3.1.68 \(\int \frac {\sqrt {x}}{(a+b \text {sech}(c+d \sqrt {x}))^2} \, dx\) [68]

3.1.68.1 Optimal result

Integrand size = 22, antiderivative size = 1027 \[ \int \frac {\sqrt {x}}{\left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=\frac {2 b^2 x}{a^2 \left (a^2-b^2\right ) d}+\frac {2 x^{3/2}}{3 a^2}-\frac {4 b^2 \sqrt {x} \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \left (a^2-b^2\right ) d^2}+\frac {2 b^3 x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d}-\frac {4 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d}-\frac {4 b^2 \sqrt {x} \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \left (a^2-b^2\right ) d^2}-\frac {2 b^3 x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d}+\frac {4 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d}-\frac {4 b^2 \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \left (a^2-b^2\right ) d^3}+\frac {4 b^3 \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d^2}-\frac {8 b \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d^2}-\frac {4 b^2 \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \left (a^2-b^2\right ) d^3}-\frac {4 b^3 \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d^2}+\frac {8 b \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d^2}-\frac {4 b^3 \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d^3}+\frac {8 b \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d^3}+\frac {4 b^3 \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d^3}-\frac {8 b \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d^3}+\frac {2 b^2 x \sinh \left (c+d \sqrt {x}\right )}{a \left (a^2-b^2\right ) d \left (b+a \cosh \left (c+d \sqrt {x}\right )\right )} \]

output

2*b^2*x/a^2/(a^2-b^2)/d+2/3*x^(3/2)/a^2+2*b^3*x*ln(1+a*exp(c+d*x^(1/2))/(b 
-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d-2*b^3*x*ln(1+a*exp(c+d*x^(1/2)) 
/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d-4*b^2*polylog(2,-a*exp(c+d*x 
^(1/2))/(b-(-a^2+b^2)^(1/2)))/a^2/(a^2-b^2)/d^3-4*b^2*polylog(2,-a*exp(c+d 
*x^(1/2))/(b+(-a^2+b^2)^(1/2)))/a^2/(a^2-b^2)/d^3-4*b^3*polylog(3,-a*exp(c 
+d*x^(1/2))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^3+4*b^3*polylog(3 
,-a*exp(c+d*x^(1/2))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^3+2*b^2* 
x*sinh(c+d*x^(1/2))/a/(a^2-b^2)/d/(b+a*cosh(c+d*x^(1/2)))-4*b*x*ln(1+a*exp 
(c+d*x^(1/2))/(b-(-a^2+b^2)^(1/2)))/a^2/d/(-a^2+b^2)^(1/2)+4*b*x*ln(1+a*ex 
p(c+d*x^(1/2))/(b+(-a^2+b^2)^(1/2)))/a^2/d/(-a^2+b^2)^(1/2)+8*b*polylog(3, 
-a*exp(c+d*x^(1/2))/(b-(-a^2+b^2)^(1/2)))/a^2/d^3/(-a^2+b^2)^(1/2)-8*b*pol 
ylog(3,-a*exp(c+d*x^(1/2))/(b+(-a^2+b^2)^(1/2)))/a^2/d^3/(-a^2+b^2)^(1/2)- 
4*b^2*ln(1+a*exp(c+d*x^(1/2))/(b-(-a^2+b^2)^(1/2)))*x^(1/2)/a^2/(a^2-b^2)/ 
d^2-4*b^2*ln(1+a*exp(c+d*x^(1/2))/(b+(-a^2+b^2)^(1/2)))*x^(1/2)/a^2/(a^2-b 
^2)/d^2+4*b^3*polylog(2,-a*exp(c+d*x^(1/2))/(b-(-a^2+b^2)^(1/2)))*x^(1/2)/ 
a^2/(-a^2+b^2)^(3/2)/d^2-4*b^3*polylog(2,-a*exp(c+d*x^(1/2))/(b+(-a^2+b^2) 
^(1/2)))*x^(1/2)/a^2/(-a^2+b^2)^(3/2)/d^2-8*b*polylog(2,-a*exp(c+d*x^(1/2) 
)/(b-(-a^2+b^2)^(1/2)))*x^(1/2)/a^2/d^2/(-a^2+b^2)^(1/2)+8*b*polylog(2,-a* 
exp(c+d*x^(1/2))/(b+(-a^2+b^2)^(1/2)))*x^(1/2)/a^2/d^2/(-a^2+b^2)^(1/2)
 

