Integrand size = 20, antiderivative size = 317 \[ \int x^3 \text {csch}^3(a+b x) \text {sech}^2(a+b x) \, dx=\frac {6 x^2 \arctan \left (e^{a+b x}\right )}{b^2}-\frac {6 x \text {arctanh}\left (e^{a+b x}\right )}{b^3}+\frac {3 x^3 \text {arctanh}\left (e^{a+b x}\right )}{b}-\frac {3 x^2 \text {csch}(a+b x)}{2 b^2}-\frac {3 \operatorname {PolyLog}\left (2,-e^{a+b x}\right )}{b^4}+\frac {9 x^2 \operatorname {PolyLog}\left (2,-e^{a+b x}\right )}{2 b^2}-\frac {6 i x \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}+\frac {6 i x \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac {3 \operatorname {PolyLog}\left (2,e^{a+b x}\right )}{b^4}-\frac {9 x^2 \operatorname {PolyLog}\left (2,e^{a+b x}\right )}{2 b^2}-\frac {9 x \operatorname {PolyLog}\left (3,-e^{a+b x}\right )}{b^3}+\frac {6 i \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}-\frac {6 i \operatorname {PolyLog}\left (3,i e^{a+b x}\right )}{b^4}+\frac {9 x \operatorname {PolyLog}\left (3,e^{a+b x}\right )}{b^3}+\frac {9 \operatorname {PolyLog}\left (4,-e^{a+b x}\right )}{b^4}-\frac {9 \operatorname {PolyLog}\left (4,e^{a+b x}\right )}{b^4}-\frac {3 x^3 \text {sech}(a+b x)}{2 b}-\frac {x^3 \text {csch}^2(a+b x) \text {sech}(a+b x)}{2 b} \]
6*x^2*arctan(exp(b*x+a))/b^2-6*x*arctanh(exp(b*x+a))/b^3+3*x^3*arctanh(exp (b*x+a))/b-3/2*x^2*csch(b*x+a)/b^2-3*polylog(2,-exp(b*x+a))/b^4+9/2*x^2*po lylog(2,-exp(b*x+a))/b^2+6*I*polylog(3,-I*exp(b*x+a))/b^4-6*I*x*polylog(2, -I*exp(b*x+a))/b^3+3*polylog(2,exp(b*x+a))/b^4-9/2*x^2*polylog(2,exp(b*x+a ))/b^2-9*x*polylog(3,-exp(b*x+a))/b^3+6*I*x*polylog(2,I*exp(b*x+a))/b^3-6* I*polylog(3,I*exp(b*x+a))/b^4+9*x*polylog(3,exp(b*x+a))/b^3+9*polylog(4,-e xp(b*x+a))/b^4-9*polylog(4,exp(b*x+a))/b^4-3/2*x^3*sech(b*x+a)/b-1/2*x^3*c sch(b*x+a)^2*sech(b*x+a)/b
Time = 6.44 (sec) , antiderivative size = 466, normalized size of antiderivative = 1.47 \[ \int x^3 \text {csch}^3(a+b x) \text {sech}^2(a+b x) \, dx=-\frac {3 x^2 \text {csch}(a)}{2 b^2}-\frac {x^3 \text {csch}^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b}+\frac {3 i \left (b^2 x^2 \log \left (1-i e^{a+b x}\right )-b^2 x^2 \log \left (1+i e^{a+b x}\right )-2 b x \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )+2 b x \operatorname {PolyLog}\left (2,i e^{a+b x}\right )+2 \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )-2 \operatorname {PolyLog}\left (3,i e^{a+b x}\right )\right )}{b^4}-\frac {3 \left (-\frac {x \log \left (1-e^{a+b x}\right )}{b}+\frac {1}{2} b x^3 \log \left (1-e^{a+b x}\right )+\frac {x \log \left (1+e^{a+b x}\right )}{b}-\frac {1}{2} b x^3 \log \left (1+e^{a+b x}\right )+\frac {\operatorname {PolyLog}\left (2,-e^{a+b x}\right )}{b^2}-\frac {3}{2} x^2 \operatorname {PolyLog}\left (2,-e^{a+b x}\right )-\frac {\operatorname {PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac {3}{2} x^2 \operatorname {PolyLog}\left (2,e^{a+b x}\right )+\frac {3 x \operatorname {PolyLog}\left (3,-e^{a+b x}\right )}{b}-\frac {3 x \operatorname {PolyLog}\left (3,e^{a+b x}\right )}{b}-\frac {3 \operatorname {PolyLog}\left (4,-e^{a+b x}\right )}{b^2}+\frac {3 \operatorname {PolyLog}\left (4,e^{a+b x}\right )}{b^2}\right )}{b^2}-\frac {x^3 \text {sech}^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b}-\frac {x^3 \text {sech}(a+b x)}{b}+\frac {3 x^2 \text {csch}\left (\frac {a}{2}\right ) \text {csch}\left (\frac {a}{2}+\frac {b x}{2}\right ) \sinh \left (\frac {b x}{2}\right )}{4 b^2}+\frac {3 x^2 \text {sech}\left (\frac {a}{2}\right ) \text {sech}\left (\frac {a}{2}+\frac {b x}{2}\right ) \sinh \left (\frac {b x}{2}\right )}{4 b^2} \]
