Optimal. Leaf size=109 \[ -\frac {2 b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {b^2 \tanh (c+d x)}{a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))}+\frac {x}{a^2} \]
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Rubi [A] time = 0.16, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3785, 3919, 3831, 2659, 208} \[ -\frac {2 b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {b^2 \tanh (c+d x)}{a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))}+\frac {x}{a^2} \]
Antiderivative was successfully verified.
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Rule 208
Rule 2659
Rule 3785
Rule 3831
Rule 3919
Rubi steps
\begin {align*} \int \frac {1}{(a+b \text {sech}(c+d x))^2} \, dx &=\frac {b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))}-\frac {\int \frac {-a^2+b^2+a b \text {sech}(c+d x)}{a+b \text {sech}(c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {x}{a^2}+\frac {b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))}-\frac {\left (b \left (2 a^2-b^2\right )\right ) \int \frac {\text {sech}(c+d x)}{a+b \text {sech}(c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac {x}{a^2}+\frac {b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))}-\frac {\left (2 a^2-b^2\right ) \int \frac {1}{1+\frac {a \cosh (c+d x)}{b}} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac {x}{a^2}+\frac {b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))}+\frac {\left (2 i \left (2 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{a^2 \left (a^2-b^2\right ) d}\\ &=\frac {x}{a^2}-\frac {2 b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.41, size = 203, normalized size = 1.86 \[ \frac {b \left (\left (a^2-b^2\right )^{3/2} (c+d x)+a b \sqrt {a^2-b^2} \sinh (c+d x)+\left (4 a^2 b-2 b^3\right ) \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )\right )+a \cosh (c+d x) \left (\left (a^2-b^2\right )^{3/2} (c+d x)+\left (4 a^2 b-2 b^3\right ) \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )\right )}{a^2 d (a-b) (a+b) \sqrt {a^2-b^2} (a \cosh (c+d x)+b)} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.42, size = 1207, normalized size = 11.07 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.12, size = 134, normalized size = 1.23 \[ -\frac {\frac {2 \, {\left (2 \, a^{2} b - b^{3}\right )} \arctan \left (\frac {a e^{\left (d x + c\right )} + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (b^{3} e^{\left (d x + c\right )} + a b^{2}\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} {\left (a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b e^{\left (d x + c\right )} + a\right )}} - \frac {d x + c}{a^{2}}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.20, size = 221, normalized size = 2.03 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}+\frac {2 b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (a^{2}-b^{2}\right ) \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {4 b \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{d \left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 b^{3} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{d \,a^{2} \left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.85, size = 296, normalized size = 2.72 \[ \frac {\frac {2\,b^2}{d\,\left (a\,b^2-a^3\right )}+\frac {2\,b^3\,{\mathrm {e}}^{c+d\,x}}{a\,d\,\left (a\,b^2-a^3\right )}}{a+2\,b\,{\mathrm {e}}^{c+d\,x}+a\,{\mathrm {e}}^{2\,c+2\,d\,x}}+\frac {x}{a^2}+\frac {b\,\ln \left (\frac {2\,{\mathrm {e}}^{c+d\,x}\,\left (2\,a^2\,b-b^3\right )}{a^3\,\left (a^2-b^2\right )}-\frac {2\,b\,\left (2\,a^2-b^2\right )\,\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}{a^3\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}\right )\,\left (2\,a^2-b^2\right )}{a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}-\frac {b\,\ln \left (\frac {2\,{\mathrm {e}}^{c+d\,x}\,\left (2\,a^2\,b-b^3\right )}{a^3\,\left (a^2-b^2\right )}+\frac {2\,b\,\left (2\,a^2-b^2\right )\,\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}{a^3\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}\right )\,\left (2\,a^2-b^2\right )}{a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}} \]
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 \[ \int \frac {1}{\left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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