### 3.1006 $$\int e^{n \sinh (\frac{a}{2}+\frac{b x}{2})} \sinh (a+b x) \, dx$$

Optimal. Leaf size=64 $\frac{4 \sinh \left (\frac{a}{2}+\frac{b x}{2}\right ) e^{n \sinh \left (\frac{a}{2}+\frac{b x}{2}\right )}}{b n}-\frac{4 e^{n \sinh \left (\frac{a}{2}+\frac{b x}{2}\right )}}{b n^2}$

[Out]

(-4*E^(n*Sinh[a/2 + (b*x)/2]))/(b*n^2) + (4*E^(n*Sinh[a/2 + (b*x)/2])*Sinh[a/2 + (b*x)/2])/(b*n)

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Rubi [A]  time = 0.038549, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {12, 2176, 2194} $\frac{4 \sinh \left (\frac{a}{2}+\frac{b x}{2}\right ) e^{n \sinh \left (\frac{a}{2}+\frac{b x}{2}\right )}}{b n}-\frac{4 e^{n \sinh \left (\frac{a}{2}+\frac{b x}{2}\right )}}{b n^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(n*Sinh[a/2 + (b*x)/2])*Sinh[a + b*x],x]

[Out]

(-4*E^(n*Sinh[a/2 + (b*x)/2]))/(b*n^2) + (4*E^(n*Sinh[a/2 + (b*x)/2])*Sinh[a/2 + (b*x)/2])/(b*n)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2176

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] - Dist[(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] &&  !\$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin{align*} \int e^{n \sinh \left (\frac{a}{2}+\frac{b x}{2}\right )} \sinh (a+b x) \, dx &=\frac{2 \operatorname{Subst}\left (\int 2 e^{n x} x \, dx,x,\sinh \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}{b}\\ &=\frac{4 \operatorname{Subst}\left (\int e^{n x} x \, dx,x,\sinh \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}{b}\\ &=\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}-\frac{4 \operatorname{Subst}\left (\int e^{n x} \, dx,x,\sinh \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}{b n}\\ &=-\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}\\ \end{align*}

Mathematica [A]  time = 0.0603013, size = 36, normalized size = 0.56 $\frac{4 e^{n \sinh \left (\frac{1}{2} (a+b x)\right )} \left (n \sinh \left (\frac{1}{2} (a+b x)\right )-1\right )}{b n^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(n*Sinh[a/2 + (b*x)/2])*Sinh[a + b*x],x]

[Out]

(4*E^(n*Sinh[(a + b*x)/2])*(-1 + n*Sinh[(a + b*x)/2]))/(b*n^2)

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Maple [A]  time = 0.07, size = 65, normalized size = 1. \begin{align*} 2\,{\frac{ \left ( n{{\rm e}^{bx+a}}-n-2\,{{\rm e}^{1/2\,bx+a/2}} \right ){{\rm e}^{-1/2\,bx-a/2+1/2\,n{{\rm e}^{1/2\,bx+a/2}}-1/2\,n{{\rm e}^{-1/2\,bx-a/2}}}}}{{n}^{2}b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*sinh(1/2*b*x+1/2*a))*sinh(b*x+a),x)

[Out]

2*(n*exp(b*x+a)-n-2*exp(1/2*b*x+1/2*a))/b/n^2*exp(-1/2*b*x-1/2*a+1/2*n*exp(1/2*b*x+1/2*a)-1/2*n*exp(-1/2*b*x-1
/2*a))

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Maxima [B]  time = 1.41199, size = 158, normalized size = 2.47 \begin{align*} \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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*sinh(1/2*a+1/2*b*x))*sinh(b*x+a),x, algorithm="maxima")

[Out]

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)

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Fricas [A]  time = 2.12055, size = 258, normalized size = 4.03 \begin{align*} \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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*sinh(1/2*a+1/2*b*x))*sinh(b*x+a),x, algorithm="fricas")

[Out]

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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{n \sinh{\left (\frac{a}{2} + \frac{b x}{2} \right )}} \sinh{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*sinh(1/2*a+1/2*b*x))*sinh(b*x+a),x)

[Out]

Integral(exp(n*sinh(a/2 + b*x/2))*sinh(a + b*x), x)

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Giac [B]  time = 1.23528, size = 344, normalized size = 5.38 \begin{align*} \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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*sinh(1/2*a+1/2*b*x))*sinh(b*x+a),x, algorithm="giac")

[Out]

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)