Optimal. Leaf size=361 \[ \frac {4 b \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {4 b \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {4 b \sqrt {x} \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {4 b \sqrt {x} \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {2 b x \log \left (\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}+1\right )}{a d \sqrt {b^2-a^2}}+\frac {2 b x \log \left (\frac {a e^{c+d \sqrt {x}}}{\sqrt {b^2-a^2}+b}+1\right )}{a d \sqrt {b^2-a^2}}+\frac {2 x^{3/2}}{3 a} \]
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Rubi [A] time = 0.78, antiderivative size = 361, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5436, 4191, 3320, 2264, 2190, 2531, 2282, 6589} \[ -\frac {4 b \sqrt {x} \text {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {4 b \sqrt {x} \text {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{\sqrt {b^2-a^2}+b}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {4 b \text {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {4 b \text {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{\sqrt {b^2-a^2}+b}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {2 b x \log \left (\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}+1\right )}{a d \sqrt {b^2-a^2}}+\frac {2 b x \log \left (\frac {a e^{c+d \sqrt {x}}}{\sqrt {b^2-a^2}+b}+1\right )}{a d \sqrt {b^2-a^2}}+\frac {2 x^{3/2}}{3 a} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2282
Rule 2531
Rule 3320
Rule 4191
Rule 5436
Rule 6589
Rubi steps
\begin {align*} \int \frac {\sqrt {x}}{a+b \text {sech}\left (c+d \sqrt {x}\right )} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^2}{a+b \text {sech}(c+d x)} \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (\frac {x^2}{a}-\frac {b x^2}{a (b+a \cosh (c+d x))}\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {2 x^{3/2}}{3 a}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {x^2}{b+a \cosh (c+d x)} \, dx,x,\sqrt {x}\right )}{a}\\ &=\frac {2 x^{3/2}}{3 a}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {e^{c+d x} x^2}{a+2 b e^{c+d x}+a e^{2 (c+d x)}} \, dx,x,\sqrt {x}\right )}{a}\\ &=\frac {2 x^{3/2}}{3 a}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {e^{c+d x} x^2}{2 b-2 \sqrt {-a^2+b^2}+2 a e^{c+d x}} \, dx,x,\sqrt {x}\right )}{\sqrt {-a^2+b^2}}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {e^{c+d x} x^2}{2 b+2 \sqrt {-a^2+b^2}+2 a e^{c+d x}} \, dx,x,\sqrt {x}\right )}{\sqrt {-a^2+b^2}}\\ &=\frac {2 x^{3/2}}{3 a}-\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {(4 b) \operatorname {Subst}\left (\int x \log \left (1+\frac {2 a e^{c+d x}}{2 b-2 \sqrt {-a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {-a^2+b^2} d}-\frac {(4 b) \operatorname {Subst}\left (\int x \log \left (1+\frac {2 a e^{c+d x}}{2 b+2 \sqrt {-a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {-a^2+b^2} d}\\ &=\frac {2 x^{3/2}}{3 a}-\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {4 b \sqrt {x} \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {4 b \sqrt {x} \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {(4 b) \operatorname {Subst}\left (\int \text {Li}_2\left (-\frac {2 a e^{c+d x}}{2 b-2 \sqrt {-a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {(4 b) \operatorname {Subst}\left (\int \text {Li}_2\left (-\frac {2 a e^{c+d x}}{2 b+2 \sqrt {-a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {-a^2+b^2} d^2}\\ &=\frac {2 x^{3/2}}{3 a}-\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {4 b \sqrt {x} \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {4 b \sqrt {x} \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {a x}{-b+\sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {a x}{b+\sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{a \sqrt {-a^2+b^2} d^3}\\ &=\frac {2 x^{3/2}}{3 a}-\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {4 b \sqrt {x} \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {4 b \sqrt {x} \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {4 b \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {4 b \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}\\ \end {align*}
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Mathematica [A] time = 8.03, size = 390, normalized size = 1.08 \[ \frac {2 \left (d^3 x^{3/2} \sqrt {e^{2 c} \left (b^2-a^2\right )}-3 b e^c d^2 x \log \left (\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {e^{2 c} \left (b^2-a^2\right )}}+1\right )+3 b e^c d^2 x \log \left (\frac {a e^{2 c+d \sqrt {x}}}{\sqrt {e^{2 c} \left (b^2-a^2\right )}+b e^c}+1\right )-6 b e^c d \sqrt {x} \text {Li}_2\left (-\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (b^2-a^2\right ) e^{2 c}}}\right )+6 b e^c d \sqrt {x} \text {Li}_2\left (-\frac {a e^{2 c+d \sqrt {x}}}{e^c b+\sqrt {\left (b^2-a^2\right ) e^{2 c}}}\right )+6 b e^c \text {Li}_3\left (-\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (b^2-a^2\right ) e^{2 c}}}\right )-6 b e^c \text {Li}_3\left (-\frac {a e^{2 c+d \sqrt {x}}}{e^c b+\sqrt {\left (b^2-a^2\right ) e^{2 c}}}\right )\right )}{3 a d^3 \sqrt {e^{2 c} \left (b^2-a^2\right )}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x}}{b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a}, x\right ) \]
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 \[ \int \frac {\sqrt {x}}{b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.59, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x}}{a +b \,\mathrm {sech}\left (c +d \sqrt {x}\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
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 \[ \int \frac {\sqrt {x}}{a+\frac {b}{\mathrm {cosh}\left (c+d\,\sqrt {x}\right )}} \,d x \]
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 \[ \int \frac {\sqrt {x}}{a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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