3.1.81 \(\int e^{a x} \sin (b x) \, dx\) [81]

Optimal. Leaf size=42 \[ -\frac {b e^{a x} \cos (b x)}{a^2+b^2}+\frac {a e^{a x} \sin (b x)}{a^2+b^2} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 42, 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 {a e^{a x} \sin (b x)}{a^2+b^2}-\frac {b e^{a x} \cos (b x)}{a^2+b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*E^(a*x)*Cos[b*x])/(a^2 + b^2)) + (a*E^(a*x)*Sin[b*x])/(a^2 + 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^{a x} \sin (b x) \, dx &=-\frac {b e^{a x} \cos (b x)}{a^2+b^2}+\frac {a e^{a x} \sin (b x)}{a^2+b^2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 29, normalized size = 0.69 \begin {gather*} \frac {e^{a x} (-b \cos (b x)+a \sin (b x))}{a^2+b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 2.91, size = 129, normalized size = 3.07 \begin {gather*} \text {Piecewise}\left [\left \{\left \{0,a\text {==}0\text {\&\&}b\text {==}0\right \},\left \{\frac {\left (b x \left (\text {Sin}\left [b x\right ]-I \text {Cos}\left [b x\right ]\right )-\text {Cos}\left [b x\right ]\right ) E^{-I b x}}{2 b},a\text {==}-I b\right \},\left \{\frac {\left (b x \left (\text {Sin}\left [b x\right ]+I \text {Cos}\left [b x\right ]\right )-\text {Cos}\left [b x\right ]\right ) E^{I b x}}{2 b},a\text {==}I b\right \}\right \},\frac {a E^{a x} \text {Sin}\left [b x\right ]}{a^2+b^2}-\frac {b \text {Cos}\left [b x\right ] E^{a x}}{a^2+b^2}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[E^(a*x)*Sin[b*x],x]')

[Out]

Piecewise[{{0, a == 0 && b == 0}, {(b x (Sin[b x] - I Cos[b x]) - Cos[b x]) E ^ (-I b x) / (2 b), a == -I b},
{(b x (Sin[b x] + I Cos[b x]) - Cos[b x]) E ^ (I b x) / (2 b), a == I b}}, a E ^ (a x) Sin[b x] / (a ^ 2 + b ^
2) - b Cos[b x] E ^ (a x) / (a ^ 2 + b ^ 2)]

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Maple [A]
time = 0.02, size = 41, normalized size = 0.98

method	result	size
default	\(-\frac {b \,{\mathrm e}^{a x} \cos \left (b x \right )}{a^{2}+b^{2}}+\frac {a \,{\mathrm e}^{a x} \sin \left (b x \right )}{a^{2}+b^{2}}\)	\(41\)
risch	\(-\frac {i {\mathrm e}^{x \left (i b +a \right )}}{2 \left (i b +a \right )}+\frac {i {\mathrm e}^{x \left (-i b +a \right )}}{-2 i b +2 a}\)	\(42\)
norman	\(\frac {\frac {b \,{\mathrm e}^{a x} \left (\tan ^{2}\left (\frac {b x}{2}\right )\right )}{a^{2}+b^{2}}-\frac {b \,{\mathrm e}^{a x}}{a^{2}+b^{2}}+\frac {2 a \,{\mathrm e}^{a x} \tan \left (\frac {b x}{2}\right )}{a^{2}+b^{2}}}{1+\tan ^{2}\left (\frac {b x}{2}\right )}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 29, normalized size = 0.69 \begin {gather*} -\frac {{\left (b \cos \left (b x\right ) - a \sin \left (b x\right )\right )} e^{\left (a x\right )}}{a^{2} + b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 33, normalized size = 0.79 \begin {gather*} -\frac {b \cos \left (b x\right ) e^{\left (a x\right )} - a e^{\left (a x\right )} \sin \left (b x\right )}{a^{2} + b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.29, size = 136, normalized size = 3.24 \begin {gather*} \begin {cases} 0 & \text {for}\: a = 0 \wedge b = 0 \\\frac {x e^{- i b x} \sin {\left (b x \right )}}{2} - \frac {i x e^{- i b x} \cos {\left (b x \right )}}{2} - \frac {e^{- i b x} \cos {\left (b x \right )}}{2 b} & \text {for}\: a = - i b \\\frac {x e^{i b x} \sin {\left (b x \right )}}{2} + \frac {i x e^{i b x} \cos {\left (b x \right )}}{2} - \frac {e^{i b x} \cos {\left (b x \right )}}{2 b} & \text {for}\: a = i b \\\frac {a e^{a x} \sin {\left (b x \right )}}{a^{2} + b^{2}} - \frac {b e^{a x} \cos {\left (b x \right )}}{a^{2} + b^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((0, Eq(a, 0) & Eq(b, 0)), (x*exp(-I*b*x)*sin(b*x)/2 - I*x*exp(-I*b*x)*cos(b*x)/2 - exp(-I*b*x)*cos(b
*x)/(2*b), Eq(a, -I*b)), (x*exp(I*b*x)*sin(b*x)/2 + I*x*exp(I*b*x)*cos(b*x)/2 - exp(I*b*x)*cos(b*x)/(2*b), Eq(
a, I*b)), (a*exp(a*x)*sin(b*x)/(a**2 + b**2) - b*exp(a*x)*cos(b*x)/(a**2 + b**2), True))

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Giac [A]
time = 0.00, size = 35, normalized size = 0.83 \begin {gather*} \mathrm {e}^{a x} \left (-\frac {b \cos \left (b x\right )}{a^{2}+b^{2}}+\frac {a \sin \left (b x\right )}{a^{2}+b^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.02, size = 29, normalized size = 0.69 \begin {gather*} -\frac {{\mathrm {e}}^{a\,x}\,\left (b\,\cos \left (b\,x\right )-a\,\sin \left (b\,x\right )\right )}{a^2+b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a*x)*sin(b*x),x)

[Out]

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

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