3.1.73 \(\int e^x \sin (a+b x) \, dx\) [73]

Optimal. Leaf size=37 \[ -\frac {b e^x \cos (a+b x)}{1+b^2}+\frac {e^x \sin (a+b x)}{1+b^2} \]

[Out]

-b*exp(x)*cos(b*x+a)/(b^2+1)+exp(x)*sin(b*x+a)/(b^2+1)

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Rubi [A]
time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4517} \begin {gather*} \frac {e^x \sin (a+b x)}{b^2+1}-\frac {b e^x \cos (a+b x)}{b^2+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^x*Sin[a + b*x],x]

[Out]

-((b*E^x*Cos[a + b*x])/(1 + b^2)) + (E^x*Sin[a + b*x])/(1 + b^2)

Rule 4517

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(S
in[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x] - Simp[e*F^(c*(a + b*x))*(Cos[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rubi steps

\begin {align*} \int e^x \sin (a+b x) \, dx &=-\frac {b e^x \cos (a+b x)}{1+b^2}+\frac {e^x \sin (a+b x)}{1+b^2}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 27, normalized size = 0.73 \begin {gather*} \frac {e^x (-b \cos (a+b x)+\sin (a+b x))}{1+b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sin[a + b*x],x]

[Out]

(E^x*(-(b*Cos[a + b*x]) + Sin[a + b*x]))/(1 + b^2)

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Maple [A]
time = 0.06, size = 36, normalized size = 0.97

method	result	size
default	\(-\frac {b \,{\mathrm e}^{x} \cos \left (b x +a \right )}{b^{2}+1}+\frac {{\mathrm e}^{x} \sin \left (b x +a \right )}{b^{2}+1}\)	\(36\)
risch	\(-\frac {{\mathrm e}^{x} {\mathrm e}^{i b x} {\mathrm e}^{i a}}{2 \left (b -i\right )}-\frac {{\mathrm e}^{x} {\mathrm e}^{-i b x} {\mathrm e}^{-i a}}{2 \left (i+b \right )}\)	\(44\)
norman	\(\frac {\frac {b \,{\mathrm e}^{x} \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b^{2}+1}-\frac {b \,{\mathrm e}^{x}}{b^{2}+1}+\frac {2 \,{\mathrm e}^{x} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{b^{2}+1}}{1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*exp(x)*cos(b*x+a)/(b^2+1)+exp(x)*sin(b*x+a)/(b^2+1)

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Maxima [A]
time = 0.27, size = 28, normalized size = 0.76 \begin {gather*} -\frac {{\left (b \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} e^{x}}{b^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.00, size = 30, normalized size = 0.81 \begin {gather*} -\frac {b \cos \left (b x + a\right ) e^{x} - e^{x} \sin \left (b x + a\right )}{b^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.25, size = 116, normalized size = 3.14 \begin {gather*} \begin {cases} \frac {x e^{x} \sin {\left (a - i x \right )}}{2} + \frac {i x e^{x} \cos {\left (a - i x \right )}}{2} - \frac {i e^{x} \cos {\left (a - i x \right )}}{2} & \text {for}\: b = - i \\\frac {x e^{x} \sin {\left (a + i x \right )}}{2} - \frac {i x e^{x} \cos {\left (a + i x \right )}}{2} + \frac {i e^{x} \cos {\left (a + i x \right )}}{2} & \text {for}\: b = i \\- \frac {b e^{x} \cos {\left (a + b x \right )}}{b^{2} + 1} + \frac {e^{x} \sin {\left (a + b x \right )}}{b^{2} + 1} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*exp(x)*sin(a - I*x)/2 + I*x*exp(x)*cos(a - I*x)/2 - I*exp(x)*cos(a - I*x)/2, Eq(b, -I)), (x*exp(x
)*sin(a + I*x)/2 - I*x*exp(x)*cos(a + I*x)/2 + I*exp(x)*cos(a + I*x)/2, Eq(b, I)), (-b*exp(x)*cos(a + b*x)/(b*
*2 + 1) + exp(x)*sin(a + b*x)/(b**2 + 1), True))

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Giac [A]
time = 0.40, size = 35, normalized size = 0.95 \begin {gather*} -{\left (\frac {b \cos \left (b x + a\right )}{b^{2} + 1} - \frac {\sin \left (b x + a\right )}{b^{2} + 1}\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.10, size = 26, normalized size = 0.70 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (\sin \left (a+b\,x\right )-b\,\cos \left (a+b\,x\right )\right )}{b^2+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*sin(a + b*x),x)

[Out]

(exp(x)*(sin(a + b*x) - b*cos(a + b*x)))/(b^2 + 1)

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