Optimal. Leaf size=224 \[ 4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {PolyLog}\left (2,1-\frac {2}{1-c \sqrt {x}}\right )+3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {PolyLog}\left (2,-1+\frac {2}{1-c \sqrt {x}}\right )+3 b^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {PolyLog}\left (3,1-\frac {2}{1-c \sqrt {x}}\right )-3 b^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {PolyLog}\left (3,-1+\frac {2}{1-c \sqrt {x}}\right )-\frac {3}{2} b^3 \text {PolyLog}\left (4,1-\frac {2}{1-c \sqrt {x}}\right )+\frac {3}{2} b^3 \text {PolyLog}\left (4,-1+\frac {2}{1-c \sqrt {x}}\right ) \]
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Rubi [A]
time = 0.36, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {6035, 6033,
6199, 6095, 6205, 6209, 6745} \begin {gather*} 3 b^2 \text {Li}_3\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-3 b^2 \text {Li}_3\left (\frac {2}{1-c \sqrt {x}}-1\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-3 b \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+3 b \text {Li}_2\left (\frac {2}{1-c \sqrt {x}}-1\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-\frac {3}{2} b^3 \text {Li}_4\left (1-\frac {2}{1-c \sqrt {x}}\right )+\frac {3}{2} b^3 \text {Li}_4\left (\frac {2}{1-c \sqrt {x}}-1\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 6033
Rule 6035
Rule 6095
Rule 6199
Rule 6205
Rule 6209
Rule 6745
Rubi steps
\begin {align*} \int \frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3}{x} \, dx &=2 \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{x} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-(12 b c) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3+(6 b c) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )-(6 b c) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right )+3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c \sqrt {x}}\right )+\left (6 b^2 c\right ) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )-\left (6 b^2 c\right ) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right )+3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c \sqrt {x}}\right )+3 b^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {Li}_3\left (1-\frac {2}{1-c \sqrt {x}}\right )-3 b^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {Li}_3\left (-1+\frac {2}{1-c \sqrt {x}}\right )-\left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\text {Li}_3\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )+\left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\text {Li}_3\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right )+3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c \sqrt {x}}\right )+3 b^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {Li}_3\left (1-\frac {2}{1-c \sqrt {x}}\right )-3 b^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {Li}_3\left (-1+\frac {2}{1-c \sqrt {x}}\right )-\frac {3}{2} b^3 \text {Li}_4\left (1-\frac {2}{1-c \sqrt {x}}\right )+\frac {3}{2} b^3 \text {Li}_4\left (-1+\frac {2}{1-c \sqrt {x}}\right )\\ \end {align*}
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Mathematica [A]
time = 0.14, size = 248, normalized size = 1.11 \begin {gather*} 4 \tanh ^{-1}\left (1+\frac {2}{-1+c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3+\frac {3}{2} b \left (2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {PolyLog}\left (2,\frac {1+c \sqrt {x}}{1-c \sqrt {x}}\right )-2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {PolyLog}\left (2,\frac {1+c \sqrt {x}}{-1+c \sqrt {x}}\right )+b \left (-2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {PolyLog}\left (3,\frac {1+c \sqrt {x}}{1-c \sqrt {x}}\right )+2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {PolyLog}\left (3,\frac {1+c \sqrt {x}}{-1+c \sqrt {x}}\right )+b \left (\text {PolyLog}\left (4,\frac {1+c \sqrt {x}}{1-c \sqrt {x}}\right )-\text {PolyLog}\left (4,\frac {1+c \sqrt {x}}{-1+c \sqrt {x}}\right )\right )\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 5.51, size = 1542, normalized size = 6.88
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1542\) |
default | \(\text {Expression too large to display}\) | \(1542\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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 {\left (a + b \operatorname {atanh}{\left (c \sqrt {x} \right )}\right )^{3}}{x}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )\right )}^3}{x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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