3.1.38 \(\int x^2 (a+b \text {sech}(c+d \sqrt {x}))^2 \, dx\) [38]

3.1.38.1 Optimal result

Integrand size = 20, antiderivative size = 497 \[ \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}+\frac {8 a b x^{5/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 b^2 x^2 \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 i a b x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}-\frac {20 b^2 x^{3/2} \operatorname {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 i a b x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 i a b x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}+\frac {30 b^2 x \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 i a b x \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 i a b x \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}-\frac {30 b^2 \sqrt {x} \operatorname {PolyLog}\left (4,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 i a b \sqrt {x} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 i a b \sqrt {x} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}+\frac {15 b^2 \operatorname {PolyLog}\left (5,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {480 i a b \operatorname {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {480 i a b \operatorname {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {2 b^2 x^{5/2} \tanh \left (c+d \sqrt {x}\right )}{d} \]

output

2*b^2*x^(5/2)/d+1/3*a^2*x^3+8*a*b*x^(5/2)*arctan(exp(c+d*x^(1/2)))/d-10*b^ 
2*x^2*ln(1+exp(2*c+2*d*x^(1/2)))/d^2-20*I*a*b*x^2*polylog(2,-I*exp(c+d*x^( 
1/2)))/d^2+480*I*a*b*polylog(6,I*exp(c+d*x^(1/2)))/d^6-20*b^2*x^(3/2)*poly 
log(2,-exp(2*c+2*d*x^(1/2)))/d^3-480*I*a*b*polylog(6,-I*exp(c+d*x^(1/2)))/ 
d^6-480*I*a*b*polylog(5,I*exp(c+d*x^(1/2)))*x^(1/2)/d^5+30*b^2*x*polylog(3 
,-exp(2*c+2*d*x^(1/2)))/d^4+80*I*a*b*x^(3/2)*polylog(3,-I*exp(c+d*x^(1/2)) 
)/d^3+20*I*a*b*x^2*polylog(2,I*exp(c+d*x^(1/2)))/d^2+15*b^2*polylog(5,-exp 
(2*c+2*d*x^(1/2)))/d^6-80*I*a*b*x^(3/2)*polylog(3,I*exp(c+d*x^(1/2)))/d^3+ 
240*I*a*b*x*polylog(4,I*exp(c+d*x^(1/2)))/d^4-30*b^2*polylog(4,-exp(2*c+2* 
d*x^(1/2)))*x^(1/2)/d^5+480*I*a*b*polylog(5,-I*exp(c+d*x^(1/2)))*x^(1/2)/d 
^5-240*I*a*b*x*polylog(4,-I*exp(c+d*x^(1/2)))/d^4+2*b^2*x^(5/2)*tanh(c+d*x 
^(1/2))/d
 

3.1.38.2 Mathematica [A] (verified)

Time = 7.13 (sec) , antiderivative size = 582, normalized size of antiderivative = 1.17 \[ \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {\cosh \left (c+d \sqrt {x}\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \left (\frac {12 b^2 e^{2 c} x^{5/2} \cosh \left (c+d \sqrt {x}\right )}{d \left (1+e^{2 c}\right )}+a^2 x^3 \cosh \left (c+d \sqrt {x}\right )+\frac {3 i b \cosh \left (c+d \sqrt {x}\right ) \left (4 a d^5 x^{5/2} \log \left (1-i e^{c+d \sqrt {x}}\right )-4 a d^5 x^{5/2} \log \left (1+i e^{c+d \sqrt {x}}\right )+10 i b d^4 x^2 \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )-20 a d^4 x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )+20 a d^4 x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )+20 i b d^3 x^{3/2} \operatorname {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )+80 a d^3 x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )-80 a d^3 x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )-30 i b d^2 x \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )-240 a d^2 x \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )+240 a d^2 x \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )+30 i b d \sqrt {x} \operatorname {PolyLog}\left (4,-e^{2 \left (c+d \sqrt {x}\right )}\right )+480 a d \sqrt {x} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )-480 a d \sqrt {x} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )-15 i b \operatorname {PolyLog}\left (5,-e^{2 \left (c+d \sqrt {x}\right )}\right )-480 a \operatorname {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )+480 a \operatorname {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )\right )}{d^6}+\frac {6 b^2 x^{5/2} \text {sech}(c) \sinh \left (d \sqrt {x}\right )}{d}\right )}{3 \left (b+a \cosh \left (c+d \sqrt {x}\right )\right )^2} \]

input

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

output

(Cosh[c + d*Sqrt[x]]*(a + b*Sech[c + d*Sqrt[x]])^2*((12*b^2*E^(2*c)*x^(5/2 
)*Cosh[c + d*Sqrt[x]])/(d*(1 + E^(2*c))) + a^2*x^3*Cosh[c + d*Sqrt[x]] + ( 
(3*I)*b*Cosh[c + d*Sqrt[x]]*(4*a*d^5*x^(5/2)*Log[1 - I*E^(c + d*Sqrt[x])] 
- 4*a*d^5*x^(5/2)*Log[1 + I*E^(c + d*Sqrt[x])] + (10*I)*b*d^4*x^2*Log[1 + 
E^(2*(c + d*Sqrt[x]))] - 20*a*d^4*x^2*PolyLog[2, (-I)*E^(c + d*Sqrt[x])] + 
 20*a*d^4*x^2*PolyLog[2, I*E^(c + d*Sqrt[x])] + (20*I)*b*d^3*x^(3/2)*PolyL 
og[2, -E^(2*(c + d*Sqrt[x]))] + 80*a*d^3*x^(3/2)*PolyLog[3, (-I)*E^(c + d* 
Sqrt[x])] - 80*a*d^3*x^(3/2)*PolyLog[3, I*E^(c + d*Sqrt[x])] - (30*I)*b*d^ 
2*x*PolyLog[3, -E^(2*(c + d*Sqrt[x]))] - 240*a*d^2*x*PolyLog[4, (-I)*E^(c 
+ d*Sqrt[x])] + 240*a*d^2*x*PolyLog[4, I*E^(c + d*Sqrt[x])] + (30*I)*b*d*S 
qrt[x]*PolyLog[4, -E^(2*(c + d*Sqrt[x]))] + 480*a*d*Sqrt[x]*PolyLog[5, (-I 
)*E^(c + d*Sqrt[x])] - 480*a*d*Sqrt[x]*PolyLog[5, I*E^(c + d*Sqrt[x])] - ( 
15*I)*b*PolyLog[5, -E^(2*(c + d*Sqrt[x]))] - 480*a*PolyLog[6, (-I)*E^(c + 
d*Sqrt[x])] + 480*a*PolyLog[6, I*E^(c + d*Sqrt[x])]))/d^6 + (6*b^2*x^(5/2) 
*Sech[c]*Sinh[d*Sqrt[x]])/d))/(3*(b + a*Cosh[c + d*Sqrt[x]])^2)
 

