Integrand size = 14, antiderivative size = 53 \[ \int e^x \coth (2 x) \text {csch}^2(2 x) \, dx=-\frac {e^{5 x}}{\left (1-e^{4 x}\right )^2}+\frac {e^x}{4 \left (1-e^{4 x}\right )}-\frac {\arctan \left (e^x\right )}{8}-\frac {\text {arctanh}\left (e^x\right )}{8} \]
Time = 0.06 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.02 \[ \int e^x \coth (2 x) \text {csch}^2(2 x) \, dx=-\frac {-2 e^x+10 e^{5 x}+\left (-1+e^{4 x}\right )^2 \arctan \left (e^x\right )+\left (-1+e^{4 x}\right )^2 \text {arctanh}\left (e^x\right )}{8 \left (-1+e^{4 x}\right )^2} \]
-1/8*(-2*E^x + 10*E^(5*x) + (-1 + E^(4*x))^2*ArcTan[E^x] + (-1 + E^(4*x))^ 2*ArcTanh[E^x])/(-1 + E^(4*x))^2
Time = 0.22 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.26, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2720, 27, 957, 817, 756, 216, 219}
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.
\(\displaystyle \int e^x \coth (2 x) \text {csch}^2(2 x) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \int -\frac {4 e^{4 x} \left (e^{4 x}+1\right )}{\left (1-e^{4 x}\right )^3}de^x\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -4 \int \frac {e^{4 x} \left (1+e^{4 x}\right )}{\left (1-e^{4 x}\right )^3}de^x\) |
\(\Big \downarrow \) 957 |
\(\displaystyle -4 \left (\frac {e^{5 x}}{4 \left (1-e^{4 x}\right )^2}-\frac {1}{4} \int \frac {e^{4 x}}{\left (1-e^{4 x}\right )^2}de^x\right )\) |
\(\Big \downarrow \) 817 |
\(\displaystyle -4 \left (\frac {1}{4} \left (\frac {1}{4} \int \frac {1}{1-e^{4 x}}de^x-\frac {e^x}{4 \left (1-e^{4 x}\right )}\right )+\frac {e^{5 x}}{4 \left (1-e^{4 x}\right )^2}\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle -4 \left (\frac {1}{4} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1}{1-e^{2 x}}de^x+\frac {1}{2} \int \frac {1}{1+e^{2 x}}de^x\right )-\frac {e^x}{4 \left (1-e^{4 x}\right )}\right )+\frac {e^{5 x}}{4 \left (1-e^{4 x}\right )^2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -4 \left (\frac {1}{4} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1}{1-e^{2 x}}de^x+\frac {\arctan \left (e^x\right )}{2}\right )-\frac {e^x}{4 \left (1-e^{4 x}\right )}\right )+\frac {e^{5 x}}{4 \left (1-e^{4 x}\right )^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -4 \left (\frac {1}{4} \left (\frac {1}{4} \left (\frac {\arctan \left (e^x\right )}{2}+\frac {\text {arctanh}\left (e^x\right )}{2}\right )-\frac {e^x}{4 \left (1-e^{4 x}\right )}\right )+\frac {e^{5 x}}{4 \left (1-e^{4 x}\right )^2}\right )\) |
-4*(E^(5*x)/(4*(1 - E^(4*x))^2) + (-1/4*E^x/(1 - E^(4*x)) + (ArcTan[E^x]/2 + ArcTanh[E^x]/2)/4)/4)
3.10.42.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a *b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* (p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
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 = 0.53 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.02
method | result | size |
risch | \(-\frac {{\mathrm e}^{x} \left (5 \,{\mathrm e}^{4 x}-1\right )}{4 \left ({\mathrm e}^{4 x}-1\right )^{2}}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{16}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{16}-\frac {\ln \left (1+{\mathrm e}^{x}\right )}{16}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{16}\) | \(54\) |
default | \(-\frac {\coth \left (x \right ) \operatorname {csch}\left (x \right )}{8}-\frac {\operatorname {arctanh}\left ({\mathrm e}^{x}\right )}{8}+\frac {1}{16 \sinh \left (x \right )^{2} \cosh \left (x \right )}+\frac {3}{16 \cosh \left (x \right )}-\frac {1}{4 \sinh \left (x \right )}-\frac {\arctan \left ({\mathrm e}^{x}\right )}{8}+\frac {1}{8 \sinh \left (x \right ) \cosh \left (x \right )^{2}}+\frac {3 \,\operatorname {sech}\left (x \right ) \tanh \left (x \right )}{16}\) | \(56\) |
-1/4*exp(x)*(5*exp(4*x)-1)/(exp(4*x)-1)^2+1/16*I*ln(exp(x)-I)-1/16*I*ln(ex p(x)+I)-1/16*ln(1+exp(x))+1/16*ln(exp(x)-1)
Leaf count of result is larger than twice the leaf count of optimal. 522 vs. \(2 (37) = 74\).
