Integrand size = 29, antiderivative size = 87 \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^2} \, dx=\frac {2 e^{2 a-2 d} (f-g)}{b (f+g)^2 \left (f-g-e^{2 d+2 b x} (f+g)\right )}+\frac {2 e^{2 a-2 d} \log \left (f-g-e^{2 d+2 b x} (f+g)\right )}{b (f+g)^2} \] Output:
2*exp(2*a-2*d)*(f-g)/b/(f+g)^2/(f-g-exp(2*b*x+2*d)*(f+g))+2*exp(2*a-2*d)*l n(f-g-exp(2*b*x+2*d)*(f+g))/b/(f+g)^2
Time = 0.42 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77 \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^2} \, dx=\frac {2 e^{2 a-2 d} \left (\frac {f-g}{f-g-e^{2 (d+b x)} (f+g)}+\log \left (f-g-e^{2 (d+b x)} (f+g)\right )\right )}{b (f+g)^2} \] Input:
Integrate[E^(2*(a + b*x))/(g*Cosh[d + b*x] + f*Sinh[d + b*x])^2,x]
Output:
(2*E^(2*a - 2*d)*((f - g)/(f - g - E^(2*(d + b*x))*(f + g)) + Log[f - g - E^(2*(d + b*x))*(f + g)]))/(b*(f + g)^2)
Time = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2720, 27, 243, 49, 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.
\(\displaystyle \int \frac {e^{2 (a+b x)}}{(f \sinh (b x+d)+g \cosh (b x+d))^2} \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {4 e^{2 a+3 b x}}{\left (f-g-e^{2 b x} (f+g)\right )^2}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 e^{2 a} \int \frac {e^{3 b x}}{\left (f-g-e^{2 b x} (f+g)\right )^2}de^{b x}}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {2 e^{2 a} \int \frac {e^{2 b x}}{\left (f-g-e^{2 b x} (f+g)\right )^2}de^{2 b x}}{b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {2 e^{2 a} \int \left (\frac {f-g}{(f+g) \left (f-g-e^{2 b x} (f+g)\right )^2}+\frac {1}{(-f-g) \left (f-g-e^{2 b x} (f+g)\right )}\right )de^{2 b x}}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 e^{2 a} \left (\frac {f-g}{(f+g)^2 \left (-e^{2 b x} (f+g)+f-g\right )}+\frac {\log \left (-e^{2 b x} (f+g)+f-g\right )}{(f+g)^2}\right )}{b}\) |
Input:
Int[E^(2*(a + b*x))/(g*Cosh[d + b*x] + f*Sinh[d + b*x])^2,x]
Output:
(2*E^(2*a)*((f - g)/((f + g)^2*(f - g - E^(2*b*x)*(f + g))) + Log[f - g - E^(2*b*x)*(f + g)]/(f + g)^2))/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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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]]]
Leaf count of result is larger than twice the leaf count of optimal. \(231\) vs. \(2(83)=166\).
