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

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

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Rubi [A]  time = 0.04, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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]

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*}

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Mathematica [A]  time = 0.18, 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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{m x} \sin ^3(x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 1.12, size = 65, normalized size = 0.79 \[ \frac {{\left (m^{3} - {\left (m^{3} + m\right )} \cos \relax (x)^{2} + 7 \, m\right )} e^{\left (m x\right )} \sin \relax (x) + 3 \, {\left ({\left (m^{2} + 1\right )} \cos \relax (x)^{3} - {\left (m^{2} + 3\right )} \cos \relax (x)\right )} e^{\left (m x\right )}}{m^{4} + 10 \, m^{2} + 9} \]

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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giac [A]  time = 0.61, size = 63, normalized size = 0.77 \[ -\frac {1}{4} \, {\left (\frac {m \sin \left (3 \, x\right )}{m^{2} + 9} - \frac {3 \, \cos \left (3 \, x\right )}{m^{2} + 9}\right )} e^{\left (m x\right )} + \frac {3}{4} \, {\left (\frac {m \sin \relax (x)}{m^{2} + 1} - \frac {\cos \relax (x)}{m^{2} + 1}\right )} e^{\left (m x\right )} \]

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)

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maple [C]  time = 0.11, size = 66, normalized size = 0.80

method	result	size
risch	\(\frac {i {\mathrm e}^{\left (3 i+m \right ) x}}{24 i+8 m}-\frac {3 i {\mathrm e}^{\left (i+m \right ) x}}{8 \left (i+m \right )}+\frac {3 i {\mathrm e}^{x \left (m -i\right )}}{8 \left (m -i\right )}-\frac {i {\mathrm e}^{x \left (m -3 i\right )}}{8 \left (m -3 i\right )}\)	\(66\)
norman	\(\frac {-\frac {6 \,{\mathrm e}^{m x}}{m^{4}+10 m^{2}+9}+\frac {6 \,{\mathrm e}^{m x} \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{m^{4}+10 m^{2}+9}+\frac {12 m \,{\mathrm e}^{m x} \tan \left (\frac {x}{2}\right )}{m^{4}+10 m^{2}+9}+\frac {12 m \,{\mathrm e}^{m x} \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{m^{4}+10 m^{2}+9}-\frac {6 \left (2 m^{2}+3\right ) {\mathrm e}^{m x} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{m^{4}+10 m^{2}+9}+\frac {6 \left (2 m^{2}+3\right ) {\mathrm e}^{m x} \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{m^{4}+10 m^{2}+9}+\frac {8 m \left (m^{2}+4\right ) {\mathrm e}^{m x} \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{m^{4}+10 m^{2}+9}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{3}}\)	\(195\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(m*x)*sin(x)^3,x,method=_RETURNVERBOSE)

[Out]

1/8*I/(3*I+m)*exp((3*I+m)*x)-3/8*I/(I+m)*exp((I+m)*x)+3/8*I/(m-I)*exp(x*(m-I))-1/8*I/(m-3*I)*exp(x*(m-3*I))

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maxima [A]  time = 0.55, size = 73, normalized size = 0.89 \[ \frac {3 \, {\left (m^{2} + 1\right )} \cos \left (3 \, x\right ) e^{\left (m x\right )} - 3 \, {\left (m^{2} + 9\right )} \cos \relax (x) e^{\left (m x\right )} - {\left (m^{3} + m\right )} e^{\left (m x\right )} \sin \left (3 \, x\right ) + 3 \, {\left (m^{3} + 9 \, m\right )} e^{\left (m x\right )} \sin \relax (x)}{4 \, {\left (m^{4} + 10 \, m^{2} + 9\right )}} \]

