Integrand size = 18, antiderivative size = 186 \[ \int e^{\frac {5}{3} (a+b x)} \coth (d+b x) \, dx=\frac {3 e^{\frac {5 (a-d)}{3}+\frac {5}{3} (d+b x)}}{5 b}-\frac {\sqrt {3} e^{\frac {5 (a-d)}{3}} \arctan \left (\frac {1-2 e^{\frac {1}{3} (d+b x)}}{\sqrt {3}}\right )}{b}+\frac {\sqrt {3} e^{\frac {5 (a-d)}{3}} \arctan \left (\frac {1+2 e^{\frac {1}{3} (d+b x)}}{\sqrt {3}}\right )}{b}-\frac {2 e^{\frac {5 (a-d)}{3}} \text {arctanh}\left (e^{\frac {1}{3} (d+b x)}\right )}{b}-\frac {e^{\frac {5 (a-d)}{3}} \text {arctanh}\left (\frac {e^{\frac {1}{3} (d+b x)}}{1+e^{\frac {2}{3} (d+b x)}}\right )}{b} \] Output:
3/5*exp(5/3*b*x+5/3*a)/b-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+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-2*exp(5/3*a-5/3*d)*arctanh(exp(1/3*b*x+1/3*d))/b-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 = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.47 \[ \int e^{\frac {5}{3} (a+b x)} \coth (d+b x) \, dx=\frac {e^{5 a/3} \left (9 e^{\frac {5 b x}{3}}-5 \text {RootSum}\left [-\cosh (d)+\sinh (d)+\cosh (d) \text {$\#$1}^6+\sinh (d) \text {$\#$1}^6\&,\frac {b x-3 \log \left (e^{\frac {b x}{3}}-\text {$\#$1}\right )}{\text {$\#$1}}\&\right ] (\cosh (2 d)-\sinh (2 d))\right )}{15 b} \] Input:
Integrate[E^((5*(a + b*x))/3)*Coth[d + b*x],x]
Output:
(E^((5*a)/3)*(9*E^((5*b*x)/3) - 5*RootSum[-Cosh[d] + Sinh[d] + Cosh[d]*#1^ 6 + Sinh[d]*#1^6 & , (b*x - 3*Log[E^((b*x)/3) - #1])/#1 & ]*(Cosh[2*d] - S inh[2*d])))/(15*b)
Time = 0.34 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.81, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2720, 25, 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.
| \(\displaystyle \int e^{\frac {5}{3} (a+b x)} \coth (b x+d) \, dx\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {3 \int -\frac {e^{\frac {5 a}{3}+\frac {4 b x}{3}} \left (1+e^{2 b x}\right )}{1-e^{2 b x}}de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {3 \int \frac {e^{\frac {5 a}{3}+\frac {4 b x}{3}} \left (1+e^{2 b x}\right )}{1-e^{2 b x}}de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {3 e^{5 a/3} \int \frac {e^{\frac {4 b x}{3}} \left (1+e^{2 b x}\right )}{1-e^{2 b x}}de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle -\frac {3 e^{5 a/3} \left (2 \int \frac {e^{\frac {4 b x}{3}}}{1-e^{2 b x}}de^{\frac {b x}{3}}-\frac {1}{5} e^{\frac {5 b x}{3}}\right )}{b}\) |
| \(\Big \downarrow \) 825 |
| \(\displaystyle -\frac {3 e^{5 a/3} \left (2 \left (\frac {1}{3} \int \frac {1}{1-e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}+\frac {1}{3} \int -\frac {1+e^{\frac {b x}{3}}}{2 \left (1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}\right )}de^{\frac {b x}{3}}+\frac {1}{3} \int -\frac {1-e^{\frac {b x}{3}}}{2 \left (1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}\right )}de^{\frac {b x}{3}}\right )-\frac {1}{5} e^{\frac {5 b x}{3}}\right )}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {3 e^{5 a/3} \left (2 \left (\frac {1}{3} \int \frac {1}{1-e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1+e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1-e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )-\frac {1}{5} e^{\frac {5 b x}{3}}\right )}{b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {3 e^{5 a/3} \left (2 \left (-\frac {1}{6} \int \frac {1+e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1-e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )-\frac {1}{5} e^{\frac {5 b x}{3}}\right )}{b}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle -\frac {3 e^{5 a/3} \left (2 \left (\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{2} \int -\frac {1-2 e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1+2 e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {3}{2} \int \frac {1}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )-\frac {1}{5} e^{\frac {5 b x}{3}}\right )}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {3 e^{5 a/3} \left (2 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1-2 e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {3}{2} \int \frac {1}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1+2 e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {3}{2} \int \frac {1}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )-\frac {1}{5} e^{\frac {5 b x}{3}}\right )}{b}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {3 e^{5 a/3} \left (2 \left (\frac {1}{6} \left (3 \int \frac {1}{-3-e^{\frac {2 b x}{3}}}d\left (-1+2 e^{\frac {b x}{3}}\right )+\frac {1}{2} \int \frac {1-2 e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (3 \int \frac {1}{-3-e^{\frac {2 b x}{3}}}d\left (1+2 e^{\frac {b x}{3}}\right )+\frac {1}{2} \int \frac {1+2 e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )-\frac {1}{5} e^{\frac {5 b x}{3}}\right )}{b}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {3 e^{5 a/3} \left (2 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1-2 e^{\frac {b x}{3}}}{1-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\sqrt {3} \arctan \left (\frac {2 e^{\frac {b x}{3}}-1}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1+2 e^{\frac {b x}{3}}}{1+e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\sqrt {3} \arctan \left (\frac {2 e^{\frac {b x}{3}}+1}{\sqrt {3}}\right )\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )-\frac {1}{5} e^{\frac {5 b x}{3}}\right )}{b}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle -\frac {3 e^{5 a/3} \left (2 \left (\frac {1}{6} \left (-\sqrt {3} \arctan \left (\frac {2 e^{\frac {b x}{3}}-1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (-e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}+1\right )\right )+\frac {1}{6} \left (\frac {1}{2} \log \left (e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}+1\right )-\sqrt {3} \arctan \left (\frac {2 e^{\frac {b x}{3}}+1}{\sqrt {3}}\right )\right )+\frac {1}{3} \text {arctanh}\left (e^{\frac {b x}{3}}\right )\right )-\frac {1}{5} e^{\frac {5 b x}{3}}\right )}{b}\) |
Input:
Int[E^((5*(a + b*x))/3)*Coth[d + b*x],x]
Output:
(-3*E^((5*a)/3)*(-1/5*E^((5*b*x)/3) + 2*(ArcTanh[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)))/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[((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 = 0.28 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.77
| method | result | size |
| risch | \(\frac {3 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}}}{5 b}+\frac {\ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-1\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{b}-\frac {\ln \left (1+{\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{b}-\frac {\ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{2 b}+\frac {i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{2 b}-\frac {\ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{2 b}-\frac {i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{2 b}+\frac {\ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{2 b}+\frac {i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{2 b}+\frac {\ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{2 b}-\frac {i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{2 b}\) | \(329\) |
Input:
int(exp(5/3*b*x+5/3*a)*coth(b*x+d),x,method=_RETURNVERBOSE)
Output:
3/5*exp(5/3*b*x+5/3*a)/b+1/b*ln(exp(1/3*b*x+1/3*d)-1)*exp(5/3*a-5/3*d)-1/b *ln(1+exp(1/3*b*x+1/3*d))*exp(5/3*a-5/3*d)-1/2/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)+1/2*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)-1/2/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)-1/2*I/b*ln(exp(1/3*b*x+1/3*d)+1/2-1/2*I*3^(1/2))*ex p(5/3*a-5/3*d)*3^(1/2)+1/2/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)+1/2*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)+1/2/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)-1/2*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. 622 vs. \(2 (144) = 288\).
