Integrand size = 16, antiderivative size = 81 \[ \int e^{a+b x} \coth ^3(a+b x) \, dx=\frac {e^{a+b x}}{b}-\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {3 \text {arctanh}\left (e^{a+b x}\right )}{b} \]
exp(b*x+a)/b-2*exp(b*x+a)/b/(1-exp(2*b*x+2*a))^2+3*exp(b*x+a)/b/(1-exp(2*b *x+2*a))-3*arctanh(exp(b*x+a))/b
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 2.48 (sec) , antiderivative size = 286, normalized size of antiderivative = 3.53 \[ \int e^{a+b x} \coth ^3(a+b x) \, dx=-\frac {e^{-5 (a+b x)} \left (-21 \left (252105+507305 e^{2 (a+b x)}+173916 e^{4 (a+b x)}-154296 e^{6 (a+b x)}-73885 e^{8 (a+b x)}+4887 e^{10 (a+b x)}\right )-\frac {315 \left (-16807-28218 e^{2 (a+b x)}+1173 e^{4 (a+b x)}+17748 e^{6 (a+b x)}+4299 e^{8 (a+b x)}-1434 e^{10 (a+b x)}+7 e^{12 (a+b x)}\right ) \text {arctanh}\left (\sqrt {e^{2 (a+b x)}}\right )}{\sqrt {e^{2 (a+b x)}}}+384 e^{8 (a+b x)} \left (1+e^{2 (a+b x)}\right )^2 \left (7+5 e^{2 (a+b x)}\right ) \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {11}{2};e^{2 (a+b x)}\right )+256 e^{8 (a+b x)} \left (1+e^{2 (a+b x)}\right )^3 \, _6F_5\left (\frac {3}{2},2,2,2,2,2;1,1,1,1,\frac {11}{2};e^{2 (a+b x)}\right )\right )}{60480 b} \]
-1/60480*(-21*(252105 + 507305*E^(2*(a + b*x)) + 173916*E^(4*(a + b*x)) - 154296*E^(6*(a + b*x)) - 73885*E^(8*(a + b*x)) + 4887*E^(10*(a + b*x))) - (315*(-16807 - 28218*E^(2*(a + b*x)) + 1173*E^(4*(a + b*x)) + 17748*E^(6*( a + b*x)) + 4299*E^(8*(a + b*x)) - 1434*E^(10*(a + b*x)) + 7*E^(12*(a + b* x)))*ArcTanh[Sqrt[E^(2*(a + b*x))]])/Sqrt[E^(2*(a + b*x))] + 384*E^(8*(a + b*x))*(1 + E^(2*(a + b*x)))^2*(7 + 5*E^(2*(a + b*x)))*HypergeometricPFQ[{ 3/2, 2, 2, 2, 2}, {1, 1, 1, 11/2}, E^(2*(a + b*x))] + 256*E^(8*(a + b*x))* (1 + E^(2*(a + b*x)))^3*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 11/2}, E^(2*(a + b*x))])/(b*E^(5*(a + b*x)))
Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2720, 25, 300, 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.
\(\displaystyle \int e^{a+b x} \coth ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {\left (1+e^{2 a+2 b x}\right )^3}{\left (1-e^{2 a+2 b x}\right )^3}de^{a+b x}}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\left (1+e^{2 a+2 b x}\right )^3}{\left (1-e^{2 a+2 b x}\right )^3}de^{a+b x}}{b}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle -\frac {\int \left (\frac {2 \left (1+3 e^{4 a+4 b x}\right )}{\left (1-e^{2 a+2 b x}\right )^3}-1\right )de^{a+b x}}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-3 \text {arctanh}\left (e^{a+b x}\right )+e^{a+b x}+\frac {3 e^{a+b x}}{1-e^{2 a+2 b x}}-\frac {2 e^{a+b x}}{\left (1-e^{2 a+2 b x}\right )^2}}{b}\) |
(E^(a + b*x) - (2*E^(a + b*x))/(1 - E^(2*a + 2*b*x))^2 + (3*E^(a + b*x))/( 1 - E^(2*a + 2*b*x)) - 3*ArcTanh[E^(a + b*x)])/b
3.3.12.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
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]]]
Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {{\mathrm e}^{b x +a}}{b}-\frac {{\mathrm e}^{b x +a} \left (3 \,{\mathrm e}^{2 b x +2 a}-1\right )}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{b x +a}+1\right )}{2 b}+\frac {3 \ln \left ({\mathrm e}^{b x +a}-1\right )}{2 b}\) | \(77\) |
derivativedivides | \(\frac {\frac {\cosh \left (b x +a \right )^{2}}{\sinh \left (b x +a \right )}-\frac {2}{\sinh \left (b x +a \right )}+\frac {\cosh \left (b x +a \right )^{3}}{\sinh \left (b x +a \right )^{2}}-\frac {3 \cosh \left (b x +a \right )}{\sinh \left (b x +a \right )^{2}}+\frac {3 \,\operatorname {csch}\left (b x +a \right ) \coth \left (b x +a \right )}{2}-3 \,\operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )}{b}\) | \(89\) |
