3.5.74 \(\int x^3 \text {csch}(a+b x) \text {sech}^2(a+b x) \, dx\) [474]

3.5.74.1 Optimal result

Integrand size = 18, antiderivative size = 226 \[ \int x^3 \text {csch}(a+b x) \text {sech}^2(a+b x) \, dx=-\frac {6 x^2 \arctan \left (e^{a+b x}\right )}{b^2}-\frac {2 x^3 \text {arctanh}\left (e^{a+b x}\right )}{b}-\frac {3 x^2 \operatorname {PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac {6 i x \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac {6 i x \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac {3 x^2 \operatorname {PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac {6 x \operatorname {PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac {6 i \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}+\frac {6 i \operatorname {PolyLog}\left (3,i e^{a+b x}\right )}{b^4}-\frac {6 x \operatorname {PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac {6 \operatorname {PolyLog}\left (4,-e^{a+b x}\right )}{b^4}+\frac {6 \operatorname {PolyLog}\left (4,e^{a+b x}\right )}{b^4}+\frac {x^3 \text {sech}(a+b x)}{b} \]

output

-6*x^2*arctan(exp(b*x+a))/b^2-2*x^3*arctanh(exp(b*x+a))/b-3*x^2*polylog(2, 
-exp(b*x+a))/b^2+6*I*x*polylog(2,-I*exp(b*x+a))/b^3-6*I*x*polylog(2,I*exp( 
b*x+a))/b^3+3*x^2*polylog(2,exp(b*x+a))/b^2+6*x*polylog(3,-exp(b*x+a))/b^3 
-6*I*polylog(3,-I*exp(b*x+a))/b^4+6*I*polylog(3,I*exp(b*x+a))/b^4-6*x*poly 
log(3,exp(b*x+a))/b^3-6*polylog(4,-exp(b*x+a))/b^4+6*polylog(4,exp(b*x+a)) 
/b^4+x^3*sech(b*x+a)/b
 

3.5.74.2 Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.16 \[ \int x^3 \text {csch}(a+b x) \text {sech}^2(a+b x) \, dx=\frac {b^3 x^3 \log \left (1-e^{a+b x}\right )-3 i b^2 x^2 \log \left (1-i e^{a+b x}\right )+3 i b^2 x^2 \log \left (1+i e^{a+b x}\right )-b^3 x^3 \log \left (1+e^{a+b x}\right )-3 b^2 x^2 \operatorname {PolyLog}\left (2,-e^{a+b x}\right )+6 i b x \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )-6 i b x \operatorname {PolyLog}\left (2,i e^{a+b x}\right )+3 b^2 x^2 \operatorname {PolyLog}\left (2,e^{a+b x}\right )+6 b x \operatorname {PolyLog}\left (3,-e^{a+b x}\right )-6 i \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )+6 i \operatorname {PolyLog}\left (3,i e^{a+b x}\right )-6 b x \operatorname {PolyLog}\left (3,e^{a+b x}\right )-6 \operatorname {PolyLog}\left (4,-e^{a+b x}\right )+6 \operatorname {PolyLog}\left (4,e^{a+b x}\right )+b^3 x^3 \text {sech}(a+b x)}{b^4} \]

input

Integrate[x^3*Csch[a + b*x]*Sech[a + b*x]^2,x]
 

output

(b^3*x^3*Log[1 - E^(a + b*x)] - (3*I)*b^2*x^2*Log[1 - I*E^(a + b*x)] + (3* 
I)*b^2*x^2*Log[1 + I*E^(a + b*x)] - b^3*x^3*Log[1 + E^(a + b*x)] - 3*b^2*x 
^2*PolyLog[2, -E^(a + b*x)] + (6*I)*b*x*PolyLog[2, (-I)*E^(a + b*x)] - (6* 
I)*b*x*PolyLog[2, I*E^(a + b*x)] + 3*b^2*x^2*PolyLog[2, E^(a + b*x)] + 6*b 
*x*PolyLog[3, -E^(a + b*x)] - (6*I)*PolyLog[3, (-I)*E^(a + b*x)] + (6*I)*P 
olyLog[3, I*E^(a + b*x)] - 6*b*x*PolyLog[3, E^(a + b*x)] - 6*PolyLog[4, -E 
^(a + b*x)] + 6*PolyLog[4, E^(a + b*x)] + b^3*x^3*Sech[a + b*x])/b^4
 

3.5.74.3 Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5985, 25, 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.

