Integrand size = 12, antiderivative size = 107 \[ \int (a+b \text {sech}(c+d x))^4 \, dx=a^4 x+\frac {2 a b \left (2 a^2+b^2\right ) \arctan (\sinh (c+d x))}{d}+\frac {b^2 \left (17 a^2+2 b^2\right ) \tanh (c+d x)}{3 d}+\frac {4 a b^3 \text {sech}(c+d x) \tanh (c+d x)}{3 d}+\frac {b^2 (a+b \text {sech}(c+d x))^2 \tanh (c+d x)}{3 d} \]
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Time = 0.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3867, 4133, 3855, 3852, 8} \[ \int (a+b \text {sech}(c+d x))^4 \, dx=a^4 x+\frac {2 a b \left (2 a^2+b^2\right ) \arctan (\sinh (c+d x))}{d}+\frac {b^2 \left (17 a^2+2 b^2\right ) \tanh (c+d x)}{3 d}+\frac {4 a b^3 \tanh (c+d x) \text {sech}(c+d x)}{3 d}+\frac {b^2 \tanh (c+d x) (a+b \text {sech}(c+d x))^2}{3 d} \]
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Rule 8
Rule 3852
Rule 3855
Rule 3867
Rule 4133
Rubi steps \begin{align*} \text {integral}& = \frac {b^2 (a+b \text {sech}(c+d x))^2 \tanh (c+d x)}{3 d}+\frac {1}{3} \int (a+b \text {sech}(c+d x)) \left (3 a^3+b \left (9 a^2+2 b^2\right ) \text {sech}(c+d x)+8 a b^2 \text {sech}^2(c+d x)\right ) \, dx \\ & = \frac {4 a b^3 \text {sech}(c+d x) \tanh (c+d x)}{3 d}+\frac {b^2 (a+b \text {sech}(c+d x))^2 \tanh (c+d x)}{3 d}+\frac {1}{6} \int \left (6 a^4+12 a b \left (2 a^2+b^2\right ) \text {sech}(c+d x)+2 b^2 \left (17 a^2+2 b^2\right ) \text {sech}^2(c+d x)\right ) \, dx \\ & = a^4 x+\frac {4 a b^3 \text {sech}(c+d x) \tanh (c+d x)}{3 d}+\frac {b^2 (a+b \text {sech}(c+d x))^2 \tanh (c+d x)}{3 d}+\left (2 a b \left (2 a^2+b^2\right )\right ) \int \text {sech}(c+d x) \, dx+\frac {1}{3} \left (b^2 \left (17 a^2+2 b^2\right )\right ) \int \text {sech}^2(c+d x) \, dx \\ & = a^4 x+\frac {2 a b \left (2 a^2+b^2\right ) \arctan (\sinh (c+d x))}{d}+\frac {4 a b^3 \text {sech}(c+d x) \tanh (c+d x)}{3 d}+\frac {b^2 (a+b \text {sech}(c+d x))^2 \tanh (c+d x)}{3 d}+\frac {\left (i b^2 \left (17 a^2+2 b^2\right )\right ) \text {Subst}(\int 1 \, dx,x,-i \tanh (c+d x))}{3 d} \\ & = a^4 x+\frac {2 a b \left (2 a^2+b^2\right ) \arctan (\sinh (c+d x))}{d}+\frac {b^2 \left (17 a^2+2 b^2\right ) \tanh (c+d x)}{3 d}+\frac {4 a b^3 \text {sech}(c+d x) \tanh (c+d x)}{3 d}+\frac {b^2 (a+b \text {sech}(c+d x))^2 \tanh (c+d x)}{3 d} \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.73 \[ \int (a+b \text {sech}(c+d x))^4 \, dx=\frac {3 a^4 d x+6 a b \left (2 a^2+b^2\right ) \arctan (\sinh (c+d x))+3 b^2 \left (6 a^2+b^2+2 a b \text {sech}(c+d x)\right ) \tanh (c+d x)-b^4 \tanh ^3(c+d x)}{3 d} \]
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Time = 1.41 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(\frac {a^{4} \left (d x +c \right )+8 a^{3} b \arctan \left ({\mathrm e}^{d x +c}\right )+6 a^{2} b^{2} \tanh \left (d x +c \right )+4 a \,b^{3} \left (\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )+b^{4} \left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{d}\) | \(92\) |
default | \(\frac {a^{4} \left (d x +c \right )+8 a^{3} b \arctan \left ({\mathrm e}^{d x +c}\right )+6 a^{2} b^{2} \tanh \left (d x +c \right )+4 a \,b^{3} \left (\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )+b^{4} \left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{d}\) | \(92\) |
parts | \(x \,a^{4}+\frac {b^{4} \left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{d}+\frac {4 a \,b^{3} \left (\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )}{d}+\frac {6 a^{2} b^{2} \tanh \left (d x +c \right )}{d}+\frac {4 a^{3} b \arctan \left (\sinh \left (d x +c \right )\right )}{d}\) | \(96\) |