Time = 90.38 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.67
method | result | size |
risch | \(-\frac {4 \,{\mathrm e}^{2 a -2 d} a}{b \left (f^{2}+2 f g +g^{2}\right )}+\frac {2 \,{\mathrm e}^{4 a -2 d} f}{\left (-f \,{\mathrm e}^{2 b x +2 a +2 d}-g \,{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a} f -{\mathrm e}^{2 a} g \right ) b \left (f^{2}+2 f g +g^{2}\right )}-\frac {2 \,{\mathrm e}^{4 a -2 d} g}{\left (-f \,{\mathrm e}^{2 b x +2 a +2 d}-g \,{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a} f -{\mathrm e}^{2 a} g \right ) b \left (f^{2}+2 f g +g^{2}\right )}+\frac {2 \ln \left ({\mathrm e}^{2 b x +2 a}-\frac {\left (f -g \right ) {\mathrm e}^{2 a -2 d}}{f +g}\right ) {\mathrm e}^{2 a -2 d}}{b \left (f^{2}+2 f g +g^{2}\right )}\) | \(232\) |
Input:
int(exp(2*b*x+2*a)/(g*cosh(b*x+d)+f*sinh(b*x+d))^2,x,method=_RETURNVERBOSE )
Output:
-4/b/(f^2+2*f*g+g^2)*exp(2*a-2*d)*a+2/(-f*exp(2*b*x+2*a+2*d)-g*exp(2*b*x+2 *a+2*d)+exp(2*a)*f-exp(2*a)*g)/b/(f^2+2*f*g+g^2)*exp(4*a-2*d)*f-2/(-f*exp( 2*b*x+2*a+2*d)-g*exp(2*b*x+2*a+2*d)+exp(2*a)*f-exp(2*a)*g)/b/(f^2+2*f*g+g^ 2)*exp(4*a-2*d)*g+2*ln(exp(2*b*x+2*a)-1/(f+g)*(f-g)*exp(2*a-2*d))/b/(f^2+2 *f*g+g^2)*exp(2*a-2*d)
Leaf count of result is larger than twice the leaf count of optimal. 351 vs. \(2 (82) = 164\).
Time = 0.09 (sec) , antiderivative size = 351, normalized size of antiderivative = 4.03 \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^2} \, dx=\frac {2 \, {\left ({\left (f - g\right )} \cosh \left (-2 \, a + 2 \, d\right ) - {\left ({\left (f + g\right )} \cosh \left (b x + d\right )^{2} \cosh \left (-2 \, a + 2 \, d\right ) + {\left ({\left (f + g\right )} \cosh \left (-2 \, a + 2 \, d\right ) - {\left (f + g\right )} \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{2} - {\left (f - g\right )} \cosh \left (-2 \, a + 2 \, d\right ) + 2 \, {\left ({\left (f + g\right )} \cosh \left (b x + d\right ) \cosh \left (-2 \, a + 2 \, d\right ) - {\left (f + g\right )} \cosh \left (b x + d\right ) \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right ) - {\left ({\left (f + g\right )} \cosh \left (b x + d\right )^{2} - f + g\right )} \sinh \left (-2 \, a + 2 \, d\right )\right )} \log \left (\frac {2 \, {\left (g \cosh \left (b x + d\right ) + f \sinh \left (b x + d\right )\right )}}{\cosh \left (b x + d\right ) - \sinh \left (b x + d\right )}\right ) - {\left (f - g\right )} \sinh \left (-2 \, a + 2 \, d\right )\right )}}{b f^{3} + b f^{2} g - b f g^{2} - b g^{3} - {\left (b f^{3} + 3 \, b f^{2} g + 3 \, b f g^{2} + b g^{3}\right )} \cosh \left (b x + d\right )^{2} - 2 \, {\left (b f^{3} + 3 \, b f^{2} g + 3 \, b f g^{2} + b g^{3}\right )} \cosh \left (b x + d\right ) \sinh \left (b x + d\right ) - {\left (b f^{3} + 3 \, b f^{2} g + 3 \, b f g^{2} + b g^{3}\right )} \sinh \left (b x + d\right )^{2}} \] Input:
