Integrand size = 20, antiderivative size = 296 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^4(d+b x) \, dx=\frac {8 e^{\frac {5 (a-d)}{3}+\frac {11}{3} (d+b x)}}{3 b \left (1-e^{2 (d+b x)}\right )^3}-\frac {22 e^{\frac {5 (a-d)}{3}+\frac {5}{3} (d+b x)}}{9 b \left (1-e^{2 (d+b x)}\right )^2}+\frac {55 e^{\frac {5 (a-d)}{3}+\frac {5}{3} (d+b x)}}{27 b \left (1-e^{2 (d+b x)}\right )}+\frac {55 e^{\frac {5 (a-d)}{3}} \arctan \left (\frac {1-2 e^{\frac {1}{3} (d+b x)}}{\sqrt {3}}\right )}{54 \sqrt {3} b}-\frac {55 e^{\frac {5 (a-d)}{3}} \arctan \left (\frac {1+2 e^{\frac {1}{3} (d+b x)}}{\sqrt {3}}\right )}{54 \sqrt {3} b}+\frac {55 e^{\frac {5 (a-d)}{3}} \text {arctanh}\left (e^{\frac {1}{3} (d+b x)}\right )}{81 b}+\frac {55 e^{\frac {5 (a-d)}{3}} \text {arctanh}\left (\frac {e^{\frac {1}{3} (d+b x)}}{1+e^{\frac {2}{3} (d+b x)}}\right )}{162 b} \] Output:
8/3*exp(5/3*a+2*d+11/3*b*x)/b/(1-exp(2*b*x+2*d))^3-22/9*exp(5/3*b*x+5/3*a) /b/(1-exp(2*b*x+2*d))^2+55/27*exp(5/3*b*x+5/3*a)/b/(1-exp(2*b*x+2*d))+55/1 62*3^(1/2)*exp(5/3*a-5/3*d)*arctan(1/3*(1-2*exp(1/3*b*x+1/3*d))*3^(1/2))/b -55/162*3^(1/2)*exp(5/3*a-5/3*d)*arctan(1/3*(1+2*exp(1/3*b*x+1/3*d))*3^(1/ 2))/b+55/81*exp(5/3*a-5/3*d)*arctanh(exp(1/3*b*x+1/3*d))/b+55/162*exp(5/3* a-5/3*d)*arctanh(exp(1/3*b*x+1/3*d)/(1+exp(2/3*b*x+2/3*d)))/b
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.52 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.53 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^4(d+b x) \, dx=\frac {e^{5 a/3} (\cosh (d)-\sinh (d)) \left (55 \text {RootSum}\left [-\cosh (d)+\sinh (d)+\cosh (d) \text {$\#$1}^6+\sinh (d) \text {$\#$1}^6\&,\frac {b x-3 \log \left (e^{\frac {b x}{3}}-\text {$\#$1}\right )}{\text {$\#$1}}\&\right ] (\cosh (d)-\sinh (d))-\frac {18 e^{\frac {5 b x}{3}} \left (28 e^{2 b x}+11 \left (-1+5 e^{4 b x}\right ) \cosh (2 d)+11 \left (1+5 e^{4 b x}\right ) \sinh (2 d)\right )}{\left (\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)\right )^3}\right )}{486 b} \] Input:
Integrate[E^((5*(a + b*x))/3)*Csch[d + b*x]^4,x]
Output:
(E^((5*a)/3)*(Cosh[d] - Sinh[d])*(55*RootSum[-Cosh[d] + Sinh[d] + Cosh[d]* #1^6 + Sinh[d]*#1^6 & , (b*x - 3*Log[E^((b*x)/3) - #1])/#1 & ]*(Cosh[d] - Sinh[d]) - (18*E^((5*b*x)/3)*(28*E^(2*b*x) + 11*(-1 + 5*E^(4*b*x))*Cosh[2* d] + 11*(1 + 5*E^(4*b*x))*Sinh[2*d]))/((-1 + E^(2*b*x))*Cosh[d] + (1 + E^( 2*b*x))*Sinh[d])^3))/(486*b)
Time = 0.44 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.75, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.650, Rules used = {2720, 27, 817, 817, 819, 825, 27, 219, 1142, 25, 1083, 217, 1103}
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.
