### 3.1004 $$\int e^{n \sinh (a+b x)} \sinh (2 a+2 b x) \, dx$$

Optimal. Leaf size=43 $\frac{2 \sinh (a+b x) e^{n \sinh (a+b x)}}{b n}-\frac{2 e^{n \sinh (a+b x)}}{b n^2}$

[Out]

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

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Rubi [A]  time = 0.039057, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.15, Rules used = {12, 2176, 2194} $\frac{2 \sinh (a+b x) e^{n \sinh (a+b x)}}{b n}-\frac{2 e^{n \sinh (a+b x)}}{b n^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0557636, size = 28, normalized size = 0.65 $\frac{2 e^{n \sinh (a+b x)} (n \sinh (a+b x)-1)}{b n^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.071, size = 61, normalized size = 1.4 \begin{align*}{\frac{n{{\rm e}^{2\,bx+2\,a}}-n-2\,{{\rm e}^{bx+a}}}{{n}^{2}b}{{\rm e}^{-bx-a+{\frac{n{{\rm e}^{bx+a}}}{2}}-{\frac{n{{\rm e}^{-bx-a}}}{2}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.32767, size = 140, normalized size = 3.26 \begin{align*} \frac{e^{\left (b x + \frac{1}{2} \, n e^{\left (b x + a\right )} - \frac{1}{2} \, n e^{\left (-b x - a\right )} + a\right )}}{b n} - \frac{e^{\left (-b x + \frac{1}{2} \, n e^{\left (b x + a\right )} - \frac{1}{2} \, n e^{\left (-b x - a\right )} - a\right )}}{b n} - \frac{2 \, e^{\left (\frac{1}{2} \, n e^{\left (b x + a\right )} - \frac{1}{2} \, n e^{\left (-b x - a\right )}\right )}}{b n^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.03058, size = 193, normalized size = 4.49 \begin{align*} \frac{2 \,{\left ({\left (n \sinh \left (b x + a\right ) - 1\right )} \cosh \left (n \sinh \left (b x + a\right )\right ) +{\left (n \sinh \left (b x + a\right ) - 1\right )} \sinh \left (n \sinh \left (b x + a\right )\right )\right )}}{b n^{2} \cosh \left (b x + a\right )^{2} - b n^{2} \sinh \left (b x + a\right )^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(e^(n*sinh(b*x + a))*sinh(2*b*x + 2*a), x)