Optimal. Leaf size=441 \[ \frac {a^2 x^3}{3}-\frac {480 a b \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {480 a b \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 a b x \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {15 b^2 \text {Li}_5\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {30 b^2 \sqrt {x} \text {Li}_4\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {30 b^2 x \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {20 b^2 x^{3/2} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}-\frac {2 b^2 x^{5/2}}{d} \]
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Rubi [A] time = 0.63, antiderivative size = 441, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5437, 4190, 4182, 2531, 6609, 2282, 6589, 4184, 3716, 2190} \[ -\frac {20 a b x^2 \text {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {80 a b x^{3/2} \text {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {240 a b x \text {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {480 a b \sqrt {x} \text {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \text {PolyLog}\left (6,-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {480 a b \text {PolyLog}\left (6,e^{c+d \sqrt {x}}\right )}{d^6}+\frac {20 b^2 x^{3/2} \text {PolyLog}\left (2,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {30 b^2 x \text {PolyLog}\left (3,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {30 b^2 \sqrt {x} \text {PolyLog}\left (4,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {15 b^2 \text {PolyLog}\left (5,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}-\frac {2 b^2 x^{5/2}}{d} \]
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^2 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \operatorname {Subst}\left (\int x^5 (a+b \text {csch}(c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (a^2 x^5+2 a b x^5 \text {csch}(c+d x)+b^2 x^5 \text {csch}^2(c+d x)\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^3}{3}+(4 a b) \operatorname {Subst}\left (\int x^5 \text {csch}(c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \operatorname {Subst}\left (\int x^5 \text {csch}^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}-\frac {(20 a b) \operatorname {Subst}\left (\int x^4 \log \left (1-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(20 a b) \operatorname {Subst}\left (\int x^4 \log \left (1+e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {\left (10 b^2\right ) \operatorname {Subst}\left (\int x^4 \coth (c+d x) \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {(80 a b) \operatorname {Subst}\left (\int x^3 \text {Li}_2\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {(80 a b) \operatorname {Subst}\left (\int x^3 \text {Li}_2\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {\left (20 b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 (c+d x)} x^4}{1-e^{2 (c+d x)}} \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {(240 a b) \operatorname {Subst}\left (\int x^2 \text {Li}_3\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {(240 a b) \operatorname {Subst}\left (\int x^2 \text {Li}_3\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {\left (40 b^2\right ) \operatorname {Subst}\left (\int x^3 \log \left (1-e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 b^2 x^{3/2} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {240 a b x \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(480 a b) \operatorname {Subst}\left (\int x \text {Li}_4\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(480 a b) \operatorname {Subst}\left (\int x \text {Li}_4\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {\left (60 b^2\right ) \operatorname {Subst}\left (\int x^2 \text {Li}_2\left (e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 b^2 x^{3/2} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {30 b^2 x \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 a b x \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(480 a b) \operatorname {Subst}\left (\int \text {Li}_5\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {(480 a b) \operatorname {Subst}\left (\int \text {Li}_5\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {\left (60 b^2\right ) \operatorname {Subst}\left (\int x \text {Li}_3\left (e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 b^2 x^{3/2} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {30 b^2 x \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 a b x \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {30 b^2 \sqrt {x} \text {Li}_4\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(480 a b) \operatorname {Subst}\left (\int \frac {\text {Li}_5(-x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^6}+\frac {(480 a b) \operatorname {Subst}\left (\int \frac {\text {Li}_5(x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^6}-\frac {\left (30 b^2\right ) \operatorname {Subst}\left (\int \text {Li}_4\left (e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 b^2 x^{3/2} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {30 b^2 x \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 a b x \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {30 b^2 \sqrt {x} \text {Li}_4\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {480 a b \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}-\frac {\left (15 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_4(x)}{x} \, dx,x,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^6}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 b^2 x^{3/2} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {30 b^2 x \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 a b x \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {30 b^2 \sqrt {x} \text {Li}_4\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {15 b^2 \text {Li}_5\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {480 a b \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {480 a b \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}\\ \end {align*}
