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

Optimal. Leaf size=17 \[ \frac {e^{n \sinh (a+b x)}}{b n} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4336, 2194} \[ \frac {e^{n \sinh (a+b x)}}{b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

E^(n*Sinh[a + b*x])/(b*n)

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]

Rule 4336

Int[Cosh[(c_.)*((a_.) + (b_.)*(x_))]*(u_), x_Symbol] :> With[{d = FreeFactors[Sinh[c*(a + b*x)], x]}, Dist[d/(
b*c), Subst[Int[SubstFor[1, Sinh[c*(a + b*x)]/d, u, x], x], x, Sinh[c*(a + b*x)]/d], x] /; FunctionOfQ[Sinh[c*
(a + b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 17, normalized size = 1.00 \[ \frac {e^{n \sinh (a+b x)}}{b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

E^(n*Sinh[a + b*x])/(b*n)

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fricas [A]  time = 0.43, size = 26, normalized size = 1.53 \[ \frac {\cosh \left (n \sinh \left (b x + a\right )\right ) + \sinh \left (n \sinh \left (b x + a\right )\right )}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cosh(n*sinh(b*x + a)) + sinh(n*sinh(b*x + a)))/(b*n)

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giac [A]  time = 0.15, size = 30, normalized size = 1.76 \[ \frac {e^{\left (\frac {1}{2} \, n e^{\left (b x + a\right )} - \frac {1}{2} \, n e^{\left (-b x - a\right )}\right )}}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 17, normalized size = 1.00 \[ \frac {{\mathrm e}^{n \sinh \left (b x +a \right )}}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.30, size = 16, normalized size = 0.94 \[ \frac {e^{\left (n \sinh \left (b x + a\right )\right )}}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(n*sinh(b*x + a))/(b*n)

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mupad [B]  time = 0.07, size = 16, normalized size = 0.94 \[ \frac {{\mathrm {e}}^{n\,\mathrm {sinh}\left (a+b\,x\right )}}{b\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(n*sinh(a + b*x))/(b*n)

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sympy [A]  time = 0.42, size = 36, normalized size = 2.12 \[ \begin {cases} x \cosh {\relax (a )} & \text {for}\: b = 0 \wedge n = 0 \\x e^{n \sinh {\relax (a )}} \cosh {\relax (a )} & \text {for}\: b = 0 \\\frac {\sinh {\left (a + b x \right )}}{b} & \text {for}\: n = 0 \\\frac {e^{n \sinh {\left (a + b x \right )}}}{b n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*cosh(a), Eq(b, 0) & Eq(n, 0)), (x*exp(n*sinh(a))*cosh(a), Eq(b, 0)), (sinh(a + b*x)/b, Eq(n, 0)),
(exp(n*sinh(a + b*x))/(b*n), True))

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