Time = 0.27 (sec) , antiderivative size = 522, normalized size of antiderivative = 9.85 \[ \int e^x \coth (2 x) \text {csch}^2(2 x) \, dx=\text {Too large to display} \]
-1/16*(20*cosh(x)^5 + 200*cosh(x)^3*sinh(x)^2 + 200*cosh(x)^2*sinh(x)^3 + 100*cosh(x)*sinh(x)^4 + 20*sinh(x)^5 + 2*(cosh(x)^8 + 56*cosh(x)^3*sinh(x) ^5 + 28*cosh(x)^2*sinh(x)^6 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 2*(35*cosh (x)^4 - 1)*sinh(x)^4 - 2*cosh(x)^4 + 8*(7*cosh(x)^5 - cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 - 3*cosh(x)^2)*sinh(x)^2 + 8*(cosh(x)^7 - cosh(x)^3)*sinh( x) + 1)*arctan(cosh(x) + sinh(x)) + (cosh(x)^8 + 56*cosh(x)^3*sinh(x)^5 + 28*cosh(x)^2*sinh(x)^6 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 2*(35*cosh(x)^4 - 1)*sinh(x)^4 - 2*cosh(x)^4 + 8*(7*cosh(x)^5 - cosh(x))*sinh(x)^3 + 4*(7 *cosh(x)^6 - 3*cosh(x)^2)*sinh(x)^2 + 8*(cosh(x)^7 - cosh(x)^3)*sinh(x) + 1)*log(cosh(x) + sinh(x) + 1) - (cosh(x)^8 + 56*cosh(x)^3*sinh(x)^5 + 28*c osh(x)^2*sinh(x)^6 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 2*(35*cosh(x)^4 - 1 )*sinh(x)^4 - 2*cosh(x)^4 + 8*(7*cosh(x)^5 - cosh(x))*sinh(x)^3 + 4*(7*cos h(x)^6 - 3*cosh(x)^2)*sinh(x)^2 + 8*(cosh(x)^7 - cosh(x)^3)*sinh(x) + 1)*l og(cosh(x) + sinh(x) - 1) + 4*(25*cosh(x)^4 - 1)*sinh(x) - 4*cosh(x))/(cos h(x)^8 + 56*cosh(x)^3*sinh(x)^5 + 28*cosh(x)^2*sinh(x)^6 + 8*cosh(x)*sinh( x)^7 + sinh(x)^8 + 2*(35*cosh(x)^4 - 1)*sinh(x)^4 - 2*cosh(x)^4 + 8*(7*cos h(x)^5 - cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 - 3*cosh(x)^2)*sinh(x)^2 + 8* (cosh(x)^7 - cosh(x)^3)*sinh(x) + 1)
\[ \int e^x \coth (2 x) \text {csch}^2(2 x) \, dx=\int e^{x} \coth {\left (2 x \right )} \operatorname {csch}^{2}{\left (2 x \right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.89 \[ \int e^x \coth (2 x) \text {csch}^2(2 x) \, dx=-\frac {5 \, e^{\left (5 \, x\right )} - e^{x}}{4 \, {\left (e^{\left (8 \, x\right )} - 2 \, e^{\left (4 \, x\right )} + 1\right )}} - \frac {1}{8} \, \arctan \left (e^{x}\right ) - \frac {1}{16} \, \log \left (e^{x} + 1\right ) + \frac {1}{16} \, \log \left (e^{x} - 1\right ) \]
-1/4*(5*e^(5*x) - e^x)/(e^(8*x) - 2*e^(4*x) + 1) - 1/8*arctan(e^x) - 1/16* log(e^x + 1) + 1/16*log(e^x - 1)
Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79 \[ \int e^x \coth (2 x) \text {csch}^2(2 x) \, dx=-\frac {5 \, e^{\left (5 \, x\right )} - e^{x}}{4 \, {\left (e^{\left (4 \, x\right )} - 1\right )}^{2}} - \frac {1}{8} \, \arctan \left (e^{x}\right ) - \frac {1}{16} \, \log \left (e^{x} + 1\right ) + \frac {1}{16} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]
-1/4*(5*e^(5*x) - e^x)/(e^(4*x) - 1)^2 - 1/8*arctan(e^x) - 1/16*log(e^x + 1) + 1/16*log(abs(e^x - 1))
Time = 2.71 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.51 \[ \int e^x \coth (2 x) \text {csch}^2(2 x) \, dx=\frac {\ln \left (\frac {1}{4}-\frac {{\mathrm {e}}^x}{4}\right )}{16}-\frac {\ln \left (\frac {{\mathrm {e}}^x}{4}+\frac {1}{4}\right )}{16}-\frac {\mathrm {atan}\left ({\mathrm {e}}^x\right )}{8}-\frac {{\mathrm {e}}^{5\,x}}{2\,\left ({\mathrm {e}}^{8\,x}-2\,{\mathrm {e}}^{4\,x}+1\right )}-\frac {3\,{\mathrm {e}}^x}{4\,\left ({\mathrm {e}}^{4\,x}-1\right )}-\frac {{\mathrm {e}}^x}{2\,\left ({\mathrm {e}}^{8\,x}-2\,{\mathrm {e}}^{4\,x}+1\right )} \]