Integrand size = 10, antiderivative size = 134 \[ \int e^x \coth ^2(4 x) \, dx=e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}-\frac {\arctan \left (e^x\right )}{8}+\frac {\arctan \left (1-\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {\text {arctanh}\left (e^x\right )}{8}+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}} \]
exp(x)+1/2*exp(x)/(1-exp(8*x))-1/8*arctan(exp(x))-1/8*arctanh(exp(x))-1/16 *arctan(-1+exp(x)*2^(1/2))*2^(1/2)-1/16*arctan(1+exp(x)*2^(1/2))*2^(1/2)+1 /32*ln(1+exp(2*x)-exp(x)*2^(1/2))*2^(1/2)-1/32*ln(1+exp(2*x)+exp(x)*2^(1/2 ))*2^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 2.00 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.84 \[ \int e^x \coth ^2(4 x) \, dx=\frac {e^{-15 x} \left (-44217-80225 e^{8 x}-15127 e^{16 x}+9361 e^{24 x}+9 \left (4913+8368 e^{8 x}+1486 e^{16 x}-1456 e^{24 x}+e^{32 x}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{8},1,\frac {9}{8},e^{8 x}\right )\right )}{9216}+\frac {64 e^{9 x} \left (1+e^{8 x}\right )^2 \, _4F_3\left (\frac {9}{8},2,2,2;1,1,\frac {33}{8};e^{8 x}\right )}{3825} \]
(-44217 - 80225*E^(8*x) - 15127*E^(16*x) + 9361*E^(24*x) + 9*(4913 + 8368* E^(8*x) + 1486*E^(16*x) - 1456*E^(24*x) + E^(32*x))*Hypergeometric2F1[1/8, 1, 9/8, E^(8*x)])/(9216*E^(15*x)) + (64*E^(9*x)*(1 + E^(8*x))^2*Hypergeom etricPFQ[{9/8, 2, 2, 2}, {1, 1, 33/8}, E^(8*x)])/3825
Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2720, 915, 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.
\(\displaystyle \int e^x \coth ^2(4 x) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \int \frac {\left (e^{8 x}+1\right )^2}{\left (1-e^{8 x}\right )^2}de^x\) |
\(\Big \downarrow \) 915 |
\(\displaystyle \int \left (\frac {4 e^{8 x}}{\left (1-e^{8 x}\right )^2}+1\right )de^x\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{8} \arctan \left (e^x\right )+\frac {\arctan \left (1-\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {\arctan \left (\sqrt {2} e^x+1\right )}{8 \sqrt {2}}-\frac {\text {arctanh}\left (e^x\right )}{8}+e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}+\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{16 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{16 \sqrt {2}}\) |
E^x + E^x/(2*(1 - E^(8*x))) - ArcTan[E^x]/8 + ArcTan[1 - Sqrt[2]*E^x]/(8*S qrt[2]) - ArcTan[1 + Sqrt[2]*E^x]/(8*Sqrt[2]) - ArcTanh[E^x]/8 + Log[1 - S qrt[2]*E^x + E^(2*x)]/(16*Sqrt[2]) - Log[1 + Sqrt[2]*E^x + E^(2*x)]/(16*Sq rt[2])
3.3.25.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a , b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 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]]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.51
method | result | size |
risch | \({\mathrm e}^{x}-\frac {{\mathrm e}^{x}}{2 \left ({\mathrm e}^{8 x}-1\right )}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{16}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{16}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{16}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{16}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (65536 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}-16 \textit {\_R} \right )\right )\) | \(68\) |
exp(x)-1/2*exp(x)/(exp(8*x)-1)+1/16*I*ln(exp(x)-I)-1/16*I*ln(exp(x)+I)-1/1 6*ln(exp(x)+1)+1/16*ln(exp(x)-1)+sum(_R*ln(exp(x)-16*_R),_R=RootOf(65536*_ Z^4+1))
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 939, normalized size of antiderivative = 7.01 \[ \int e^x \coth ^2(4 x) \, dx=\text {Too large to display} \]