3.1.68.2 Mathematica [A] (verified)

Time = 5.15 (sec) , antiderivative size = 986, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {x}}{\left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=\frac {2 \left (b+a \cosh \left (c+d \sqrt {x}\right )\right ) \text {sech}^2\left (c+d \sqrt {x}\right ) \left (x^{3/2} \left (b+a \cosh \left (c+d \sqrt {x}\right )\right )+\frac {3 b e^c \left (b+a \cosh \left (c+d \sqrt {x}\right )\right ) \left (2 b e^c x-\frac {e^{-c} \left (1+e^{2 c}\right ) \left (2 b d \sqrt {\left (-a^2+b^2\right ) e^{2 c}} \sqrt {x} \log \left (1+\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (-a^2+b^2\right ) e^{2 c}}}\right )+2 a^2 d^2 e^c x \log \left (1+\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (-a^2+b^2\right ) e^{2 c}}}\right )-b^2 d^2 e^c x \log \left (1+\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (-a^2+b^2\right ) e^{2 c}}}\right )+2 b d \sqrt {\left (-a^2+b^2\right ) e^{2 c}} \sqrt {x} \log \left (1+\frac {a e^{2 c+d \sqrt {x}}}{b e^c+\sqrt {\left (-a^2+b^2\right ) e^{2 c}}}\right )-2 a^2 d^2 e^c x \log \left (1+\frac {a e^{2 c+d \sqrt {x}}}{b e^c+\sqrt {\left (-a^2+b^2\right ) e^{2 c}}}\right )+b^2 d^2 e^c x \log \left (1+\frac {a e^{2 c+d \sqrt {x}}}{b e^c+\sqrt {\left (-a^2+b^2\right ) e^{2 c}}}\right )+2 \left (b \sqrt {\left (-a^2+b^2\right ) e^{2 c}}+2 a^2 d e^c \sqrt {x}-b^2 d e^c \sqrt {x}\right ) \operatorname {PolyLog}\left (2,-\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (-a^2+b^2\right ) e^{2 c}}}\right )+2 \left (b \sqrt {\left (-a^2+b^2\right ) e^{2 c}}-2 a^2 d e^c \sqrt {x}+b^2 d e^c \sqrt {x}\right ) \operatorname {PolyLog}\left (2,-\frac {a e^{2 c+d \sqrt {x}}}{b e^c+\sqrt {\left (-a^2+b^2\right ) e^{2 c}}}\right )-4 a^2 e^c \operatorname {PolyLog}\left (3,-\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (-a^2+b^2\right ) e^{2 c}}}\right )+2 b^2 e^c \operatorname {PolyLog}\left (3,-\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (-a^2+b^2\right ) e^{2 c}}}\right )+4 a^2 e^c \operatorname {PolyLog}\left (3,-\frac {a e^{2 c+d \sqrt {x}}}{b e^c+\sqrt {\left (-a^2+b^2\right ) e^{2 c}}}\right )-2 b^2 e^c \operatorname {PolyLog}\left (3,-\frac {a e^{2 c+d \sqrt {x}}}{b e^c+\sqrt {\left (-a^2+b^2\right ) e^{2 c}}}\right )\right )}{d^2 \sqrt {\left (-a^2+b^2\right ) e^{2 c}}}\right )}{\left (a^2-b^2\right ) d \left (1+e^{2 c}\right )}+\frac {3 b^2 x \text {sech}(c) \left (b \sinh (c)-a \sinh \left (d \sqrt {x}\right )\right )}{\left (-a^2+b^2\right ) d}\right )}{3 a^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \]

input

Integrate[Sqrt[x]/(a + b*Sech[c + d*Sqrt[x]])^2,x]
 