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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mupad [B]  time = 0.06, size = 47, normalized size = 0.57 \[ -\frac {{\mathrm {e}}^{m\,x}\,\left (\frac {3\,\left (\cos \relax (x)-m\,\sin \relax (x)\right )}{m^2+1}-\frac {3\,\cos \left (3\,x\right )-m\,\sin \left (3\,x\right )}{m^2+9}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 11.65, size = 638, normalized size = 7.78 \[ \begin {cases} \frac {x e^{- 3 i x} \sin ^{3}{\relax (x )}}{8} - \frac {3 i x e^{- 3 i x} \sin ^{2}{\relax (x )} \cos {\relax (x )}}{8} - \frac {3 x e^{- 3 i x} \sin {\relax (x )} \cos ^{2}{\relax (x )}}{8} + \frac {i x e^{- 3 i x} \cos ^{3}{\relax (x )}}{8} + \frac {7 i e^{- 3 i x} \sin ^{3}{\relax (x )}}{24} + \frac {i e^{- 3 i x} \sin {\relax (x )} \cos ^{2}{\relax (x )}}{4} + \frac {e^{- 3 i x} \cos ^{3}{\relax (x )}}{8} & \text {for}\: m = - 3 i \\\frac {3 x e^{- i x} \sin ^{3}{\relax (x )}}{8} - \frac {3 i x e^{- i x} \sin ^{2}{\relax (x )} \cos {\relax (x )}}{8} + \frac {3 x e^{- i x} \sin {\relax (x )} \cos ^{2}{\relax (x )}}{8} - \frac {3 i x e^{- i x} \cos ^{3}{\relax (x )}}{8} + \frac {5 i e^{- i x} \sin ^{3}{\relax (x )}}{8} + \frac {3 i e^{- i x} \sin {\relax (x )} \cos ^{2}{\relax (x )}}{4} + \frac {3 e^{- i x} \cos ^{3}{\relax (x )}}{8} & \text {for}\: m = - i \\\frac {3 x e^{i x} \sin ^{3}{\relax (x )}}{8} + \frac {3 i x e^{i x} \sin ^{2}{\relax (x )} \cos {\relax (x )}}{8} + \frac {3 x e^{i x} \sin {\relax (x )} \cos ^{2}{\relax (x )}}{8} + \frac {3 i x e^{i x} \cos ^{3}{\relax (x )}}{8} - \frac {5 i e^{i x} \sin ^{3}{\relax (x )}}{8} - \frac {3 i e^{i x} \sin {\relax (x )} \cos ^{2}{\relax (x )}}{4} + \frac {3 e^{i x} \cos ^{3}{\relax (x )}}{8} & \text {for}\: m = i \\\frac {x e^{3 i x} \sin ^{3}{\relax (x )}}{8} + \frac {3 i x e^{3 i x} \sin ^{2}{\relax (x )} \cos {\relax (x )}}{8} - \frac {3 x e^{3 i x} \sin {\relax (x )} \cos ^{2}{\relax (x )}}{8} - \frac {i x e^{3 i x} \cos ^{3}{\relax (x )}}{8} - \frac {7 i e^{3 i x} \sin ^{3}{\relax (x )}}{24} - \frac {i e^{3 i x} \sin {\relax (x )} \cos ^{2}{\relax (x )}}{4} + \frac {e^{3 i x} \cos ^{3}{\relax (x )}}{8} & \text {for}\: m = 3 i \\\frac {m^{3} e^{m x} \sin ^{3}{\relax (x )}}{m^{4} + 10 m^{2} + 9} - \frac {3 m^{2} e^{m x} \sin ^{2}{\relax (x )} \cos {\relax (x )}}{m^{4} + 10 m^{2} + 9} + \frac {7 m e^{m x} \sin ^{3}{\relax (x )}}{m^{4} + 10 m^{2} + 9} + \frac {6 m e^{m x} \sin {\relax (x )} \cos ^{2}{\relax (x )}}{m^{4} + 10 m^{2} + 9} - \frac {9 e^{m x} \sin ^{2}{\relax (x )} \cos {\relax (x )}}{m^{4} + 10 m^{2} + 9} - \frac {6 e^{m x} \cos ^{3}{\relax (x )}}{m^{4} + 10 m^{2} + 9} & \text {otherwise} \end {cases} \]

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*I*exp(-3*I*x)*sin(x)**3/24 + I*exp(-3*I*x)*sin(x)*cos(x)**2/4 + exp(-3*I*
x)*cos(x)**3/8, 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*exp(-I*x)*
sin(x)*cos(x)**2/8 - 3*I*x*exp(-I*x)*cos(x)**3/8 + 5*I*exp(-I*x)*sin(x)**3/8 + 3*I*exp(-I*x)*sin(x)*cos(x)**2/
4 + 3*exp(-I*x)*cos(x)**3/8, 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*e
xp(I*x)*sin(x)*cos(x)**2/8 + 3*I*x*exp(I*x)*cos(x)**3/8 - 5*I*exp(I*x)*sin(x)**3/8 - 3*I*exp(I*x)*sin(x)*cos(x
)**2/4 + 3*exp(I*x)*cos(x)**3/8, 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*I*exp(3*I*x)*sin(x)**3/24 - I*exp(3*I*x)*si
n(x)*cos(x)**2/4 + exp(3*I*x)*cos(x)**3/8, Eq(m, 3*I)), (m**3*exp(m*x)*sin(x)**3/(m**4 + 10*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)*s
in(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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