| \(\displaystyle \int e^{\frac {5}{3} (a+b x)} \text {csch}^4(b x+d) \, dx\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {3 \int \frac {16 e^{\frac {5 a}{3}+\frac {16 b x}{3}}}{\left (1-e^{2 b x}\right )^4}de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {48 e^{5 a/3} \int \frac {e^{\frac {16 b x}{3}}}{\left (1-e^{2 b x}\right )^4}de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {e^{\frac {11 b x}{3}}}{18 \left (1-e^{2 b x}\right )^3}-\frac {11}{18} \int \frac {e^{\frac {10 b x}{3}}}{\left (1-e^{2 b x}\right )^3}de^{\frac {b x}{3}}\right )}{b}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {e^{\frac {11 b x}{3}}}{18 \left (1-e^{2 b x}\right )^3}-\frac {11}{18} \left (\frac {e^{\frac {5 b x}{3}}}{12 \left (1-e^{2 b x}\right )^2}-\frac {5}{12} \int \frac {e^{\frac {4 b x}{3}}}{\left (1-e^{2 b x}\right )^2}de^{\frac {b x}{3}}\right )\right )}{b}\) |
| \(\Big \downarrow \) 819 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {e^{\frac {11 b x}{3}}}{18 \left (1-e^{2 b x}\right )^3}-\frac {11}{18} \left (\frac {e^{\frac {5 b x}{3}}}{12 \left (1-e^{2 b x}\right )^2}-\frac {5}{12} \left (\frac {1}{6} \int \frac {e^{\frac {4 b x}{3}}}{1-e^{2 b x}}de^{\frac {b x}{3}}+\frac {e^{\frac {5 b x}{3}}}{6 \left (1-e^{2 b x}\right )}\right )\right )\right )}{b}\) |
| \(\Big \downarrow \) 825 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {e^{\frac {11 b x}{3}}}{18 \left (1-e^{2 b x}\right )^3}-\frac {11}{18} \left (\frac {e^{\frac {5 b x}{3}}}{12 \left (1-e^{2 b x}\right )^2}-\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{3} \int \frac {1}{1-e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}+\frac {1}{3} \int -\frac {1+e^{\frac {b x}{3}}}{2 \left (1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}\right )}de^{\frac {b x}{3}}+\frac {1}{3} \int -\frac {1-e^{\frac {b x}{3}}}{2 \left (1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}\right )}de^{\frac {b x}{3}}\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (1-e^{2 b x}\right )}\right )\right )\right )}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {e^{\frac {11 b x}{3}}}{18 \left (1-e^{2 b x}\right )^3}-\frac {11}{18} \left (\frac {e^{\frac {5 b x}{3}}}{12 \left (1-e^{2 b x}\right )^2}-\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{3} \int \frac {1}{1-e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1+e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1-e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (1-e^{2 b x}\right )}\right )\right )\right )}{b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {e^{\frac {11 b x}{3}}}{18 \left (1-e^{2 b x}\right )^3}-\frac {11}{18} \left (\frac {e^{\frac {5 b x}{3}}}{12 \left (1-e^{2 b x}\right )^2}-\frac {5}{12} \left (\frac {1}{6} \left (-\frac {1}{6} \int \frac {1+e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1-e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (1-e^{2 b x}\right )}\right )\right )\right )}{b}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {e^{\frac {11 b x}{3}}}{18 \left (1-e^{2 b x}\right )^3}-\frac {11}{18} \left (\frac {e^{\frac {5 b x}{3}}}{12 \left (1-e^{2 b x}\right )^2}-\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{2} \int -\frac {1-2 e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1+2 e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {3}{2} \int \frac {1}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (1-e^{2 b x}\right )}\right )\right )\right )}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {e^{\frac {11 b x}{3}}}{18 \left (1-e^{2 b x}\right )^3}-\frac {11}{18} \left (\frac {e^{\frac {5 b x}{3}}}{12 \left (1-e^{2 b x}\right )^2}-\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1-2 e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {3}{2} \int \frac {1}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1+2 e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {3}{2} \int \frac {1}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (1-e^{2 b x}\right )}\right )\right )\right )}{b}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {e^{\frac {11 b x}{3}}}{18 \left (1-e^{2 b x}\right )^3}-\frac {11}{18} \left (\frac {e^{\frac {5 b x}{3}}}{12 \left (1-e^{2 b x}\right )^2}-\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{6} \left (3 \int \frac {1}{-3-e^{\frac {2 b x}{3}}}d\left (-1+2 e^{\frac {b x}{3}}\right )+\frac {1}{2} \int \frac {1-2 e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (3 \int \frac {1}{-3-e^{\frac {2 b x}{3}}}d\left (1+2 e^{\frac {b x}{3}}\right )+\frac {1}{2} \int \frac {1+2 e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (1-e^{2 b x}\right )}\right )\right )\right )}{b}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {e^{\frac {11 b x}{3}}}{18 \left (1-e^{2 b x}\right )^3}-\frac {11}{18} \left (\frac {e^{\frac {5 b x}{3}}}{12 \left (1-e^{2 b x}\right )^2}-\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1-2 e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\sqrt {3} \arctan \left (\frac {2 e^{\frac {b x}{3}}-1}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1+2 e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\sqrt {3} \arctan \left (\frac {2 e^{\frac {b x}{3}}+1}{\sqrt {3}}\right )\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (1-e^{2 b x}\right )}\right )\right )\right )}{b}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {e^{\frac {11 b x}{3}}}{18 \left (1-e^{2 b x}\right )^3}-\frac {11}{18} \left (\frac {e^{\frac {5 b x}{3}}}{12 \left (1-e^{2 b x}\right )^2}-\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{6} \left (-\sqrt {3} \arctan \left (\frac {2 e^{\frac {b x}{3}}-1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}+1\right )\right )+\frac {1}{6} \left (\frac {1}{2} \log \left (e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}+1\right )-\sqrt {3} \arctan \left (\frac {2 e^{\frac {b x}{3}}+1}{\sqrt {3}}\right )\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (1-e^{2 b x}\right )}\right )\right )\right )}{b}\) |
Input:
Int[E^((5*(a + b*x))/3)*Csch[d + b*x]^4,x]
Output:
(48*E^((5*a)/3)*(E^((11*b*x)/3)/(18*(1 - E^(2*b*x))^3) - (11*(E^((5*b*x)/3 )/(12*(1 - E^(2*b*x))^2) - (5*(E^((5*b*x)/3)/(6*(1 - E^(2*b*x))) + (ArcTan h[E^((b*x)/3)]/3 + (-(Sqrt[3]*ArcTan[(-1 + 2*E^((b*x)/3))/Sqrt[3]]) - Log[ 1 - E^((b*x)/3) + E^((2*b*x)/3)]/2)/6 + (-(Sqrt[3]*ArcTan[(1 + 2*E^((b*x)/ 3))/Sqrt[3]]) + Log[1 + E^((b*x)/3) + E^((2*b*x)/3)]/2)/6)/6))/12))/18))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator [Rt[-a/b, n]], s = Denominator[Rt[-a/b, n]], k, u}, Simp[u = Int[(r*Cos[2*k *m*(Pi/n)] - s*Cos[2*k*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[2*k*(Pi/n)]*x + s^2*x^2), x] + Int[(r*Cos[2*k*m*(Pi/n)] + s*Cos[2*k*(m + 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[2*k*(Pi/n)]*x + s^2*x^2), x]; 2*(r^(m + 2)/(a*n*s^m)) Int[1/ (r^2 - s^2*x^2), x] + 2*(r^(m + 1)/(a*n*s^m)) Sum[u, {k, 1, (n - 2)/4}], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1 ] && NegQ[a/b]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
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]]]
Result contains complex when optimal does not.
