Integrand size = 20, antiderivative size = 232 \[ \int e^{\frac {5}{3} (a+b x)} \coth ^2(d+b x) \, dx=\frac {3 e^{\frac {5 (a-d)}{3}+\frac {5}{3} (d+b x)}}{5 b}+\frac {2 e^{\frac {5 (a-d)}{3}+\frac {5}{3} (d+b x)}}{b \left (1-e^{2 (d+b x)}\right )}-\frac {5 e^{\frac {5 (a-d)}{3}} \arctan \left (\frac {1-2 e^{\frac {1}{3} (d+b x)}}{\sqrt {3}}\right )}{\sqrt {3} b}+\frac {5 e^{\frac {5 (a-d)}{3}} \arctan \left (\frac {1+2 e^{\frac {1}{3} (d+b x)}}{\sqrt {3}}\right )}{\sqrt {3} b}-\frac {10 e^{\frac {5 (a-d)}{3}} \text {arctanh}\left (e^{\frac {1}{3} (d+b x)}\right )}{3 b}-\frac {5 e^{\frac {5 (a-d)}{3}} \text {arctanh}\left (\frac {e^{\frac {1}{3} (d+b x)}}{1+e^{\frac {2}{3} (d+b x)}}\right )}{3 b} \] Output:
3/5*exp(5/3*b*x+5/3*a)/b+2*exp(5/3*b*x+5/3*a)/b/(1-exp(2*b*x+2*d))-5/3*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+5/3*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-10/ 3*exp(5/3*a-5/3*d)*arctanh(exp(1/3*b*x+1/3*d))/b-5/3*exp(5/3*a-5/3*d)*arct anh(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 = 0.27 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.55 \[ \int e^{\frac {5}{3} (a+b x)} \coth ^2(d+b x) \, dx=\frac {e^{5 a/3} \left (27 e^{\frac {5 b x}{3}}-25 \text {RootSum}\left [-\cosh (d)+\sinh (d)+\cosh (d) \text {$\#$1}^6+\sinh (d) \text {$\#$1}^6\&,\frac {b x-3 \log \left (e^{\frac {b x}{3}}-\text {$\#$1}\right )}{\text {$\#$1}}\&\right ] (\cosh (d)-\sinh (d))^2-\frac {90 e^{\frac {5 b x}{3}} (\cosh (d)-\sinh (d))}{\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)}\right )}{45 b} \] Input:
Integrate[E^((5*(a + b*x))/3)*Coth[d + b*x]^2,x]
Output:
(E^((5*a)/3)*(27*E^((5*b*x)/3) - 25*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] - S inh[d])^2 - (90*E^((5*b*x)/3)*(Cosh[d] - Sinh[d]))/((-1 + E^(2*b*x))*Cosh[ d] + (1 + E^(2*b*x))*Sinh[d])))/(45*b)
Time = 0.37 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.78, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.650, Rules used = {2720, 27, 963, 27, 959, 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.
| \(\displaystyle \int e^{\frac {5}{3} (a+b x)} \coth ^2(b x+d) \, dx\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {3 \int \frac {e^{\frac {5 a}{3}+\frac {4 b x}{3}} \left (1+e^{2 b x}\right )^2}{\left (1-e^{2 b x}\right )^2}de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 e^{5 a/3} \int \frac {e^{\frac {4 b x}{3}} \left (1+e^{2 b x}\right )^2}{\left (1-e^{2 b x}\right )^2}de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 963 |
| \(\displaystyle \frac {3 e^{5 a/3} \left (\frac {2 e^{\frac {5 b x}{3}}}{3 \left (1-e^{2 b x}\right )}-\frac {1}{6} \int \frac {2 e^{\frac {4 b x}{3}} \left (7+3 e^{2 b x}\right )}{1-e^{2 b x}}de^{\frac {b x}{3}}\right )}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 e^{5 a/3} \left (\frac {2 e^{\frac {5 b x}{3}}}{3 \left (1-e^{2 b x}\right )}-\frac {1}{3} \int \frac {e^{\frac {4 b x}{3}} \left (7+3 e^{2 b x}\right )}{1-e^{2 b x}}de^{\frac {b x}{3}}\right )}{b}\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle \frac {3 e^{5 a/3} \left (\frac {1}{3} \left (\frac {3}{5} e^{\frac {5 b x}{3}}-10 \int \frac {e^{\frac {4 b x}{3}}}{1-e^{2 b x}}de^{\frac {b x}{3}}\right )+\frac {2 e^{\frac {5 b x}{3}}}{3 \left (1-e^{2 b x}\right )}\right )}{b}\) |
| \(\Big \downarrow \) 825 |
| \(\displaystyle \frac {3 e^{5 a/3} \left (\frac {1}{3} \left (\frac {3}{5} e^{\frac {5 b x}{3}}-10 \left (\frac {1}{3} \int \frac {1}{1-e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}+\frac {1}{3} \int -\frac {1+e^{\frac {b x}{3}}}{2 \left (1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}\right )}de^{\frac {b x}{3}}+\frac {1}{3} \int -\frac {1-e^{\frac {b x}{3}}}{2 \left (1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}\right )}de^{\frac {b x}{3}}\right )\right )+\frac {2 e^{\frac {5 b x}{3}}}{3 \left (1-e^{2 b x}\right )}\right )}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 