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

Optimal result

Integrand size = 10, antiderivative size = 37 \[ \int e^x \sin (a+b x) \, dx=-\frac {b e^x \cos (a+b x)}{1+b^2}+\frac {e^x \sin (a+b x)}{1+b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4517} \[ \int e^x \sin (a+b x) \, dx=\frac {e^x \sin (a+b x)}{b^2+1}-\frac {b e^x \cos (a+b x)}{b^2+1} \]

[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*} \text {integral}& = -\frac {b e^x \cos (a+b x)}{1+b^2}+\frac {e^x \sin (a+b x)}{1+b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int e^x \sin (a+b x) \, dx=\frac {e^x (-b \cos (a+b x)+\sin (a+b x))}{1+b^2} \]

[In]

Integrate[E^x*Sin[a + b*x],x]

[Out]

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

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{x} \left (-b \cos \left (x b +a \right )+\sin \left (x b +a \right )\right )}{b^{2}+1}\)	\(27\)
default	\(-\frac {b \,{\mathrm e}^{x} \cos \left (x b +a \right )}{b^{2}+1}+\frac {{\mathrm e}^{x} \sin \left (x b +a \right )}{b^{2}+1}\)	\(36\)
risch	\(\frac {{\mathrm e}^{x} \left (2 b \cos \left (x b +a \right )-2 \sin \left (x b +a \right )\right )}{2 \left (-b +i\right ) \left (i+b \right )}\)	\(37\)
norman	\(\frac {\frac {b \,{\mathrm e}^{x} \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{b^{2}+1}-\frac {b \,{\mathrm e}^{x}}{b^{2}+1}+\frac {2 \,{\mathrm e}^{x} \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{b^{2}+1}}{1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}\)	\(72\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int e^x \sin (a+b x) \, dx=-\frac {b \cos \left (b x + a\right ) e^{x} - e^{x} \sin \left (b x + a\right )}{b^{2} + 1} \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 114, normalized size of antiderivative = 3.08 \[ \int e^x \sin (a+b x) \, dx=\begin {cases} \frac {x e^{x} \sin {\left (a - i x \right )}}{2} + \frac {i x e^{x} \cos {\left (a - i x \right )}}{2} + \frac {e^{x} \sin {\left (a - i x \right )}}{2} & \text {for}\: b = - i \\\frac {x e^{x} \sin {\left (a + i x \right )}}{2} - \frac {i x e^{x} \cos {\left (a + i x \right )}}{2} + \frac {i e^{x} \cos {\left (a + i x \right )}}{2} & \text {for}\: b = i \\- \frac {b e^{x} \cos {\left (a + b x \right )}}{b^{2} + 1} + \frac {e^{x} \sin {\left (a + b x \right )}}{b^{2} + 1} & \text {otherwise} \end {cases} \]

[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 + exp(x)*sin(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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int e^x \sin (a+b x) \, dx=-\frac {{\left (b \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} e^{x}}{b^{2} + 1} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int e^x \sin (a+b x) \, dx=-{\left (\frac {b \cos \left (b x + a\right )}{b^{2} + 1} - \frac {\sin \left (b x + a\right )}{b^{2} + 1}\right )} e^{x} \]

[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

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int e^x \sin (a+b x) \, dx=\frac {{\mathrm {e}}^x\,\left (\sin \left (a+b\,x\right )-b\,\cos \left (a+b\,x\right )\right )}{b^2+1} \]

[In]

int(exp(x)*sin(a + b*x),x)

[Out]

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