(-3*x^2*Csch[a])/(2*b^2) - (x^3*Csch[a/2 + (b*x)/2]^2)/(8*b) + ((3*I)*(b^2 *x^2*Log[1 - I*E^(a + b*x)] - b^2*x^2*Log[1 + I*E^(a + b*x)] - 2*b*x*PolyL og[2, (-I)*E^(a + b*x)] + 2*b*x*PolyLog[2, I*E^(a + b*x)] + 2*PolyLog[3, ( -I)*E^(a + b*x)] - 2*PolyLog[3, I*E^(a + b*x)]))/b^4 - (3*(-((x*Log[1 - E^ (a + b*x)])/b) + (b*x^3*Log[1 - E^(a + b*x)])/2 + (x*Log[1 + E^(a + b*x)]) /b - (b*x^3*Log[1 + E^(a + b*x)])/2 + PolyLog[2, -E^(a + b*x)]/b^2 - (3*x^ 2*PolyLog[2, -E^(a + b*x)])/2 - PolyLog[2, E^(a + b*x)]/b^2 + (3*x^2*PolyL og[2, E^(a + b*x)])/2 + (3*x*PolyLog[3, -E^(a + b*x)])/b - (3*x*PolyLog[3, E^(a + b*x)])/b - (3*PolyLog[4, -E^(a + b*x)])/b^2 + (3*PolyLog[4, E^(a + b*x)])/b^2))/b^2 - (x^3*Sech[a/2 + (b*x)/2]^2)/(8*b) - (x^3*Sech[a + b*x] )/b + (3*x^2*Csch[a/2]*Csch[a/2 + (b*x)/2]*Sinh[(b*x)/2])/(4*b^2) + (3*x^2 *Sech[a/2]*Sech[a/2 + (b*x)/2]*Sinh[(b*x)/2])/(4*b^2)
Time = 1.33 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5985, 27, 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.
\(\displaystyle \int x^3 \text {csch}^3(a+b x) \text {sech}^2(a+b x) \, dx\) |
\(\Big \downarrow \) 5985 |
\(\displaystyle -3 \int \frac {1}{2} x^2 \left (-\frac {\text {sech}(a+b x) \text {csch}^2(a+b x)}{b}+\frac {3 \text {arctanh}(\cosh (a+b x))}{b}-\frac {3 \text {sech}(a+b x)}{b}\right )dx+\frac {3 x^3 \text {arctanh}(\cosh (a+b x))}{2 b}-\frac {3 x^3 \text {sech}(a+b x)}{2 b}-\frac {x^3 \text {csch}^2(a+b x) \text {sech}(a+b x)}{2 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3}{2} \int x^2 \left (-\frac {\text {sech}(a+b x) \text {csch}^2(a+b x)}{b}+\frac {3 \text {arctanh}(\cosh (a+b x))}{b}-\frac {3 \text {sech}(a+b x)}{b}\right )dx+\frac {3 x^3 \text {arctanh}(\cosh (a+b x))}{2 b}-\frac {3 x^3 \text {sech}(a+b x)}{2 b}-\frac {x^3 \text {csch}^2(a+b x) \text {sech}(a+b x)}{2 b}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {3}{2} \int \left (\frac {3 x^2 \text {arctanh}(\cosh (a+b x))}{b}-\frac {x^2 \left (\text {csch}^2(a+b x)+3\right ) \text {sech}(a+b x)}{b}\right )dx+\frac {3 x^3 \text {arctanh}(\cosh (a+b x))}{2 b}-\frac {3 x^3 \text {sech}(a+b x)}{2 b}-\frac {x^3 \text {csch}^2(a+b x) \text {sech}(a+b x)}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3}{2} \left (-\frac {4 x^2 \arctan \left (e^{a+b x}\right )}{b^2}+\frac {4 x \text {arctanh}\left (e^{a+b x}\right )}{b^3}-\frac {2 x^3 \text {arctanh}\left (e^{a+b x}\right )}{b}+\frac {x^3 \text {arctanh}(\cosh (a+b x))}{b}+\frac {2 \operatorname {PolyLog}\left (2,-e^{a+b x}\right )}{b^4}-\frac {2 \operatorname {PolyLog}\left (2,e^{a+b x}\right )}{b^4}-\frac {4 i \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}+\frac {4 i \operatorname {PolyLog}\left (3,i e^{a+b x}\right )}{b^4}-\frac {6 \operatorname {PolyLog}\left (4,-e^{a+b x}\right )}{b^4}+\frac {6 \operatorname {PolyLog}\left (4,e^{a+b x}\right )}{b^4}+\frac {4 i x \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac {4 i x \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac {6 x \operatorname {PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac {6 x \operatorname {PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac {3 x^2 \operatorname {PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac {3 x^2 \operatorname {PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac {x^2 \text {csch}(a+b x)}{b^2}\right )+\frac {3 x^3 \text {arctanh}(\cosh (a+b x))}{2 b}-\frac {3 x^3 \text {sech}(a+b x)}{2 b}-\frac {x^3 \text {csch}^2(a+b x) \text {sech}(a+b x)}{2 b}\) |
(3*x^3*ArcTanh[Cosh[a + b*x]])/(2*b) - (3*((-4*x^2*ArcTan[E^(a + b*x)])/b^ 2 + (4*x*ArcTanh[E^(a + b*x)])/b^3 - (2*x^3*ArcTanh[E^(a + b*x)])/b + (x^3 *ArcTanh[Cosh[a + b*x]])/b + (x^2*Csch[a + b*x])/b^2 + (2*PolyLog[2, -E^(a + b*x)])/b^4 - (3*x^2*PolyLog[2, -E^(a + b*x)])/b^2 + ((4*I)*x*PolyLog[2, (-I)*E^(a + b*x)])/b^3 - ((4*I)*x*PolyLog[2, I*E^(a + b*x)])/b^3 - (2*Pol yLog[2, E^(a + b*x)])/b^4 + (3*x^2*PolyLog[2, E^(a + b*x)])/b^2 + (6*x*Pol yLog[3, -E^(a + b*x)])/b^3 - ((4*I)*PolyLog[3, (-I)*E^(a + b*x)])/b^4 + (( 4*I)*PolyLog[3, I*E^(a + b*x)])/b^4 - (6*x*PolyLog[3, E^(a + b*x)])/b^3 - (6*PolyLog[4, -E^(a + b*x)])/b^4 + (6*PolyLog[4, E^(a + b*x)])/b^4))/2 - ( 3*x^3*Sech[a + b*x])/(2*b) - (x^3*Csch[a + b*x]^2*Sech[a + b*x])/(2*b)
3.6.15.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]]
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]
\[\int x^{3} \operatorname {csch}\left (b x +a \right )^{3} \operatorname {sech}\left (b x +a \right )^{2}d x\]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 5356 vs. \(2 (270) = 540\).
Time = 0.33 (sec) , antiderivative size = 5356, normalized size of antiderivative = 16.90 \[ \int x^3 \text {csch}^3(a+b x) \text {sech}^2(a+b x) \, dx=\text {Too large to display} \]
\[ \int x^3 \text {csch}^3(a+b x) \text {sech}^2(a+b x) \, dx=\int x^{3} \operatorname {csch}^{3}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]
\[ \int x^3 \text {csch}^3(a+b x) \text {sech}^2(a+b x) \, dx=\int { x^{3} \operatorname {csch}\left (b x + a\right )^{3} \operatorname {sech}\left (b x + a\right )^{2} \,d x } \]
(2*b*x^3*e^(3*b*x + 3*a) - 3*(b*x^3*e^(5*a) + x^2*e^(5*a))*e^(5*b*x) - 3*( b*x^3*e^a - x^2*e^a)*e^(b*x))/(b^2*e^(6*b*x + 6*a) - b^2*e^(4*b*x + 4*a) - b^2*e^(2*b*x + 2*a) + b^2) + 3/2*(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 - 3/2*(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 - 3*(b*x*log(e^(b*x + a) + 1) + dilog(-e^(b*x + a)))/b^4 + 3*(b*x*log(-e^( b*x + a) + 1) + dilog(e^(b*x + a)))/b^4 + 96*integrate(1/16*x^2*e^(b*x + a )/(b*e^(2*b*x + 2*a) + b), x)
\[ \int x^3 \text {csch}^3(a+b x) \text {sech}^2(a+b x) \, dx=\int { x^{3} \operatorname {csch}\left (b x + a\right )^{3} \operatorname {sech}\left (b x + a\right )^{2} \,d x } \]
Timed out. \[ \int x^3 \text {csch}^3(a+b x) \text {sech}^2(a+b x) \, dx=\int \frac {x^3}{{\mathrm {cosh}\left (a+b\,x\right )}^2\,{\mathrm {sinh}\left (a+b\,x\right )}^3} \,d x \]