Optimal. Leaf size=65 \[ -\frac {32}{35} b^3 x^{7/2}+\frac {16}{5} b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3 \]
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Rubi [A]
time = 0.02, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps
used = 4, 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 {16}{5} b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3-\frac {32}{35} b^3 x^{7/2} \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))^3}{\sqrt {x}} \, dx &=2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3-(6 b) \int \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3+\left (8 b^2\right ) \int x^{3/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=\frac {16}{5} b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3-\frac {1}{5} \left (16 b^3\right ) \int x^{5/2} \, dx\\ &=-\frac {32}{35} b^3 x^{7/2}+\frac {16}{5} b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 57, normalized size = 0.88 \begin {gather*} \frac {2}{35} \sqrt {x} \left (-16 b^3 x^3+56 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-70 b x \tanh ^{-1}(\tanh (a+b x))^2+35 \tanh ^{-1}(\tanh (a+b x))^3\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.32, size = 69, normalized size = 1.06
method | result | size |
derivativedivides | \(\frac {2 b^{3} x^{\frac {7}{2}}}{7}+\frac {6 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b^{2} x^{\frac {5}{2}}}{5}+2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} b \,x^{\frac {3}{2}}+2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}\) | \(69\) |
default | \(\frac {2 b^{3} x^{\frac {7}{2}}}{7}+\frac {6 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b^{2} x^{\frac {5}{2}}}{5}+2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} b \,x^{\frac {3}{2}}+2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}\) | \(69\) |
risch | \(\text {Expression too large to display}\) | \(7814\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 55, normalized size = 0.85 \begin {gather*} -4 \, b x^{\frac {3}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + 2 \, \sqrt {x} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {16}{35} \, {\left (2 \, b^{2} x^{\frac {7}{2}} - 7 \, b x^{\frac {5}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.35, size = 35, normalized size = 0.54 \begin {gather*} \frac {2}{35} \, {\left (5 \, b^{3} x^{3} + 21 \, a b^{2} x^{2} + 35 \, a^{2} b x + 35 \, a^{3}\right )} \sqrt {x} \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*} \int \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{\sqrt {x}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 35, normalized size = 0.54 \begin {gather*} \frac {2}{7} \, b^{3} x^{\frac {7}{2}} + \frac {6}{5} \, a b^{2} x^{\frac {5}{2}} + 2 \, a^{2} b x^{\frac {3}{2}} + 2 \, a^{3} \sqrt {x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.19, size = 182, normalized size = 2.80 \begin {gather*} \frac {2\,b^3\,x^{7/2}}{7}-\frac {\sqrt {x}\,{\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}{4}+\frac {b\,x^{3/2}\,{\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}{2}-\frac {3\,b^2\,x^{5/2}\,\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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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