\(\int e^{n \sin (a+b x)} \cos (a+b x) \, dx\) [684]

Optimal result

Integrand size = 17, antiderivative size = 17 \[ \int e^{n \sin (a+b x)} \cos (a+b x) \, dx=\frac {e^{n \sin (a+b x)}}{b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4419, 2225} \[ \int e^{n \sin (a+b x)} \cos (a+b x) \, dx=\frac {e^{n \sin (a+b x)}}{b n} \]

[In]

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

[Out]

E^(n*Sin[a + b*x])/(b*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+b x)\right )}{b} \\ & = \frac {e^{n \sin (a+b x)}}{b n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int e^{n \sin (a+b x)} \cos (a+b x) \, dx=\frac {e^{n \sin (a+b x)}}{b n} \]

[In]

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

[Out]

E^(n*Sin[a + b*x])/(b*n)

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {{\mathrm e}^{n \sin \left (x b +a \right )}}{b n}\)	\(17\)
default	\(\frac {{\mathrm e}^{n \sin \left (x b +a \right )}}{b n}\)	\(17\)
risch	\(\frac {{\mathrm e}^{n \sin \left (x b +a \right )}}{b n}\)	\(17\)
parallelrisch	\(\frac {{\mathrm e}^{n \sin \left (x b +a \right )}}{b n}\)	\(17\)
norman	\(\frac {\frac {{\mathrm e}^{\frac {2 n \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}}}{n b}+\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2} {\mathrm e}^{\frac {2 n \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}}}{n b}}{1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}\)	\(99\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int e^{n \sin (a+b x)} \cos (a+b x) \, dx=\frac {e^{\left (n \sin \left (b x + a\right )\right )}}{b n} \]

[In]

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

[Out]

e^(n*sin(b*x + a))/(b*n)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (12) = 24\).

Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 2.12 \[ \int e^{n \sin (a+b x)} \cos (a+b x) \, dx=\begin {cases} x \cos {\left (a \right )} & \text {for}\: b = 0 \wedge n = 0 \\x e^{n \sin {\left (a \right )}} \cos {\left (a \right )} & \text {for}\: b = 0 \\\frac {\sin {\left (a + b x \right )}}{b} & \text {for}\: n = 0 \\\frac {e^{n \sin {\left (a + b x \right )}}}{b n} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int e^{n \sin (a+b x)} \cos (a+b x) \, dx=\frac {e^{\left (n \sin \left (b x + a\right )\right )}}{b n} \]

[In]

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

[Out]

e^(n*sin(b*x + a))/(b*n)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int e^{n \sin (a+b x)} \cos (a+b x) \, dx=\frac {e^{\left (n \sin \left (b x + a\right )\right )}}{b n} \]

[In]

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

[Out]

e^(n*sin(b*x + a))/(b*n)

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int e^{n \sin (a+b x)} \cos (a+b x) \, dx=\frac {{\mathrm {e}}^{n\,\sin \left (a+b\,x\right )}}{b\,n} \]

[In]

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

[Out]

exp(n*sin(a + b*x))/(b*n)