\(\int x^{3/2} (a+b \text {sech}(c+d \sqrt {x}))^2 \, dx\) [64]

Optimal result

Integrand size = 22, antiderivative size = 407 \[ \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}+\frac {8 a b x^2 \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {8 b^2 x^{3/2} \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 i a b x^{3/2} \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}-\frac {12 b^2 x \operatorname {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 i a b x \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 i a b x \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 i a b \sqrt {x} \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}-\frac {6 b^2 \operatorname {PolyLog}\left (4,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {96 i a b \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {96 i a b \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}+\frac {2 b^2 x^2 \tanh \left (c+d \sqrt {x}\right )}{d} \] Output:

2*b^2*x^2/d+2/5*a^2*x^(5/2)+8*a*b*x^2*arctan(exp(c+d*x^(1/2)))/d-8*b^2*x^( 
3/2)*ln(1+exp(2*c+2*d*x^(1/2)))/d^2+48*I*a*b*x*polylog(3,-I*exp(c+d*x^(1/2 
)))/d^3-96*I*a*b*polylog(5,I*exp(c+d*x^(1/2)))/d^5-12*b^2*x*polylog(2,-exp 
(2*c+2*d*x^(1/2)))/d^3-48*I*a*b*x*polylog(3,I*exp(c+d*x^(1/2)))/d^3-96*I*a 
*b*x^(1/2)*polylog(4,-I*exp(c+d*x^(1/2)))/d^4+12*b^2*x^(1/2)*polylog(3,-ex 
p(2*c+2*d*x^(1/2)))/d^4+96*I*a*b*polylog(5,-I*exp(c+d*x^(1/2)))/d^5-16*I*a 
*b*x^(3/2)*polylog(2,-I*exp(c+d*x^(1/2)))/d^2-6*b^2*polylog(4,-exp(2*c+2*d 
*x^(1/2)))/d^5+16*I*a*b*x^(3/2)*polylog(2,I*exp(c+d*x^(1/2)))/d^2+96*I*a*b 
*x^(1/2)*polylog(4,I*exp(c+d*x^(1/2)))/d^4+2*b^2*x^2*tanh(c+d*x^(1/2))/d
 

Mathematica [A] (verified)

Time = 3.74 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.22 \[ \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {2 \cosh \left (c+d \sqrt {x}\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \left (\frac {10 b^2 e^{2 c} x^2 \cosh \left (c+d \sqrt {x}\right )}{d \left (1+e^{2 c}\right )}+a^2 x^{5/2} \cosh \left (c+d \sqrt {x}\right )+\frac {5 i b \cosh \left (c+d \sqrt {x}\right ) \left (2 a d^4 x^2 \log \left (1-i e^{c+d \sqrt {x}}\right )-2 a d^4 x^2 \log \left (1+i e^{c+d \sqrt {x}}\right )+4 i b d^3 x^{3/2} \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )-8 a d^3 x^{3/2} \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )+8 a d^3 x^{3/2} \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )+6 i b d^2 x \operatorname {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )+24 a d^2 x \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )-24 a d^2 x \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )-6 i b d \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )-48 a d \sqrt {x} \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )+48 a d \sqrt {x} \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )+3 i b \operatorname {PolyLog}\left (4,-e^{2 \left (c+d \sqrt {x}\right )}\right )+48 a \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )-48 a \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )\right )}{d^5}+\frac {5 b^2 x^2 \text {sech}(c) \sinh \left (d \sqrt {x}\right )}{d}\right )}{5 \left (b+a \cosh \left (c+d \sqrt {x}\right )\right )^2} \] Input:

Integrate[x^(3/2)*(a + b*Sech[c + d*Sqrt[x]])^2,x]
 

Output:

(2*Cosh[c + d*Sqrt[x]]*(a + b*Sech[c + d*Sqrt[x]])^2*((10*b^2*E^(2*c)*x^2* 
Cosh[c + d*Sqrt[x]])/(d*(1 + E^(2*c))) + a^2*x^(5/2)*Cosh[c + d*Sqrt[x]] + 
 ((5*I)*b*Cosh[c + d*Sqrt[x]]*(2*a*d^4*x^2*Log[1 - I*E^(c + d*Sqrt[x])] - 
2*a*d^4*x^2*Log[1 + I*E^(c + d*Sqrt[x])] + (4*I)*b*d^3*x^(3/2)*Log[1 + E^( 
2*(c + d*Sqrt[x]))] - 8*a*d^3*x^(3/2)*PolyLog[2, (-I)*E^(c + d*Sqrt[x])] + 
 8*a*d^3*x^(3/2)*PolyLog[2, I*E^(c + d*Sqrt[x])] + (6*I)*b*d^2*x*PolyLog[2 
, -E^(2*(c + d*Sqrt[x]))] + 24*a*d^2*x*PolyLog[3, (-I)*E^(c + d*Sqrt[x])] 
- 24*a*d^2*x*PolyLog[3, I*E^(c + d*Sqrt[x])] - (6*I)*b*d*Sqrt[x]*PolyLog[3 
, -E^(2*(c + d*Sqrt[x]))] - 48*a*d*Sqrt[x]*PolyLog[4, (-I)*E^(c + d*Sqrt[x 
])] + 48*a*d*Sqrt[x]*PolyLog[4, I*E^(c + d*Sqrt[x])] + (3*I)*b*PolyLog[4, 
-E^(2*(c + d*Sqrt[x]))] + 48*a*PolyLog[5, (-I)*E^(c + d*Sqrt[x])] - 48*a*P 
olyLog[5, I*E^(c + d*Sqrt[x])]))/d^5 + (5*b^2*x^2*Sech[c]*Sinh[d*Sqrt[x]]) 
/d))/(5*(b + a*Cosh[c + d*Sqrt[x]])^2)
 

Rubi [A] (verified)

Time = 1.34 (sec) , antiderivative size = 407, 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, 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.

\(\displaystyle \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx\)

\(\Big \downarrow \) 5959

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4678

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

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {1}{5} a^2 x^{5/2}+\frac {4 a b x^2 \arctan \left (e^{c+d \sqrt {x}}\right )}{d}+\frac {48 i a b \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {48 i a b \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {48 i a b \sqrt {x} \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {48 i a b \sqrt {x} \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {24 i a b x \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 i a b x \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {8 i a b x^{3/2} \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {8 i a b x^{3/2} \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}-\frac {3 b^2 \operatorname {PolyLog}\left (4,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {6 b^2 \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 b^2 x \operatorname {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {4 b^2 x^{3/2} \log \left (e^{2 \left (c+d \sqrt {x}\right )}+1\right )}{d^2}+\frac {b^2 x^2 \tanh \left (c+d \sqrt {x}\right )}{d}+\frac {b^2 x^2}{d}\right )\)

Input:

Int[x^(3/2)*(a + b*Sech[c + d*Sqrt[x]])^2,x]
 

Output:

2*((b^2*x^2)/d + (a^2*x^(5/2))/5 + (4*a*b*x^2*ArcTan[E^(c + d*Sqrt[x])])/d 
 - (4*b^2*x^(3/2)*Log[1 + E^(2*(c + d*Sqrt[x]))])/d^2 - ((8*I)*a*b*x^(3/2) 
*PolyLog[2, (-I)*E^(c + d*Sqrt[x])])/d^2 + ((8*I)*a*b*x^(3/2)*PolyLog[2, I 
*E^(c + d*Sqrt[x])])/d^2 - (6*b^2*x*PolyLog[2, -E^(2*(c + d*Sqrt[x]))])/d^ 
3 + ((24*I)*a*b*x*PolyLog[3, (-I)*E^(c + d*Sqrt[x])])/d^3 - ((24*I)*a*b*x* 
PolyLog[3, I*E^(c + d*Sqrt[x])])/d^3 + (6*b^2*Sqrt[x]*PolyLog[3, -E^(2*(c 
+ d*Sqrt[x]))])/d^4 - ((48*I)*a*b*Sqrt[x]*PolyLog[4, (-I)*E^(c + d*Sqrt[x] 
)])/d^4 + ((48*I)*a*b*Sqrt[x]*PolyLog[4, I*E^(c + d*Sqrt[x])])/d^4 - (3*b^ 
2*PolyLog[4, -E^(2*(c + d*Sqrt[x]))])/d^5 + ((48*I)*a*b*PolyLog[5, (-I)*E^ 
(c + d*Sqrt[x])])/d^5 - ((48*I)*a*b*PolyLog[5, I*E^(c + d*Sqrt[x])])/d^5 + 
 (b^2*x^2*Tanh[c + d*Sqrt[x]])/d)
 

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]
 

Maple [F]

\[\int x^{\frac {3}{2}} \left (a +b \,\operatorname {sech}\left (c +d \sqrt {x}\right )\right )^{2}d x\]

Input:

int(x^(3/2)*(a+b*sech(c+d*x^(1/2)))^2,x)
 

Output:

int(x^(3/2)*(a+b*sech(c+d*x^(1/2)))^2,x)
 

Fricas [F]

