\(\int x^3 \text {csch}(a+b x) \text {sech}(a+b x) \, dx\) [161]

Optimal result

Integrand size = 16, antiderivative size = 148 \[ \int x^3 \text {csch}(a+b x) \text {sech}(a+b x) \, dx=-\frac {2 x^3 \text {arctanh}\left (e^{2 a+2 b x}\right )}{b}-\frac {3 x^2 \operatorname {PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}+\frac {3 x^2 \operatorname {PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}+\frac {3 x \operatorname {PolyLog}\left (3,-e^{2 a+2 b x}\right )}{2 b^3}-\frac {3 x \operatorname {PolyLog}\left (3,e^{2 a+2 b x}\right )}{2 b^3}-\frac {3 \operatorname {PolyLog}\left (4,-e^{2 a+2 b x}\right )}{4 b^4}+\frac {3 \operatorname {PolyLog}\left (4,e^{2 a+2 b x}\right )}{4 b^4} \] Output:

-2*x^3*arctanh(exp(2*b*x+2*a))/b-3/2*x^2*polylog(2,-exp(2*b*x+2*a))/b^2+3/ 
2*x^2*polylog(2,exp(2*b*x+2*a))/b^2+3/2*x*polylog(3,-exp(2*b*x+2*a))/b^3-3 
/2*x*polylog(3,exp(2*b*x+2*a))/b^3-3/4*polylog(4,-exp(2*b*x+2*a))/b^4+3/4* 
polylog(4,exp(2*b*x+2*a))/b^4
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.01 \[ \int x^3 \text {csch}(a+b x) \text {sech}(a+b x) \, dx=\frac {4 b^3 x^3 \log \left (1-e^{2 (a+b x)}\right )-4 b^3 x^3 \log \left (1+e^{2 (a+b x)}\right )-6 b^2 x^2 \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (2,e^{2 (a+b x)}\right )+6 b x \operatorname {PolyLog}\left (3,-e^{2 (a+b x)}\right )-6 b x \operatorname {PolyLog}\left (3,e^{2 (a+b x)}\right )-3 \operatorname {PolyLog}\left (4,-e^{2 (a+b x)}\right )+3 \operatorname {PolyLog}\left (4,e^{2 (a+b x)}\right )}{4 b^4} \] Input:

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

Output:

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

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 1.11 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.24, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5984, 3042, 26, 4670, 3011, 7163, 2720, 7143}

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.

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

\(\Big \downarrow \) 5984

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

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int i x^3 \csc (2 i a+2 i b x)dx\)

\(\Big \downarrow \) 26

\(\displaystyle 2 i \int x^3 \csc (2 i a+2 i b x)dx\)

\(\Big \downarrow \) 4670

\(\displaystyle 2 i \left (\frac {3 i \int x^2 \log \left (1-e^{2 a+2 b x}\right )dx}{2 b}-\frac {3 i \int x^2 \log \left (1+e^{2 a+2 b x}\right )dx}{2 b}+\frac {i x^3 \text {arctanh}\left (e^{2 a+2 b x}\right )}{b}\right )\)

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

(2*I)*((I*x^3*ArcTanh[E^(2*a + 2*b*x)])/b - (((3*I)/2)*(-1/2*(x^2*PolyLog[ 
2, -E^(2*a + 2*b*x)])/b + ((x*PolyLog[3, -E^(2*a + 2*b*x)])/(2*b) - PolyLo 
g[4, -E^(2*a + 2*b*x)]/(4*b^2))/b))/b + (((3*I)/2)*(-1/2*(x^2*PolyLog[2, E 
^(2*a + 2*b*x)])/b + ((x*PolyLog[3, E^(2*a + 2*b*x)])/(2*b) - PolyLog[4, E 
^(2*a + 2*b*x)]/(4*b^2))/b))/b)
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4670

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x 
_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)]/(f*fz*I)), x] 
 + (-Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[1 - E^((-I)*e + f*fz*x 
)], x], x] + Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e 
+ f*fz*x)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
 

rule 5984

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + 
(b_.)*(x_)]^(n_.), x_Symbol] :> Simp[2^n   Int[(c + d*x)^m*Csch[2*a + 2*b*x 
]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