Time = 0.99 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.24
| method | result | size |
| risch | \(-\frac {\left (55 \,{\mathrm e}^{4 b x +4 d}+28 \,{\mathrm e}^{2 b x +2 d}-11\right ) {\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}}}{27 \left ({\mathrm e}^{2 b x +2 d}-1\right )^{3} b}+\frac {55 \ln \left (1+{\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{162 b}-\frac {55 \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-1\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{162 b}+\frac {55 \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{324 b}+\frac {55 i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{324 b}+\frac {55 \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{324 b}-\frac {55 i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{324 b}-\frac {55 \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{324 b}+\frac {55 i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{324 b}-\frac {55 \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{324 b}-\frac {55 i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{324 b}\) | \(367\) |
Input:
int(exp(5/3*b*x+5/3*a)*csch(b*x+d)^4,x,method=_RETURNVERBOSE)
Output:
-1/27/(exp(2*b*x+2*d)-1)^3/b*(55*exp(4*b*x+4*d)+28*exp(2*b*x+2*d)-11)*exp( 5/3*b*x+5/3*a)+55/162/b*ln(1+exp(1/3*b*x+1/3*d))*exp(5/3*a-5/3*d)-55/162/b *ln(exp(1/3*b*x+1/3*d)-1)*exp(5/3*a-5/3*d)+55/324/b*ln(exp(1/3*b*x+1/3*d)+ 1/2-1/2*I*3^(1/2))*exp(5/3*a-5/3*d)+55/324*I/b*ln(exp(1/3*b*x+1/3*d)+1/2-1 /2*I*3^(1/2))*exp(5/3*a-5/3*d)*3^(1/2)+55/324/b*ln(exp(1/3*b*x+1/3*d)+1/2+ 1/2*I*3^(1/2))*exp(5/3*a-5/3*d)-55/324*I/b*ln(exp(1/3*b*x+1/3*d)+1/2+1/2*I *3^(1/2))*exp(5/3*a-5/3*d)*3^(1/2)-55/324/b*ln(exp(1/3*b*x+1/3*d)-1/2-1/2* I*3^(1/2))*exp(5/3*a-5/3*d)+55/324*I/b*ln(exp(1/3*b*x+1/3*d)-1/2-1/2*I*3^( 1/2))*exp(5/3*a-5/3*d)*3^(1/2)-55/324/b*ln(exp(1/3*b*x+1/3*d)-1/2+1/2*I*3^ (1/2))*exp(5/3*a-5/3*d)-55/324*I/b*ln(exp(1/3*b*x+1/3*d)-1/2+1/2*I*3^(1/2) )*exp(5/3*a-5/3*d)*3^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 13031 vs. \(2 (215) = 430\).
