Optimal. Leaf size=60 \[ -\frac {2 A \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-\frac {B \log (a+b \sinh (x))}{a}+\frac {B \log (\sinh (x))}{a} \]
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Rubi [A] time = 0.15, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {4401, 2660, 618, 206, 2721, 36, 29, 31} \[ -\frac {2 A \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-\frac {B \log (a+b \sinh (x))}{a}+\frac {B \log (\sinh (x))}{a} \]
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 206
Rule 618
Rule 2660
Rule 2721
Rule 4401
Rubi steps
\begin {align*} \int \frac {A+B \coth (x)}{a+b \sinh (x)} \, dx &=\int \left (\frac {A}{a+b \sinh (x)}+\frac {B \coth (x)}{a+b \sinh (x)}\right ) \, dx\\ &=A \int \frac {1}{a+b \sinh (x)} \, dx+B \int \frac {\coth (x)}{a+b \sinh (x)} \, dx\\ &=(2 A) \operatorname {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )+B \operatorname {Subst}\left (\int \frac {1}{x (a+x)} \, dx,x,b \sinh (x)\right )\\ &=-\left ((4 A) \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 )\right )+\frac {B \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,b \sinh (x)\right )}{a}-\frac {B \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \sinh (x)\right )}{a}\\ &=-\frac {2 A \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}+\frac {B \log (\sinh (x))}{a}-\frac {B \log (a+b \sinh (x))}{a}\\ \end {align*}
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Mathematica [A] time = 0.17, size = 65, normalized size = 1.08 \[ \frac {2 A \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+\frac {B (\log (\sinh (x))-\log (a+b \sinh (x)))}{a} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.63, size = 183, normalized size = 3.05 \[ \frac {\sqrt {a^{2} + b^{2}} A a \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) - b}\right ) - {\left (B a^{2} + B b^{2}\right )} \log \left (\frac {2 \, {\left (b \sinh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + {\left (B a^{2} + B b^{2}\right )} \log \left (\frac {2 \, \sinh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{3} + a b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.36, size = 102, normalized size = 1.70 \[ \frac {A \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}}} + \frac {B \log \left (e^{x} + 1\right )}{a} - \frac {B \log \left ({\left | b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b \right |}\right )}{a} + \frac {B \log \left ({\left | e^{x} - 1 \right |}\right )}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 73, normalized size = 1.22 \[ -\frac {B \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a \right )}{a}+\frac {2 \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right ) A}{\sqrt {a^{2}+b^{2}}}+\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.48, size = 106, normalized size = 1.77 \[ -B {\left (\frac {\log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a} - \frac {\log \left (e^{\left (-x\right )} + 1\right )}{a} - \frac {\log \left (e^{\left (-x\right )} - 1\right )}{a}\right )} + \frac {A \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 11.21, size = 164, normalized size = 2.73 \[ \frac {B\,\ln \left (16\,B^2\,a^2+16\,B^2\,b^2-16\,B^2\,a^2\,{\mathrm {e}}^{2\,x}-16\,B^2\,b^2\,{\mathrm {e}}^{2\,x}\right )}{a}-\frac {2\,\mathrm {atan}\left (\frac {A^2\,b^2\,{\mathrm {e}}^x\,\sqrt {-a^2-b^2}+A^2\,a\,b\,\sqrt {-a^2-b^2}}{A\,b\,\sqrt {A^2}\,\left (a^2+b^2\right )}\right )\,\sqrt {A^2}}{\sqrt {-a^2-b^2}}-\frac {B\,\ln \left (32\,B^2\,a\,{\mathrm {e}}^x-16\,B^2\,b+16\,B^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a} \]
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 \[ \int \frac {A + B \coth {\relax (x )}}{a + b \sinh {\relax (x )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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