Optimal. Leaf size=31 \[ \frac {\tanh ^3(a+b x)}{3 b}-\frac {\tanh ^5(a+b x)}{5 b} \]
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Rubi [A] time = 0.03, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2607, 14} \[ \frac {\tanh ^3(a+b x)}{3 b}-\frac {\tanh ^5(a+b x)}{5 b} \]
Antiderivative was successfully verified.
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Rule 14
Rule 2607
Rubi steps
\begin {align*} \int \text {sech}^4(a+b x) \tanh ^2(a+b x) \, dx &=\frac {i \operatorname {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=\frac {i \operatorname {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=\frac {\tanh ^3(a+b x)}{3 b}-\frac {\tanh ^5(a+b x)}{5 b}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 56, normalized size = 1.81 \[ \frac {2 \tanh (a+b x)}{15 b}-\frac {\tanh (a+b x) \text {sech}^4(a+b x)}{5 b}+\frac {\tanh (a+b x) \text {sech}^2(a+b x)}{15 b} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.44, size = 304, normalized size = 9.81 \[ -\frac {8 \, {\left (8 \, \cosh \left (b x + a\right )^{3} + 24 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 7 \, \sinh \left (b x + a\right )^{3} + {\left (21 \, \cosh \left (b x + a\right )^{2} - 5\right )} \sinh \left (b x + a\right )\right )}}{15 \, {\left (b \cosh \left (b x + a\right )^{7} + 7 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{6} + b \sinh \left (b x + a\right )^{7} + 5 \, b \cosh \left (b x + a\right )^{5} + {\left (21 \, b \cosh \left (b x + a\right )^{2} + 5 \, b\right )} \sinh \left (b x + a\right )^{5} + 5 \, {\left (7 \, b \cosh \left (b x + a\right )^{3} + 5 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{4} + 11 \, b \cosh \left (b x + a\right )^{3} + {\left (35 \, b \cosh \left (b x + a\right )^{4} + 50 \, b \cosh \left (b x + a\right )^{2} + 9 \, b\right )} \sinh \left (b x + a\right )^{3} + {\left (21 \, b \cosh \left (b x + a\right )^{5} + 50 \, b \cosh \left (b x + a\right )^{3} + 33 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 15 \, b \cosh \left (b x + a\right ) + {\left (7 \, b \cosh \left (b x + a\right )^{6} + 25 \, b \cosh \left (b x + a\right )^{4} + 27 \, b \cosh \left (b x + a\right )^{2} + 5 \, b\right )} \sinh \left (b x + a\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 53, normalized size = 1.71 \[ -\frac {4 \, {\left (15 \, e^{\left (6 \, b x + 6 \, a\right )} - 5 \, e^{\left (4 \, b x + 4 \, a\right )} + 5 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}{15 \, b {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.32, size = 52, normalized size = 1.68 \[ \frac {-\frac {\sinh \left (b x +a \right )}{4 \cosh \left (b x +a \right )^{5}}+\frac {\left (\frac {8}{15}+\frac {\mathrm {sech}\left (b x +a \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (b x +a \right )^{2}}{15}\right ) \tanh \left (b x +a \right )}{4}}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.39, size = 276, normalized size = 8.90 \[ \frac {4 \, e^{\left (-2 \, b x - 2 \, a\right )}}{3 \, b {\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} - \frac {4 \, e^{\left (-4 \, b x - 4 \, a\right )}}{3 \, b {\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} + \frac {4 \, e^{\left (-6 \, b x - 6 \, a\right )}}{b {\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} + \frac {4}{15 \, b {\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.13, size = 270, normalized size = 8.71 \[ \frac {\frac {8}{15\,b}-\frac {4\,{\mathrm {e}}^{2\,a+2\,b\,x}}{5\,b}}{3\,{\mathrm {e}}^{2\,a+2\,b\,x}+3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}+1}-\frac {\frac {2}{5\,b}-\frac {8\,{\mathrm {e}}^{2\,a+2\,b\,x}}{5\,b}+\frac {6\,{\mathrm {e}}^{4\,a+4\,b\,x}}{5\,b}}{4\,{\mathrm {e}}^{2\,a+2\,b\,x}+6\,{\mathrm {e}}^{4\,a+4\,b\,x}+4\,{\mathrm {e}}^{6\,a+6\,b\,x}+{\mathrm {e}}^{8\,a+8\,b\,x}+1}-\frac {\frac {8\,{\mathrm {e}}^{2\,a+2\,b\,x}}{5\,b}-\frac {16\,{\mathrm {e}}^{4\,a+4\,b\,x}}{5\,b}+\frac {8\,{\mathrm {e}}^{6\,a+6\,b\,x}}{5\,b}}{5\,{\mathrm {e}}^{2\,a+2\,b\,x}+10\,{\mathrm {e}}^{4\,a+4\,b\,x}+10\,{\mathrm {e}}^{6\,a+6\,b\,x}+5\,{\mathrm {e}}^{8\,a+8\,b\,x}+{\mathrm {e}}^{10\,a+10\,b\,x}+1}-\frac {2}{5\,b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )} \]
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 \[ \int \tanh ^{2}{\left (a + b x \right )} \operatorname {sech}^{4}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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