Time = 0.10 (sec) , antiderivative size = 622, normalized size of antiderivative = 3.34 \[ \int e^{\frac {5}{3} (a+b x)} \coth (d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(5/3*b*x+5/3*a)*coth(b*x+d),x, algorithm="fricas")
Output:
1/10*(6*cosh(1/3*b*x + 1/3*d)^5*cosh(-5/3*a + 5/3*d) + 6*(cosh(-5/3*a + 5/ 3*d) - sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^5 - 6*cosh(1/3*b*x + 1/ 3*d)^5*sinh(-5/3*a + 5/3*d) + 30*(cosh(1/3*b*x + 1/3*d)*cosh(-5/3*a + 5/3* d) - cosh(1/3*b*x + 1/3*d)*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^4 + 60*(cosh(1/3*b*x + 1/3*d)^2*cosh(-5/3*a + 5/3*d) - cosh(1/3*b*x + 1/3*d)^ 2*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^3 + 60*(cosh(1/3*b*x + 1/3*d )^3*cosh(-5/3*a + 5/3*d) - cosh(1/3*b*x + 1/3*d)^3*sinh(-5/3*a + 5/3*d))*s inh(1/3*b*x + 1/3*d)^2 + 10*(sqrt(3)*cosh(-5/3*a + 5/3*d) - sqrt(3)*sinh(- 5/3*a + 5/3*d))*arctan(2/3*sqrt(3)*cosh(1/3*b*x + 1/3*d) + 2/3*sqrt(3)*sin h(1/3*b*x + 1/3*d) + 1/3*sqrt(3)) + 10*(sqrt(3)*cosh(-5/3*a + 5/3*d) - sqr t(3)*sinh(-5/3*a + 5/3*d))*arctan(2/3*sqrt(3)*cosh(1/3*b*x + 1/3*d) + 2/3* sqrt(3)*sinh(1/3*b*x + 1/3*d) - 1/3*sqrt(3)) - 5*(cosh(-5/3*a + 5/3*d) - s inh(-5/3*a + 5/3*d))*log((2*cosh(1/3*b*x + 1/3*d) + 1)/(cosh(1/3*b*x + 1/3 *d) - sinh(1/3*b*x + 1/3*d))) + 5*(cosh(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/ 3*d))*log((2*cosh(1/3*b*x + 1/3*d) - 1)/(cosh(1/3*b*x + 1/3*d) - sinh(1/3* b*x + 1/3*d))) - 10*(cosh(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))*log(cosh (1/3*b*x + 1/3*d) + sinh(1/3*b*x + 1/3*d) + 1) + 10*(cosh(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))*log(cosh(1/3*b*x + 1/3*d) + sinh(1/3*b*x + 1/3*d) - 1) + 30*(cosh(1/3*b*x + 1/3*d)^4*cosh(-5/3*a + 5/3*d) - cosh(1/3*b*x + 1 /3*d)^4*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d))/b
\[ \int e^{\frac {5}{3} (a+b x)} \coth (d+b x) \, dx=e^{\frac {5 a}{3}} \int e^{\frac {5 b x}{3}} \coth {\left (b x + d \right )}\, dx \] Input:
integrate(exp(5/3*b*x+5/3*a)*coth(b*x+d),x)
Output:
exp(5*a/3)*Integral(exp(5*b*x/3)*coth(b*x + d), x)
Time = 0.12 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.10 \[ \int e^{\frac {5}{3} (a+b x)} \coth (d+b x) \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + 1\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{b} - \frac {\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} - 1\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{b} - \frac {e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )}{2 \, b} - \frac {e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + 1\right )}{b} + \frac {e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} - 1\right )}{b} + \frac {e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (-e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )}{2 \, b} + \frac {3 \, e^{\left (\frac {5}{3} \, b x + \frac {5}{3} \, a\right )}}{5 \, b} \] Input:
integrate(exp(5/3*b*x+5/3*a)*coth(b*x+d),x, algorithm="maxima")
Output:
-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 - 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 - 1/2*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 - e^(5/3*a - 5/3*d)*log(e^(-1/3*b*x - 1/3*d) + 1)/b + e^(5/3 *a - 5/3*d)*log(e^(-1/3*b*x - 1/3*d) - 1)/b + 1/2*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 + 3/5*e^(5/3*b*x + 5/3*a )/b
