3.545 \(\int e^{m x} \sin ^3(x) \, dx\)

Optimal. Leaf size=82 \[ \frac{m e^{m x} \sin ^3(x)}{m^2+9}+\frac{6 m e^{m x} \sin (x)}{m^4+10 m^2+9}-\frac{6 e^{m x} \cos (x)}{m^4+10 m^2+9}-\frac{3 e^{m x} \sin ^2(x) \cos (x)}{m^2+9} \]

[Out]

(-6*E^(m*x)*Cos[x])/(9 + 10*m^2 + m^4) + (6*E^(m*x)*m*Sin[x])/(9 + 10*m^2 + m^4) - (3*E^(m*x)*Cos[x]*Sin[x]^2)
/(9 + m^2) + (E^(m*x)*m*Sin[x]^3)/(9 + m^2)

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Rubi [A]  time = 0.0353972, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4434, 4432} \[ \frac{m e^{m x} \sin ^3(x)}{m^2+9}+\frac{6 m e^{m x} \sin (x)}{m^4+10 m^2+9}-\frac{6 e^{m x} \cos (x)}{m^4+10 m^2+9}-\frac{3 e^{m x} \sin ^2(x) \cos (x)}{m^2+9} \]

Antiderivative was successfully verified.

[In]

Int[E^(m*x)*Sin[x]^3,x]

[Out]

(-6*E^(m*x)*Cos[x])/(9 + 10*m^2 + m^4) + (6*E^(m*x)*m*Sin[x])/(9 + 10*m^2 + m^4) - (3*E^(m*x)*Cos[x]*Sin[x]^2)
/(9 + m^2) + (E^(m*x)*m*Sin[x]^3)/(9 + m^2)

Rule 4434

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*
x))*Sin[d + e*x]^n)/(e^2*n^2 + b^2*c^2*Log[F]^2), x] + (Dist[(n*(n - 1)*e^2)/(e^2*n^2 + b^2*c^2*Log[F]^2), Int
[F^(c*(a + b*x))*Sin[d + e*x]^(n - 2), x], x] - Simp[(e*n*F^(c*(a + b*x))*Cos[d + e*x]*Sin[d + e*x]^(n - 1))/(
e^2*n^2 + b^2*c^2*Log[F]^2), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*n^2 + b^2*c^2*Log[F]^2, 0] && GtQ[
n, 1]

Rule 4432

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^{m x} \sin ^3(x) \, dx &=-\frac{3 e^{m x} \cos (x) \sin ^2(x)}{9+m^2}+\frac{e^{m x} m \sin ^3(x)}{9+m^2}+\frac{6 \int e^{m x} \sin (x) \, dx}{9+m^2}\\ &=-\frac{6 e^{m x} \cos (x)}{9+10 m^2+m^4}+\frac{6 e^{m x} m \sin (x)}{9+10 m^2+m^4}-\frac{3 e^{m x} \cos (x) \sin ^2(x)}{9+m^2}+\frac{e^{m x} m \sin ^3(x)}{9+m^2}\\ \end{align*}

Mathematica [A]  time = 0.166801, size = 64, normalized size = 0.78 \[ \frac{e^{m x} \left (-3 \left (m^2+9\right ) \cos (x)+3 \left (m^2+1\right ) \cos (3 x)-2 m \sin (x) \left (\left (m^2+1\right ) \cos (2 x)-m^2-13\right )\right )}{4 \left (m^4+10 m^2+9\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(m*x)*Sin[x]^3,x]

[Out]

(E^(m*x)*(-3*(9 + m^2)*Cos[x] + 3*(1 + m^2)*Cos[3*x] - 2*m*(-13 - m^2 + (1 + m^2)*Cos[2*x])*Sin[x]))/(4*(9 + 1
0*m^2 + m^4))

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Maple [A]  time = 0.023, size = 68, normalized size = 0.8 \begin{align*} -{\frac{3\,{{\rm e}^{mx}}\cos \left ( x \right ) }{4\,{m}^{2}+4}}+{\frac{3\,m{{\rm e}^{mx}}\sin \left ( x \right ) }{4\,{m}^{2}+4}}+{\frac{3\,{{\rm e}^{mx}}\cos \left ( 3\,x \right ) }{4\,{m}^{2}+36}}-{\frac{m{{\rm e}^{mx}}\sin \left ( 3\,x \right ) }{4\,{m}^{2}+36}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(m*x)*sin(x)^3,x)

[Out]

-3/4/(m^2+1)*exp(m*x)*cos(x)+3/4*m/(m^2+1)*exp(m*x)*sin(x)+3/4/(m^2+9)*exp(m*x)*cos(3*x)-1/4*m/(m^2+9)*exp(m*x
)*sin(3*x)

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Maxima [A]  time = 0.976098, size = 99, normalized size = 1.21 \begin{align*} \frac{3 \,{\left (m^{2} + 1\right )} \cos \left (3 \, x\right ) e^{\left (m x\right )} - 3 \,{\left (m^{2} + 9\right )} \cos \left (x\right ) e^{\left (m x\right )} -{\left (m^{3} + m\right )} e^{\left (m x\right )} \sin \left (3 \, x\right ) + 3 \,{\left (m^{3} + 9 \, m\right )} e^{\left (m x\right )} \sin \left (x\right )}{4 \,{\left (m^{4} + 10 \, m^{2} + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(m*x)*sin(x)^3,x, algorithm="maxima")

