Optimal. Leaf size=53 \[ -\frac {2 b (a+b) \tanh ^3(c+d x)}{3 d}+\frac {(a+b)^2 \tanh (c+d x)}{d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \]
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Rubi [A] time = 0.07, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {4146, 194} \[ -\frac {2 b (a+b) \tanh ^3(c+d x)}{3 d}+\frac {(a+b)^2 \tanh (c+d x)}{d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \]
Antiderivative was successfully verified.
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Rule 194
Rule 4146
Rubi steps
\begin {align*} \int \text {sech}^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b-b x^2\right )^2 \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (a^2 \left (1+\frac {b (2 a+b)}{a^2}\right )-2 a b \left (1+\frac {b}{a}\right ) x^2+b^2 x^4\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {(a+b)^2 \tanh (c+d x)}{d}-\frac {2 b (a+b) \tanh ^3(c+d x)}{3 d}+\frac {b^2 \tanh ^5(c+d x)}{5 d}\\ \end {align*}
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Mathematica [A] time = 0.03, size = 93, normalized size = 1.75 \[ \frac {a^2 \tanh (c+d x)}{d}-\frac {2 a b \tanh ^3(c+d x)}{3 d}+\frac {2 a b \tanh (c+d x)}{d}+\frac {b^2 \tanh ^5(c+d x)}{5 d}-\frac {2 b^2 \tanh ^3(c+d x)}{3 d}+\frac {b^2 \tanh (c+d x)}{d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.41, size = 404, normalized size = 7.62 \[ -\frac {4 \, {\left ({\left (15 \, a^{2} + 10 \, a b + 4 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} - 8 \, {\left (5 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (15 \, a^{2} + 10 \, a b + 4 \, b^{2}\right )} \sinh \left (d x + c\right )^{4} + 20 \, {\left (3 \, a^{2} + 4 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (15 \, a^{2} + 10 \, a b + 4 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 30 \, a^{2} + 40 \, a b + 10 \, b^{2}\right )} \sinh \left (d x + c\right )^{2} + 45 \, a^{2} + 70 \, a b + 40 \, b^{2} - 8 \, {\left ({\left (5 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (a b + b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}}{15 \, {\left (d \cosh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + d \sinh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right )^{4} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (5 \, d \cosh \left (d x + c\right )^{3} + 4 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 15 \, d \cosh \left (d x + c\right )^{2} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{4} + 12 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{5} + 8 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 10 \, d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.17, size = 156, normalized size = 2.94 \[ -\frac {2 \, {\left (15 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 60 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 140 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 80 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 100 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 40 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} + 20 \, a b + 8 \, b^{2}\right )}}{15 \, d {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.42, size = 70, normalized size = 1.32 \[ \frac {a^{2} \tanh \left (d x +c \right )+2 a b \left (\frac {2}{3}+\frac {\mathrm {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )+b^{2} \left (\frac {8}{15}+\frac {\mathrm {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.33, size = 324, normalized size = 6.11 \[ \frac {16}{15} \, b^{2} {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac {8}{3} \, a b {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + \frac {2 \, a^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.46, size = 452, normalized size = 8.53 \[ -\frac {\frac {2\,a\,\left (a+2\,b\right )}{5\,d}+\frac {2\,a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,a^2}{5\,d}+\frac {2\,a^2\,{\mathrm {e}}^{8\,c+8\,d\,x}}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2+8\,a\,b+8\,b^2\right )}{5\,d}+\frac {8\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+2\,b\right )}{5\,d}+\frac {8\,a\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a+2\,b\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {2\,a\,\left (a+2\,b\right )}{5\,d}+\frac {2\,a^2\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2+8\,a\,b+8\,b^2\right )}{5\,d}+\frac {6\,a\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a+2\,b\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2+8\,a\,b+8\,b^2\right )}{15\,d}+\frac {2\,a^2\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}+\frac {4\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+2\,b\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {2\,a^2}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{2} \operatorname {sech}^{2}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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