3.1.38.3 Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 499, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5959, 3042, 4678, 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.

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

\(\Big \downarrow \) 5959

\(\displaystyle 2 \int x^{5/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2d\sqrt {x}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int x^{5/2} \left (a+b \csc \left (i c+i d \sqrt {x}+\frac {\pi }{2}\right )\right )^2d\sqrt {x}\)

\(\Big \downarrow \) 4678

\(\displaystyle 2 \int \left (a^2 x^{5/2}+b^2 \text {sech}^2\left (c+d \sqrt {x}\right ) x^{5/2}+2 a b \text {sech}\left (c+d \sqrt {x}\right ) x^{5/2}\right )d\sqrt {x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {a^2 x^3}{6}+\frac {4 a b x^{5/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {240 i a b \operatorname {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 i a b \operatorname {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 i a b \sqrt {x} \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i a b \sqrt {x} \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {120 i a b x \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i a b x \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {40 i a b x^{3/2} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i a b x^{3/2} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {10 i a b x^2 \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i a b x^2 \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {15 b^2 \operatorname {PolyLog}\left (5,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {15 b^2 \sqrt {x} \operatorname {PolyLog}\left (4,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {15 b^2 x \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {10 b^2 x^{3/2} \operatorname {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {5 b^2 x^2 \log \left (e^{2 \left (c+d \sqrt {x}\right )}+1\right )}{d^2}+\frac {b^2 x^{5/2} \tanh \left (c+d \sqrt {x}\right )}{d}+\frac {b^2 x^{5/2}}{d}\right )\)

input

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

output

2*((b^2*x^(5/2))/d + (a^2*x^3)/6 + (4*a*b*x^(5/2)*ArcTan[E^(c + d*Sqrt[x]) 
])/d - (5*b^2*x^2*Log[1 + E^(2*(c + d*Sqrt[x]))])/d^2 - ((10*I)*a*b*x^2*Po 
lyLog[2, (-I)*E^(c + d*Sqrt[x])])/d^2 + ((10*I)*a*b*x^2*PolyLog[2, I*E^(c 
+ d*Sqrt[x])])/d^2 - (10*b^2*x^(3/2)*PolyLog[2, -E^(2*(c + d*Sqrt[x]))])/d 
^3 + ((40*I)*a*b*x^(3/2)*PolyLog[3, (-I)*E^(c + d*Sqrt[x])])/d^3 - ((40*I) 
*a*b*x^(3/2)*PolyLog[3, I*E^(c + d*Sqrt[x])])/d^3 + (15*b^2*x*PolyLog[3, - 
E^(2*(c + d*Sqrt[x]))])/d^4 - ((120*I)*a*b*x*PolyLog[4, (-I)*E^(c + d*Sqrt 
[x])])/d^4 + ((120*I)*a*b*x*PolyLog[4, I*E^(c + d*Sqrt[x])])/d^4 - (15*b^2 
*Sqrt[x]*PolyLog[4, -E^(2*(c + d*Sqrt[x]))])/d^5 + ((240*I)*a*b*Sqrt[x]*Po 
lyLog[5, (-I)*E^(c + d*Sqrt[x])])/d^5 - ((240*I)*a*b*Sqrt[x]*PolyLog[5, I* 
E^(c + d*Sqrt[x])])/d^5 + (15*b^2*PolyLog[5, -E^(2*(c + d*Sqrt[x]))])/(2*d 
^6) - ((240*I)*a*b*PolyLog[6, (-I)*E^(c + d*Sqrt[x])])/d^6 + ((240*I)*a*b* 
PolyLog[6, I*E^(c + d*Sqrt[x])])/d^6 + (b^2*x^(5/2)*Tanh[c + d*Sqrt[x]])/d 
)
 

3.1.38.3.1 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 4678

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

3.1.38.4 Maple [F]

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

input

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

output

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

3.1.38.5 Fricas [F]

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

input

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

output

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

3.1.38.6 Sympy [F]

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

input

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

output

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

3.1.38.7 Maxima [F]

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

input

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

output

1/3*(a^2*d*x^3*e^(2*d*sqrt(x) + 2*c) + a^2*d*x^3 - 12*b^2*x^(5/2))/(d*e^(2 
*d*sqrt(x) + 2*c) + d) + integrate(2*(2*a*b*d*x^2*e^(d*sqrt(x) + c) + 5*b^ 
2*x^(3/2))/(d*e^(2*d*sqrt(x) + 2*c) + d), x)
 

3.1.38.8 Giac [F]

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

input

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

output

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

3.1.38.9 Mupad [F(-1)]

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

input

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

output

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