3.661 \(\int e^{n \cos (c (a+b x))} \sin (a c+b c x) \, dx\)

Optimal. Leaf size=24 \[ -\frac {e^{n \cos (a c+b c x)}}{b c n} \]

[Out]

-exp(n*cos(b*c*x+a*c))/b/c/n

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Rubi [A]  time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4335, 2194} \[ -\frac {e^{n \cos (a c+b c x)}}{b c n} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*Cos[c*(a + b*x)])*Sin[a*c + b*c*x],x]

[Out]

-(E^(n*Cos[a*c + b*c*x])/(b*c*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 4335

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dist[d/(
b*c), Subst[Int[SubstFor[1, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a
+ b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 23, normalized size = 0.96 \[ -\frac {e^{n \cos (c (a+b x))}}{b c n} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*Cos[c*(a + b*x)])*Sin[a*c + b*c*x],x]

[Out]

-(E^(n*Cos[c*(a + b*x)])/(b*c*n))

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fricas [A]  time = 0.90, size = 23, normalized size = 0.96 \[ -\frac {e^{\left (n \cos \left (b c x + a c\right )\right )}}{b c n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*cos(c*(b*x+a)))*sin(b*c*x+a*c),x, algorithm="fricas")

[Out]

-e^(n*cos(b*c*x + a*c))/(b*c*n)

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giac [A]  time = 0.15, size = 23, normalized size = 0.96 \[ -\frac {e^{\left (n \cos \left (b c x + a c\right )\right )}}{b c n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*cos(c*(b*x+a)))*sin(b*c*x+a*c),x, algorithm="giac")

[Out]

-e^(n*cos(b*c*x + a*c))/(b*c*n)

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maple [A]  time = 0.03, size = 24, normalized size = 1.00 \[ -\frac {{\mathrm e}^{n \cos \left (b c x +a c \right )}}{b c n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*cos(c*(b*x+a)))*sin(b*c*x+a*c),x)

[Out]

-exp(n*cos(b*c*x+a*c))/b/c/n

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maxima [A]  time = 0.32, size = 23, normalized size = 0.96 \[ -\frac {e^{\left (n \cos \left (b c x + a c\right )\right )}}{b c n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*cos(c*(b*x+a)))*sin(b*c*x+a*c),x, algorithm="maxima")

[Out]

-e^(n*cos(b*c*x + a*c))/(b*c*n)

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mupad [B]  time = 3.02, size = 23, normalized size = 0.96 \[ -\frac {{\mathrm {e}}^{n\,\cos \left (a\,c+b\,c\,x\right )}}{b\,c\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*cos(c*(a + b*x)))*sin(a*c + b*c*x),x)

[Out]

-exp(n*cos(a*c + b*c*x))/(b*c*n)

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sympy [A]  time = 2.23, size = 54, normalized size = 2.25 \[ \begin {cases} 0 & \text {for}\: c = 0 \wedge \left (b = 0 \vee c = 0\right ) \wedge \left (c = 0 \vee n = 0\right ) \\x e^{n \cos {\left (a c \right )}} \sin {\left (a c \right )} & \text {for}\: b = 0 \\- \frac {\cos {\left (a c + b c x \right )}}{b c} & \text {for}\: n = 0 \\- \frac {e^{n \cos {\left (a c + b c x \right )}}}{b c n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*cos(c*(b*x+a)))*sin(b*c*x+a*c),x)

[Out]

Piecewise((0, Eq(c, 0) & (Eq(b, 0) | Eq(c, 0)) & (Eq(c, 0) | Eq(n, 0))), (x*exp(n*cos(a*c))*sin(a*c), Eq(b, 0)
), (-cos(a*c + b*c*x)/(b*c), Eq(n, 0)), (-exp(n*cos(a*c + b*c*x))/(b*c*n), True))

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