[Out]

1/4*(3*(m^2 + 1)*cos(3*x)*e^(m*x) - 3*(m^2 + 9)*cos(x)*e^(m*x) - (m^3 + m)*e^(m*x)*sin(3*x) + 3*(m^3 + 9*m)*e^
(m*x)*sin(x))/(m^4 + 10*m^2 + 9)

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Fricas [A]  time = 1.89389, size = 165, normalized size = 2.01 \begin{align*} \frac{{\left (m^{3} -{\left (m^{3} + m\right )} \cos \left (x\right )^{2} + 7 \, m\right )} e^{\left (m x\right )} \sin \left (x\right ) + 3 \,{\left ({\left (m^{2} + 1\right )} \cos \left (x\right )^{3} -{\left (m^{2} + 3\right )} \cos \left (x\right )\right )} e^{\left (m x\right )}}{m^{4} + 10 \, m^{2} + 9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(m*x)*sin(x)^3,x, algorithm="fricas")

[Out]

((m^3 - (m^3 + m)*cos(x)^2 + 7*m)*e^(m*x)*sin(x) + 3*((m^2 + 1)*cos(x)^3 - (m^2 + 3)*cos(x))*e^(m*x))/(m^4 + 1
0*m^2 + 9)

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Sympy [A]  time = 14.6775, size = 644, normalized size = 7.85 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(m*x)*sin(x)**3,x)

[Out]

Piecewise((x*exp(-3*I*x)*sin(x)**3/8 - 3*I*x*exp(-3*I*x)*sin(x)**2*cos(x)/8 - 3*x*exp(-3*I*x)*sin(x)*cos(x)**2
/8 + I*x*exp(-3*I*x)*cos(x)**3/8 - 7*exp(-3*I*x)*sin(x)**2*cos(x)/8 + 9*I*exp(-3*I*x)*sin(x)*cos(x)**2/8 + 5*e
xp(-3*I*x)*cos(x)**3/12, Eq(m, -3*I)), (3*x*exp(-I*x)*sin(x)**3/8 - 3*I*x*exp(-I*x)*sin(x)**2*cos(x)/8 + 3*x*e
xp(-I*x)*sin(x)*cos(x)**2/8 - 3*I*x*exp(-I*x)*cos(x)**3/8 - 5*exp(-I*x)*sin(x)**2*cos(x)/8 + I*exp(-I*x)*sin(x
)*cos(x)**2/8 - exp(-I*x)*cos(x)**3/4, Eq(m, -I)), (3*x*exp(I*x)*sin(x)**3/8 + 3*I*x*exp(I*x)*sin(x)**2*cos(x)
/8 + 3*x*exp(I*x)*sin(x)*cos(x)**2/8 + 3*I*x*exp(I*x)*cos(x)**3/8 - 5*exp(I*x)*sin(x)**2*cos(x)/8 - I*exp(I*x)
*sin(x)*cos(x)**2/8 - exp(I*x)*cos(x)**3/4, Eq(m, I)), (x*exp(3*I*x)*sin(x)**3/8 + 3*I*x*exp(3*I*x)*sin(x)**2*
cos(x)/8 - 3*x*exp(3*I*x)*sin(x)*cos(x)**2/8 - I*x*exp(3*I*x)*cos(x)**3/8 - 7*exp(3*I*x)*sin(x)**2*cos(x)/8 -
9*I*exp(3*I*x)*sin(x)*cos(x)**2/8 + 5*exp(3*I*x)*cos(x)**3/12, Eq(m, 3*I)), (m**3*exp(m*x)*sin(x)**3/(m**4 + 1
0*m**2 + 9) - 3*m**2*exp(m*x)*sin(x)**2*cos(x)/(m**4 + 10*m**2 + 9) + 7*m*exp(m*x)*sin(x)**3/(m**4 + 10*m**2 +
9) + 6*m*exp(m*x)*sin(x)*cos(x)**2/(m**4 + 10*m**2 + 9) - 9*exp(m*x)*sin(x)**2*cos(x)/(m**4 + 10*m**2 + 9) -
6*exp(m*x)*cos(x)**3/(m**4 + 10*m**2 + 9), True))

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Giac [A]  time = 1.10856, size = 85, normalized size = 1.04 \begin{align*} -\frac{1}{4} \,{\left (\frac{m \sin \left (3 \, x\right )}{m^{2} + 9} - \frac{3 \, \cos \left (3 \, x\right )}{m^{2} + 9}\right )} e^{\left (m x\right )} + \frac{3}{4} \,{\left (\frac{m \sin \left (x\right )}{m^{2} + 1} - \frac{\cos \left (x\right )}{m^{2} + 1}\right )} e^{\left (m x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(m*x)*sin(x)^3,x, algorithm="giac")

[Out]

-1/4*(m*sin(3*x)/(m^2 + 9) - 3*cos(3*x)/(m^2 + 9))*e^(m*x) + 3/4*(m*sin(x)/(m^2 + 1) - cos(x)/(m^2 + 1))*e^(m*
x)