output

(2*(b + a*Cosh[c + d*Sqrt[x]])*Sech[c + d*Sqrt[x]]^2*(x^(3/2)*(b + a*Cosh[ 
c + d*Sqrt[x]]) + (3*b*E^c*(b + a*Cosh[c + d*Sqrt[x]])*(2*b*E^c*x - ((1 + 
E^(2*c))*(2*b*d*Sqrt[(-a^2 + b^2)*E^(2*c)]*Sqrt[x]*Log[1 + (a*E^(2*c + d*S 
qrt[x]))/(b*E^c - Sqrt[(-a^2 + b^2)*E^(2*c)])] + 2*a^2*d^2*E^c*x*Log[1 + ( 
a*E^(2*c + d*Sqrt[x]))/(b*E^c - Sqrt[(-a^2 + b^2)*E^(2*c)])] - b^2*d^2*E^c 
*x*Log[1 + (a*E^(2*c + d*Sqrt[x]))/(b*E^c - Sqrt[(-a^2 + b^2)*E^(2*c)])] + 
 2*b*d*Sqrt[(-a^2 + b^2)*E^(2*c)]*Sqrt[x]*Log[1 + (a*E^(2*c + d*Sqrt[x]))/ 
(b*E^c + Sqrt[(-a^2 + b^2)*E^(2*c)])] - 2*a^2*d^2*E^c*x*Log[1 + (a*E^(2*c 
+ d*Sqrt[x]))/(b*E^c + Sqrt[(-a^2 + b^2)*E^(2*c)])] + b^2*d^2*E^c*x*Log[1 
+ (a*E^(2*c + d*Sqrt[x]))/(b*E^c + Sqrt[(-a^2 + b^2)*E^(2*c)])] + 2*(b*Sqr 
t[(-a^2 + b^2)*E^(2*c)] + 2*a^2*d*E^c*Sqrt[x] - b^2*d*E^c*Sqrt[x])*PolyLog 
[2, -((a*E^(2*c + d*Sqrt[x]))/(b*E^c - Sqrt[(-a^2 + b^2)*E^(2*c)]))] + 2*( 
b*Sqrt[(-a^2 + b^2)*E^(2*c)] - 2*a^2*d*E^c*Sqrt[x] + b^2*d*E^c*Sqrt[x])*Po 
lyLog[2, -((a*E^(2*c + d*Sqrt[x]))/(b*E^c + Sqrt[(-a^2 + b^2)*E^(2*c)]))] 
- 4*a^2*E^c*PolyLog[3, -((a*E^(2*c + d*Sqrt[x]))/(b*E^c - Sqrt[(-a^2 + b^2 
)*E^(2*c)]))] + 2*b^2*E^c*PolyLog[3, -((a*E^(2*c + d*Sqrt[x]))/(b*E^c - Sq 
rt[(-a^2 + b^2)*E^(2*c)]))] + 4*a^2*E^c*PolyLog[3, -((a*E^(2*c + d*Sqrt[x] 
))/(b*E^c + Sqrt[(-a^2 + b^2)*E^(2*c)]))] - 2*b^2*E^c*PolyLog[3, -((a*E^(2 
*c + d*Sqrt[x]))/(b*E^c + Sqrt[(-a^2 + b^2)*E^(2*c)]))]))/(d^2*E^c*Sqrt[(- 
a^2 + b^2)*E^(2*c)])))/((a^2 - b^2)*d*(1 + E^(2*c))) + (3*b^2*x*Sech[c]...
 

3.1.68.3 Rubi [A] (verified)

Time = 2.15 (sec) , antiderivative size = 1026, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5959, 3042, 4679, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sqrt {x}}{\left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx\)

\(\Big \downarrow \) 5959

\(\displaystyle 2 \int \frac {x}{\left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2}d\sqrt {x}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int \frac {x}{\left (a+b \csc \left (i c+i d \sqrt {x}+\frac {\pi }{2}\right )\right )^2}d\sqrt {x}\)

\(\Big \downarrow \) 4679

\(\displaystyle 2 \int \left (\frac {x b^2}{a^2 \left (b+a \cosh \left (c+d \sqrt {x}\right )\right )^2}-\frac {2 x b}{a^2 \left (b+a \cosh \left (c+d \sqrt {x}\right )\right )}+\frac {x}{a^2}\right )d\sqrt {x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {x \log \left (\frac {e^{c+d \sqrt {x}} a}{b-\sqrt {b^2-a^2}}+1\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d}-\frac {x \log \left (\frac {e^{c+d \sqrt {x}} a}{b+\sqrt {b^2-a^2}}+1\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d}+\frac {2 \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^2}-\frac {2 \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^2}-\frac {2 \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^3}+\frac {2 \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^3}+\frac {x b^2}{a^2 \left (a^2-b^2\right ) d}-\frac {2 \sqrt {x} \log \left (\frac {e^{c+d \sqrt {x}} a}{b-\sqrt {b^2-a^2}}+1\right ) b^2}{a^2 \left (a^2-b^2\right ) d^2}-\frac {2 \sqrt {x} \log \left (\frac {e^{c+d \sqrt {x}} a}{b+\sqrt {b^2-a^2}}+1\right ) b^2}{a^2 \left (a^2-b^2\right ) d^2}-\frac {2 \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}\right ) b^2}{a^2 \left (a^2-b^2\right ) d^3}-\frac {2 \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {b^2-a^2}}\right ) b^2}{a^2 \left (a^2-b^2\right ) d^3}+\frac {x \sinh \left (c+d \sqrt {x}\right ) b^2}{a \left (a^2-b^2\right ) d \left (b+a \cosh \left (c+d \sqrt {x}\right )\right )}-\frac {2 x \log \left (\frac {e^{c+d \sqrt {x}} a}{b-\sqrt {b^2-a^2}}+1\right ) b}{a^2 \sqrt {b^2-a^2} d}+\frac {2 x \log \left (\frac {e^{c+d \sqrt {x}} a}{b+\sqrt {b^2-a^2}}+1\right ) b}{a^2 \sqrt {b^2-a^2} d}-\frac {4 \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^2}+\frac {4 \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^2}+\frac {4 \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^3}-\frac {4 \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^3}+\frac {x^{3/2}}{3 a^2}\right )\)

