Integrand size = 18, antiderivative size = 260 \[ \int x^2 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^3}{3}-\frac {4 b x^{5/2} \text {arctanh}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 b x^2 \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 b x^2 \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 b x \operatorname {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 b x \operatorname {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 b \sqrt {x} \operatorname {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 b \sqrt {x} \operatorname {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 b \operatorname {PolyLog}\left (6,-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 b \operatorname {PolyLog}\left (6,e^{c+d \sqrt {x}}\right )}{d^6} \]
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Time = 0.20 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {14, 5545, 4267, 2611, 6744, 2320, 6724} \[ \int x^2 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^3}{3}-\frac {4 b x^{5/2} \text {arctanh}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {240 b \operatorname {PolyLog}\left (6,-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 b \operatorname {PolyLog}\left (6,e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 b \sqrt {x} \operatorname {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 b \sqrt {x} \operatorname {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )}{d^5}-\frac {120 b x \operatorname {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 b x \operatorname {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {10 b x^2 \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 b x^2 \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2} \]
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Rule 14
Rule 2320
Rule 2611
Rule 4267
Rule 5545
Rule 6724
Rule 6744
Rubi steps \begin{align*} \text {integral}& = \int \left (a x^2+b x^2 \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx \\ & = \frac {a x^3}{3}+b \int x^2 \text {csch}\left (c+d \sqrt {x}\right ) \, dx \\ & = \frac {a x^3}{3}+(2 b) \text {Subst}\left (\int x^5 \text {csch}(c+d x) \, dx,x,\sqrt {x}\right ) \\ & = \frac {a x^3}{3}-\frac {4 b x^{5/2} \text {arctanh}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {(10 b) \text {Subst}\left (\int x^4 \log \left (1-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(10 b) \text {Subst}\left (\int x^4 \log \left (1+e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d} \\ & = \frac {a x^3}{3}-\frac {4 b x^{5/2} \text {arctanh}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 b x^2 \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 b x^2 \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {(40 b) \text {Subst}\left (\int x^3 \operatorname {PolyLog}\left (2,-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {(40 b) \text {Subst}\left (\int x^3 \operatorname {PolyLog}\left (2,e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2} \\ & = \frac {a x^3}{3}-\frac {4 b x^{5/2} \text {arctanh}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 b x^2 \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 b x^2 \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {(120 b) \text {Subst}\left (\int x^2 \operatorname {PolyLog}\left (3,-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {(120 b) \text {Subst}\left (\int x^2 \operatorname {PolyLog}\left (3,e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3} \\ & = \frac {a x^3}{3}-\frac {4 b x^{5/2} \text {arctanh}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 b x^2 \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 b x^2 \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 b x \operatorname {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 b x \operatorname {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(240 b) \text {Subst}\left (\int x \operatorname {PolyLog}\left (4,-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(240 b) \text {Subst}\left (\int x \operatorname {PolyLog}\left (4,e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4} \\ & = \frac {a x^3}{3}-\frac {4 b x^{5/2} \text {arctanh}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 b x^2 \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 b x^2 \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 b x \operatorname {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 b x \operatorname {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 b \sqrt {x} \operatorname {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 b \sqrt {x} \operatorname {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(240 b) \text {Subst}\left (\int \operatorname {PolyLog}\left (5,-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {(240 b) \text {Subst}\left (\int \operatorname {PolyLog}\left (5,e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5} \\ & = \frac {a x^3}{3}-\frac {4 b x^{5/2} \text {arctanh}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 b x^2 \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 b x^2 \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 b x \operatorname {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 b x \operatorname {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 b \sqrt {x} \operatorname {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 b \sqrt {x} \operatorname {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(240 b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(5,-x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^6}+\frac {(240 b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(5,x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^6} \\ & = \frac {a x^3}{3}-\frac {4 b x^{5/2} \text {arctanh}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 b x^2 \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 b x^2 \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 b x^{3/2} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 b x \operatorname {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 b x \operatorname {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 b \sqrt {x} \operatorname {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 b \sqrt {x} \operatorname {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 b \operatorname {PolyLog}\left (6,-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 b \operatorname {PolyLog}\left (6,e^{c+d \sqrt {x}}\right )}{d^6} \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.05 \[ \int x^2 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^3}{3}+\frac {2 b \left (d^5 x^{5/2} \log \left (1-e^{c+d \sqrt {x}}\right )-d^5 x^{5/2} \log \left (1+e^{c+d \sqrt {x}}\right )-5 d^4 x^2 \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )+5 d^4 x^2 \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )+20 d^3 x^{3/2} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )-20 d^3 x^{3/2} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )-60 d^2 x \operatorname {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )+60 d^2 x \operatorname {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )+120 d \sqrt {x} \operatorname {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )-120 d \sqrt {x} \operatorname {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )-120 \operatorname {PolyLog}\left (6,-e^{c+d \sqrt {x}}\right )+120 \operatorname {PolyLog}\left (6,e^{c+d \sqrt {x}}\right )\right )}{d^6} \]
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\[\int x^{2} \left (a +b \,\operatorname {csch}\left (c +d \sqrt {x}\right )\right )d x\]
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\[ \int x^2 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \operatorname {csch}\left (d \sqrt {x} + c\right ) + a\right )} x^{2} \,d x } \]
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\[ \int x^2 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\int x^{2} \left (a + b \operatorname {csch}{\left (c + d \sqrt {x} \right )}\right )\, dx \]
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Time = 0.38 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {1}{3} \, a x^{3} - \frac {2 \, {\left (\log \left (e^{\left (d \sqrt {x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{5} + 5 \, {\rm Li}_2\left (-e^{\left (d \sqrt {x} + c\right )}\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{4} - 20 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{3} {\rm Li}_{3}(-e^{\left (d \sqrt {x} + c\right )}) + 60 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{2} {\rm Li}_{4}(-e^{\left (d \sqrt {x} + c\right )}) - 120 \, \log \left (e^{\left (d \sqrt {x}\right )}\right ) {\rm Li}_{5}(-e^{\left (d \sqrt {x} + c\right )}) + 120 \, {\rm Li}_{6}(-e^{\left (d \sqrt {x} + c\right )})\right )} b}{d^{6}} + \frac {2 \, {\left (\log \left (-e^{\left (d \sqrt {x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{5} + 5 \, {\rm Li}_2\left (e^{\left (d \sqrt {x} + c\right )}\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{4} - 20 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{3} {\rm Li}_{3}(e^{\left (d \sqrt {x} + c\right )}) + 60 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{2} {\rm Li}_{4}(e^{\left (d \sqrt {x} + c\right )}) - 120 \, \log \left (e^{\left (d \sqrt {x}\right )}\right ) {\rm Li}_{5}(e^{\left (d \sqrt {x} + c\right )}) + 120 \, {\rm Li}_{6}(e^{\left (d \sqrt {x} + c\right )})\right )} b}{d^{6}} \]
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\[ \int x^2 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \operatorname {csch}\left (d \sqrt {x} + c\right ) + a\right )} x^{2} \,d x } \]
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Timed out. \[ \int x^2 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\int x^2\,\left (a+\frac {b}{\mathrm {sinh}\left (c+d\,\sqrt {x}\right )}\right ) \,d x \]
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