default | \(\frac {\frac {\cosh \left (b x +a \right )^{2}}{\sinh \left (b x +a \right )}-\frac {2}{\sinh \left (b x +a \right )}+\frac {\cosh \left (b x +a \right )^{3}}{\sinh \left (b x +a \right )^{2}}-\frac {3 \cosh \left (b x +a \right )}{\sinh \left (b x +a \right )^{2}}+\frac {3 \,\operatorname {csch}\left (b x +a \right ) \coth \left (b x +a \right )}{2}-3 \,\operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )}{b}\) | \(89\) |
exp(b*x+a)/b-exp(b*x+a)*(3*exp(2*b*x+2*a)-1)/b/(exp(2*b*x+2*a)-1)^2-3/2/b* ln(exp(b*x+a)+1)+3/2/b*ln(exp(b*x+a)-1)
Leaf count of result is larger than twice the leaf count of optimal. 459 vs. \(2 (71) = 142\).
Time = 0.26 (sec) , antiderivative size = 459, normalized size of antiderivative = 5.67 \[ \int e^{a+b x} \coth ^3(a+b x) \, dx=\frac {2 \, \cosh \left (b x + a\right )^{5} + 10 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 2 \, \sinh \left (b x + a\right )^{5} + 10 \, {\left (2 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{3} - 10 \, \cosh \left (b x + a\right )^{3} + 10 \, {\left (2 \, \cosh \left (b x + a\right )^{3} - 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 3 \, {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 3 \, {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \, {\left (5 \, \cosh \left (b x + a\right )^{4} - 15 \, \cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right ) + 4 \, \cosh \left (b x + a\right )}{2 \, {\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \]
1/2*(2*cosh(b*x + a)^5 + 10*cosh(b*x + a)*sinh(b*x + a)^4 + 2*sinh(b*x + a )^5 + 10*(2*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^3 - 10*cosh(b*x + a)^3 + 10 *(2*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^2 - 3*(cosh(b*x + a)^ 4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a) ^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b* x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 3*(cos h(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*co sh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^ 3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 2*(5*cosh(b*x + a)^4 - 15*cosh(b*x + a)^2 + 2)*sinh(b*x + a) + 4*cosh (b*x + a))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh (b*x + a)^4 - 2*b*cosh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^2 + 4*(b*cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a) + b)
\[ \int e^{a+b x} \coth ^3(a+b x) \, dx=e^{a} \int e^{b x} \coth ^{3}{\left (a + b x \right )}\, dx \]
Time = 0.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.09 \[ \int e^{a+b x} \coth ^3(a+b x) \, dx=\frac {e^{\left (b x + a\right )}}{b} - \frac {3 \, \log \left (e^{\left (b x + a\right )} + 1\right )}{2 \, b} + \frac {3 \, \log \left (e^{\left (b x + a\right )} - 1\right )}{2 \, b} - \frac {3 \, e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}}{b {\left (e^{\left (4 \, b x + 4 \, a\right )} - 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \]
e^(b*x + a)/b - 3/2*log(e^(b*x + a) + 1)/b + 3/2*log(e^(b*x + a) - 1)/b - (3*e^(3*b*x + 3*a) - e^(b*x + a))/(b*(e^(4*b*x + 4*a) - 2*e^(2*b*x + 2*a) + 1))
Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89 \[ \int e^{a+b x} \coth ^3(a+b x) \, dx=-\frac {\frac {2 \, {\left (3 \, e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - 2 \, e^{\left (b x + a\right )} + 3 \, \log \left (e^{\left (b x + a\right )} + 1\right ) - 3 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{2 \, b} \]
-1/2*(2*(3*e^(3*b*x + 3*a) - e^(b*x + a))/(e^(2*b*x + 2*a) - 1)^2 - 2*e^(b *x + a) + 3*log(e^(b*x + a) + 1) - 3*log(abs(e^(b*x + a) - 1)))/b
Time = 1.72 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.20 \[ \int e^{a+b x} \coth ^3(a+b x) \, dx=\frac {{\mathrm {e}}^{a+b\,x}}{b}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {3\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]