Time = 0.13 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.05 \[ \int e^{\frac {5}{3} (a+b x)} \coth (d+b x) \, dx=\frac {10 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (\frac {1}{3} \, b x\right )} + e^{\left (-\frac {1}{3} \, d\right )}\right )} e^{\left (\frac {1}{3} \, d\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} + 10 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (\frac {1}{3} \, b x\right )} - e^{\left (-\frac {1}{3} \, d\right )}\right )} e^{\left (\frac {1}{3} \, d\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} - 5 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) + 5 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (-e^{\left (\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) - 10 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (\frac {1}{3} \, b x\right )} + e^{\left (-\frac {1}{3} \, d\right )}\right ) + 10 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left ({\left | e^{\left (\frac {1}{3} \, b x\right )} - e^{\left (-\frac {1}{3} \, d\right )} \right |}\right ) + 6 \, e^{\left (\frac {5}{3} \, b x + \frac {5}{3} \, a\right )}}{10 \, b} \] Input:
integrate(exp(5/3*b*x+5/3*a)*coth(b*x+d),x, algorithm="giac")
Output:
1/10*(10*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) + 10*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) - 5*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)) + 5*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)) - 10*e^(5/3*a - 5/3*d)*log(e^(1/3*b*x ) + e^(-1/3*d)) + 10*e^(5/3*a - 5/3*d)*log(abs(e^(1/3*b*x) - e^(-1/3*d))) + 6*e^(5/3*b*x + 5/3*a))/b
Time = 4.42 (sec) , antiderivative size = 418, normalized size of antiderivative = 2.25 \[ \int e^{\frac {5}{3} (a+b x)} \coth (d+b x) \, dx =\text {Too large to display} \] Input:
int(coth(d + b*x)*exp((5*a)/3 + (5*b*x)/3),x)
Output:
(3*exp((5*a)/3 + (5*b*x)/3))/(5*b) - (exp(10*a - 10*d)^(1/6)*log(- 4*exp(( 10*a)/3)*exp(-(10*d)/3) - 4*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/ 3)*(exp(10*a)*exp(-10*d))^(1/6)))/b + (exp(10*a - 10*d)^(1/6)*log(4*exp((5 *a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*(exp(10*a)*exp(-10*d))^(1/6) - 4*exp((10*a)/3)*exp(-(10*d)/3)))/b - (log(- 4*exp((10*a)/3)*exp(-(10*d)/3) - 4*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))*exp(10*a - 10*d)^(1/6)*((3^(1/2)*1i)/2 - 1/2))/b + (log(4*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) - 4*exp((10*a)/3)*exp(-(10*d)/ 3))*exp(10*a - 10*d)^(1/6)*((3^(1/2)*1i)/2 - 1/2))/b - (log(- 4*exp((10*a) /3)*exp(-(10*d)/3) - 4*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))*exp(10*a - 10*d)^(1/6)* ((3^(1/2)*1i)/2 + 1/2))/b + (log(4*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) - 4*exp((10* a)/3)*exp(-(10*d)/3))*exp(10*a - 10*d)^(1/6)*((3^(1/2)*1i)/2 + 1/2))/b
\[ \int e^{\frac {5}{3} (a+b x)} \coth (d+b x) \, dx=\int e^{\frac {5 b x}{3}+\frac {5 a}{3}} \coth \left (b x +d \right )d x \] Input:
int(exp(5/3*b*x+5/3*a)*coth(b*x+d),x)
Output:
int(e**((5*a + 5*b*x)/3)*coth(b*x + d),x)