Optimal. Leaf size=45 \[ \frac {e^{a c+b c x} \coth ^{-1}(\tanh (c (a+b x)))}{b c}-\frac {e^{a c+b c x}}{b c} \]
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Rubi [A] time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2194, 6276} \[ \frac {e^{a c+b c x} \coth ^{-1}(\tanh (c (a+b x)))}{b c}-\frac {e^{a c+b c x}}{b c} \]
Antiderivative was successfully verified.
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Rule 2194
Rule 6276
Rubi steps
\begin {align*} \int e^{c (a+b x)} \coth ^{-1}(\tanh (a c+b c x)) \, dx &=\frac {\operatorname {Subst}\left (\int e^x \coth ^{-1}(\tanh (x)) \, dx,x,a c+b c x\right )}{b c}\\ &=\frac {e^{a c+b c x} \coth ^{-1}(\tanh (c (a+b x)))}{b c}-\frac {\operatorname {Subst}\left (\int e^x \, dx,x,a c+b c x\right )}{b c}\\ &=-\frac {e^{a c+b c x}}{b c}+\frac {e^{a c+b c x} \coth ^{-1}(\tanh (c (a+b x)))}{b c}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 46, normalized size = 1.02 \[ \frac {e^{c (a+b x)} \left (\coth ^{-1}\left (\frac {e^{2 c (a+b x)}-1}{e^{2 c (a+b x)}+1}\right )-1\right )}{b c} \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 0.58, size = 25, normalized size = 0.56 \[ \frac {{\left (b c x + a c - 1\right )} e^{\left (b c x + a c\right )}}{b c} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {arcoth}\left (\tanh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.38, size = 349, normalized size = 7.76 \[ \frac {{\mathrm e}^{c \left (b x +a \right )} \ln \left ({\mathrm e}^{c \left (b x +a \right )}\right )}{b c}-\frac {i \left (-2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{c \left (b x +a \right )}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{c \left (b x +a \right )}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )^{3}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{3}-4 i+2 \pi \right ) {\mathrm e}^{c \left (b x +a \right )}}{4 b c} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 43, normalized size = 0.96 \[ \frac {\operatorname {arcoth}\left (\tanh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} - \frac {e^{\left (b c x + a c\right )}}{b c} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.10, size = 28, normalized size = 0.62 \[ \frac {{\mathrm {e}}^{a\,c+b\,c\,x}\,\left (\mathrm {acoth}\left (\mathrm {tanh}\left (a\,c+b\,c\,x\right )\right )-1\right )}{b\,c} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.12, size = 66, normalized size = 1.47 \[ \begin {cases} \frac {i \pi x}{2} & \text {for}\: b = 0 \wedge c = 0 \\x e^{a c} \operatorname {acoth}{\left (\tanh {\left (a c \right )} \right )} & \text {for}\: b = 0 \\\frac {i \pi x}{2} & \text {for}\: c = 0 \\\frac {e^{a c} e^{b c x} \operatorname {acoth}{\left (\tanh {\left (a c + b c x \right )} \right )}}{b c} - \frac {e^{a c} e^{b c x}}{b c} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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