input

Int[Sqrt[x]/(a + b*Sech[c + d*Sqrt[x]])^2,x]
 

output

2*((b^2*x)/(a^2*(a^2 - b^2)*d) + x^(3/2)/(3*a^2) - (2*b^2*Sqrt[x]*Log[1 + 
(a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2])])/(a^2*(a^2 - b^2)*d^2) + (b^ 
3*x*Log[1 + (a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^ 
2)^(3/2)*d) - (2*b*x*Log[1 + (a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2])] 
)/(a^2*Sqrt[-a^2 + b^2]*d) - (2*b^2*Sqrt[x]*Log[1 + (a*E^(c + d*Sqrt[x]))/ 
(b + Sqrt[-a^2 + b^2])])/(a^2*(a^2 - b^2)*d^2) - (b^3*x*Log[1 + (a*E^(c + 
d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d) + (2*b*x*L 
og[1 + (a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2 
]*d) - (2*b^2*PolyLog[2, -((a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2]))]) 
/(a^2*(a^2 - b^2)*d^3) + (2*b^3*Sqrt[x]*PolyLog[2, -((a*E^(c + d*Sqrt[x])) 
/(b - Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b^2)^(3/2)*d^2) - (4*b*Sqrt[x]*Pol 
yLog[2, -((a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2]))])/(a^2*Sqrt[-a^2 + 
 b^2]*d^2) - (2*b^2*PolyLog[2, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^ 
2]))])/(a^2*(a^2 - b^2)*d^3) - (2*b^3*Sqrt[x]*PolyLog[2, -((a*E^(c + d*Sqr 
t[x]))/(b + Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b^2)^(3/2)*d^2) + (4*b*Sqrt[ 
x]*PolyLog[2, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2]))])/(a^2*Sqrt[ 
-a^2 + b^2]*d^2) - (2*b^3*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^ 
2 + b^2]))])/(a^2*(-a^2 + b^2)^(3/2)*d^3) + (4*b*PolyLog[3, -((a*E^(c + d* 
Sqrt[x]))/(b - Sqrt[-a^2 + b^2]))])/(a^2*Sqrt[-a^2 + b^2]*d^3) + (2*b^3*Po 
lyLog[3, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 +...
 

3.1.68.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4679

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si 
n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt 
Q[m, 0]
 

rule 5959

Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbo 
l] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sech[c + d*x] 
)^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m 
 + 1)/n], 0] && IntegerQ[p]
 

3.1.68.4 Maple [F]

\[\int \frac {\sqrt {x}}{\left (a +b \,\operatorname {sech}\left (c +d \sqrt {x}\right )\right )^{2}}d x\]

input

int(x^(1/2)/(a+b*sech(c+d*x^(1/2)))^2,x)
 

output

int(x^(1/2)/(a+b*sech(c+d*x^(1/2)))^2,x)
 

3.1.68.5 Fricas [F]

\[ \int \frac {\sqrt {x}}{\left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=\int { \frac {\sqrt {x}}{{\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )}^{2}} \,d x } \]

input

integrate(x^(1/2)/(a+b*sech(c+d*x^(1/2)))^2,x, algorithm="fricas")
 

output

integral(sqrt(x)/(b^2*sech(d*sqrt(x) + c)^2 + 2*a*b*sech(d*sqrt(x) + c) + 
a^2), x)
 

3.1.68.6 Sympy [F]

\[ \int \frac {\sqrt {x}}{\left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=\int \frac {\sqrt {x}}{\left (a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}\right )^{2}}\, dx \]

input

integrate(x**(1/2)/(a+b*sech(c+d*x**(1/2)))**2,x)
 

output

Integral(sqrt(x)/(a + b*sech(c + d*sqrt(x)))**2, x)
 

3.1.68.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {x}}{\left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=\text {Exception raised: ValueError} \]

input

integrate(x^(1/2)/(a+b*sech(c+d*x^(1/2)))^2,x, algorithm="maxima")
 

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a-b>0)', see `assume?` for more 
details)Is
 

3.1.68.8 Giac [F]

\[ \int \frac {\sqrt {x}}{\left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=\int { \frac {\sqrt {x}}{{\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )}^{2}} \,d x } \]

input

integrate(x^(1/2)/(a+b*sech(c+d*x^(1/2)))^2,x, algorithm="giac")
 

output

integrate(sqrt(x)/(b*sech(d*sqrt(x) + c) + a)^2, x)
 

3.1.68.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{\left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=\int \frac {\sqrt {x}}{{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,\sqrt {x}\right )}\right )}^2} \,d x \]

input

int(x^(1/2)/(a + b/cosh(c + d*x^(1/2)))^2,x)
 

output

int(x^(1/2)/(a + b/cosh(c + d*x^(1/2)))^2, x)