Optimal. Leaf size=100 \[ \frac {\text {sech}^3(x) (a \sinh (x)+b)}{3 \left (a^2+b^2\right )}-\frac {2 b^4 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac {\text {sech}(x) \left (a \left (2 a^2+5 b^2\right ) \sinh (x)+3 b^3\right )}{3 \left (a^2+b^2\right )^2} \]
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Rubi [A] time = 0.21, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2696, 2866, 12, 2660, 618, 206} \[ -\frac {2 b^4 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac {\text {sech}^3(x) (a \sinh (x)+b)}{3 \left (a^2+b^2\right )}+\frac {\text {sech}(x) \left (a \left (2 a^2+5 b^2\right ) \sinh (x)+3 b^3\right )}{3 \left (a^2+b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 206
Rule 618
Rule 2660
Rule 2696
Rule 2866
Rubi steps
\begin {align*} \int \frac {\text {sech}^4(x)}{a+b \sinh (x)} \, dx &=\frac {\text {sech}^3(x) (b+a \sinh (x))}{3 \left (a^2+b^2\right )}-\frac {\int \frac {\text {sech}^2(x) \left (-2 a^2-3 b^2-2 a b \sinh (x)\right )}{a+b \sinh (x)} \, dx}{3 \left (a^2+b^2\right )}\\ &=\frac {\text {sech}^3(x) (b+a \sinh (x))}{3 \left (a^2+b^2\right )}+\frac {\text {sech}(x) \left (3 b^3+a \left (2 a^2+5 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}+\frac {\int \frac {3 b^4}{a+b \sinh (x)} \, dx}{3 \left (a^2+b^2\right )^2}\\ &=\frac {\text {sech}^3(x) (b+a \sinh (x))}{3 \left (a^2+b^2\right )}+\frac {\text {sech}(x) \left (3 b^3+a \left (2 a^2+5 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}+\frac {b^4 \int \frac {1}{a+b \sinh (x)} \, dx}{\left (a^2+b^2\right )^2}\\ &=\frac {\text {sech}^3(x) (b+a \sinh (x))}{3 \left (a^2+b^2\right )}+\frac {\text {sech}(x) \left (3 b^3+a \left (2 a^2+5 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}+\frac {\left (2 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 )^2}\\ &=\frac {\text {sech}^3(x) (b+a \sinh (x))}{3 \left (a^2+b^2\right )}+\frac {\text {sech}(x) \left (3 b^3+a \left (2 a^2+5 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}-\frac {\left (4 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 )^2}\\ &=-\frac {2 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 )^{5/2}}+\frac {\text {sech}^3(x) (b+a \sinh (x))}{3 \left (a^2+b^2\right )}+\frac {\text {sech}(x) \left (3 b^3+a \left (2 a^2+5 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}\\ \end {align*}
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Mathematica [A] time = 0.34, size = 102, normalized size = 1.02 \[ \frac {a \left (2 a^2+5 b^2\right ) \tanh (x)+\left (a^2+b^2\right ) \text {sech}^3(x) (a \sinh (x)+b)+\frac {6 b^4 \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+3 b^3 \text {sech}(x)}{3 \left (a^2+b^2\right )^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.61, size = 1142, normalized size = 11.42 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.47, size = 180, normalized size = 1.80 \[ \frac {b^{4} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (3 \, b^{3} e^{\left (5 \, x\right )} - 3 \, a b^{2} e^{\left (4 \, x\right )} + 4 \, a^{2} b e^{\left (3 \, x\right )} + 10 \, b^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} - 12 \, a b^{2} e^{\left (2 \, x\right )} + 3 \, b^{3} e^{x} - 2 \, a^{3} - 5 \, a b^{2}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.06, size = 182, normalized size = 1.82 \[ \frac {2 b^{4} \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}}-\frac {2 \left (\left (-a^{3}-2 a \,b^{2}\right ) \left (\tanh ^{5}\left (\frac {x}{2}\right )\right )+\left (-a^{2} b -2 b^{3}\right ) \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )+\left (-\frac {2}{3} a^{3}-\frac {8}{3} a \,b^{2}\right ) \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )-2 b^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+\left (-a^{3}-2 a \,b^{2}\right ) \tanh \left (\frac {x}{2}\right )-\frac {a^{2} b}{3}-\frac {4 b^{3}}{3}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.44, size = 230, normalized size = 2.30 \[ \frac {b^{4} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (3 \, b^{3} e^{\left (-x\right )} + 3 \, a b^{2} e^{\left (-4 \, x\right )} + 3 \, b^{3} e^{\left (-5 \, x\right )} + 2 \, a^{3} + 5 \, a b^{2} + 6 \, {\left (a^{3} + 2 \, a b^{2}\right )} e^{\left (-2 \, x\right )} + 2 \, {\left (2 \, a^{2} b + 5 \, b^{3}\right )} e^{\left (-3 \, x\right )}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + 3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-4 \, x\right )} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-6 \, x\right )}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.61, size = 634, normalized size = 6.34 \[ \frac {\frac {2\,b^3\,{\mathrm {e}}^x}{{\left (a^2+b^2\right )}^2}-\frac {2\,a\,b^2}{{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,x}+1}-\frac {\frac {4\,\left (a^3+a\,b^2\right )}{{\left (a^2+b^2\right )}^2}-\frac {8\,{\mathrm {e}}^x\,\left (a^2\,b+b^3\right )}{3\,{\left (a^2+b^2\right )}^2}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {\frac {8\,a}{3\,\left (a^2+b^2\right )}-\frac {8\,b\,{\mathrm {e}}^x}{3\,\left (a^2+b^2\right )}}{3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1}-\frac {2\,\mathrm {atan}\left (\left ({\mathrm {e}}^x\,\left (\frac {2\,b^2}{\sqrt {b^8}\,{\left (a^2+b^2\right )}^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {2\,a\,\left (a^5\,\sqrt {b^8}+2\,a^3\,b^2\,\sqrt {b^8}+a\,b^4\,\sqrt {b^8}\right )}{b^6\,\sqrt {-{\left (a^2+b^2\right )}^5}\,\left (a^4+2\,a^2\,b^2+b^4\right )\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}}\right )-\frac {2\,a\,\left (b^5\,\sqrt {b^8}+2\,a^2\,b^3\,\sqrt {b^8}+a^4\,b\,\sqrt {b^8}\right )}{b^6\,\sqrt {-{\left (a^2+b^2\right )}^5}\,\left (a^4+2\,a^2\,b^2+b^4\right )\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}}\right )\,\left (\frac {b^5\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}}{2}+\frac {a^4\,b\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}}{2}+a^2\,b^3\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}\right )\right )\,\sqrt {b^8}}{\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}} \]
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 \[ \int \frac {\operatorname {sech}^{4}{\relax (x )}}{a + b \sinh {\relax (x )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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