Optimal. Leaf size=144 \[ -\frac {b \text {sech}^3(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}^3(x) \left (\left (a^2-4 b^2\right ) \sinh (x)+5 a b\right )}{3 \left (a^2+b^2\right )^2}-\frac {10 a b^4 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}+\frac {\text {sech}(x) \left (\left (2 a^4+9 a^2 b^2-8 b^4\right ) \sinh (x)+15 a b^3\right )}{3 \left (a^2+b^2\right )^3} \]
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Rubi [A] time = 0.31, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2694, 2866, 12, 2660, 618, 206} \[ -\frac {10 a b^4 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}-\frac {b \text {sech}^3(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}^3(x) \left (\left (a^2-4 b^2\right ) \sinh (x)+5 a b\right )}{3 \left (a^2+b^2\right )^2}+\frac {\text {sech}(x) \left (\left (9 a^2 b^2+2 a^4-8 b^4\right ) \sinh (x)+15 a b^3\right )}{3 \left (a^2+b^2\right )^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 206
Rule 618
Rule 2660
Rule 2694
Rule 2866
Rubi steps
\begin {align*} \int \frac {\text {sech}^4(x)}{(a+b \sinh (x))^2} \, dx &=-\frac {b \text {sech}^3(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {\int \frac {\text {sech}^4(x) (-a+4 b \sinh (x))}{a+b \sinh (x)} \, dx}{a^2+b^2}\\ &=-\frac {b \text {sech}^3(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}^3(x) \left (5 a b+\left (a^2-4 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}+\frac {\int \frac {\text {sech}^2(x) \left (a \left (2 a^2+7 b^2\right )+2 b \left (a^2-4 b^2\right ) \sinh (x)\right )}{a+b \sinh (x)} \, dx}{3 \left (a^2+b^2\right )^2}\\ &=-\frac {b \text {sech}^3(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}^3(x) \left (5 a b+\left (a^2-4 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}+\frac {\text {sech}(x) \left (15 a b^3+\left (2 a^4+9 a^2 b^2-8 b^4\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^3}-\frac {\int -\frac {15 a b^4}{a+b \sinh (x)} \, dx}{3 \left (a^2+b^2\right )^3}\\ &=-\frac {b \text {sech}^3(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}^3(x) \left (5 a b+\left (a^2-4 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}+\frac {\text {sech}(x) \left (15 a b^3+\left (2 a^4+9 a^2 b^2-8 b^4\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^3}+\frac {\left (5 a b^4\right ) \int \frac {1}{a+b \sinh (x)} \, dx}{\left (a^2+b^2\right )^3}\\ &=-\frac {b \text {sech}^3(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}^3(x) \left (5 a b+\left (a^2-4 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}+\frac {\text {sech}(x) \left (15 a b^3+\left (2 a^4+9 a^2 b^2-8 b^4\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^3}+\frac {\left (10 a b^4\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{\left (a^2+b^2\right )^3}\\ &=-\frac {b \text {sech}^3(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}^3(x) \left (5 a b+\left (a^2-4 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}+\frac {\text {sech}(x) \left (15 a b^3+\left (2 a^4+9 a^2 b^2-8 b^4\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^3}-\frac {\left (20 a b^4\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{\left (a^2+b^2\right )^3}\\ &=-\frac {10 a b^4 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}-\frac {b \text {sech}^3(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}^3(x) \left (5 a b+\left (a^2-4 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}+\frac {\text {sech}(x) \left (15 a b^3+\left (2 a^4+9 a^2 b^2-8 b^4\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^3}\\ \end {align*}
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Mathematica [A] time = 0.44, size = 137, normalized size = 0.95 \[ \frac {\left (a^2+b^2\right ) \text {sech}^3(x) \left (\left (a^2-b^2\right ) \sinh (x)+2 a b\right )+\frac {30 a b^4 \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+\left (2 a^4+9 a^2 b^2-5 b^4\right ) \tanh (x)-\frac {3 b^5 \cosh (x)}{a+b \sinh (x)}+12 a b^3 \text {sech}(x)}{3 \left (a^2+b^2\right )^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.00, size = 3044, normalized size = 21.14 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.26, size = 287, normalized size = 1.99 \[ \frac {5 \, a b^{4} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (a b^{4} e^{x} - b^{5}\right )}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )}} + \frac {2 \, {\left (12 \, a b^{3} e^{\left (5 \, x\right )} - 9 \, a^{2} b^{2} e^{\left (4 \, x\right )} + 3 \, b^{4} e^{\left (4 \, x\right )} + 8 \, a^{3} b e^{\left (3 \, x\right )} + 32 \, a b^{3} e^{\left (3 \, x\right )} - 6 \, a^{4} e^{\left (2 \, x\right )} - 18 \, a^{2} b^{2} e^{\left (2 \, x\right )} + 12 \, b^{4} e^{\left (2 \, x\right )} + 12 \, a b^{3} e^{x} - 2 \, a^{4} - 9 \, a^{2} b^{2} + 5 \, b^{4}\right )}}{3 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 266, normalized size = 1.85 \[ -\frac {2 b^{4} \left (\frac {-\frac {b^{2} \tanh \left (\frac {x}{2}\right )}{a}-b}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a}-\frac {5 a \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right )}-\frac {2 \left (\left (-a^{4}-3 a^{2} b^{2}+2 b^{4}\right ) \left (\tanh ^{5}\left (\frac {x}{2}\right )\right )+\left (-2 a^{3} b -6 a \,b^{3}\right ) \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )+\left (-\frac {2}{3} a^{4}-6 a^{2} b^{2}+\frac {8}{3} b^{4}\right ) \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )-8 a \,b^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+\left (-a^{4}-3 a^{2} b^{2}+2 b^{4}\right ) \tanh \left (\frac {x}{2}\right )-\frac {2 a^{3} b}{3}-\frac {14 a \,b^{3}}{3}\right )}{\left (a^{2}+b^{2}\right ) \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.42, size = 490, normalized size = 3.40 \[ \frac {5 \, a b^{4} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (15 \, a^{2} b^{3} e^{\left (-6 \, x\right )} - 15 \, a b^{4} e^{\left (-7 \, x\right )} + 2 \, a^{4} b + 9 \, a^{2} b^{3} - 8 \, b^{5} + {\left (4 \, a^{5} + 18 \, a^{3} b^{2} - a b^{4}\right )} e^{\left (-x\right )} + {\left (4 \, a^{4} b + 33 \, a^{2} b^{3} - 16 \, b^{5}\right )} e^{\left (-2 \, x\right )} + {\left (12 \, a^{5} + 44 \, a^{3} b^{2} - 13 \, a b^{4}\right )} e^{\left (-3 \, x\right )} + 5 \, {\left (2 \, a^{4} b + 11 \, a^{2} b^{3}\right )} e^{\left (-4 \, x\right )} + 5 \, {\left (2 \, a^{3} b^{2} - 7 \, a b^{4}\right )} e^{\left (-5 \, x\right )}\right )}}{3 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7} + 2 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} e^{\left (-x\right )} + 2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} e^{\left (-2 \, x\right )} + 6 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} e^{\left (-3 \, x\right )} + 6 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} e^{\left (-5 \, x\right )} - 2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} e^{\left (-6 \, x\right )} + 2 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} e^{\left (-7 \, x\right )} - {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} e^{\left (-8 \, x\right )}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.06, size = 476, normalized size = 3.31 \[ \frac {\frac {8\,\left (a^2-b^2\right )}{3\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {16\,a\,b\,{\mathrm {e}}^x}{3\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1}-\frac {\frac {4\,\left (a^6+a^4\,b^2-a^2\,b^4-b^6\right )}{{\left (a^4+2\,a^2\,b^2+b^4\right )}^2}-\frac {16\,{\mathrm {e}}^x\,\left (a^5\,b+2\,a^3\,b^3+a\,b^5\right )}{3\,{\left (a^4+2\,a^2\,b^2+b^4\right )}^2}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}-\frac {\frac {2\,\left (3\,a^4\,b^2+2\,a^2\,b^4-b^6\right )}{{\left (a^4+2\,a^2\,b^2+b^4\right )}^2}-\frac {8\,{\mathrm {e}}^x\,\left (a^3\,b^3+a\,b^5\right )}{{\left (a^4+2\,a^2\,b^2+b^4\right )}^2}}{{\mathrm {e}}^{2\,x}+1}-\frac {\frac {2\,\left (a^2\,b^9+b^{11}\right )}{b^3\,\left (a^2\,b+b^3\right )\,{\left (a^2+b^2\right )}^3}-\frac {2\,{\mathrm {e}}^x\,\left (a^3\,b^9+a\,b^{11}\right )}{b^4\,\left (a^2\,b+b^3\right )\,{\left (a^2+b^2\right )}^3}}{2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}}-\frac {5\,a\,b^4\,\ln \left (-\frac {10\,a\,b^3\,\left (b-a\,{\mathrm {e}}^x\right )}{{\left (a^2+b^2\right )}^{7/2}}-\frac {10\,a\,b^3\,{\mathrm {e}}^x}{{\left (a^2+b^2\right )}^3}\right )}{{\left (a^2+b^2\right )}^{7/2}}+\frac {5\,a\,b^4\,\ln \left (\frac {10\,a\,b^3\,\left (b-a\,{\mathrm {e}}^x\right )}{{\left (a^2+b^2\right )}^{7/2}}-\frac {10\,a\,b^3\,{\mathrm {e}}^x}{{\left (a^2+b^2\right )}^3}\right )}{{\left (a^2+b^2\right )}^{7/2}} \]
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 \[ \int \frac {\operatorname {sech}^{4}{\relax (x )}}{\left (a + b \sinh {\relax (x )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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