e^{5 a/3} \left (\frac {1}{3} \left (\frac {3}{5} e^{\frac {5 b x}{3}}-10 \left (\frac {1}{3} \int \frac {1}{1-e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1+e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1-e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )\right )+\frac {2 e^{\frac {5 b x}{3}}}{3 \left (1-e^{2 b x}\right )}\right )}{b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3 e^{5 a/3} \left (\frac {1}{3} \left (\frac {3}{5} e^{\frac {5 b x}{3}}-10 \left (-\frac {1}{6} \int \frac {1+e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1-e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )\right )+\frac {2 e^{\frac {5 b x}{3}}}{3 \left (1-e^{2 b x}\right )}\right )}{b}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {3 e^{5 a/3} \left (\frac {1}{3} \left (\frac {3}{5} e^{\frac {5 b x}{3}}-10 \left (\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{2} \int -\frac {1-2 e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1+2 e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {3}{2} \int \frac {1}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )\right )+\frac {2 e^{\frac {5 b x}{3}}}{3 \left (1-e^{2 b x}\right )}\right )}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {3 e^{5 a/3} \left (\frac {1}{3} \left (\frac {3}{5} e^{\frac {5 b x}{3}}-10 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1-2 e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {3}{2} \int \frac {1}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1+2 e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {3}{2} \int \frac {1}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )\right )+\frac {2 e^{\frac {5 b x}{3}}}{3 \left (1-e^{2 b x}\right )}\right )}{b}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {3 e^{5 a/3} \left (\frac {1}{3} \left (\frac {3}{5} e^{\frac {5 b x}{3}}-10 \left (\frac {1}{6} \left (3 \int \frac {1}{-3-e^{\frac {2 b x}{3}}}d\left (-1+2 e^{\frac {b x}{3}}\right )+\frac {1}{2} \int \frac {1-2 e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (3 \int \frac {1}{-3-e^{\frac {2 b x}{3}}}d\left (1+2 e^{\frac {b x}{3}}\right )+\frac {1}{2} \int \frac {1+2 e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )\right )+\frac {2 e^{\frac {5 b x}{3}}}{3 \left (1-e^{2 b x}\right )}\right )}{b}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {3 e^{5 a/3} \left (\frac {1}{3} \left (\frac {3}{5} e^{\frac {5 b x}{3}}-10 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1-2 e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\sqrt {3} \arctan \left (\frac {2 e^{\frac {b x}{3}}-1}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1+2 e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\sqrt {3} \arctan \left (\frac {2 e^{\frac {b x}{3}}+1}{\sqrt {3}}\right )\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )\right )+\frac {2 e^{\frac {5 b x}{3}}}{3 \left (1-e^{2 b x}\right )}\right )}{b}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {3 e^{5 a/3} \left (\frac {1}{3} \left (\frac {3}{5} e^{\frac {5 b x}{3}}-10 \left (\frac {1}{6} \left (-\sqrt {3} \arctan \left (\frac {2 e^{\frac {b x}{3}}-1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}+1\right )\right )+\frac {1}{6} \left (\frac {1}{2} \log \left (e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}+1\right )-\sqrt {3} \arctan \left (\frac {2 e^{\frac {b x}{3}}+1}{\sqrt {3}}\right )\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )\right )+\frac {2 e^{\frac {5 b x}{3}}}{3 \left (1-e^{2 b x}\right )}\right )}{b}\) |
Input:
Int[E^((5*(a + b*x))/3)*Coth[d + b*x]^2,x]
Output:
(3*E^((5*a)/3)*((2*E^((5*b*x)/3))/(3*(1 - E^(2*b*x))) + ((3*E^((5*b*x)/3)) /5 - 10*(ArcTanh[E^((b*x)/3)]/3 + (-(Sqrt[3]*ArcTan[(-1 + 2*E^((b*x)/3))/S qrt[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)) /3))/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[(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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^n)^(p + 1) /(a*b^2*e*n*(p + 1))), x] + Simp[1/(a*b^2*n*(p + 1)) Int[(e*x)^m*(a + b*x ^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]
