 c}-5 b \,{\mathrm e}^{4 d x +4 c}+25 \,{\mathrm e}^{2 d x +2 c} a +5 b \,{\mathrm e}^{2 d x +2 c}+5 a +b \right )}{15 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\) | \(96\) |
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Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (44) = 88\).
Time = 0.26 (sec) , antiderivative size = 345, normalized size of antiderivative = 7.19 \[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {8 \, {\left (2 \, {\left (5 \, a + 4 \, b\right )} \cosh \left (d x + c\right )^{3} + 6 \, {\left (5 \, a + 4 \, b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (5 \, a + 7 \, b\right )} \sinh \left (d x + c\right )^{3} + 30 \, a \cosh \left (d x + c\right ) + {\left (3 \, {\left (5 \, a + 7 \, b\right )} \cosh \left (d x + c\right )^{2} + 5 \, a - 5 \, b\right )} \sinh \left (d x + c\right )\right )}}{15 \, {\left (d \cosh \left (d x + c\right )^{7} + 7 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} + d \sinh \left (d x + c\right )^{7} + 5 \, d \cosh \left (d x + c\right )^{5} + {\left (21 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (7 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 11 \, d \cosh \left (d x + c\right )^{3} + {\left (35 \, d \cosh \left (d x + c\right )^{4} + 50 \, d \cosh \left (d x + c\right )^{2} + 9 \, d\right )} \sinh \left (d x + c\right )^{3} + {\left (21 \, d \cosh \left (d x + c\right )^{5} + 50 \, d \cosh \left (d x + c\right )^{3} + 33 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 15 \, d \cosh \left (d x + c\right ) + {\left (7 \, d \cosh \left (d x + c\right )^{6} + 25 \, d \cosh \left (d x + c\right )^{4} + 27 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )\right )}} \]
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\[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \operatorname {sech}^{4}{\left (c + d x \right )}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (44) = 88\).
Time = 0.20 (sec) , antiderivative size = 371, normalized size of antiderivative = 7.73 \[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {4}{15} \, b {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} - \frac {5 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {15 \, e^{\left (-6 \, d x - 6 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac {4}{3} \, a {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (44) = 88\).
Time = 0.31 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.98 \[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (15 \, a e^{\left (6 \, d x + 6 \, c\right )} + 15 \, b e^{\left (6 \, d x + 6 \, c\right )} + 35 \, a e^{\left (4 \, d x + 4 \, c\right )} - 5 \, b e^{\left (4 \, d x + 4 \, c\right )} + 25 \, a e^{\left (2 \, d x + 2 \, c\right )} + 5 \, b e^{\left (2 \, d x + 2 \, c\right )} + 5 \, a + b\right )}}{15 \, d {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \]
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Time = 0.15 (sec) , antiderivative size = 304, normalized size of antiderivative = 6.33 \[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {\frac {8\,\left (a-b\right )}{15\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {16\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a-b\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {2\,\left (a+b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a-b\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {2\,\left (a+b\right )}{5\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \]
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