Optimal. Leaf size=287 \[ \frac {a^2 x^2}{2}-\frac {24 a b \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {24 a b \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {24 a b \sqrt {x} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 a b \sqrt {x} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 a b x \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 a b x \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}-\frac {8 a b x^{3/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {3 b^2 \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {6 b^2 \sqrt {x} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 b^2 x \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b^2 x^{3/2} \coth \left (c+d \sqrt {x}\right )}{d}-\frac {2 b^2 x^{3/2}}{d} \]
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Rubi [A] time = 0.45, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {5437, 4190, 4182, 2531, 6609, 2282, 6589, 4184, 3716, 2190} \[ -\frac {12 a b x \text {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 a b x \text {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {24 a b \sqrt {x} \text {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 a b \sqrt {x} \text {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 a b \text {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {24 a b \text {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {6 b^2 \sqrt {x} \text {PolyLog}\left (2,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {3 b^2 \text {PolyLog}\left (3,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {a^2 x^2}{2}-\frac {8 a b x^{3/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}+\frac {6 b^2 x \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b^2 x^{3/2} \coth \left (c+d \sqrt {x}\right )}{d}-\frac {2 b^2 x^{3/2}}{d} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3716
Rule 4182
Rule 4184
Rule 4190
Rule 5437
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \operatorname {Subst}\left (\int x^3 (a+b \text {csch}(c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (a^2 x^3+2 a b x^3 \text {csch}(c+d x)+b^2 x^3 \text {csch}^2(c+d x)\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^2}{2}+(4 a b) \operatorname {Subst}\left (\int x^3 \text {csch}(c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \operatorname {Subst}\left (\int x^3 \text {csch}^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^2}{2}-\frac {8 a b x^{3/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{3/2} \coth \left (c+d \sqrt {x}\right )}{d}-\frac {(12 a b) \operatorname {Subst}\left (\int x^2 \log \left (1-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(12 a b) \operatorname {Subst}\left (\int x^2 \log \left (1+e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int x^2 \coth (c+d x) \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}-\frac {8 a b x^{3/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{3/2} \coth \left (c+d \sqrt {x}\right )}{d}-\frac {12 a b x \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 a b x \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {(24 a b) \operatorname {Subst}\left (\int x \text {Li}_2\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {(24 a b) \operatorname {Subst}\left (\int x \text {Li}_2\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {\left (12 b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 (c+d x)} x^2}{1-e^{2 (c+d x)}} \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}-\frac {8 a b x^{3/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{3/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {6 b^2 x \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 a b x \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 a b x \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {24 a b \sqrt {x} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 a b \sqrt {x} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {(24 a b) \operatorname {Subst}\left (\int \text {Li}_3\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {(24 a b) \operatorname {Subst}\left (\int \text {Li}_3\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {\left (12 b^2\right ) \operatorname {Subst}\left (\int x \log \left (1-e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=-\frac {2 b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}-\frac {8 a b x^{3/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{3/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {6 b^2 x \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 a b x \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 a b x \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {6 b^2 \sqrt {x} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 a b \sqrt {x} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 a b \sqrt {x} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {(24 a b) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(24 a b) \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^4}-\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=-\frac {2 b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}-\frac {8 a b x^{3/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{3/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {6 b^2 x \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 a b x \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 a b x \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {6 b^2 \sqrt {x} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 a b \sqrt {x} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 a b \sqrt {x} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 a b \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {24 a b \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}\\ &=-\frac {2 b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}-\frac {8 a b x^{3/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{3/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {6 b^2 x \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 a b x \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 a b x \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {6 b^2 \sqrt {x} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 a b \sqrt {x} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 a b \sqrt {x} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {3 b^2 \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {24 a b \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {24 a b \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}\\ \end {align*}
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Mathematica [B] time = 11.51, size = 616, normalized size = 2.15 \[ \frac {a^2 x^2 \sinh ^2\left (c+d \sqrt {x}\right ) \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )^2}{2 \left (a \sinh \left (c+d \sqrt {x}\right )+b\right )^2}+\frac {b^2 x^{3/2} \text {csch}\left (\frac {c}{2}\right ) \sinh \left (\frac {d \sqrt {x}}{2}\right ) \sinh ^2\left (c+d \sqrt {x}\right ) \text {csch}\left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right ) \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )^2}{d \left (a \sinh \left (c+d \sqrt {x}\right )+b\right )^2}-\frac {b^2 x^{3/2} \text {sech}\left (\frac {c}{2}\right ) \sinh \left (\frac {d \sqrt {x}}{2}\right ) \sinh ^2\left (c+d \sqrt {x}\right ) \text {sech}\left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right ) \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )^2}{d \left (a \sinh \left (c+d \sqrt {x}\right )+b\right )^2}+\frac {2 b \sinh ^2\left (c+d \sqrt {x}\right ) \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )^2 \left (6 \left (a d^2 x-b d \sqrt {x}\right ) \text {Li}_2\left (-e^{-c-d \sqrt {x}}\right )-6 \left (a d^2 x+b d \sqrt {x}\right ) \text {Li}_2\left (e^{-c-d \sqrt {x}}\right )+2 a d^3 x^{3/2} \log \left (1-e^{-c-d \sqrt {x}}\right )-2 a d^3 x^{3/2} \log \left (e^{-c-d \sqrt {x}}+1\right )+12 a d \sqrt {x} \text {Li}_3\left (-e^{-c-d \sqrt {x}}\right )-12 a d \sqrt {x} \text {Li}_3\left (e^{-c-d \sqrt {x}}\right )+12 a \text {Li}_4\left (-e^{-c-d \sqrt {x}}\right )-12 a \text {Li}_4\left (e^{-c-d \sqrt {x}}\right )-\frac {2 b d^3 x^{3/2}}{e^{2 c}-1}+3 b d^2 x \log \left (1-e^{-c-d \sqrt {x}}\right )+3 b d^2 x \log \left (e^{-c-d \sqrt {x}}+1\right )-6 b \text {Li}_3\left (-e^{-c-d \sqrt {x}}\right )-6 b \text {Li}_3\left (e^{-c-d \sqrt {x}}\right )\right )}{d^4 \left (a \sinh \left (c+d \sqrt {x}\right )+b\right )^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x \operatorname {csch}\left (d \sqrt {x} + c\right )^{2} + 2 \, a b x \operatorname {csch}\left (d \sqrt {x} + c\right ) + a^{2} x, x\right ) \]
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 \[ \int {\left (b \operatorname {csch}\left (d \sqrt {x} + c\right ) + a\right )}^{2} x\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.74, size = 0, normalized size = 0.00 \[ \int x \left (a +b \,\mathrm {csch}\left (c +d \sqrt {x}\right )\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.80, size = 343, normalized size = 1.20 \[ \frac {1}{2} \, a^{2} x^{2} - \frac {4 \, b^{2} x^{\frac {3}{2}}}{d e^{\left (2 \, d \sqrt {x} + 2 \, c\right )} - d} - \frac {4 \, {\left (d^{3} x^{\frac {3}{2}} \log \left (e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 3 \, d^{2} x {\rm Li}_2\left (-e^{\left (d \sqrt {x} + c\right )}\right ) - 6 \, d \sqrt {x} {\rm Li}_{3}(-e^{\left (d \sqrt {x} + c\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (d \sqrt {x} + c\right )})\right )} a b}{d^{4}} + \frac {4 \, {\left (d^{3} x^{\frac {3}{2}} \log \left (-e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 3 \, d^{2} x {\rm Li}_2\left (e^{\left (d \sqrt {x} + c\right )}\right ) - 6 \, d \sqrt {x} {\rm Li}_{3}(e^{\left (d \sqrt {x} + c\right )}) + 6 \, {\rm Li}_{4}(e^{\left (d \sqrt {x} + c\right )})\right )} a b}{d^{4}} + \frac {6 \, {\left (d^{2} x \log \left (e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 2 \, d \sqrt {x} {\rm Li}_2\left (-e^{\left (d \sqrt {x} + c\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (d \sqrt {x} + c\right )})\right )} b^{2}}{d^{4}} + \frac {6 \, {\left (d^{2} x \log \left (-e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 2 \, d \sqrt {x} {\rm Li}_2\left (e^{\left (d \sqrt {x} + c\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (d \sqrt {x} + c\right )})\right )} b^{2}}{d^{4}} - \frac {a b d^{4} x^{2} + 2 \, b^{2} d^{3} x^{\frac {3}{2}}}{d^{4}} + \frac {a b d^{4} x^{2} - 2 \, b^{2} d^{3} x^{\frac {3}{2}}}{d^{4}} \]
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 \[ \int x\,{\left (a+\frac {b}{\mathrm {sinh}\left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]
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 \[ \int x \left (a + b \operatorname {csch}{\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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