risch | \(x \,a^{4}-\frac {4 b^{2} \left (-3 a b \,{\mathrm e}^{5 d x +5 c}+9 a^{2} {\mathrm e}^{4 d x +4 c}+18 a^{2} {\mathrm e}^{2 d x +2 c}+3 \,{\mathrm e}^{2 d x +2 c} b^{2}+3 \,{\mathrm e}^{d x +c} a b +9 a^{2}+b^{2}\right )}{3 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3}}+\frac {4 i a^{3} b \ln \left ({\mathrm e}^{d x +c}+i\right )}{d}+\frac {2 i a \,b^{3} \ln \left ({\mathrm e}^{d x +c}+i\right )}{d}-\frac {4 i a^{3} b \ln \left ({\mathrm e}^{d x +c}-i\right )}{d}-\frac {2 i a \,b^{3} \ln \left ({\mathrm e}^{d x +c}-i\right )}{d}\) | \(182\) |
parallelrisch | \(\frac {-36 i \left (\frac {\cosh \left (3 d x +3 c \right )}{3}+\cosh \left (d x +c \right )\right ) b a \left (a^{2}+\frac {b^{2}}{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )+36 i \left (\frac {\cosh \left (3 d x +3 c \right )}{3}+\cosh \left (d x +c \right )\right ) b a \left (a^{2}+\frac {b^{2}}{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )+3 a^{4} d x \cosh \left (3 d x +3 c \right )+2 \left (9 a^{2} b^{2}+b^{4}\right ) \sinh \left (3 d x +3 c \right )+12 a \,b^{3} \sinh \left (2 d x +2 c \right )+9 a^{4} d x \cosh \left (d x +c \right )+6 \left (3 a^{2} b^{2}+b^{4}\right ) \sinh \left (d x +c \right )}{3 d \left (\cosh \left (3 d x +3 c \right )+3 \cosh \left (d x +c \right )\right )}\) | \(204\) |
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Leaf count of result is larger than twice the leaf count of optimal. 1028 vs. \(2 (101) = 202\).
Time = 0.27 (sec) , antiderivative size = 1028, normalized size of antiderivative = 9.61 \[ \int (a+b \text {sech}(c+d x))^4 \, dx=\text {Too large to display} \]
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\[ \int (a+b \text {sech}(c+d x))^4 \, dx=\int \left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{4}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (101) = 202\).
Time = 0.28 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.97 \[ \int (a+b \text {sech}(c+d x))^4 \, dx=a^{4} x - 4 \, a b^{3} {\left (\frac {\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + \frac {4}{3} \, b^{4} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + \frac {4 \, a^{3} b \arctan \left (\sinh \left (d x + c\right )\right )}{d} + \frac {12 \, a^{2} b^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.32 \[ \int (a+b \text {sech}(c+d x))^4 \, dx=\frac {3 \, {\left (d x + c\right )} a^{4} + 12 \, {\left (2 \, a^{3} b + a b^{3}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) + \frac {4 \, {\left (3 \, a b^{3} e^{\left (5 \, d x + 5 \, c\right )} - 9 \, a^{2} b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 18 \, a^{2} b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 3 \, b^{4} e^{\left (2 \, d x + 2 \, c\right )} - 3 \, a b^{3} e^{\left (d x + c\right )} - 9 \, a^{2} b^{2} - b^{4}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{3 \, d} \]
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Time = 2.11 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.18 \[ \int (a+b \text {sech}(c+d x))^4 \, dx=a^4\,x-\frac {\frac {12\,a^2\,b^2}{d}-\frac {4\,a\,b^3\,{\mathrm {e}}^{c+d\,x}}{d}}{{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {4\,b^4}{d}+\frac {8\,a\,b^3\,{\mathrm {e}}^{c+d\,x}}{d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+\frac {8\,b^4}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}+\frac {4\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (a\,b^3\,\sqrt {d^2}+2\,a^3\,b\,\sqrt {d^2}\right )}{d\,\sqrt {4\,a^6\,b^2+4\,a^4\,b^4+a^2\,b^6}}\right )\,\sqrt {4\,a^6\,b^2+4\,a^4\,b^4+a^2\,b^6}}{\sqrt {d^2}} \]
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