\(\int \frac {\sqrt {x}}{a+b \text {sech}(c+d \sqrt {x})} \, dx\) [70]

Optimal result

Integrand size = 22, antiderivative size = 361 \[ \int \frac {\sqrt {x}}{a+b \text {sech}\left (c+d \sqrt {x}\right )} \, dx=\frac {2 x^{3/2}}{3 a}-\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {4 b \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {4 b \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {4 b \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {4 b \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3} \] Output:

2/3*x^(3/2)/a-2*b*x*ln(1+a*exp(c+d*x^(1/2))/(b-(-a^2+b^2)^(1/2)))/a/(-a^2+ 
b^2)^(1/2)/d+2*b*x*ln(1+a*exp(c+d*x^(1/2))/(b+(-a^2+b^2)^(1/2)))/a/(-a^2+b 
^2)^(1/2)/d-4*b*x^(1/2)*polylog(2,-a*exp(c+d*x^(1/2))/(b-(-a^2+b^2)^(1/2)) 
)/a/(-a^2+b^2)^(1/2)/d^2+4*b*x^(1/2)*polylog(2,-a*exp(c+d*x^(1/2))/(b+(-a^ 
2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^2+4*b*polylog(3,-a*exp(c+d*x^(1/2))/(b 
-(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^3-4*b*polylog(3,-a*exp(c+d*x^(1/2 
))/(b+(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^3
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {x}}{a+b \text {sech}\left (c+d \sqrt {x}\right )} \, dx=\frac {2 \left (\sqrt {-a^2+b^2} d^3 x^{3/2}-3 b d^2 x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )+3 b d^2 x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )-6 b d \sqrt {x} \operatorname {PolyLog}\left (2,\frac {a e^{c+d \sqrt {x}}}{-b+\sqrt {-a^2+b^2}}\right )+6 b d \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )+6 b \operatorname {PolyLog}\left (3,\frac {a e^{c+d \sqrt {x}}}{-b+\sqrt {-a^2+b^2}}\right )-6 b \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )\right )}{3 a \sqrt {-a^2+b^2} d^3} \] Input:

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

Output:

(2*(Sqrt[-a^2 + b^2]*d^3*x^(3/2) - 3*b*d^2*x*Log[1 + (a*E^(c + d*Sqrt[x])) 
/(b - Sqrt[-a^2 + b^2])] + 3*b*d^2*x*Log[1 + (a*E^(c + d*Sqrt[x]))/(b + Sq 
rt[-a^2 + b^2])] - 6*b*d*Sqrt[x]*PolyLog[2, (a*E^(c + d*Sqrt[x]))/(-b + Sq 
rt[-a^2 + b^2])] + 6*b*d*Sqrt[x]*PolyLog[2, -((a*E^(c + d*Sqrt[x]))/(b + S 
qrt[-a^2 + b^2]))] + 6*b*PolyLog[3, (a*E^(c + d*Sqrt[x]))/(-b + Sqrt[-a^2 
+ b^2])] - 6*b*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2]))] 
))/(3*a*Sqrt[-a^2 + b^2]*d^3)
 

Rubi [A] (verified)

Time = 1.09 (sec) , antiderivative size = 362, 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.

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

\(\Big \downarrow \) 5959

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

2*(x^(3/2)/(3*a) - (b*x*Log[1 + (a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2 
])])/(a*Sqrt[-a^2 + b^2]*d) + (b*x*Log[1 + (a*E^(c + d*Sqrt[x]))/(b + Sqrt 
[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) - (2*b*Sqrt[x]*PolyLog[2, -((a*E^(c 
 + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + (2*b*S 
qrt[x]*PolyLog[2, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqr 
t[-a^2 + b^2]*d^2) + (2*b*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^ 
2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3) - (2*b*PolyLog[3, -((a*E^(c + d*Sqrt 
[x]))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3))
 

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

e**(2*c)*int((sqrt(x)*e**(2*sqrt(x)*d))/(e**(2*sqrt(x)*d + 2*c)*a + 2*e**( 
sqrt(x)*d + c)*b + a),x) + int(sqrt(x)/(e**(2*sqrt(x)*d + 2*c)*a + 2*e**(s 
qrt(x)*d + c)*b + a),x)