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

integrate(x^(3/2)*(a+b*sech(c+d*x^(1/2)))^2,x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(x**(3/2)*(a+b*sech(c+d*x**(1/2)))**2,x)
 

Output:

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

Maxima [F]

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

integrate(x^(3/2)*(a+b*sech(c+d*x^(1/2)))^2,x, algorithm="maxima")
 

Output:

2/5*(a^2*d*x^(5/2)*e^(2*d*sqrt(x) + 2*c) + a^2*d*x^(5/2) - 10*b^2*x^2)/(d* 
e^(2*d*sqrt(x) + 2*c) + d) + integrate(4*(a*b*d*x^(5/2)*e^(d*sqrt(x) + c) 
+ 2*b^2*x^2)/(d*x*e^(2*d*sqrt(x) + 2*c) + d*x), x)
 

Giac [F]

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

integrate(x^(3/2)*(a+b*sech(c+d*x^(1/2)))^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(x^(3/2)*(a + b/cosh(c + d*x^(1/2)))^2,x)
 

Output:

int(x^(3/2)*(a + b/cosh(c + d*x^(1/2)))^2, x)
 

Reduce [F]

\[ \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx =\text {Too large to display} \] Input:

int(x^(3/2)*(a+b*sech(c+d*x^(1/2)))^2,x)
 

Output:

(2*(480*e**(2*sqrt(x)*d + 2*c)*atan(e**(sqrt(x)*d + c))*a*b + 480*atan(e** 
(sqrt(x)*d + c))*a*b + 480*e**(2*sqrt(x)*d + 3*c)*int(e**(sqrt(x)*d)/(e**( 
4*sqrt(x)*d + 4*c) + 2*e**(2*sqrt(x)*d + 2*c) + 1),x)*a*b*d**2 + 80*e**(2* 
sqrt(x)*d + 3*c)*int((e**(sqrt(x)*d)*x)/(e**(4*sqrt(x)*d + 4*c) + 2*e**(2* 
sqrt(x)*d + 2*c) + 1),x)*a*b*d**4 + 20*e**(2*sqrt(x)*d + 3*c)*int((sqrt(x) 
*e**(sqrt(x)*d)*x)/(e**(4*sqrt(x)*d + 4*c) + 2*e**(2*sqrt(x)*d + 2*c) + 1) 
,x)*a*b*d**5 + 240*e**(2*sqrt(x)*d + 3*c)*int((sqrt(x)*e**(sqrt(x)*d))/(e* 
*(4*sqrt(x)*d + 4*c) + 2*e**(2*sqrt(x)*d + 2*c) + 1),x)*a*b*d**3 + sqrt(x) 
*e**(2*sqrt(x)*d + 2*c)*a**2*d**5*x**2 + 30*sqrt(x)*e**(2*sqrt(x)*d + 2*c) 
*b**2*d + 30*e**(2*sqrt(x)*d + 2*c)*int(sqrt(x)/(e**(4*sqrt(x)*d + 4*c) + 
2*e**(2*sqrt(x)*d + 2*c) + 1),x)*b**2*d**3 + 20*e**(2*sqrt(x)*d + 2*c)*int 
(x/(e**(4*sqrt(x)*d + 4*c) + 2*e**(2*sqrt(x)*d + 2*c) + 1),x)*b**2*d**4 + 
30*e**(2*sqrt(x)*d + 2*c)*int(1/(e**(4*sqrt(x)*d + 4*c) + 2*e**(2*sqrt(x)* 
d + 2*c) + 1),x)*b**2*d**2 - 15*e**(2*sqrt(x)*d + 2*c)*log(e**(2*sqrt(x)*d 
 + 2*c) + 1)*b**2 - 80*sqrt(x)*e**(sqrt(x)*d + c)*a*b*d**3*x - 480*sqrt(x) 
*e**(sqrt(x)*d + c)*a*b*d - 20*e**(sqrt(x)*d + c)*a*b*d**4*x**2 - 240*e**( 
sqrt(x)*d + c)*a*b*d**2*x + 480*e**c*int(e**(sqrt(x)*d)/(e**(4*sqrt(x)*d + 
 4*c) + 2*e**(2*sqrt(x)*d + 2*c) + 1),x)*a*b*d**2 + 80*e**c*int((e**(sqrt( 
x)*d)*x)/(e**(4*sqrt(x)*d + 4*c) + 2*e**(2*sqrt(x)*d + 2*c) + 1),x)*a*b*d* 
*4 + 20*e**c*int((sqrt(x)*e**(sqrt(x)*d)*x)/(e**(4*sqrt(x)*d + 4*c) + 2...