Integrand size = 21, antiderivative size = 63 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {1}{8} (3 a-b) x+\frac {(3 a-b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d} \]
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Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3756, 393, 205, 212} \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {(a+b) \sinh (c+d x) \cosh ^3(c+d x)}{4 d}+\frac {(3 a-b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac {1}{8} x (3 a-b) \]
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Rule 205
Rule 212
Rule 393
Rule 3756
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {a+b x^2}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = \frac {(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac {(3 a-b) \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d} \\ & = \frac {(3 a-b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac {(3 a-b) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d} \\ & = \frac {1}{8} (3 a-b) x+\frac {(3 a-b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {-4 b d x+12 a (c+d x)+8 a \sinh (2 (c+d x))+(a+b) \sinh (4 (c+d x))}{32 d} \]
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Time = 4.87 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.30
method | result | size |
derivativedivides | \(\frac {a \left (\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )}{d}\) | \(82\) |
default | \(\frac {a \left (\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )}{d}\) | \(82\) |
risch | \(\frac {3 a x}{8}-\frac {b x}{8}+\frac {{\mathrm e}^{4 d x +4 c} a}{64 d}+\frac {{\mathrm e}^{4 d x +4 c} b}{64 d}+\frac {a \,{\mathrm e}^{2 d x +2 c}}{8 d}-\frac {a \,{\mathrm e}^{-2 d x -2 c}}{8 d}-\frac {{\mathrm e}^{-4 d x -4 c} a}{64 d}-\frac {{\mathrm e}^{-4 d x -4 c} b}{64 d}\) | \(100\) |
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Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {{\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (3 \, a - b\right )} d x + {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{3} + 4 \, a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{8 \, d} \]
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\[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \cosh ^{4}{\left (c + d x \right )}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.65 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {1}{64} \, a {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} - \frac {1}{64} \, b {\left (\frac {8 \, {\left (d x + c\right )}}{d} - \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} \]
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Time = 0.32 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.65 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {8 \, {\left (d x + c\right )} {\left (3 \, a - b\right )} + a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a e^{\left (2 \, d x + 2 \, c\right )} - {\left (18 \, a e^{\left (4 \, d x + 4 \, c\right )} - 6 \, b e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} \]
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Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.17 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=x\,\left (\frac {3\,a}{8}-\frac {b}{8}\right )-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}\,\left (a+b\right )}{64\,d}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a+b\right )}{64\,d}-\frac {a\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}+\frac {a\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d} \]
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