Optimal. Leaf size=76 \[ \frac {\tanh (a+b x)}{5 b \text {sech}^2(a+b x)^{5/2}}+\frac {4 \tanh (a+b x)}{15 b \text {sech}^2(a+b x)^{3/2}}+\frac {8 \tanh (a+b x)}{15 b \sqrt {\text {sech}^2(a+b x)}} \]
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Rubi [A]
time = 0.02, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4207, 198, 197}
\begin {gather*} \frac {8 \tanh (a+b x)}{15 b \sqrt {\text {sech}^2(a+b x)}}+\frac {4 \tanh (a+b x)}{15 b \text {sech}^2(a+b x)^{3/2}}+\frac {\tanh (a+b x)}{5 b \text {sech}^2(a+b x)^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 197
Rule 198
Rule 4207
Rubi steps
\begin {align*} \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^{7/2}} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac {\tanh (a+b x)}{5 b \text {sech}^2(a+b x)^{5/2}}+\frac {4 \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^{5/2}} \, dx,x,\tanh (a+b x)\right )}{5 b}\\ &=\frac {\tanh (a+b x)}{5 b \text {sech}^2(a+b x)^{5/2}}+\frac {4 \tanh (a+b x)}{15 b \text {sech}^2(a+b x)^{3/2}}+\frac {8 \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^{3/2}} \, dx,x,\tanh (a+b x)\right )}{15 b}\\ &=\frac {\tanh (a+b x)}{5 b \text {sech}^2(a+b x)^{5/2}}+\frac {4 \tanh (a+b x)}{15 b \text {sech}^2(a+b x)^{3/2}}+\frac {8 \tanh (a+b x)}{15 b \sqrt {\text {sech}^2(a+b x)}}\\ \end {align*}
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Mathematica [A]
time = 0.07, size = 54, normalized size = 0.71 \begin {gather*} \frac {\text {sech}(a+b x) (150 \sinh (a+b x)+25 \sinh (3 (a+b x))+3 \sinh (5 (a+b x)))}{240 b \sqrt {\text {sech}^2(a+b x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(304\) vs.
\(2(64)=128\).
time = 2.50, size = 305, normalized size = 4.01
method | result | size |
risch | \(\frac {{\mathrm e}^{6 b x +6 a}}{160 b \left ({\mathrm e}^{2 b x +2 a}+1\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}}}+\frac {5 \,{\mathrm e}^{4 b x +4 a}}{96 b \left ({\mathrm e}^{2 b x +2 a}+1\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}}}+\frac {5 \,{\mathrm e}^{2 b x +2 a}}{16 b \left ({\mathrm e}^{2 b x +2 a}+1\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}}}-\frac {5}{16 b \left ({\mathrm e}^{2 b x +2 a}+1\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}}}-\frac {5 \,{\mathrm e}^{-2 b x -2 a}}{96 b \left ({\mathrm e}^{2 b x +2 a}+1\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}}}-\frac {{\mathrm e}^{-4 b x -4 a}}{160 b \left ({\mathrm e}^{2 b x +2 a}+1\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}}}\) | \(305\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.27, size = 82, normalized size = 1.08 \begin {gather*} \frac {e^{\left (5 \, b x + 5 \, a\right )}}{160 \, b} + \frac {5 \, e^{\left (3 \, b x + 3 \, a\right )}}{96 \, b} + \frac {5 \, e^{\left (b x + a\right )}}{16 \, b} - \frac {5 \, e^{\left (-b x - a\right )}}{16 \, b} - \frac {5 \, e^{\left (-3 \, b x - 3 \, a\right )}}{96 \, b} - \frac {e^{\left (-5 \, b x - 5 \, a\right )}}{160 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 66, normalized size = 0.87 \begin {gather*} \frac {3 \, \sinh \left (b x + a\right )^{5} + 5 \, {\left (6 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right )^{3} + 15 \, {\left (\cosh \left (b x + a\right )^{4} + 5 \, \cosh \left (b x + a\right )^{2} + 10\right )} \sinh \left (b x + a\right )}{240 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 14.51, size = 80, normalized size = 1.05 \begin {gather*} \begin {cases} \frac {8 \tanh ^{5}{\left (a + b x \right )}}{15 b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {5}{2}}} - \frac {4 \tanh ^{3}{\left (a + b x \right )}}{3 b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {5}{2}}} + \frac {\tanh {\left (a + b x \right )}}{b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {5}{2}}} & \text {for}\: b \neq 0 \\\frac {x}{\left (\operatorname {sech}^{2}{\left (a \right )}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.39, size = 70, normalized size = 0.92 \begin {gather*} -\frac {{\left (150 \, e^{\left (4 \, b x + 4 \, a\right )} + 25 \, e^{\left (2 \, b x + 2 \, a\right )} + 3\right )} e^{\left (-5 \, b x - 5 \, a\right )} - 3 \, e^{\left (5 \, b x + 5 \, a\right )} - 25 \, e^{\left (3 \, b x + 3 \, a\right )} - 150 \, e^{\left (b x + a\right )}}{480 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (\frac {1}{{\mathrm {cosh}\left (a+b\,x\right )}^2}\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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