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 = 0.31 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.54
| method | result | size |
| risch | \(\frac {3 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}}}{5 b}-\frac {2 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}}}{\left ({\mathrm e}^{2 b x +2 d}-1\right ) b}-\frac {5 \ln \left (1+{\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{3 b}+\frac {5 \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-1\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{3 b}+\frac {5 \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{6 b}+\frac {5 i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{6 b}+\frac {5 \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{6 b}-\frac {5 i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{6 b}-\frac {5 \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{6 b}+\frac {5 i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{6 b}-\frac {5 \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{6 b}-\frac {5 i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{6 b}\) | \(357\) |
Input:
int(exp(5/3*b*x+5/3*a)*coth(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
3/5*exp(5/3*b*x+5/3*a)/b-2/(exp(2*b*x+2*d)-1)/b*exp(5/3*b*x+5/3*a)-5/3/b*l n(1+exp(1/3*b*x+1/3*d))*exp(5/3*a-5/3*d)+5/3/b*ln(exp(1/3*b*x+1/3*d)-1)*ex p(5/3*a-5/3*d)+5/6/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)+5/6*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)+5/6/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)-5/6*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)-5/6/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)+5/6*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)-5/6/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)-5/6*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. 3400 vs. \(2 (172) = 344\).
Time = 0.11 (sec) , antiderivative size = 3400, normalized size of antiderivative = 14.66 \[ \int e^{\frac {5}{3} (a+b x)} \coth ^2(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(5/3*b*x+5/3*a)*coth(b*x+d)^2,x, algorithm="fricas")
Output:
Too large to include
\[ \int e^{\frac {5}{3} (a+b x)} \coth ^2(d+b x) \, dx=e^{\frac {5 a}{3}} \int e^{\frac {5 b x}{3}} \coth ^{2}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(5/3*b*x+5/3*a)*coth(b*x+d)**2,x)
Output:
exp(5*a/3)*Integral(exp(5*b*x/3)*coth(b*x + d)**2, x)
Time = 0.13 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.03 \[ \int e^{\frac {5}{3} (a+b x)} \coth ^2(d+b x) \, dx=-\frac {5 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + 1\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{3 \, b} - \frac {5 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} - 1\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{3 \, b} - \frac {5 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )}{6 \, b} - \frac {5 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + 1\right )}{3 \, b} + \frac {5 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} - 1\right )}{3 \, b} + \frac {5 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (-e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )}{6 \, b} - \frac {{\left (13 \, e^{\left (-2 \, b x - 2 \, d\right )} - 3\right )} e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{5 \, b {\left (e^{\left (-\frac {5}{3} \, b x - \frac {5}{3} \, d\right )} - e^{\left (-\frac {11}{3} \, b x - \frac {11}{3} \, d\right )}\right )}} \] Input:
integrate(exp(5/3*b*x+5/3*a)*coth(b*x+d)^2,x, algorithm="maxima")
Output:
