\(\int \frac {x^5}{a+b \text {sech}(c+d x^2)} \, dx\) [20]

Optimal result

Integrand size = 18, antiderivative size = 349 \[ \int \frac {x^5}{a+b \text {sech}\left (c+d x^2\right )} \, dx=\frac {x^6}{6 a}-\frac {b x^4 \log \left (1+\frac {a e^{c+d x^2}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {b x^4 \log \left (1+\frac {a e^{c+d x^2}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^2}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^2}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {b \operatorname {PolyLog}\left (3,-\frac {a e^{c+d x^2}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {b \operatorname {PolyLog}\left (3,-\frac {a e^{c+d x^2}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3} \] Output:

1/6*x^6/a-1/2*b*x^4*ln(1+a*exp(d*x^2+c)/(b-(-a^2+b^2)^(1/2)))/a/(-a^2+b^2) 
^(1/2)/d+1/2*b*x^4*ln(1+a*exp(d*x^2+c)/(b+(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^ 
(1/2)/d-b*x^2*polylog(2,-a*exp(d*x^2+c)/(b-(-a^2+b^2)^(1/2)))/a/(-a^2+b^2) 
^(1/2)/d^2+b*x^2*polylog(2,-a*exp(d*x^2+c)/(b+(-a^2+b^2)^(1/2)))/a/(-a^2+b 
^2)^(1/2)/d^2+b*polylog(3,-a*exp(d*x^2+c)/(b-(-a^2+b^2)^(1/2)))/a/(-a^2+b^ 
2)^(1/2)/d^3-b*polylog(3,-a*exp(d*x^2+c)/(b+(-a^2+b^2)^(1/2)))/a/(-a^2+b^2 
)^(1/2)/d^3
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.78 \[ \int \frac {x^5}{a+b \text {sech}\left (c+d x^2\right )} \, dx=\frac {\sqrt {-a^2+b^2} d^3 x^6-3 b d^2 x^4 \log \left (1+\frac {a e^{c+d x^2}}{b-\sqrt {-a^2+b^2}}\right )+3 b d^2 x^4 \log \left (1+\frac {a e^{c+d x^2}}{b+\sqrt {-a^2+b^2}}\right )-6 b d x^2 \operatorname {PolyLog}\left (2,\frac {a e^{c+d x^2}}{-b+\sqrt {-a^2+b^2}}\right )+6 b d x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^2}}{b+\sqrt {-a^2+b^2}}\right )+6 b \operatorname {PolyLog}\left (3,\frac {a e^{c+d x^2}}{-b+\sqrt {-a^2+b^2}}\right )-6 b \operatorname {PolyLog}\left (3,-\frac {a e^{c+d x^2}}{b+\sqrt {-a^2+b^2}}\right )}{6 a \sqrt {-a^2+b^2} d^3} \] Input:

Integrate[x^5/(a + b*Sech[c + d*x^2]),x]
 

Output:

(Sqrt[-a^2 + b^2]*d^3*x^6 - 3*b*d^2*x^4*Log[1 + (a*E^(c + d*x^2))/(b - Sqr 
t[-a^2 + b^2])] + 3*b*d^2*x^4*Log[1 + (a*E^(c + d*x^2))/(b + Sqrt[-a^2 + b 
^2])] - 6*b*d*x^2*PolyLog[2, (a*E^(c + d*x^2))/(-b + Sqrt[-a^2 + b^2])] + 
6*b*d*x^2*PolyLog[2, -((a*E^(c + d*x^2))/(b + Sqrt[-a^2 + b^2]))] + 6*b*Po 
lyLog[3, (a*E^(c + d*x^2))/(-b + Sqrt[-a^2 + b^2])] - 6*b*PolyLog[3, -((a* 
E^(c + d*x^2))/(b + Sqrt[-a^2 + b^2]))])/(6*a*Sqrt[-a^2 + b^2]*d^3)
 

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, 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.

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

\(\Big \downarrow \) 5959

\(\displaystyle \frac {1}{2} \int \frac {x^4}{a+b \text {sech}\left (d x^2+c\right )}dx^2\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \frac {x^4}{a+b \csc \left (i d x^2+i c+\frac {\pi }{2}\right )}dx^2\)

\(\Big \downarrow \) 4679

\(\displaystyle \frac {1}{2} \int \left (\frac {x^4}{a}-\frac {b x^4}{a \left (b+a \cosh \left (d x^2+c\right )\right )}\right )dx^2\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {2 b \operatorname {PolyLog}\left (3,-\frac {a e^{d x^2+c}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {2 b \operatorname {PolyLog}\left (3,-\frac {a e^{d x^2+c}}{b+\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {2 b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{d x^2+c}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {2 b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{d x^2+c}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {b x^4 \log \left (\frac {a e^{c+d x^2}}{b-\sqrt {b^2-a^2}}+1\right )}{a d \sqrt {b^2-a^2}}+\frac {b x^4 \log \left (\frac {a e^{c+d x^2}}{\sqrt {b^2-a^2}+b}+1\right )}{a d \sqrt {b^2-a^2}}+\frac {x^6}{3 a}\right )\)

Input:

Int[x^5/(a + b*Sech[c + d*x^2]),x]
 

Output:

(x^6/(3*a) - (b*x^4*Log[1 + (a*E^(c + d*x^2))/(b - Sqrt[-a^2 + b^2])])/(a* 
Sqrt[-a^2 + b^2]*d) + (b*x^4*Log[1 + (a*E^(c + d*x^2))/(b + Sqrt[-a^2 + b^ 
2])])/(a*Sqrt[-a^2 + b^2]*d) - (2*b*x^2*PolyLog[2, -((a*E^(c + d*x^2))/(b 
- Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + (2*b*x^2*PolyLog[2, -((a 
*E^(c + d*x^2))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + (2*b* 
PolyLog[3, -((a*E^(c + d*x^2))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^ 
2]*d^3) - (2*b*PolyLog[3, -((a*E^(c + d*x^2))/(b + Sqrt[-a^2 + b^2]))])/(a 
*Sqrt[-a^2 + b^2]*d^3))/2
 

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]

\[\int \frac {x^{5}}{a +b \,\operatorname {sech}\left (d \,x^{2}+c \right )}d x\]

Input:

int(x^5/(a+b*sech(d*x^2+c)),x)
 

Output:

int(x^5/(a+b*sech(d*x^2+c)),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 731 vs. \(2 (311) = 622\).