\(\displaystyle \int x^3 \text {csch}(a+b x) \text {sech}^2(a+b x) \, dx\)

\(\Big \downarrow \) 5985

\(\displaystyle -3 \int -x^2 \left (\frac {\text {arctanh}(\cosh (a+b x))}{b}-\frac {\text {sech}(a+b x)}{b}\right )dx-\frac {x^3 \text {arctanh}(\cosh (a+b x))}{b}+\frac {x^3 \text {sech}(a+b x)}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle 3 \int x^2 \left (\frac {\text {arctanh}(\cosh (a+b x))}{b}-\frac {\text {sech}(a+b x)}{b}\right )dx-\frac {x^3 \text {arctanh}(\cosh (a+b x))}{b}+\frac {x^3 \text {sech}(a+b x)}{b}\)

\(\Big \downarrow \) 2010

\(\displaystyle 3 \int \left (\frac {x^2 \text {arctanh}(\cosh (a+b x))}{b}-\frac {x^2 \text {sech}(a+b x)}{b}\right )dx-\frac {x^3 \text {arctanh}(\cosh (a+b x))}{b}+\frac {x^3 \text {sech}(a+b x)}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle 3 \left (-\frac {2 x^2 \arctan \left (e^{a+b x}\right )}{b^2}-\frac {2 x^3 \text {arctanh}\left (e^{a+b x}\right )}{3 b}+\frac {x^3 \text {arctanh}(\cosh (a+b x))}{3 b}-\frac {2 i \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}+\frac {2 i \operatorname {PolyLog}\left (3,i e^{a+b x}\right )}{b^4}-\frac {2 \operatorname {PolyLog}\left (4,-e^{a+b x}\right )}{b^4}+\frac {2 \operatorname {PolyLog}\left (4,e^{a+b x}\right )}{b^4}+\frac {2 i x \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac {2 i x \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac {2 x \operatorname {PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac {2 x \operatorname {PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac {x^2 \operatorname {PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac {x^2 \operatorname {PolyLog}\left (2,e^{a+b x}\right )}{b^2}\right )-\frac {x^3 \text {arctanh}(\cosh (a+b x))}{b}+\frac {x^3 \text {sech}(a+b x)}{b}\)

input

Int[x^3*Csch[a + b*x]*Sech[a + b*x]^2,x]
 

output

-((x^3*ArcTanh[Cosh[a + b*x]])/b) + 3*((-2*x^2*ArcTan[E^(a + b*x)])/b^2 - 
(2*x^3*ArcTanh[E^(a + b*x)])/(3*b) + (x^3*ArcTanh[Cosh[a + b*x]])/(3*b) - 
(x^2*PolyLog[2, -E^(a + b*x)])/b^2 + ((2*I)*x*PolyLog[2, (-I)*E^(a + b*x)] 
)/b^3 - ((2*I)*x*PolyLog[2, I*E^(a + b*x)])/b^3 + (x^2*PolyLog[2, E^(a + b 
*x)])/b^2 + (2*x*PolyLog[3, -E^(a + b*x)])/b^3 - ((2*I)*PolyLog[3, (-I)*E^ 
(a + b*x)])/b^4 + ((2*I)*PolyLog[3, I*E^(a + b*x)])/b^4 - (2*x*PolyLog[3, 
E^(a + b*x)])/b^3 - (2*PolyLog[4, -E^(a + b*x)])/b^4 + (2*PolyLog[4, E^(a 
+ b*x)])/b^4) + (x^3*Sech[a + b*x])/b
 

3.5.74.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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]]
 

rule 5985

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + 
(b_.)*(x_)]^(p_.), x_Symbol] :> With[{u = IntHide[Csch[a + b*x]^n*Sech[a + 
b*x]^p, x]}, Simp[(c + d*x)^m   u, x] - Simp[d*m   Int[(c + d*x)^(m - 1)*u, 
 x], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n 
, p]
 

3.5.74.4 Maple [F]

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

input

int(x^3*csch(b*x+a)*sech(b*x+a)^2,x)
 

output

int(x^3*csch(b*x+a)*sech(b*x+a)^2,x)
 

3.5.74.5 Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1280 vs. \(2 (194) = 388\).

Time = 0.30 (sec) , antiderivative size = 1280, normalized size of antiderivative = 5.66 \[ \int x^3 \text {csch}(a+b x) \text {sech}^2(a+b x) \, dx=\text {Too large to display} \]

input

integrate(x^3*csch(b*x+a)*sech(b*x+a)^2,x, algorithm="fricas")
 