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Mathematica [A] time = 12.07, size = 833, normalized size = 1.89 \[ \frac {a^2 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )^2 \sinh ^2\left (c+d \sqrt {x}\right ) x^3}{3 \left (b+a \sinh \left (c+d \sqrt {x}\right )\right )^2}+\frac {b^2 \text {csch}\left (\frac {c}{2}\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 \sinh ^2\left (c+d \sqrt {x}\right ) \sinh \left (\frac {d \sqrt {x}}{2}\right ) x^{5/2}}{d \left (b+a \sinh \left (c+d \sqrt {x}\right )\right )^2}-\frac {b^2 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )^2 \text {sech}\left (\frac {c}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right ) \sinh ^2\left (c+d \sqrt {x}\right ) \sinh \left (\frac {d \sqrt {x}}{2}\right ) x^{5/2}}{d \left (b+a \sinh \left (c+d \sqrt {x}\right )\right )^2}+\frac {2 b \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )^2 \left (-\frac {2 b x^{5/2} d^5}{-1+e^{2 c}}+2 a x^{5/2} \log \left (1-e^{-c-d \sqrt {x}}\right ) d^5-2 a x^{5/2} \log \left (1+e^{-c-d \sqrt {x}}\right ) d^5+5 b x^2 \log \left (1-e^{-c-d \sqrt {x}}\right ) d^4+5 b x^2 \log \left (1+e^{-c-d \sqrt {x}}\right ) d^4+40 a x^{3/2} \text {Li}_3\left (-e^{-c-d \sqrt {x}}\right ) d^3-40 a x^{3/2} \text {Li}_3\left (e^{-c-d \sqrt {x}}\right ) d^3-60 b x \text {Li}_3\left (-e^{-c-d \sqrt {x}}\right ) d^2-60 b x \text {Li}_3\left (e^{-c-d \sqrt {x}}\right ) d^2+120 a x \text {Li}_4\left (-e^{-c-d \sqrt {x}}\right ) d^2-120 a x \text {Li}_4\left (e^{-c-d \sqrt {x}}\right ) d^2-120 b \sqrt {x} \text {Li}_4\left (-e^{-c-d \sqrt {x}}\right ) d-120 b \sqrt {x} \text {Li}_4\left (e^{-c-d \sqrt {x}}\right ) d+240 a \sqrt {x} \text {Li}_5\left (-e^{-c-d \sqrt {x}}\right ) d-240 a \sqrt {x} \text {Li}_5\left (e^{-c-d \sqrt {x}}\right ) d+10 \left (a d^4 x^2-2 b d^3 x^{3/2}\right ) \text {Li}_2\left (-e^{-c-d \sqrt {x}}\right )-10 \left (a x^2 d^4+2 b x^{3/2} d^3\right ) \text {Li}_2\left (e^{-c-d \sqrt {x}}\right )-120 b \text {Li}_5\left (-e^{-c-d \sqrt {x}}\right )-120 b \text {Li}_5\left (e^{-c-d \sqrt {x}}\right )+240 a \text {Li}_6\left (-e^{-c-d \sqrt {x}}\right )-240 a \text {Li}_6\left (e^{-c-d \sqrt {x}}\right )\right ) \sinh ^2\left (c+d \sqrt {x}\right )}{d^6 \left (b+a \sinh \left (c+d \sqrt {x}\right )\right )^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x^{2} \operatorname {csch}\left (d \sqrt {x} + c\right )^{2} + 2 \, a b x^{2} \operatorname {csch}\left (d \sqrt {x} + c\right ) + a^{2} x^{2}, x\right ) \]
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 \[ \int {\left (b \operatorname {csch}\left (d \sqrt {x} + c\right ) + a\right )}^{2} x^{2}\,{d x} \]
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 \[ \int x^{2} \left (a +b \,\mathrm {csch}\left (c +d \sqrt {x}\right )\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.78, size = 496, normalized size = 1.12 \[ \frac {1}{3} \, a^{2} x^{3} - \frac {4 \, b^{2} x^{\frac {5}{2}}}{d e^{\left (2 \, d \sqrt {x} + 2 \, c\right )} - d} - \frac {4 \, {\left (d^{5} x^{\frac {5}{2}} \log \left (e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 5 \, d^{4} x^{2} {\rm Li}_2\left (-e^{\left (d \sqrt {x} + c\right )}\right ) - 20 \, d^{3} x^{\frac {3}{2}} {\rm Li}_{3}(-e^{\left (d \sqrt {x} + c\right )}) + 60 \, d^{2} x {\rm Li}_{4}(-e^{\left (d \sqrt {x} + c\right )}) - 120 \, d \sqrt {x} {\rm Li}_{5}(-e^{\left (d \sqrt {x} + c\right )}) + 120 \, {\rm Li}_{6}(-e^{\left (d \sqrt {x} + c\right )})\right )} a b}{d^{6}} + \frac {4 \, {\left (d^{5} x^{\frac {5}{2}} \log \left (-e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 5 \, d^{4} x^{2} {\rm Li}_2\left (e^{\left (d \sqrt {x} + c\right )}\right ) - 20 \, d^{3} x^{\frac {3}{2}} {\rm Li}_{3}(e^{\left (d \sqrt {x} + c\right )}) + 60 \, d^{2} x {\rm Li}_{4}(e^{\left (d \sqrt {x} + c\right )}) - 120 \, d \sqrt {x} {\rm Li}_{5}(e^{\left (d \sqrt {x} + c\right )}) + 120 \, {\rm Li}_{6}(e^{\left (d \sqrt {x} + c\right )})\right )} a b}{d^{6}} + \frac {10 \, {\left (d^{4} x^{2} \log \left (e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 4 \, d^{3} x^{\frac {3}{2}} {\rm Li}_2\left (-e^{\left (d \sqrt {x} + c\right )}\right ) - 12 \, d^{2} x {\rm Li}_{3}(-e^{\left (d \sqrt {x} + c\right )}) + 24 \, d \sqrt {x} {\rm Li}_{4}(-e^{\left (d \sqrt {x} + c\right )}) - 24 \, {\rm Li}_{5}(-e^{\left (d \sqrt {x} + c\right )})\right )} b^{2}}{d^{6}} + \frac {10 \, {\left (d^{4} x^{2} \log \left (-e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 4 \, d^{3} x^{\frac {3}{2}} {\rm Li}_2\left (e^{\left (d \sqrt {x} + c\right )}\right ) - 12 \, d^{2} x {\rm Li}_{3}(e^{\left (d \sqrt {x} + c\right )}) + 24 \, d \sqrt {x} {\rm Li}_{4}(e^{\left (d \sqrt {x} + c\right )}) - 24 \, {\rm Li}_{5}(e^{\left (d \sqrt {x} + c\right )})\right )} b^{2}}{d^{6}} - \frac {2 \, {\left (a b d^{6} x^{3} + 3 \, b^{2} d^{5} x^{\frac {5}{2}}\right )}}{3 \, d^{6}} + \frac {2 \, {\left (a b d^{6} x^{3} - 3 \, b^{2} d^{5} x^{\frac {5}{2}}\right )}}{3 \, d^{6}} \]
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 \[ \int x^2\,{\left (a+\frac {b}{\mathrm {sinh}\left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]
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 \[ \int x^{2} \left (a + b \operatorname {csch}{\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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