Optimal. Leaf size=48 \[ -\frac {16 b^2}{15 \sqrt {x}}-\frac {8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}} \]
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Rubi [A]
time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2199, 30}
\begin {gather*} -\frac {8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}-\frac {16 b^2}{15 \sqrt {x}} \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2199
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{x^{7/2}} \, dx &=-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}+\frac {1}{5} (4 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))}{x^{5/2}} \, dx\\ &=-\frac {8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}+\frac {1}{15} \left (8 b^2\right ) \int \frac {1}{x^{3/2}} \, dx\\ &=-\frac {16 b^2}{15 \sqrt {x}}-\frac {8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}\\ \end {align*}
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Mathematica [A]
time = 0.03, size = 40, normalized size = 0.83 \begin {gather*} -\frac {2 \left (8 b^2 x^2+4 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2\right )}{15 x^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.29, size = 38, normalized size = 0.79
method | result | size |
derivativedivides | \(-\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{5 x^{\frac {5}{2}}}+\frac {8 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{3 x^{\frac {3}{2}}}-\frac {2 b}{3 \sqrt {x}}\right )}{5}\) | \(38\) |
default | \(-\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{5 x^{\frac {5}{2}}}+\frac {8 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{3 x^{\frac {3}{2}}}-\frac {2 b}{3 \sqrt {x}}\right )}{5}\) | \(38\) |
risch | \(\text {Expression too large to display}\) | \(1978\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 36, normalized size = 0.75 \begin {gather*} -\frac {16 \, b^{2}}{15 \, \sqrt {x}} - \frac {8 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{15 \, x^{\frac {3}{2}}} - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{5 \, x^{\frac {5}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.35, size = 24, normalized size = 0.50 \begin {gather*} -\frac {2 \, {\left (15 \, b^{2} x^{2} + 10 \, a b x + 3 \, a^{2}\right )}}{15 \, x^{\frac {5}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 20.26, size = 49, normalized size = 1.02 \begin {gather*} - \frac {16 b^{2}}{15 \sqrt {x}} - \frac {8 b \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{15 x^{\frac {3}{2}}} - \frac {2 \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{\frac {5}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.38, size = 24, normalized size = 0.50 \begin {gather*} -\frac {2 \, {\left (15 \, b^{2} x^{2} + 10 \, a b x + 3 \, a^{2}\right )}}{15 \, x^{\frac {5}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.13, size = 122, normalized size = 2.54 \begin {gather*} \frac {2\,b\,\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\,x^{3/2}}-\frac {2\,b^2}{\sqrt {x}}-\frac {{\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}{10\,x^{5/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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