output

(2*b^3*x^3*cosh(b*x + a) + 2*b^3*x^3*sinh(b*x + a) + 3*(b^2*x^2*cosh(b*x + 
 a)^2 + 2*b^2*x^2*cosh(b*x + a)*sinh(b*x + a) + b^2*x^2*sinh(b*x + a)^2 + 
b^2*x^2)*dilog(cosh(b*x + a) + sinh(b*x + a)) - 6*(I*b*x*cosh(b*x + a)^2 + 
 2*I*b*x*cosh(b*x + a)*sinh(b*x + a) + I*b*x*sinh(b*x + a)^2 + I*b*x)*dilo 
g(I*cosh(b*x + a) + I*sinh(b*x + a)) - 6*(-I*b*x*cosh(b*x + a)^2 - 2*I*b*x 
*cosh(b*x + a)*sinh(b*x + a) - I*b*x*sinh(b*x + a)^2 - I*b*x)*dilog(-I*cos 
h(b*x + a) - I*sinh(b*x + a)) - 3*(b^2*x^2*cosh(b*x + a)^2 + 2*b^2*x^2*cos 
h(b*x + a)*sinh(b*x + a) + b^2*x^2*sinh(b*x + a)^2 + b^2*x^2)*dilog(-cosh( 
b*x + a) - sinh(b*x + a)) - (b^3*x^3*cosh(b*x + a)^2 + 2*b^3*x^3*cosh(b*x 
+ a)*sinh(b*x + a) + b^3*x^3*sinh(b*x + a)^2 + b^3*x^3)*log(cosh(b*x + a) 
+ sinh(b*x + a) + 1) - 3*(I*a^2*cosh(b*x + a)^2 + 2*I*a^2*cosh(b*x + a)*si 
nh(b*x + a) + I*a^2*sinh(b*x + a)^2 + I*a^2)*log(cosh(b*x + a) + sinh(b*x 
+ a) + I) - 3*(-I*a^2*cosh(b*x + a)^2 - 2*I*a^2*cosh(b*x + a)*sinh(b*x + a 
) - I*a^2*sinh(b*x + a)^2 - I*a^2)*log(cosh(b*x + a) + sinh(b*x + a) - I) 
- (a^3*cosh(b*x + a)^2 + 2*a^3*cosh(b*x + a)*sinh(b*x + a) + a^3*sinh(b*x 
+ a)^2 + a^3)*log(cosh(b*x + a) + sinh(b*x + a) - 1) - 3*(-I*b^2*x^2 + (-I 
*b^2*x^2 + I*a^2)*cosh(b*x + a)^2 + 2*(-I*b^2*x^2 + I*a^2)*cosh(b*x + a)*s 
inh(b*x + a) + (-I*b^2*x^2 + I*a^2)*sinh(b*x + a)^2 + I*a^2)*log(I*cosh(b* 
x + a) + I*sinh(b*x + a) + 1) - 3*(I*b^2*x^2 + (I*b^2*x^2 - I*a^2)*cosh(b* 
x + a)^2 + 2*(I*b^2*x^2 - I*a^2)*cosh(b*x + a)*sinh(b*x + a) + (I*b^2*x...
 

3.5.74.6 Sympy [F]

\[ \int x^3 \text {csch}(a+b x) \text {sech}^2(a+b x) \, dx=\int x^{3} \operatorname {csch}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]

input

integrate(x**3*csch(b*x+a)*sech(b*x+a)**2,x)
 

output

Integral(x**3*csch(a + b*x)*sech(a + b*x)**2, x)
 

3.5.74.7 Maxima [F]

\[ \int x^3 \text {csch}(a+b x) \text {sech}^2(a+b x) \, dx=\int { x^{3} \operatorname {csch}\left (b x + a\right ) \operatorname {sech}\left (b x + a\right )^{2} \,d x } \]

input

integrate(x^3*csch(b*x+a)*sech(b*x+a)^2,x, algorithm="maxima")
 

output

2*x^3*e^(b*x + a)/(b*e^(2*b*x + 2*a) + b) - (b^3*x^3*log(e^(b*x + a) + 1) 
+ 3*b^2*x^2*dilog(-e^(b*x + a)) - 6*b*x*polylog(3, -e^(b*x + a)) + 6*polyl 
og(4, -e^(b*x + a)))/b^4 + (b^3*x^3*log(-e^(b*x + a) + 1) + 3*b^2*x^2*dilo 
g(e^(b*x + a)) - 6*b*x*polylog(3, e^(b*x + a)) + 6*polylog(4, e^(b*x + a)) 
)/b^4 - 24*integrate(1/4*x^2*e^(b*x + a)/(b*e^(2*b*x + 2*a) + b), x)
 

3.5.74.8 Giac [F(-1)]

Timed out. \[ \int x^3 \text {csch}(a+b x) \text {sech}^2(a+b x) \, dx=\text {Timed out} \]

input

integrate(x^3*csch(b*x+a)*sech(b*x+a)^2,x, algorithm="giac")
 

output

Timed out
 

3.5.74.9 Mupad [F(-1)]

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

input

int(x^3/(cosh(a + b*x)^2*sinh(a + b*x)),x)
 

output

int(x^3/(cosh(a + b*x)^2*sinh(a + b*x)), x)