integrate(exp(2*b*x+2*a)/(g*cosh(b*x+d)+f*sinh(b*x+d))^2,x, algorithm="fri cas")
Output:
2*((f - g)*cosh(-2*a + 2*d) - ((f + g)*cosh(b*x + d)^2*cosh(-2*a + 2*d) + ((f + g)*cosh(-2*a + 2*d) - (f + g)*sinh(-2*a + 2*d))*sinh(b*x + d)^2 - (f - g)*cosh(-2*a + 2*d) + 2*((f + g)*cosh(b*x + d)*cosh(-2*a + 2*d) - (f + g)*cosh(b*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d) - ((f + g)*cosh(b*x + d)^ 2 - f + g)*sinh(-2*a + 2*d))*log(2*(g*cosh(b*x + d) + f*sinh(b*x + d))/(co sh(b*x + d) - sinh(b*x + d))) - (f - g)*sinh(-2*a + 2*d))/(b*f^3 + b*f^2*g - b*f*g^2 - b*g^3 - (b*f^3 + 3*b*f^2*g + 3*b*f*g^2 + b*g^3)*cosh(b*x + d) ^2 - 2*(b*f^3 + 3*b*f^2*g + 3*b*f*g^2 + b*g^3)*cosh(b*x + d)*sinh(b*x + d) - (b*f^3 + 3*b*f^2*g + 3*b*f*g^2 + b*g^3)*sinh(b*x + d)^2)
\[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^2} \, dx=e^{2 a} \int \frac {e^{2 b x}}{f^{2} \sinh ^{2}{\left (b x + d \right )} + 2 f g \sinh {\left (b x + d \right )} \cosh {\left (b x + d \right )} + g^{2} \cosh ^{2}{\left (b x + d \right )}}\, dx \] Input:
integrate(exp(2*b*x+2*a)/(g*cosh(b*x+d)+f*sinh(b*x+d))**2,x)
Output:
exp(2*a)*Integral(exp(2*b*x)/(f**2*sinh(b*x + d)**2 + 2*f*g*sinh(b*x + d)* cosh(b*x + d) + g**2*cosh(b*x + d)**2), x)
Time = 0.04 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.43 \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^2} \, dx=\frac {4 \, {\left (b x + d\right )} e^{\left (2 \, a - 2 \, d\right )}}{{\left (f^{2} + 2 \, f g + g^{2}\right )} b} + \frac {2 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (-{\left (f - g\right )} e^{\left (-2 \, b x - 2 \, d\right )} + f + g\right )}{{\left (f^{2} + 2 \, f g + g^{2}\right )} b} - \frac {2 \, e^{\left (2 \, a - 2 \, d\right )}}{{\left (f^{2} + 2 \, f g + g^{2} - {\left (f^{2} - g^{2}\right )} e^{\left (-2 \, b x - 2 \, d\right )}\right )} b} \] Input:
integrate(exp(2*b*x+2*a)/(g*cosh(b*x+d)+f*sinh(b*x+d))^2,x, algorithm="max ima")
Output:
4*(b*x + d)*e^(2*a - 2*d)/((f^2 + 2*f*g + g^2)*b) + 2*e^(2*a - 2*d)*log(-( f - g)*e^(-2*b*x - 2*d) + f + g)/((f^2 + 2*f*g + g^2)*b) - 2*e^(2*a - 2*d) /((f^2 + 2*f*g + g^2 - (f^2 - g^2)*e^(-2*b*x - 2*d))*b)
Time = 0.11 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.29 \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^2} \, dx=\frac {2 \, {\left (\frac {\log \left ({\left | f e^{\left (2 \, b x + 2 \, d\right )} + g e^{\left (2 \, b x + 2 \, d\right )} - f + g \right |}\right )}{f^{2} e^{\left (2 \, d\right )} + 2 \, f g e^{\left (2 \, d\right )} + g^{2} e^{\left (2 \, d\right )}} - \frac {{\left (f - g\right )} e^{\left (-2 \, d\right )}}{{\left (f e^{\left (2 \, b x + 2 \, d\right )} + g e^{\left (2 \, b x + 2 \, d\right )} - f + g\right )} {\left (f + g\right )}^{2}}\right )} e^{\left (2 \, a\right )}}{b} \] Input:
