Optimal. Leaf size=53 \[ \frac {x^2 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {x \tanh ^{-1}(\tanh (a+b x))^6}{15 b^2}+\frac {\tanh ^{-1}(\tanh (a+b x))^7}{105 b^3} \]
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Rubi [A]
time = 0.02, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2199, 2188, 30}
\begin {gather*} \frac {\tanh ^{-1}(\tanh (a+b x))^7}{105 b^3}-\frac {x \tanh ^{-1}(\tanh (a+b x))^6}{15 b^2}+\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^5}{5 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2188
Rule 2199
Rubi steps
\begin {align*} \int x^2 \tanh ^{-1}(\tanh (a+b x))^4 \, dx &=\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {2 \int x \tanh ^{-1}(\tanh (a+b x))^5 \, dx}{5 b}\\ &=\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {x \tanh ^{-1}(\tanh (a+b x))^6}{15 b^2}+\frac {\int \tanh ^{-1}(\tanh (a+b x))^6 \, dx}{15 b^2}\\ &=\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {x \tanh ^{-1}(\tanh (a+b x))^6}{15 b^2}+\frac {\text {Subst}\left (\int x^6 \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{15 b^3}\\ &=\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {x \tanh ^{-1}(\tanh (a+b x))^6}{15 b^2}+\frac {\tanh ^{-1}(\tanh (a+b x))^7}{105 b^3}\\ \end {align*}
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Mathematica [A]
time = 0.04, size = 71, normalized size = 1.34 \begin {gather*} \frac {1}{105} x^3 \left (b^4 x^4-7 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+21 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-35 b x \tanh ^{-1}(\tanh (a+b x))^3+35 \tanh ^{-1}(\tanh (a+b x))^4\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.02, size = 74, normalized size = 1.40 \[\frac {x^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{3}-\frac {4 b \left (\frac {x^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{4}-\frac {3 b \left (\frac {x^{5} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{5}-\frac {2 b \left (\frac {x^{6} \arctanh \left (\tanh \left (b x +a \right )\right )}{6}-\frac {b \,x^{7}}{42}\right )}{5}\right )}{4}\right )}{3}\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.42, size = 72, normalized size = 1.36 \begin {gather*} -\frac {1}{3} \, b x^{4} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} + \frac {1}{3} \, x^{3} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4} + \frac {1}{105} \, {\left (21 \, b x^{5} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + {\left (b^{2} x^{7} - 7 \, b x^{6} \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 = 0.33, size = 45, normalized size = 0.85 \begin {gather*} \frac {1}{7} \, b^{4} x^{7} + \frac {2}{3} \, a b^{3} x^{6} + \frac {6}{5} \, a^{2} b^{2} x^{5} + a^{3} b x^{4} + \frac {1}{3} \, a^{4} x^{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.42, size = 75, normalized size = 1.42 \begin {gather*} \frac {b^{4} x^{7}}{105} - \frac {b^{3} x^{6} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{15} + \frac {b^{2} x^{5} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{5} - \frac {b x^{4} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3} + \frac {x^{3} \operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.39, size = 45, normalized size = 0.85 \begin {gather*} \frac {1}{7} \, b^{4} x^{7} + \frac {2}{3} \, a b^{3} x^{6} + \frac {6}{5} \, a^{2} b^{2} x^{5} + a^{3} b x^{4} + \frac {1}{3} \, a^{4} x^{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.15, size = 70, normalized size = 1.32 \begin {gather*} \frac {b^4\,x^7}{105}-\frac {b^3\,x^6\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{15}+\frac {b^2\,x^5\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{5}-\frac {b\,x^4\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{3}+\frac {x^3\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4}{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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