Maple [A] (verified)

Time = 1.46 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.63

method	result	size
risch	\(\frac {6 \operatorname {polylog}\left (4, -{\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {3 \operatorname {polylog}\left (4, -{\mathrm e}^{2 b x +2 a}\right )}{4 b^{4}}+\frac {6 \operatorname {polylog}\left (4, {\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {a^{3} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{4}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{3}}{b^{4}}-\frac {x^{3} \ln \left ({\mathrm e}^{2 b x +2 a}+1\right )}{b}-\frac {6 x \operatorname {polylog}\left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) x^{3}}{b}+\frac {3 x^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {6 x \operatorname {polylog}\left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{3}}{b}+\frac {3 x^{2} \operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {3 x^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{2}}+\frac {3 x \operatorname {polylog}\left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{3}}\)	\(241\)

Input:

int(x^3*csch(b*x+a)*sech(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 448, normalized size of antiderivative = 3.03 \[ \int x^3 \text {csch}(a+b x) \text {sech}(a+b x) \, dx=\frac {b^{3} x^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 3 \, b^{2} x^{2} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 3 \, b^{2} x^{2} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) - a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - 6 \, b x {\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 6 \, b x {\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 6 \, b x {\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) - 6 \, b x {\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - {\left (b^{3} x^{3} + a^{3}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) - {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) + {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 6 \, {\rm polylog}\left (4, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 6 \, {\rm polylog}\left (4, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 6 \, {\rm polylog}\left (4, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + 6 \, {\rm polylog}\left (4, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{b^{4}} \] Input:

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

Output:

(b^3*x^3*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 3*b^2*x^2*dilog(cosh(b*x 
 + a) + sinh(b*x + a)) - 3*b^2*x^2*dilog(I*cosh(b*x + a) + I*sinh(b*x + a) 
) - 3*b^2*x^2*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + 3*b^2*x^2*dilog( 
-cosh(b*x + a) - sinh(b*x + a)) + a^3*log(cosh(b*x + a) + sinh(b*x + a) + 
I) + a^3*log(cosh(b*x + a) + sinh(b*x + a) - I) - a^3*log(cosh(b*x + a) + 
sinh(b*x + a) - 1) - 6*b*x*polylog(3, cosh(b*x + a) + sinh(b*x + a)) + 6*b 
*x*polylog(3, I*cosh(b*x + a) + I*sinh(b*x + a)) + 6*b*x*polylog(3, -I*cos 
h(b*x + a) - I*sinh(b*x + a)) - 6*b*x*polylog(3, -cosh(b*x + a) - sinh(b*x 
 + a)) - (b^3*x^3 + a^3)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) - (b^3 
*x^3 + a^3)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) + (b^3*x^3 + a^3)* 
log(-cosh(b*x + a) - sinh(b*x + a) + 1) + 6*polylog(4, cosh(b*x + a) + sin 
h(b*x + a)) - 6*polylog(4, I*cosh(b*x + a) + I*sinh(b*x + a)) - 6*polylog( 
4, -I*cosh(b*x + a) - I*sinh(b*x + a)) + 6*polylog(4, -cosh(b*x + a) - sin 
h(b*x + a)))/b^4
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.37 \[ \int x^3 \text {csch}(a+b x) \text {sech}(a+b x) \, dx=-\frac {4 \, b^{3} x^{3} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}(-e^{\left (2 \, b x + 2 \, a\right )})}{3 \, b^{4}} + \frac {b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (b x + a\right )})}{b^{4}} + \frac {b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(e^{\left (b x + a\right )})}{b^{4}} \] Input:

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

Output:

-1/3*(4*b^3*x^3*log(e^(2*b*x + 2*a) + 1) + 6*b^2*x^2*dilog(-e^(2*b*x + 2*a 
)) - 6*b*x*polylog(3, -e^(2*b*x + 2*a)) + 3*polylog(4, -e^(2*b*x + 2*a)))/ 
b^4 + (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*polylog(4, -e^(b*x + a)))/b^4 + (b^3*x^3*lo 
g(-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*polylog(4, e^(b*x + a)))/b^4
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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