3.1.70 \(\int e^{2 \sin (x)} \cos (x) \, dx\) [70]

Optimal. Leaf size=10 \[ \frac {1}{2} e^{2 \sin (x)} \]

[Out]

1/2*exp(2*sin(x))

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Rubi [A]
time = 0.01, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4419, 2225} \begin {gather*} \frac {1}{2} e^{2 \sin (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*Sin[x])*Cos[x],x]

[Out]

E^(2*Sin[x])/2

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*} \int e^{2 \sin (x)} \cos (x) \, dx &=\text {Subst}\left (\int e^{2 x} \, dx,x,\sin (x)\right )\\ &=\frac {1}{2} e^{2 \sin (x)}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 10, normalized size = 1.00 \begin {gather*} \frac {1}{2} e^{2 \sin (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*Sin[x])*Cos[x],x]

[Out]

E^(2*Sin[x])/2

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Mathics [A]
time = 1.71, size = 8, normalized size = 0.80 \begin {gather*} \frac {E^{2 \text {Sin}\left [x\right ]}}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[Cos[x]*E^(2*Sin[x]),x]')

[Out]

E ^ (2 Sin[x]) / 2

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Maple [A]
time = 0.03, size = 8, normalized size = 0.80

method	result	size
derivativedivides	\(\frac {{\mathrm e}^{2 \sin \left (x \right )}}{2}\)	\(8\)
default	\(\frac {{\mathrm e}^{2 \sin \left (x \right )}}{2}\)	\(8\)
risch	\(\frac {{\mathrm e}^{2 \sin \left (x \right )}}{2}\)	\(8\)
norman	\(\frac {\frac {\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) {\mathrm e}^{\frac {4 \tan \left (\frac {x}{2}\right )}{1+\tan ^{2}\left (\frac {x}{2}\right )}}}{2}+\frac {{\mathrm e}^{\frac {4 \tan \left (\frac {x}{2}\right )}{1+\tan ^{2}\left (\frac {x}{2}\right )}}}{2}}{1+\tan ^{2}\left (\frac {x}{2}\right )}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*sin(x))*cos(x),x,method=_RETURNVERBOSE)

[Out]

1/2*exp(2*sin(x))

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Maxima [A]
time = 0.25, size = 7, normalized size = 0.70 \begin {gather*} \frac {1}{2} \, e^{\left (2 \, \sin \left (x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*sin(x))*cos(x),x, algorithm="maxima")

[Out]

1/2*e^(2*sin(x))

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Fricas [A]
time = 0.37, size = 7, normalized size = 0.70 \begin {gather*} \frac {1}{2} \, e^{\left (2 \, \sin \left (x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*sin(x))*cos(x),x, algorithm="fricas")

[Out]

1/2*e^(2*sin(x))

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Sympy [A]
time = 0.11, size = 7, normalized size = 0.70 \begin {gather*} \frac {e^{2 \sin {\left (x \right )}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*sin(x))*cos(x),x)

[Out]

exp(2*sin(x))/2

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Giac [A]
time = 0.00, size = 8, normalized size = 0.80 \begin {gather*} \frac {\mathrm {e}^{2 \sin x}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*sin(x))*cos(x),x)

[Out]

1/2*e^(2*sin(x))

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Mupad [B]
time = 0.10, size = 7, normalized size = 0.70 \begin {gather*} \frac {{\mathrm {e}}^{2\,\sin \left (x\right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*sin(x))*cos(x),x)

[Out]

exp(2*sin(x))/2

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