-5/3*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 - 5/3*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 - 5/6*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 - 5/3*e^(5/3*a - 5/3*d)*log(e^(-1/3*b*x - 1/3*d) + 1 )/b + 5/3*e^(5/3*a - 5/3*d)*log(e^(-1/3*b*x - 1/3*d) - 1)/b + 5/6*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 - 1/5*(1 3*e^(-2*b*x - 2*d) - 3)*e^(5/3*a - 5/3*d)/(b*(e^(-5/3*b*x - 5/3*d) - e^(-1 1/3*b*x - 11/3*d)))
Time = 0.13 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.95 \[ \int e^{\frac {5}{3} (a+b x)} \coth ^2(d+b x) \, dx=\frac {50 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (\frac {1}{3} \, b x\right )} + e^{\left (-\frac {1}{3} \, d\right )}\right )} e^{\left (\frac {1}{3} \, d\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} + 50 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (\frac {1}{3} \, b x\right )} - e^{\left (-\frac {1}{3} \, d\right )}\right )} e^{\left (\frac {1}{3} \, d\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} - 25 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) + 25 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (-e^{\left (\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) - 50 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (\frac {1}{3} \, b x\right )} + e^{\left (-\frac {1}{3} \, d\right )}\right ) + 50 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left ({\left | e^{\left (\frac {1}{3} \, b x\right )} - e^{\left (-\frac {1}{3} \, d\right )} \right |}\right ) - \frac {60 \, e^{\left (\frac {5}{3} \, b x + \frac {5}{3} \, a\right )}}{e^{\left (2 \, b x + 2 \, d\right )} - 1} + 18 \, e^{\left (\frac {5}{3} \, b x + \frac {5}{3} \, a\right )}}{30 \, b} \] Input:
integrate(exp(5/3*b*x+5/3*a)*coth(b*x+d)^2,x, algorithm="giac")
Output:
1/30*(50*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(1/3*b*x) + e^(-1/3*d))*e^(1/3*d) )*e^(5/3*a - 5/3*d) + 50*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(1/3*b*x) - e^(-1 /3*d))*e^(1/3*d))*e^(5/3*a - 5/3*d) - 25*e^(5/3*a - 5/3*d)*log(e^(1/3*b*x - 1/3*d) + e^(2/3*b*x) + e^(-2/3*d)) + 25*e^(5/3*a - 5/3*d)*log(-e^(1/3*b* x - 1/3*d) + e^(2/3*b*x) + e^(-2/3*d)) - 50*e^(5/3*a - 5/3*d)*log(e^(1/3*b *x) + e^(-1/3*d)) + 50*e^(5/3*a - 5/3*d)*log(abs(e^(1/3*b*x) - e^(-1/3*d)) ) - 60*e^(5/3*b*x + 5/3*a)/(e^(2*b*x + 2*d) - 1) + 18*e^(5/3*b*x + 5/3*a)) /b
Time = 5.81 (sec) , antiderivative size = 448, normalized size of antiderivative = 1.93 \[ \int e^{\frac {5}{3} (a+b x)} \coth ^2(d+b x) \, dx =\text {Too large to display} \] Input:
int(coth(d + b*x)^2*exp((5*a)/3 + (5*b*x)/3),x)
Output:
(3*exp((5*a)/3 + (5*b*x)/3))/(5*b) - (2*exp((5*a)/3 + (5*b*x)/3))/(b*(exp( 2*d + 2*b*x) - 1)) - (5*exp(10*a - 10*d)^(1/6)*log(- (100*exp((10*a)/3)*ex p(-(10*d)/3))/9 - (100*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*(e xp(10*a)*exp(-10*d))^(1/6))/9))/(3*b) + (5*exp(10*a - 10*d)^(1/6)*log((100 *exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*(exp(10*a)*exp(-10*d))^( 1/6))/9 - (100*exp((10*a)/3)*exp(-(10*d)/3))/9))/(3*b) - (5*log(- (100*exp ((10*a)/3)*exp(-(10*d)/3))/9 - (100*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*ex p((b*x)/3)*((3^(1/2)*1i)/2 - 1/2)*(exp(10*a)*exp(-10*d))^(1/6))/9)*exp(10* a - 10*d)^(1/6)*((3^(1/2)*1i)/2 - 1/2))/(3*b) + (5*log((100*exp((5*a)/3)*e xp(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))/9 - (100*exp((10*a)/3)*exp(-(10*d)/3))/9)*exp(10*a - 10*d)^( 1/6)*((3^(1/2)*1i)/2 - 1/2))/(3*b) - (5*log(- (100*exp((10*a)/3)*exp(-(10* d)/3))/9 - (100*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))/9)*exp(10*a - 10*d)^(1/6)*((3^ (1/2)*1i)/2 + 1/2))/(3*b) + (5*log((100*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))/9 - (1 00*exp((10*a)/3)*exp(-(10*d)/3))/9)*exp(10*a - 10*d)^(1/6)*((3^(1/2)*1i)/2 + 1/2))/(3*b)
\[ \int e^{\frac {5}{3} (a+b x)} \coth ^2(d+b x) \, dx=\int e^{\frac {5 b x}{3}+\frac {5 a}{3}} \coth \left (b x +d \right )^{2}d x \] Input:
int(exp(5/3*b*x+5/3*a)*coth(b*x+d)^2,x)
Output:
int(e**((5*a + 5*b*x)/3)*coth(b*x + d)**2,x)