3.1 \(\int \sin ^3(x) (a \cos (x)+b \sin (x)) \, dx\)

Optimal. Leaf size=36 \[ \frac{1}{4} a \sin ^4(x)+\frac{3 b x}{8}-\frac{1}{4} b \sin ^3(x) \cos (x)-\frac{3}{8} b \sin (x) \cos (x) \]

[Out]

(3*b*x)/8 - (3*b*Cos[x]*Sin[x])/8 - (b*Cos[x]*Sin[x]^3)/4 + (a*Sin[x]^4)/4

________________________________________________________________________________________

Rubi [A]  time = 0.0486186, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3089, 2564, 30, 2635, 8} \[ \frac{1}{4} a \sin ^4(x)+\frac{3 b x}{8}-\frac{1}{4} b \sin ^3(x) \cos (x)-\frac{3}{8} b \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^3*(a*Cos[x] + b*Sin[x]),x]

[Out]

(3*b*x)/8 - (3*b*Cos[x]*Sin[x])/8 - (b*Cos[x]*Sin[x]^3)/4 + (a*Sin[x]^4)/4

Rule 3089

Int[sin[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[sin[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \sin ^3(x) (a \cos (x)+b \sin (x)) \, dx &=\int \left (a \cos (x) \sin ^3(x)+b \sin ^4(x)\right ) \, dx\\ &=a \int \cos (x) \sin ^3(x) \, dx+b \int \sin ^4(x) \, dx\\ &=-\frac{1}{4} b \cos (x) \sin ^3(x)+a \operatorname{Subst}\left (\int x^3 \, dx,x,\sin (x)\right )+\frac{1}{4} (3 b) \int \sin ^2(x) \, dx\\ &=-\frac{3}{8} b \cos (x) \sin (x)-\frac{1}{4} b \cos (x) \sin ^3(x)+\frac{1}{4} a \sin ^4(x)+\frac{1}{8} (3 b) \int 1 \, dx\\ &=\frac{3 b x}{8}-\frac{3}{8} b \cos (x) \sin (x)-\frac{1}{4} b \cos (x) \sin ^3(x)+\frac{1}{4} a \sin ^4(x)\\ \end{align*}

Mathematica [A]  time = 0.0063702, size = 34, normalized size = 0.94 \[ \frac{1}{4} a \sin ^4(x)+\frac{3 b x}{8}-\frac{1}{4} b \sin (2 x)+\frac{1}{32} b \sin (4 x) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^3*(a*Cos[x] + b*Sin[x]),x]

[Out]

(3*b*x)/8 + (a*Sin[x]^4)/4 - (b*Sin[2*x])/4 + (b*Sin[4*x])/32

________________________________________________________________________________________

Maple [A]  time = 0.023, size = 28, normalized size = 0.8 \begin{align*} b \left ( -{\frac{\cos \left ( x \right ) }{4} \left ( \left ( \sin \left ( x \right ) \right ) ^{3}+{\frac{3\,\sin \left ( x \right ) }{2}} \right ) }+{\frac{3\,x}{8}} \right ) +{\frac{a \left ( \sin \left ( x \right ) \right ) ^{4}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3*(a*cos(x)+b*sin(x)),x)

[Out]

b*(-1/4*(sin(x)^3+3/2*sin(x))*cos(x)+3/8*x)+1/4*a*sin(x)^4

________________________________________________________________________________________

Maxima [A]  time = 1.20377, size = 34, normalized size = 0.94 \begin{align*} \frac{1}{4} \, a \sin \left (x\right )^{4} + \frac{1}{32} \, b{\left (12 \, x + \sin \left (4 \, x\right ) - 8 \, \sin \left (2 \, x\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3*(a*cos(x)+b*sin(x)),x, algorithm="maxima")

[Out]

1/4*a*sin(x)^4 + 1/32*b*(12*x + sin(4*x) - 8*sin(2*x))

________________________________________________________________________________________

Fricas [A]  time = 0.490686, size = 113, normalized size = 3.14 \begin{align*} \frac{1}{4} \, a \cos \left (x\right )^{4} - \frac{1}{2} \, a \cos \left (x\right )^{2} + \frac{3}{8} \, b x + \frac{1}{8} \,{\left (2 \, b \cos \left (x\right )^{3} - 5 \, b \cos \left (x\right )\right )} \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3*(a*cos(x)+b*sin(x)),x, algorithm="fricas")

[Out]

1/4*a*cos(x)^4 - 1/2*a*cos(x)^2 + 3/8*b*x + 1/8*(2*b*cos(x)^3 - 5*b*cos(x))*sin(x)

________________________________________________________________________________________

Sympy [B]  time = 0.758967, size = 75, normalized size = 2.08 \begin{align*} \frac{a \sin ^{4}{\left (x \right )}}{4} + \frac{3 b x \sin ^{4}{\left (x \right )}}{8} + \frac{3 b x \sin ^{2}{\left (x \right )} \cos ^{2}{\left (x \right )}}{4} + \frac{3 b x \cos ^{4}{\left (x \right )}}{8} - \frac{5 b \sin ^{3}{\left (x \right )} \cos{\left (x \right )}}{8} - \frac{3 b \sin{\left (x \right )} \cos ^{3}{\left (x \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**3*(a*cos(x)+b*sin(x)),x)

[Out]

a*sin(x)**4/4 + 3*b*x*sin(x)**4/8 + 3*b*x*sin(x)**2*cos(x)**2/4 + 3*b*x*cos(x)**4/8 - 5*b*sin(x)**3*cos(x)/8 -
3*b*sin(x)*cos(x)**3/8

________________________________________________________________________________________

Giac [A]  time = 1.14065, size = 45, normalized size = 1.25 \begin{align*} \frac{3}{8} \, b x + \frac{1}{32} \, a \cos \left (4 \, x\right ) - \frac{1}{8} \, a \cos \left (2 \, x\right ) + \frac{1}{32} \, b \sin \left (4 \, x\right ) - \frac{1}{4} \, b \sin \left (2 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3*(a*cos(x)+b*sin(x)),x, algorithm="giac")

[Out]

3/8*b*x + 1/32*a*cos(4*x) - 1/8*a*cos(2*x) + 1/32*b*sin(4*x) - 1/4*b*sin(2*x)