\(\int \sqrt {x} (a+b \text {sech}(c+d \sqrt {x})) \, dx\) [60]

Optimal result

Integrand size = 20, antiderivative size = 140 \[ \int \sqrt {x} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {2}{3} a x^{3/2}+\frac {4 b x \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {4 i b \sqrt {x} \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {4 i b \sqrt {x} \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {4 i b \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {4 i b \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.23 \[ \int \sqrt {x} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {2 \left (a d^3 x^{3/2}+3 i b d^2 x \log \left (1-i e^{c+d \sqrt {x}}\right )-3 i b d^2 x \log \left (1+i e^{c+d \sqrt {x}}\right )-6 i b d \sqrt {x} \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )+6 i b d \sqrt {x} \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )+6 i b \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )-6 i b \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )\right )}{3 d^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2010, 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.

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

\(\Big \downarrow \) 2010

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(2*a*x^(3/2))/3 + (4*b*x*ArcTan[E^(c + d*Sqrt[x])])/d - ((4*I)*b*Sqrt[x]*P 
olyLog[2, (-I)*E^(c + d*Sqrt[x])])/d^2 + ((4*I)*b*Sqrt[x]*PolyLog[2, I*E^( 
c + d*Sqrt[x])])/d^2 + ((4*I)*b*PolyLog[3, (-I)*E^(c + d*Sqrt[x])])/d^3 - 
((4*I)*b*PolyLog[3, I*E^(c + d*Sqrt[x])])/d^3
 

Defintions of rubi rules used

rule 2009

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

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

Maple [F]

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

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]

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

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

Output:

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

Sympy [F]

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

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

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

Output:

2/3*a*x^(3/2) + 2*b*integrate(sqrt(x)*e^(d*sqrt(x) + c)/(e^(2*d*sqrt(x) + 
2*c) + 1), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

\[ \int \sqrt {x} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {2 \sqrt {x}\, a x}{3}+\left (\int \sqrt {x}\, \mathrm {sech}\left (\sqrt {x}\, d +c \right )d x \right ) b \] Input:

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

Output:

(2*sqrt(x)*a*x + 3*int(sqrt(x)*sech(sqrt(x)*d + c),x)*b)/3