Time = 0.24 (sec) , antiderivative size = 13031, normalized size of antiderivative = 44.02 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^4(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(5/3*b*x+5/3*a)*csch(b*x+d)^4,x, algorithm="fricas")
Output:
Too large to include
\[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^4(d+b x) \, dx=e^{\frac {5 a}{3}} \int e^{\frac {5 b x}{3}} \operatorname {csch}^{4}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(5/3*b*x+5/3*a)*csch(b*x+d)**4,x)
Output:
exp(5*a/3)*Integral(exp(5*b*x/3)*csch(b*x + d)**4, x)
Time = 0.13 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.92 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^4(d+b x) \, dx=\frac {55 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + 1\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{162 \, b} + \frac {55 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} - 1\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{162 \, b} + \frac {55 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )}{324 \, b} + \frac {55 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + 1\right )}{162 \, b} - \frac {55 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} - 1\right )}{162 \, b} - \frac {55 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (-e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )}{324 \, b} + \frac {{\left (55 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + 28 \, e^{\left (-\frac {7}{3} \, b x - \frac {7}{3} \, d\right )} - 11 \, e^{\left (-\frac {13}{3} \, b x - \frac {13}{3} \, d\right )}\right )} e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{27 \, b {\left (3 \, e^{\left (-2 \, b x - 2 \, d\right )} - 3 \, e^{\left (-4 \, b x - 4 \, d\right )} + e^{\left (-6 \, b x - 6 \, d\right )} - 1\right )}} \] Input:
integrate(exp(5/3*b*x+5/3*a)*csch(b*x+d)^4,x, algorithm="maxima")
Output:
55/162*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(-1/3*b*x - 1/3*d) + 1))*e^(5/3*a - 5/3*d)/b + 55/162*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(-1/3*b*x - 1/3*d) - 1) )*e^(5/3*a - 5/3*d)/b + 55/324*e^(5/3*a - 5/3*d)*log(e^(-1/3*b*x - 1/3*d) + e^(-2/3*b*x - 2/3*d) + 1)/b + 55/162*e^(5/3*a - 5/3*d)*log(e^(-1/3*b*x - 1/3*d) + 1)/b - 55/162*e^(5/3*a - 5/3*d)*log(e^(-1/3*b*x - 1/3*d) - 1)/b - 55/324*e^(5/3*a - 5/3*d)*log(-e^(-1/3*b*x - 1/3*d) + e^(-2/3*b*x - 2/3*d ) + 1)/b + 1/27*(55*e^(-1/3*b*x - 1/3*d) + 28*e^(-7/3*b*x - 7/3*d) - 11*e^ (-13/3*b*x - 13/3*d))*e^(5/3*a - 5/3*d)/(b*(3*e^(-2*b*x - 2*d) - 3*e^(-4*b *x - 4*d) + e^(-6*b*x - 6*d) - 1))
Time = 0.12 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.74 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^4(d+b x) \, dx=-\frac {{\left (110 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (\frac {1}{3} \, b x\right )} + e^{\left (-\frac {1}{3} \, d\right )}\right )} e^{\left (\frac {1}{3} \, d\right )}\right ) e^{\left (-\frac {17}{3} \, d\right )} + 110 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (\frac {1}{3} \, b x\right )} - e^{\left (-\frac {1}{3} \, d\right )}\right )} e^{\left (\frac {1}{3} \, d\right )}\right ) e^{\left (-\frac {17}{3} \, d\right )} - 55 \, e^{\left (-\frac {17}{3} \, d\right )} \log \left (e^{\left (\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) + 55 \, e^{\left (-\frac {17}{3} \, d\right )} \log \left (-e^{\left (\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) - 