3.1005 \(\int e^{n \sinh (a+b x)} \sinh (2 (a+b x)) \, dx\)

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

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Rubi [A]  time = 0.04, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, 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 + 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+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*}

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Mathematica [A]  time = 0.03, 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 + b*x)],x]

[Out]

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

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fricas [A]  time = 0.47, size = 73, normalized size = 1.70 \[ \frac {2 \, {\left ({\left (n \sinh \left (b x + a\right ) - 1\right )} \cosh \left (n \sinh \left (b x + a\right )\right ) + {\left (n \sinh \left (b x + a\right ) - 1\right )} \sinh \left (n \sinh \left (b x + a\right )\right )\right )}}{b n^{2} \cosh \left (b x + a\right )^{2} - b n^{2} \sinh \left (b x + a\right )^{2}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{\left (n \sinh \left (b x + a\right )\right )} \sinh \left (2 \, b x + 2 \, a\right )\,{d x} \]

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)

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maple [A]  time = 0.22, size = 61, normalized size = 1.42 \[ \frac {\left ({\mathrm e}^{2 b x +2 a} n -n -2 \,{\mathrm e}^{b x +a}\right ) {\mathrm e}^{-b x -a +\frac {n \,{\mathrm e}^{b x +a}}{2}-\frac {n \,{\mathrm e}^{-b x -a}}{2}}}{n^{2} b} \]

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]

1/n^2/b*(exp(2*b*x+2*a)*n-n-2*exp(b*x+a))*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 = 0.40, size = 104, normalized size = 2.42 \[ \frac {e^{\left (b x + \frac {1}{2} \, n e^{\left (b x + a\right )} - \frac {1}{2} \, n e^{\left (-b x - a\right )} + a\right )}}{b n} - \frac {e^{\left (-b x + \frac {1}{2} \, n e^{\left (b x + a\right )} - \frac {1}{2} \, n e^{\left (-b x - a\right )} - a\right )}}{b n} - \frac {2 \, e^{\left (\frac {1}{2} \, n e^{\left (b x + a\right )} - \frac {1}{2} \, n e^{\left (-b x - a\right )}\right )}}{b n^{2}} \]

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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mupad [B]  time = 0.00, size = 108, normalized size = 2.51 \[ \frac {{\mathrm {e}}^{\frac {n\,{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a}{2}}\,{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,{\mathrm {e}}^{-\frac {n\,{\mathrm {e}}^{-a}\,{\mathrm {e}}^{-b\,x}}{2}}}{b\,n}-\frac {{\mathrm {e}}^{-a}\,{\mathrm {e}}^{\frac {n\,{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a}{2}}\,{\mathrm {e}}^{-b\,x}\,{\mathrm {e}}^{-\frac {n\,{\mathrm {e}}^{-a}\,{\mathrm {e}}^{-b\,x}}{2}}}{b\,n}-\frac {2\,{\mathrm {e}}^{\frac {n\,{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a}{2}}\,{\mathrm {e}}^{-\frac {n\,{\mathrm {e}}^{-a}\,{\mathrm {e}}^{-b\,x}}{2}}}{b\,n^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp((n*exp(b*x)*exp(a))/2)*exp(b*x)*exp(a)*exp(-(n*exp(-a)*exp(-b*x))/2))/(b*n) - (exp(-a)*exp((n*exp(b*x)*ex
p(a))/2)*exp(-b*x)*exp(-(n*exp(-a)*exp(-b*x))/2))/(b*n) - (2*exp((n*exp(b*x)*exp(a))/2)*exp(-(n*exp(-a)*exp(-b
*x))/2))/(b*n^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{n \sinh {\left (a + b x \right )}} \sinh {\left (2 a + 2 b x \right )}\, dx \]

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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