Integrand size = 18, antiderivative size = 310 \[ \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^3}{3}+\frac {4 b x^{5/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 i b x \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \operatorname {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 i b \operatorname {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )}{d^6} \]
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Time = 0.21 (sec) , antiderivative size = 310, 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, 5544, 4265, 2611, 6744, 2320, 6724} \[ \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^3}{3}+\frac {4 b x^{5/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {240 i b \operatorname {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 i b \operatorname {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {120 i b x \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {10 i b x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2} \]
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Rule 14
Rule 2320
Rule 2611
Rule 4265
Rule 5544
Rule 6724
Rule 6744
Rubi steps \begin{align*} \text {integral}& = \int \left (a x^2+b x^2 \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx \\ & = \frac {a x^3}{3}+b \int x^2 \text {sech}\left (c+d \sqrt {x}\right ) \, dx \\ & = \frac {a x^3}{3}+(2 b) \text {Subst}\left (\int x^5 \text {sech}(c+d x) \, dx,x,\sqrt {x}\right ) \\ & = \frac {a x^3}{3}+\frac {4 b x^{5/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {(10 i b) \text {Subst}\left (\int x^4 \log \left (1-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(10 i b) \text {Subst}\left (\int x^4 \log \left (1+i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d} \\ & = \frac {a x^3}{3}+\frac {4 b x^{5/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {(40 i b) \text {Subst}\left (\int x^3 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {(40 i b) \text {Subst}\left (\int x^3 \operatorname {PolyLog}\left (2,i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2} \\ & = \frac {a x^3}{3}+\frac {4 b x^{5/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {(120 i b) \text {Subst}\left (\int x^2 \operatorname {PolyLog}\left (3,-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {(120 i b) \text {Subst}\left (\int x^2 \operatorname {PolyLog}\left (3,i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3} \\ & = \frac {a x^3}{3}+\frac {4 b x^{5/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 i b x \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(240 i b) \text {Subst}\left (\int x \operatorname {PolyLog}\left (4,-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(240 i b) \text {Subst}\left (\int x \operatorname {PolyLog}\left (4,i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4} \\ & = \frac {a x^3}{3}+\frac {4 b x^{5/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 i b x \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(240 i b) \text {Subst}\left (\int \operatorname {PolyLog}\left (5,-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {(240 i b) \text {Subst}\left (\int \operatorname {PolyLog}\left (5,i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5} \\ & = \frac {a x^3}{3}+\frac {4 b x^{5/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 i b x \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(240 i b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(5,-i x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^6}+\frac {(240 i b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(5,i x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^6} \\ & = \frac {a x^3}{3}+\frac {4 b x^{5/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 i b x \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \operatorname {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 i b \operatorname {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )}{d^6} \\ \end{align*}
Time = 0.52 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^3}{3}+\frac {2 i b \left (d^5 x^{5/2} \log \left (1-i e^{c+d \sqrt {x}}\right )-d^5 x^{5/2} \log \left (1+i e^{c+d \sqrt {x}}\right )-5 d^4 x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )+5 d^4 x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )+20 d^3 x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )-20 d^3 x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )-60 d^2 x \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )+60 d^2 x \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )+120 d \sqrt {x} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )-120 d \sqrt {x} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )-120 \operatorname {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )+120 \operatorname {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )\right )}{d^6} \]
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\[\int x^{2} \left (a +b \,\operatorname {sech}\left (c +d \sqrt {x}\right )\right )d x\]
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\[ \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )} x^{2} \,d x } \]
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\[ \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\int x^{2} \left (a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}\right )\, dx \]
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\[ \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )} x^{2} \,d x } \]
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\[ \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )} x^{2} \,d x } \]
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Timed out. \[ \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\int x^2\,\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,\sqrt {x}\right )}\right ) \,d x \]
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