Optimal. Leaf size=254 \[ \frac {2}{5} a x^{5/2}+\frac {48 i b \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {48 i b \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {48 i b \sqrt {x} \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {24 i b x \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 i b x \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {8 i b x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {8 i b x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {4 b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d} \]
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Rubi [A] time = 0.22, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {14, 5436, 4180, 2531, 6609, 2282, 6589} \[ -\frac {8 i b x^{3/2} \text {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {8 i b x^{3/2} \text {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {24 i b x \text {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 i b x \text {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 i b \sqrt {x} \text {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {48 i b \text {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {48 i b \text {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}+\frac {2}{5} a x^{5/2}+\frac {4 b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 14
Rule 2282
Rule 2531
Rule 4180
Rule 5436
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx &=\int \left (a x^{3/2}+b x^{3/2} \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx\\ &=\frac {2}{5} a x^{5/2}+b \int x^{3/2} \text {sech}\left (c+d \sqrt {x}\right ) \, dx\\ &=\frac {2}{5} a x^{5/2}+(2 b) \operatorname {Subst}\left (\int x^4 \text {sech}(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{5} a x^{5/2}+\frac {4 b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {(8 i b) \operatorname {Subst}\left (\int x^3 \log \left (1-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(8 i b) \operatorname {Subst}\left (\int x^3 \log \left (1+i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {2}{5} a x^{5/2}+\frac {4 b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {8 i b x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {8 i b x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {(24 i b) \operatorname {Subst}\left (\int x^2 \text {Li}_2\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {(24 i b) \operatorname {Subst}\left (\int x^2 \text {Li}_2\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=\frac {2}{5} a x^{5/2}+\frac {4 b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {8 i b x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {8 i b x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {24 i b x \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 i b x \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {(48 i b) \operatorname {Subst}\left (\int x \text {Li}_3\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {(48 i b) \operatorname {Subst}\left (\int x \text {Li}_3\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {2}{5} a x^{5/2}+\frac {4 b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {8 i b x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {8 i b x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {24 i b x \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 i b x \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 i b \sqrt {x} \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(48 i b) \operatorname {Subst}\left (\int \text {Li}_4\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(48 i b) \operatorname {Subst}\left (\int \text {Li}_4\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=\frac {2}{5} a x^{5/2}+\frac {4 b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {8 i b x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {8 i b x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {24 i b x \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 i b x \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 i b \sqrt {x} \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(48 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_4(-i x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(48 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_4(i x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^5}\\ &=\frac {2}{5} a x^{5/2}+\frac {4 b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {8 i b x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {8 i b x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {24 i b x \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 i b x \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 i b \sqrt {x} \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {48 i b \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {48 i b \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}\\ \end {align*}
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Mathematica [A] time = 2.05, size = 288, normalized size = 1.13 \[ \frac {2 \left (a d^5 x^{5/2}+5 i b d^4 x^2 \log \left (1-i e^{c+d \sqrt {x}}\right )-5 i b d^4 x^2 \log \left (1+i e^{c+d \sqrt {x}}\right )-20 i b d^3 x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )+20 i b d^3 x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )+60 i b d^2 x \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )-60 i b d^2 x \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )-120 i b d \sqrt {x} \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )+120 i b d \sqrt {x} \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )+120 i b \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )-120 i b \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )\right )}{5 d^5} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x^{\frac {3}{2}} \operatorname {sech}\left (d \sqrt {x} + c\right ) + a x^{\frac {3}{2}}, x\right ) \]
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 \[ \int {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )} x^{\frac {3}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.54, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \left (a +b \,\mathrm {sech}\left (c +d \sqrt {x}\right )\right )\, dx \]
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 \[ \frac {2}{5} \, a x^{\frac {5}{2}} + 2 \, b \int \frac {x^{\frac {3}{2}} e^{\left (d \sqrt {x} + c\right )}}{e^{\left (2 \, d \sqrt {x} + 2 \, c\right )} + 1}\,{d x} \]
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 \[ \int x^{3/2}\,\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,\sqrt {x}\right )}\right ) \,d x \]
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 \[ \int x^{\frac {3}{2}} \left (a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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