Optimal. Leaf size=48 \[ -\frac {8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}-\frac {16 b^2}{15 \sqrt {x}} \]
[Out]
________________________________________________________________________________________
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 = {2168, 30} \[ -\frac {8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}-\frac {16 b^2}{15 \sqrt {x}} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 30
Rule 2168
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*}
________________________________________________________________________________________
Mathematica [A] time = 0.05, size = 40, normalized size = 0.83 \[ -\frac {2 \left (4 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2\right )}{15 x^{5/2}} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.53, size = 24, normalized size = 0.50 \[ -\frac {2 \, {\left (15 \, b^{2} x^{2} + 10 \, a b x + 3 \, a^{2}\right )}}{15 \, x^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.25, size = 24, normalized size = 0.50 \[ -\frac {2 \, {\left (15 \, b^{2} x^{2} + 10 \, a b x + 3 \, a^{2}\right )}}{15 \, x^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.25, size = 38, normalized size = 0.79 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{5 x^{\frac {5}{2}}}+\frac {8 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{3 x^{\frac {3}{2}}}-\frac {2 b}{3 \sqrt {x}}\right )}{5} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.35, size = 36, normalized size = 0.75 \[ -\frac {16 \, b^{2}}{15 \, \sqrt {x}} - \frac {8 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{15 \, x^{\frac {3}{2}}} - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{5 \, x^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 1.13, size = 122, normalized size = 2.54 \[ \frac {2\,b\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{3\,x^{3/2}}-\frac {2\,b^2}{\sqrt {x}}-\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{10\,x^{5/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 154.70, size = 49, normalized size = 1.02 \[ - \frac {16 b^{2}}{15 \sqrt {x}} - \frac {8 b \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{15 x^{\frac {3}{2}}} - \frac {2 \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________