110 \, e^{\left (-\frac {17}{3} \, d\right )} \log \left (e^{\left (\frac {1}{3} \, b x\right )} + e^{\left (-\frac {1}{3} \, d\right )}\right ) + 110 \, e^{\left (-\frac {17}{3} \, d\right )} \log \left ({\left | e^{\left (\frac {1}{3} \, b x\right )} - e^{\left (-\frac {1}{3} \, d\right )} \right |}\right ) - \frac {12 \, {\left (11 \, e^{\left (\frac {5}{3} \, b x\right )} - 55 \, e^{\left (\frac {17}{3} \, b x + 4 \, d\right )} - 28 \, e^{\left (\frac {11}{3} \, b x + 2 \, d\right )}\right )} e^{\left (-4 \, d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} - 1\right )}^{3}}\right )} e^{\left (\frac {5}{3} \, a + 4 \, d\right )}}{324 \, b} \] Input:
integrate(exp(5/3*b*x+5/3*a)*csch(b*x+d)^4,x, algorithm="giac")
Output:
-1/324*(110*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(1/3*b*x) + e^(-1/3*d))*e^(1/3 *d))*e^(-17/3*d) + 110*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(1/3*b*x) - e^(-1/3 *d))*e^(1/3*d))*e^(-17/3*d) - 55*e^(-17/3*d)*log(e^(1/3*b*x - 1/3*d) + e^( 2/3*b*x) + e^(-2/3*d)) + 55*e^(-17/3*d)*log(-e^(1/3*b*x - 1/3*d) + e^(2/3* b*x) + e^(-2/3*d)) - 110*e^(-17/3*d)*log(e^(1/3*b*x) + e^(-1/3*d)) + 110*e ^(-17/3*d)*log(abs(e^(1/3*b*x) - e^(-1/3*d))) - 12*(11*e^(5/3*b*x) - 55*e^ (17/3*b*x + 4*d) - 28*e^(11/3*b*x + 2*d))*e^(-4*d)/(e^(2*b*x + 2*d) - 1)^3 )*e^(5/3*a + 4*d)/b
Time = 6.22 (sec) , antiderivative size = 524, normalized size of antiderivative = 1.77 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^4(d+b x) \, dx=\text {Too large to display} \] Input:
int(exp((5*a)/3 + (5*b*x)/3)/sinh(d + b*x)^4,x)
Output:
(55*exp(10*a - 10*d)^(1/6)*log(- (3025*exp((10*a)/3)*exp(-(10*d)/3))/6561 - (3025*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*(exp(10*a)*exp(-1 0*d))^(1/6))/6561))/(162*b) - (8*exp((5*a)/3 + 2*d + (11*b*x)/3))/(3*b*(3* exp(2*d + 2*b*x) - 3*exp(4*d + 4*b*x) + exp(6*d + 6*b*x) - 1)) - (22*exp(( 5*a)/3 + (5*b*x)/3))/(9*b*(exp(4*d + 4*b*x) - 2*exp(2*d + 2*b*x) + 1)) - ( 55*exp((5*a)/3 + (5*b*x)/3))/(27*b*(exp(2*d + 2*b*x) - 1)) - (55*exp(10*a - 10*d)^(1/6)*log((3025*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*( exp(10*a)*exp(-10*d))^(1/6))/6561 - (3025*exp((10*a)/3)*exp(-(10*d)/3))/65 61))/(162*b) + (55*log(- (3025*exp((10*a)/3)*exp(-(10*d)/3))/6561 - (3025* exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*((3^(1/2)*1i)/2 - 1/2)*(e xp(10*a)*exp(-10*d))^(1/6))/6561)*exp(10*a - 10*d)^(1/6)*((3^(1/2)*1i)/2 - 1/2))/(162*b) - (55*log((3025*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b* x)/3)*((3^(1/2)*1i)/2 - 1/2)*(exp(10*a)*exp(-10*d))^(1/6))/6561 - (3025*ex p((10*a)/3)*exp(-(10*d)/3))/6561)*exp(10*a - 10*d)^(1/6)*((3^(1/2)*1i)/2 - 1/2))/(162*b) + (55*log(- (3025*exp((10*a)/3)*exp(-(10*d)/3))/6561 - (302 5*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*((3^(1/2)*1i)/2 + 1/2)* (exp(10*a)*exp(-10*d))^(1/6))/6561)*exp(10*a - 10*d)^(1/6)*((3^(1/2)*1i)/2 + 1/2))/(162*b) - (55*log((3025*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp(( b*x)/3)*((3^(1/2)*1i)/2 + 1/2)*(exp(10*a)*exp(-10*d))^(1/6))/6561 - (3025* exp((10*a)/3)*exp(-(10*d)/3))/6561)*exp(10*a - 10*d)^(1/6)*((3^(1/2)*1i...
\[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^4(d+b x) \, dx=\int e^{\frac {5 b x}{3}+\frac {5 a}{3}} \mathrm {csch}\left (b x +d \right )^{4}d x \] Input:
int(exp(5/3*b*x+5/3*a)*csch(b*x+d)^4,x)
Output:
int(e**((5*a + 5*b*x)/3)*csch(b*x + d)**4,x)