integrate(exp(2*b*x+2*a)/(g*cosh(b*x+d)+f*sinh(b*x+d))^2,x, algorithm="gia c")
Output:
2*(log(abs(f*e^(2*b*x + 2*d) + g*e^(2*b*x + 2*d) - f + g))/(f^2*e^(2*d) + 2*f*g*e^(2*d) + g^2*e^(2*d)) - (f - g)*e^(-2*d)/((f*e^(2*b*x + 2*d) + g*e^ (2*b*x + 2*d) - f + g)*(f + g)^2))*e^(2*a)/b
Time = 2.86 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.85 \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^2} \, dx=\frac {\ln \left (g-f+f\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,b\,x}+g\,{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,b\,x}\right )\,\left (2\,b\,{\mathrm {e}}^{2\,a-2\,d}\,f^2+4\,b\,{\mathrm {e}}^{2\,a-2\,d}\,f\,g+2\,b\,{\mathrm {e}}^{2\,a-2\,d}\,g^2\right )}{b^2\,f^4+4\,b^2\,f^3\,g+6\,b^2\,f^2\,g^2+4\,b^2\,f\,g^3+b^2\,g^4}-\frac {2\,{\mathrm {e}}^{2\,a-2\,d}\,\left (f-g\right )}{b\,{\left (f+g\right )}^2\,\left (g-f+{\mathrm {e}}^{2\,d+2\,b\,x}\,\left (f+g\right )\right )} \] Input:
int(exp(2*a + 2*b*x)/(g*cosh(d + b*x) + f*sinh(d + b*x))^2,x)
Output:
(log(g - f + f*exp(2*d)*exp(2*b*x) + g*exp(2*d)*exp(2*b*x))*(2*b*f^2*exp(2 *a - 2*d) + 2*b*g^2*exp(2*a - 2*d) + 4*b*f*g*exp(2*a - 2*d)))/(b^2*f^4 + b ^2*g^4 + 4*b^2*f*g^3 + 4*b^2*f^3*g + 6*b^2*f^2*g^2) - (2*exp(2*a - 2*d)*(f - g))/(b*(f + g)^2*(g - f + exp(2*d + 2*b*x)*(f + g)))
Time = 0.24 (sec) , antiderivative size = 275, normalized size of antiderivative = 3.16 \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^2} \, dx=\frac {2 e^{2 a} \left (e^{2 b x +2 d} \mathrm {log}\left (e^{2 b x +2 d} f +e^{2 b x +2 d} g -f +g \right ) f +e^{2 b x +2 d} \mathrm {log}\left (e^{2 b x +2 d} f +e^{2 b x +2 d} g -f +g \right ) g -e^{2 b x +2 d} f -e^{2 b x +2 d} g -\mathrm {log}\left (e^{2 b x +2 d} f +e^{2 b x +2 d} g -f +g \right ) f +\mathrm {log}\left (e^{2 b x +2 d} f +e^{2 b x +2 d} g -f +g \right ) g \right )}{e^{2 d} b \left (e^{2 b x +2 d} f^{3}+3 e^{2 b x +2 d} f^{2} g +3 e^{2 b x +2 d} f \,g^{2}+e^{2 b x +2 d} g^{3}-f^{3}-f^{2} g +f \,g^{2}+g^{3}\right )} \] Input:
int(exp(2*b*x+2*a)/(g*cosh(b*x+d)+f*sinh(b*x+d))^2,x)
Output:
(2*e**(2*a)*(e**(2*b*x + 2*d)*log(e**(2*b*x + 2*d)*f + e**(2*b*x + 2*d)*g - f + g)*f + e**(2*b*x + 2*d)*log(e**(2*b*x + 2*d)*f + e**(2*b*x + 2*d)*g - f + g)*g - e**(2*b*x + 2*d)*f - e**(2*b*x + 2*d)*g - log(e**(2*b*x + 2*d )*f + e**(2*b*x + 2*d)*g - f + g)*f + log(e**(2*b*x + 2*d)*f + e**(2*b*x + 2*d)*g - f + g)*g))/(e**(2*d)*b*(e**(2*b*x + 2*d)*f**3 + 3*e**(2*b*x + 2* d)*f**2*g + 3*e**(2*b*x + 2*d)*f*g**2 + e**(2*b*x + 2*d)*g**3 - f**3 - f** 2*g + f*g**2 + g**3))