\(\int x^2 (a+b \text {sech}(c+d \sqrt {x})) \, dx\) [40]

Optimal result

Integrand size = 18, antiderivative size = 310 \[ \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^3}{3}+\frac {4 b x^{5/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 i b x \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \operatorname {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 i b \operatorname {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )}{d^6} \] Output:

1/3*a*x^3+4*b*x^(5/2)*arctan(exp(c+d*x^(1/2)))/d-10*I*b*x^2*polylog(2,-I*e 
xp(c+d*x^(1/2)))/d^2+10*I*b*x^2*polylog(2,I*exp(c+d*x^(1/2)))/d^2+40*I*b*x 
^(3/2)*polylog(3,-I*exp(c+d*x^(1/2)))/d^3-40*I*b*x^(3/2)*polylog(3,I*exp(c 
+d*x^(1/2)))/d^3-120*I*b*x*polylog(4,-I*exp(c+d*x^(1/2)))/d^4+120*I*b*x*po 
lylog(4,I*exp(c+d*x^(1/2)))/d^4+240*I*b*x^(1/2)*polylog(5,-I*exp(c+d*x^(1/ 
2)))/d^5-240*I*b*x^(1/2)*polylog(5,I*exp(c+d*x^(1/2)))/d^5-240*I*b*polylog 
(6,-I*exp(c+d*x^(1/2)))/d^6+240*I*b*polylog(6,I*exp(c+d*x^(1/2)))/d^6
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^3}{3}+\frac {2 i b \left (d^5 x^{5/2} \log \left (1-i e^{c+d \sqrt {x}}\right )-d^5 x^{5/2} \log \left (1+i e^{c+d \sqrt {x}}\right )-5 d^4 x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )+5 d^4 x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )+20 d^3 x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )-20 d^3 x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )-60 d^2 x \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )+60 d^2 x \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )+120 d \sqrt {x} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )-120 d \sqrt {x} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )-120 \operatorname {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )+120 \operatorname {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )\right )}{d^6} \] Input:

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

Output:

(a*x^3)/3 + ((2*I)*b*(d^5*x^(5/2)*Log[1 - I*E^(c + d*Sqrt[x])] - d^5*x^(5/ 
2)*Log[1 + I*E^(c + d*Sqrt[x])] - 5*d^4*x^2*PolyLog[2, (-I)*E^(c + d*Sqrt[ 
x])] + 5*d^4*x^2*PolyLog[2, I*E^(c + d*Sqrt[x])] + 20*d^3*x^(3/2)*PolyLog[ 
3, (-I)*E^(c + d*Sqrt[x])] - 20*d^3*x^(3/2)*PolyLog[3, I*E^(c + d*Sqrt[x]) 
] - 60*d^2*x*PolyLog[4, (-I)*E^(c + d*Sqrt[x])] + 60*d^2*x*PolyLog[4, I*E^ 
(c + d*Sqrt[x])] + 120*d*Sqrt[x]*PolyLog[5, (-I)*E^(c + d*Sqrt[x])] - 120* 
d*Sqrt[x]*PolyLog[5, I*E^(c + d*Sqrt[x])] - 120*PolyLog[6, (-I)*E^(c + d*S 
qrt[x])] + 120*PolyLog[6, I*E^(c + d*Sqrt[x])]))/d^6
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.

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

\(\Big \downarrow \) 2010

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a x^3}{3}+\frac {4 b x^{5/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {240 i b \operatorname {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 i b \operatorname {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \sqrt {x} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {120 i b x \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {10 i b x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}\)

Input:

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

Output:

(a*x^3)/3 + (4*b*x^(5/2)*ArcTan[E^(c + d*Sqrt[x])])/d - ((10*I)*b*x^2*Poly 
Log[2, (-I)*E^(c + d*Sqrt[x])])/d^2 + ((10*I)*b*x^2*PolyLog[2, I*E^(c + d* 
Sqrt[x])])/d^2 + ((40*I)*b*x^(3/2)*PolyLog[3, (-I)*E^(c + d*Sqrt[x])])/d^3 
 - ((40*I)*b*x^(3/2)*PolyLog[3, I*E^(c + d*Sqrt[x])])/d^3 - ((120*I)*b*x*P 
olyLog[4, (-I)*E^(c + d*Sqrt[x])])/d^4 + ((120*I)*b*x*PolyLog[4, I*E^(c + 
d*Sqrt[x])])/d^4 + ((240*I)*b*Sqrt[x]*PolyLog[5, (-I)*E^(c + d*Sqrt[x])])/ 
d^5 - ((240*I)*b*Sqrt[x]*PolyLog[5, I*E^(c + d*Sqrt[x])])/d^5 - ((240*I)*b 
*PolyLog[6, (-I)*E^(c + d*Sqrt[x])])/d^6 + ((240*I)*b*PolyLog[6, I*E^(c + 
d*Sqrt[x])])/d^6
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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