Optimal. Leaf size=80 \[ \frac {b^4 x^{11}}{2310}-\frac {1}{210} b^3 x^{10} \tanh ^{-1}(\tanh (a+b x))+\frac {1}{42} b^2 x^9 \tanh ^{-1}(\tanh (a+b x))^2-\frac {1}{14} b x^8 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4 \]
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Rubi [A]
time = 0.04, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2199, 30}
\begin {gather*} -\frac {1}{210} b^3 x^{10} \tanh ^{-1}(\tanh (a+b x))+\frac {1}{42} b^2 x^9 \tanh ^{-1}(\tanh (a+b x))^2-\frac {1}{14} b x^8 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4+\frac {b^4 x^{11}}{2310} \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2199
Rubi steps
\begin {align*} \int x^6 \tanh ^{-1}(\tanh (a+b x))^4 \, dx &=\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4-\frac {1}{7} (4 b) \int x^7 \tanh ^{-1}(\tanh (a+b x))^3 \, dx\\ &=-\frac {1}{14} b x^8 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4+\frac {1}{14} \left (3 b^2\right ) \int x^8 \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=\frac {1}{42} b^2 x^9 \tanh ^{-1}(\tanh (a+b x))^2-\frac {1}{14} b x^8 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4-\frac {1}{21} b^3 \int x^9 \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=-\frac {1}{210} b^3 x^{10} \tanh ^{-1}(\tanh (a+b x))+\frac {1}{42} b^2 x^9 \tanh ^{-1}(\tanh (a+b x))^2-\frac {1}{14} b x^8 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4+\frac {1}{210} b^4 \int x^{10} \, dx\\ &=\frac {b^4 x^{11}}{2310}-\frac {1}{210} b^3 x^{10} \tanh ^{-1}(\tanh (a+b x))+\frac {1}{42} b^2 x^9 \tanh ^{-1}(\tanh (a+b x))^2-\frac {1}{14} b x^8 \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{7} x^7 \tanh ^{-1}(\tanh (a+b x))^4\\ \end {align*}
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Mathematica [A]
time = 0.04, size = 71, normalized size = 0.89 \begin {gather*} \frac {x^7 \left (b^4 x^4-11 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+55 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-165 b x \tanh ^{-1}(\tanh (a+b x))^3+330 \tanh ^{-1}(\tanh (a+b x))^4\right )}{2310} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.03, size = 74, normalized size = 0.92 \[\frac {x^{7} \arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{7}-\frac {4 b \left (\frac {x^{8} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{8}-\frac {3 b \left (\frac {x^{9} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{9}-\frac {2 b \left (\frac {x^{10} \arctanh \left (\tanh \left (b x +a \right )\right )}{10}-\frac {x^{11} b}{110}\right )}{9}\right )}{8}\right )}{7}\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.40, size = 72, normalized size = 0.90 \begin {gather*} -\frac {1}{14} \, b x^{8} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} + \frac {1}{7} \, x^{7} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4} + \frac {1}{2310} \, {\left (55 \, b x^{9} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + {\left (b^{2} x^{11} - 11 \, b x^{10} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b\right )} b \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 46, normalized size = 0.58 \begin {gather*} \frac {1}{11} \, b^{4} x^{11} + \frac {2}{5} \, a b^{3} x^{10} + \frac {2}{3} \, a^{2} b^{2} x^{9} + \frac {1}{2} \, a^{3} b x^{8} + \frac {1}{7} \, a^{4} x^{7} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 2.49, size = 75, normalized size = 0.94 \begin {gather*} \frac {b^{4} x^{11}}{2310} - \frac {b^{3} x^{10} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{210} + \frac {b^{2} x^{9} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{42} - \frac {b x^{8} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{14} + \frac {x^{7} \operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{7} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.38, size = 46, normalized size = 0.58 \begin {gather*} \frac {1}{11} \, b^{4} x^{11} + \frac {2}{5} \, a b^{3} x^{10} + \frac {2}{3} \, a^{2} b^{2} x^{9} + \frac {1}{2} \, a^{3} b x^{8} + \frac {1}{7} \, a^{4} x^{7} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.03, size = 242, normalized size = 3.02 \begin {gather*} \frac {x^7\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}{112}+\frac {b^4\,x^{11}}{11}-\frac {b\,x^8\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{16}-\frac {b^3\,x^{10}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{5}+\frac {b^2\,x^9\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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