Integrand size = 21, antiderivative size = 64 \[ \int e^{n \sinh \left (\frac {1}{2} (a+b x)\right )} \sinh (a+b x) \, dx=-\frac {4 e^{n \sinh \left (\frac {a}{2}+\frac {b x}{2}\right )}}{b n^2}+\frac {4 e^{n \sinh \left (\frac {a}{2}+\frac {b x}{2}\right )} \sinh \left (\frac {a}{2}+\frac {b x}{2}\right )}{b n} \]
Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.56 \[ \int e^{n \sinh \left (\frac {1}{2} (a+b x)\right )} \sinh (a+b x) \, dx=\frac {4 e^{n \sinh \left (\frac {1}{2} (a+b x)\right )} \left (-1+n \sinh \left (\frac {1}{2} (a+b x)\right )\right )}{b n^2} \]
Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4878, 27, 2607, 2624}
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.
\(\displaystyle \int \sinh (a+b x) e^{n \sinh \left (\frac {1}{2} (a+b x)\right )} \, dx\) |
\(\Big \downarrow \) 4878 |
\(\displaystyle \frac {2 \int 2 e^{n \sinh \left (\frac {a}{2}+\frac {b x}{2}\right )} \sinh \left (\frac {a}{2}+\frac {b x}{2}\right )d\sinh \left (\frac {a}{2}+\frac {b x}{2}\right )}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 \int e^{n \sinh \left (\frac {a}{2}+\frac {b x}{2}\right )} \sinh \left (\frac {a}{2}+\frac {b x}{2}\right )d\sinh \left (\frac {a}{2}+\frac {b x}{2}\right )}{b}\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle \frac {4 \left (\frac {\sinh \left (\frac {a}{2}+\frac {b x}{2}\right ) e^{n \sinh \left (\frac {a}{2}+\frac {b x}{2}\right )}}{n}-\frac {\int e^{n \sinh \left (\frac {a}{2}+\frac {b x}{2}\right )}d\sinh \left (\frac {a}{2}+\frac {b x}{2}\right )}{n}\right )}{b}\) |
\(\Big \downarrow \) 2624 |
\(\displaystyle \frac {4 \left (\frac {\sinh \left (\frac {a}{2}+\frac {b x}{2}\right ) e^{n \sinh \left (\frac {a}{2}+\frac {b x}{2}\right )}}{n}-\frac {e^{n \sinh \left (\frac {a}{2}+\frac {b x}{2}\right )}}{n^2}\right )}{b}\) |
3.11.7.3.1 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[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m _.), x_Symbol] :> Simp[(c + d*x)^m*((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Simp[d*(m/(f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x)))^ n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2* m] && !TrueQ[$UseGamma]
Int[((F_)^(v_))^(n_.), x_Symbol] :> Simp[(F^v)^n/(n*Log[F]*D[v, x]), x] /; FreeQ[{F, n}, x] && LinearQ[v, x]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa ctors[Sin[v], x]}, d/Coefficient[v, x, 1] Subst[Int[SubstFor[1, Sin[v]/d, u/Cos[v], x], x], x, Sin[v]/d]], x] /; !FalseQ[v] && FunctionOfQ[NonfreeF actors[Sin[v], x], u/Cos[v], x]]
Time = 0.64 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.02
method | result | size |
risch | \(\frac {2 \left (n \,{\mathrm e}^{b x +a}-n -2 \,{\mathrm e}^{\frac {a}{2}+\frac {b x}{2}}\right ) {\mathrm e}^{-\frac {a}{2}-\frac {b x}{2}-\frac {n \,{\mathrm e}^{-\frac {a}{2}-\frac {b x}{2}}}{2}+\frac {n \,{\mathrm e}^{\frac {a}{2}+\frac {b x}{2}}}{2}}}{n^{2} b}\) | \(65\) |
2/n^2/b*(n*exp(b*x+a)-n-2*exp(1/2*a+1/2*b*x))*exp(-1/2*a-1/2*b*x-1/2*n*exp (-1/2*a-1/2*b*x)+1/2*n*exp(1/2*a+1/2*b*x))
Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.42 \[ \int e^{n \sinh \left (\frac {1}{2} (a+b x)\right )} \sinh (a+b x) \, dx=\frac {4 \, {\left ({\left (n \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - 1\right )} \cosh \left (n \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right ) + {\left (n \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - 1\right )} \sinh \left (n \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right )\right )}}{b n^{2} \cosh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - b n^{2} \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2}} \]
4*((n*sinh(1/2*b*x + 1/2*a) - 1)*cosh(n*sinh(1/2*b*x + 1/2*a)) + (n*sinh(1 /2*b*x + 1/2*a) - 1)*sinh(n*sinh(1/2*b*x + 1/2*a)))/(b*n^2*cosh(1/2*b*x + 1/2*a)^2 - b*n^2*sinh(1/2*b*x + 1/2*a)^2)
\[ \int e^{n \sinh \left (\frac {1}{2} (a+b x)\right )} \sinh (a+b x) \, dx=\int e^{n \sinh {\left (\frac {a}{2} + \frac {b x}{2} \right )}} \sinh {\left (a + b x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (50) = 100\).