Time = 0.10 (sec) , antiderivative size = 731, normalized size of antiderivative = 2.09 \[ \int \frac {x^5}{a+b \text {sech}\left (c+d x^2\right )} \, dx =\text {Too large to display} \] Input:

integrate(x^5/(a+b*sech(d*x^2+c)),x, algorithm="fricas")
 

Output:

1/6*((a^2 - b^2)*d^3*x^6 + 6*a*b*d*x^2*sqrt(-(a^2 - b^2)/a^2)*dilog(-(b*co 
sh(d*x^2 + c) + b*sinh(d*x^2 + c) + (a*cosh(d*x^2 + c) + a*sinh(d*x^2 + c) 
)*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) - 6*a*b*d*x^2*sqrt(-(a^2 - b^2)/a^2)* 
dilog(-(b*cosh(d*x^2 + c) + b*sinh(d*x^2 + c) - (a*cosh(d*x^2 + c) + a*sin 
h(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) - 3*a*b*c^2*sqrt(-(a^2 - 
b^2)/a^2)*log(2*a*cosh(d*x^2 + c) + 2*a*sinh(d*x^2 + c) + 2*a*sqrt(-(a^2 - 
 b^2)/a^2) + 2*b) + 3*a*b*c^2*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cosh(d*x^2 + 
c) + 2*a*sinh(d*x^2 + c) - 2*a*sqrt(-(a^2 - b^2)/a^2) + 2*b) - 6*a*b*sqrt( 
-(a^2 - b^2)/a^2)*polylog(3, -(b*cosh(d*x^2 + c) + b*sinh(d*x^2 + c) + (a* 
cosh(d*x^2 + c) + a*sinh(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2))/a) + 6*a*b*sq 
rt(-(a^2 - b^2)/a^2)*polylog(3, -(b*cosh(d*x^2 + c) + b*sinh(d*x^2 + c) - 
(a*cosh(d*x^2 + c) + a*sinh(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2))/a) + 3*(a* 
b*d^2*x^4 - a*b*c^2)*sqrt(-(a^2 - b^2)/a^2)*log((b*cosh(d*x^2 + c) + b*sin 
h(d*x^2 + c) + (a*cosh(d*x^2 + c) + a*sinh(d*x^2 + c))*sqrt(-(a^2 - b^2)/a 
^2) + a)/a) - 3*(a*b*d^2*x^4 - a*b*c^2)*sqrt(-(a^2 - b^2)/a^2)*log((b*cosh 
(d*x^2 + c) + b*sinh(d*x^2 + c) - (a*cosh(d*x^2 + c) + a*sinh(d*x^2 + c))* 
sqrt(-(a^2 - b^2)/a^2) + a)/a))/((a^3 - a*b^2)*d^3)
 

Sympy [F]

\[ \int \frac {x^5}{a+b \text {sech}\left (c+d x^2\right )} \, dx=\int \frac {x^{5}}{a + b \operatorname {sech}{\left (c + d x^{2} \right )}}\, dx \] Input:

integrate(x**5/(a+b*sech(d*x**2+c)),x)
 

Output:

Integral(x**5/(a + b*sech(c + d*x**2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^5}{a+b \text {sech}\left (c+d x^2\right )} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^5/(a+b*sech(d*x^2+c)),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]

\[ \int \frac {x^5}{a+b \text {sech}\left (c+d x^2\right )} \, dx=\int { \frac {x^{5}}{b \operatorname {sech}\left (d x^{2} + c\right ) + a} \,d x } \] Input:

integrate(x^5/(a+b*sech(d*x^2+c)),x, algorithm="giac")
 

Output:

integrate(x^5/(b*sech(d*x^2 + c) + a), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{a+b \text {sech}\left (c+d x^2\right )} \, dx=\int \frac {x^5}{a+\frac {b}{\mathrm {cosh}\left (d\,x^2+c\right )}} \,d x \] Input:

int(x^5/(a + b/cosh(c + d*x^2)),x)
 

Output:

int(x^5/(a + b/cosh(c + d*x^2)), x)
 

Reduce [F]

\[ \int \frac {x^5}{a+b \text {sech}\left (c+d x^2\right )} \, dx=e^{2 c} \left (\int \frac {e^{2 d \,x^{2}} x^{5}}{e^{2 d \,x^{2}+2 c} a +2 e^{d \,x^{2}+c} b +a}d x \right )+\int \frac {x^{5}}{e^{2 d \,x^{2}+2 c} a +2 e^{d \,x^{2}+c} b +a}d x \] Input:

int(x^5/(a+b*sech(d*x^2+c)),x)
 

Output:

e**(2*c)*int((e**(2*d*x**2)*x**5)/(e**(2*c + 2*d*x**2)*a + 2*e**(c + d*x** 
2)*b + a),x) + int(x**5/(e**(2*c + 2*d*x**2)*a + 2*e**(c + d*x**2)*b + a), 
x)