1/32*(32*cosh(x)^9 + 1152*cosh(x)^7*sinh(x)^2 + 2688*cosh(x)^6*sinh(x)^3 + 4032*cosh(x)^5*sinh(x)^4 + 4032*cosh(x)^4*sinh(x)^5 + 2688*cosh(x)^3*sinh (x)^6 + 1152*cosh(x)^2*sinh(x)^7 + 288*cosh(x)*sinh(x)^8 + 32*sinh(x)^9 - 4*(cosh(x)^8 + 8*cosh(x)^7*sinh(x) + 28*cosh(x)^6*sinh(x)^2 + 56*cosh(x)^5 *sinh(x)^3 + 70*cosh(x)^4*sinh(x)^4 + 56*cosh(x)^3*sinh(x)^5 + 28*cosh(x)^ 2*sinh(x)^6 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 - 1)*arctan(cosh(x) + sinh(x )) + (-(I + 1)*sqrt(2)*cosh(x)^8 - (8*I + 8)*sqrt(2)*cosh(x)^7*sinh(x) - ( 28*I + 28)*sqrt(2)*cosh(x)^6*sinh(x)^2 - (56*I + 56)*sqrt(2)*cosh(x)^5*sin h(x)^3 - (70*I + 70)*sqrt(2)*cosh(x)^4*sinh(x)^4 - (56*I + 56)*sqrt(2)*cos h(x)^3*sinh(x)^5 - (28*I + 28)*sqrt(2)*cosh(x)^2*sinh(x)^6 - (8*I + 8)*sqr t(2)*cosh(x)*sinh(x)^7 - (I + 1)*sqrt(2)*sinh(x)^8 + (I + 1)*sqrt(2))*log( (I + 1)*sqrt(2) + 2*cosh(x) + 2*sinh(x)) + ((I - 1)*sqrt(2)*cosh(x)^8 + (8 *I - 8)*sqrt(2)*cosh(x)^7*sinh(x) + (28*I - 28)*sqrt(2)*cosh(x)^6*sinh(x)^ 2 + (56*I - 56)*sqrt(2)*cosh(x)^5*sinh(x)^3 + (70*I - 70)*sqrt(2)*cosh(x)^ 4*sinh(x)^4 + (56*I - 56)*sqrt(2)*cosh(x)^3*sinh(x)^5 + (28*I - 28)*sqrt(2 )*cosh(x)^2*sinh(x)^6 + (8*I - 8)*sqrt(2)*cosh(x)*sinh(x)^7 + (I - 1)*sqrt (2)*sinh(x)^8 - (I - 1)*sqrt(2))*log(-(I - 1)*sqrt(2) + 2*cosh(x) + 2*sinh (x)) + (-(I - 1)*sqrt(2)*cosh(x)^8 - (8*I - 8)*sqrt(2)*cosh(x)^7*sinh(x) - (28*I - 28)*sqrt(2)*cosh(x)^6*sinh(x)^2 - (56*I - 56)*sqrt(2)*cosh(x)^5*s inh(x)^3 - (70*I - 70)*sqrt(2)*cosh(x)^4*sinh(x)^4 - (56*I - 56)*sqrt(2...
\[ \int e^x \coth ^2(4 x) \, dx=\int e^{x} \coth ^{2}{\left (4 x \right )}\, dx \]
Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.81 \[ \int e^x \coth ^2(4 x) \, dx=-\frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{32} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{32} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {e^{x}}{2 \, {\left (e^{\left (8 \, x\right )} - 1\right )}} - \frac {1}{8} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{16} \, \log \left (e^{x} + 1\right ) + \frac {1}{16} \, \log \left (e^{x} - 1\right ) \]
-1/16*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) - 1/16*sqrt(2)*arctan( -1/2*sqrt(2)*(sqrt(2) - 2*e^x)) - 1/32*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) + 1/32*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) - 1/2*e^x/(e^(8*x) - 1) - 1/8*arctan(e^x) + e^x - 1/16*log(e^x + 1) + 1/16*log(e^x - 1)
Time = 0.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.82 \[ \int e^x \coth ^2(4 x) \, dx=-\frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{32} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{32} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {e^{x}}{2 \, {\left (e^{\left (8 \, x\right )} - 1\right )}} - \frac {1}{8} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{16} \, \log \left (e^{x} + 1\right ) + \frac {1}{16} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]
-1/16*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) - 1/16*sqrt(2)*arctan( -1/2*sqrt(2)*(sqrt(2) - 2*e^x)) - 1/32*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) + 1/32*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) - 1/2*e^x/(e^(8*x) - 1) - 1/8*arctan(e^x) + e^x - 1/16*log(e^x + 1) + 1/16*log(abs(e^x - 1))
Time = 2.02 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.91 \[ \int e^x \coth ^2(4 x) \, dx=\frac {\ln \left (\frac {1}{2}-\frac {{\mathrm {e}}^x}{2}\right )}{16}-\frac {\ln \left (-\frac {{\mathrm {e}}^x}{2}-\frac {1}{2}\right )}{16}-\frac {\mathrm {atan}\left ({\mathrm {e}}^x\right )}{8}+{\mathrm {e}}^x-\frac {{\mathrm {e}}^x}{2\,\left ({\mathrm {e}}^{8\,x}-1\right )}-\frac {\sqrt {2}\,\mathrm {atan}\left (2\,\sqrt {2}\,\left (\frac {{\mathrm {e}}^x}{2}-\frac {\sqrt {2}}{4}\right )\right )}{16}-\frac {\sqrt {2}\,\mathrm {atan}\left (2\,\sqrt {2}\,\left (\frac {{\mathrm {e}}^x}{2}+\frac {\sqrt {2}}{4}\right )\right )}{16}+\frac {\sqrt {2}\,\ln \left ({\left (\frac {{\mathrm {e}}^x}{2}-\frac {\sqrt {2}}{4}\right )}^2+\frac {1}{8}\right )}{32}-\frac {\sqrt {2}\,\ln \left ({\left (\frac {{\mathrm {e}}^x}{2}+\frac {\sqrt {2}}{4}\right )}^2+\frac {1}{8}\right )}{32} \]
log(1/2 - exp(x)/2)/16 - log(- exp(x)/2 - 1/2)/16 - atan(exp(x))/8 + exp(x ) - exp(x)/(2*(exp(8*x) - 1)) - (2^(1/2)*atan(2*2^(1/2)*(exp(x)/2 - 2^(1/2 )/4)))/16 - (2^(1/2)*atan(2*2^(1/2)*(exp(x)/2 + 2^(1/2)/4)))/16 + (2^(1/2) *log((exp(x)/2 - 2^(1/2)/4)^2 + 1/8))/32 - (2^(1/2)*log((exp(x)/2 + 2^(1/2 )/4)^2 + 1/8))/32