Time = 0.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.83 \[ \int e^{n \sinh \left (\frac {1}{2} (a+b x)\right )} \sinh (a+b x) \, dx=\frac {2 \, e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, n e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} - \frac {1}{2} \, n e^{\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a\right )} + \frac {1}{2} \, a\right )}}{b n} - \frac {2 \, e^{\left (-\frac {1}{2} \, b x + \frac {1}{2} \, n e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} - \frac {1}{2} \, n e^{\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a\right )} - \frac {1}{2} \, a\right )}}{b n} - \frac {4 \, e^{\left (\frac {1}{2} \, n e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} - \frac {1}{2} \, n e^{\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a\right )}\right )}}{b n^{2}} \]
2*e^(1/2*b*x + 1/2*n*e^(1/2*b*x + 1/2*a) - 1/2*n*e^(-1/2*b*x - 1/2*a) + 1/ 2*a)/(b*n) - 2*e^(-1/2*b*x + 1/2*n*e^(1/2*b*x + 1/2*a) - 1/2*n*e^(-1/2*b*x - 1/2*a) - 1/2*a)/(b*n) - 4*e^(1/2*n*e^(1/2*b*x + 1/2*a) - 1/2*n*e^(-1/2* b*x - 1/2*a))/(b*n^2)
Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (50) = 100\).
Time = 0.32 (sec) , antiderivative size = 255, normalized size of antiderivative = 3.98 \[ \int e^{n \sinh \left (\frac {1}{2} (a+b x)\right )} \sinh (a+b x) \, dx=\frac {2 \, {\left (n e^{\left (b x + \frac {1}{4} \, {\left (2 \, b x e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} + n e^{\left (b x + a\right )} - n\right )} e^{\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a\right )} - \frac {1}{4} \, {\left (2 \, b x e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} - n e^{\left (b x + a\right )} + n\right )} e^{\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a\right )} + a\right )} - n e^{\left (\frac {1}{4} \, {\left (2 \, b x e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} + n e^{\left (b x + a\right )} - n\right )} e^{\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a\right )} - \frac {1}{4} \, {\left (2 \, b x e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} - n e^{\left (b x + a\right )} + n\right )} e^{\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a\right )}\right )} - 2 \, e^{\left (\frac {1}{2} \, b x + \frac {1}{4} \, {\left (2 \, b x e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} + n e^{\left (b x + a\right )} - n\right )} e^{\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a\right )} - \frac {1}{4} \, {\left (2 \, b x e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} - n e^{\left (b x + a\right )} + n\right )} e^{\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a\right )} + \frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a\right )}}{b n^{2}} \]
2*(n*e^(b*x + 1/4*(2*b*x*e^(1/2*b*x + 1/2*a) + n*e^(b*x + a) - n)*e^(-1/2* b*x - 1/2*a) - 1/4*(2*b*x*e^(1/2*b*x + 1/2*a) - n*e^(b*x + a) + n)*e^(-1/2 *b*x - 1/2*a) + a) - n*e^(1/4*(2*b*x*e^(1/2*b*x + 1/2*a) + n*e^(b*x + a) - n)*e^(-1/2*b*x - 1/2*a) - 1/4*(2*b*x*e^(1/2*b*x + 1/2*a) - n*e^(b*x + a) + n)*e^(-1/2*b*x - 1/2*a)) - 2*e^(1/2*b*x + 1/4*(2*b*x*e^(1/2*b*x + 1/2*a) + n*e^(b*x + a) - n)*e^(-1/2*b*x - 1/2*a) - 1/4*(2*b*x*e^(1/2*b*x + 1/2*a ) - n*e^(b*x + a) + n)*e^(-1/2*b*x - 1/2*a) + 1/2*a))*e^(-1/2*b*x - 1/2*a) /(b*n^2)
Time = 0.00 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.98 \[ \int e^{n \sinh \left (\frac {1}{2} (a+b x)\right )} \sinh (a+b x) \, dx=\frac {2\,{\mathrm {e}}^{-\frac {a}{2}}\,{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^{-\frac {b\,x}{2}}\,{\mathrm {e}}^a\,{\mathrm {e}}^{-\frac {n\,{\mathrm {e}}^{-\frac {a}{2}}\,{\mathrm {e}}^{-\frac {b\,x}{2}}}{2}}\,{\mathrm {e}}^{\frac {n\,{\mathrm {e}}^{a/2}\,{\mathrm {e}}^{\frac {b\,x}{2}}}{2}}}{b\,n}-\frac {2\,{\mathrm {e}}^{-\frac {a}{2}}\,{\mathrm {e}}^{-\frac {b\,x}{2}}\,{\mathrm {e}}^{-\frac {n\,{\mathrm {e}}^{-\frac {a}{2}}\,{\mathrm {e}}^{-\frac {b\,x}{2}}}{2}}\,{\mathrm {e}}^{\frac {n\,{\mathrm {e}}^{a/2}\,{\mathrm {e}}^{\frac {b\,x}{2}}}{2}}}{b\,n}-\frac {4\,{\mathrm {e}}^{-\frac {n\,{\mathrm {e}}^{-\frac {a}{2}}\,{\mathrm {e}}^{-\frac {b\,x}{2}}}{2}}\,{\mathrm {e}}^{\frac {n\,{\mathrm {e}}^{a/2}\,{\mathrm {e}}^{\frac {b\,x}{2}}}{2}}}{b\,n^2} \]
(2*exp(-a/2)*exp(b*x)*exp(-(b*x)/2)*exp(a)*exp(-(n*exp(-a/2)*exp(-(b*x)/2) )/2)*exp((n*exp(a/2)*exp((b*x)/2))/2))/(b*n) - (2*exp(-a/2)*exp(-(b*x)/2)* exp(-(n*exp(-a/2)*exp(-(b*x)/2))/2)*exp((n*exp(a/2)*exp((b*x)/2))/2))/(b*n ) - (4*exp(-(n*exp(-a/2)*exp(-(b*x)/2))/2)*exp((n*exp(a/2)*exp((b*x)/2))/2 ))/(b*n^2)