\(\int e^{n \sin (c (a+b x))} \cos (a c+b c x) \, dx\) [686]

Optimal result

Integrand size = 22, antiderivative size = 23 \[ \int e^{n \sin (c (a+b x))} \cos (a c+b c x) \, dx=\frac {e^{n \sin (a c+b c x)}}{b c n} \]

[Out]

exp(n*sin(b*c*x+a*c))/b/c/n

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4419, 2225} \[ \int e^{n \sin (c (a+b x))} \cos (a c+b c x) \, dx=\frac {e^{n \sin (a c+b c x)}}{b c n} \]

[In]

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

[Out]

E^(n*Sin[a*c + b*c*x])/(b*c*n)

Rule 2225

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 4419

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int e^{n x} \, dx,x,\sin (a c+b c x)\right )}{b c} \\ & = \frac {e^{n \sin (a c+b c x)}}{b c n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int e^{n \sin (c (a+b x))} \cos (a c+b c x) \, dx=\frac {e^{n \sin (c (a+b x))}}{b c n} \]

[In]

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

[Out]

E^(n*Sin[c*(a + b*x)])/(b*c*n)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {{\mathrm e}^{n \sin \left (c \left (x b +a \right )\right )}}{b c n}\)	\(22\)
default	\(\frac {{\mathrm e}^{n \sin \left (c \left (x b +a \right )\right )}}{b c n}\)	\(22\)
risch	\(\frac {{\mathrm e}^{n \sin \left (c \left (x b +a \right )\right )}}{b c n}\)	\(22\)
parallelrisch	\(\frac {{\mathrm e}^{n \sin \left (c \left (x b +a \right )\right )}}{b c n}\)	\(22\)

[In]

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

[Out]

exp(n*sin(c*(b*x+a)))/b/c/n

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int e^{n \sin (c (a+b x))} \cos (a c+b c x) \, dx=\frac {e^{\left (n \sin \left (b c x + a c\right )\right )}}{b c n} \]

[In]

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

[Out]

e^(n*sin(b*c*x + a*c))/(b*c*n)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (17) = 34\).

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.09 \[ \int e^{n \sin (c (a+b x))} \cos (a c+b c x) \, dx=\begin {cases} x & \text {for}\: b = 0 \wedge c = 0 \wedge n = 0 \\x e^{n \sin {\left (a c \right )}} \cos {\left (a c \right )} & \text {for}\: b = 0 \\x & \text {for}\: c = 0 \\\frac {\sin {\left (a c + b c x \right )}}{b c} & \text {for}\: n = 0 \\\frac {e^{n \sin {\left (a c + b c x \right )}}}{b c n} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int e^{n \sin (c (a+b x))} \cos (a c+b c x) \, dx=\frac {e^{\left (n \sin \left (b c x + a c\right )\right )}}{b c n} \]

[In]

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

[Out]

e^(n*sin(b*c*x + a*c))/(b*c*n)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int e^{n \sin (c (a+b x))} \cos (a c+b c x) \, dx=\frac {e^{\left (n \sin \left (b c x + a c\right )\right )}}{b c n} \]

[In]

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

[Out]

e^(n*sin(b*c*x + a*c))/(b*c*n)

Mupad [B] (verification not implemented)

Time = 26.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int e^{n \sin (c (a+b x))} \cos (a c+b c x) \, dx=\frac {{\mathrm {e}}^{n\,\sin \left (a\,c+b\,c\,x\right )}}{b\,c\,n} \]

[In]

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

[Out]

exp(n*sin(a*c + b*c*x))/(b*c*n)