Optimal. Leaf size=107 \[ \frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1+e^{2 a+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1+e^{2 a+2 b x}\right )^2}+\frac {5 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}-\frac {3 \text {ArcTan}\left (e^{a+b x}\right )}{b} \]
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Rubi [A]
time = 0.05, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2320, 398,
1272, 1171, 393, 209} \begin {gather*} -\frac {3 \text {ArcTan}\left (e^{a+b x}\right )}{b}+\frac {e^{a+b x}}{b}+\frac {5 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac {14 e^{a+b x}}{3 b \left (e^{2 a+2 b x}+1\right )^2}+\frac {8 e^{a+b x}}{3 b \left (e^{2 a+2 b x}+1\right )^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 393
Rule 398
Rule 1171
Rule 1272
Rule 2320
Rubi steps
\begin {align*} \int e^{a+b x} \tanh ^4(a+b x) \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^4}{\left (1+x^2\right )^4} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\text {Subst}\left (\int \left (1-\frac {8 x^2 \left (1+x^4\right )}{\left (1+x^2\right )^4}\right ) \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}-\frac {8 \text {Subst}\left (\int \frac {x^2 \left (1+x^4\right )}{\left (1+x^2\right )^4} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1+e^{2 a+2 b x}\right )^3}+\frac {4 \text {Subst}\left (\int \frac {-2+6 x^2-6 x^4}{\left (1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{3 b}\\ &=\frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1+e^{2 a+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1+e^{2 a+2 b x}\right )^2}-\frac {\text {Subst}\left (\int \frac {-6+24 x^2}{\left (1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{3 b}\\ &=\frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1+e^{2 a+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1+e^{2 a+2 b x}\right )^2}+\frac {5 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}-\frac {3 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1+e^{2 a+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1+e^{2 a+2 b x}\right )^2}+\frac {5 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}-\frac {3 \tan ^{-1}\left (e^{a+b x}\right )}{b}\\ \end {align*}
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Mathematica [A]
time = 0.12, size = 76, normalized size = 0.71 \begin {gather*} \frac {e^{a+b x} \left (12+25 e^{2 (a+b x)}+24 e^{4 (a+b x)}+3 e^{6 (a+b x)}\right )}{3 b \left (1+e^{2 (a+b x)}\right )^3}-\frac {3 \text {ArcTan}\left (e^{a+b x}\right )}{b} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains complex when optimal does not.
time = 1.45, size = 92, normalized size = 0.86
method | result | size |
risch | \(\frac {{\mathrm e}^{b x +a}}{b}+\frac {{\mathrm e}^{b x +a} \left (15 \,{\mathrm e}^{4 b x +4 a}+16 \,{\mathrm e}^{2 b x +2 a}+9\right )}{3 b \left ({\mathrm e}^{2 b x +2 a}+1\right )^{3}}+\frac {3 i \ln \left ({\mathrm e}^{b x +a}-i\right )}{2 b}-\frac {3 i \ln \left ({\mathrm e}^{b x +a}+i\right )}{2 b}\) | \(92\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.47, size = 94, normalized size = 0.88 \begin {gather*} -\frac {3 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b} + \frac {e^{\left (b x + a\right )}}{b} + \frac {15 \, e^{\left (5 \, b x + 5 \, a\right )} + 16 \, e^{\left (3 \, b x + 3 \, a\right )} + 9 \, e^{\left (b x + a\right )}}{3 \, b {\left (e^{\left (6 \, b x + 6 \, a\right )} + 3 \, e^{\left (4 \, b x + 4 \, a\right )} + 3 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 604 vs.
\(2 (95) = 190\).
time = 0.37, size = 604, normalized size = 5.64 \begin {gather*} \frac {3 \, \cosh \left (b x + a\right )^{7} + 21 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{6} + 3 \, \sinh \left (b x + a\right )^{7} + 3 \, {\left (21 \, \cosh \left (b x + a\right )^{2} + 8\right )} \sinh \left (b x + a\right )^{5} + 24 \, \cosh \left (b x + a\right )^{5} + 15 \, {\left (7 \, \cosh \left (b x + a\right )^{3} + 8 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{4} + 5 \, {\left (21 \, \cosh \left (b x + a\right )^{4} + 48 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right )^{3} + 25 \, \cosh \left (b x + a\right )^{3} + 3 \, {\left (21 \, \cosh \left (b x + a\right )^{5} + 80 \, \cosh \left (b x + a\right )^{3} + 25 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 9 \, {\left (\cosh \left (b x + a\right )^{6} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + \sinh \left (b x + a\right )^{6} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{4} + 3 \, \cosh \left (b x + a\right )^{4} + 4 \, {\left (5 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{4} + 6 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 3 \, \cosh \left (b x + a\right )^{2} + 6 \, {\left (\cosh \left (b x + a\right )^{5} + 2 \, \cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 3 \, {\left (7 \, \cosh \left (b x + a\right )^{6} + 40 \, \cosh \left (b x + a\right )^{4} + 25 \, \cosh \left (b x + a\right )^{2} + 4\right )} \sinh \left (b x + a\right ) + 12 \, \cosh \left (b x + a\right )}{3 \, {\left (b \cosh \left (b x + a\right )^{6} + 6 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + b \sinh \left (b x + a\right )^{6} + 3 \, b \cosh \left (b x + a\right )^{4} + 3 \, {\left (5 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{4} + 4 \, {\left (5 \, b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )^{2} + 3 \, {\left (5 \, b \cosh \left (b x + a\right )^{4} + 6 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 6 \, {\left (b \cosh \left (b x + a\right )^{5} + 2 \, b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \end {gather*}
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 \begin {gather*} e^{a} \int e^{b x} \tanh ^{4}{\left (a + b x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.42, size = 68, normalized size = 0.64 \begin {gather*} \frac {\frac {15 \, e^{\left (5 \, b x + 5 \, a\right )} + 16 \, e^{\left (3 \, b x + 3 \, a\right )} + 9 \, e^{\left (b x + a\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{3}} - 9 \, \arctan \left (e^{\left (b x + a\right )}\right ) + 3 \, e^{\left (b x + a\right )}}{3 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.10, size = 155, normalized size = 1.45 \begin {gather*} \frac {{\mathrm {e}}^{a+b\,x}}{b}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}+\frac {\frac {4\,{\mathrm {e}}^{a+b\,x}}{3\,b}+\frac {4\,{\mathrm {e}}^{5\,a+5\,b\,x}}{3\,b}}{3\,{\mathrm {e}}^{2\,a+2\,b\,x}+3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}+1}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}+\frac {11\,{\mathrm {e}}^{a+b\,x}}{3\,b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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