3.3 \(\int \sin (x) (a \cos (x)+b \sin (x)) \, dx\)

Optimal. Leaf size=25 \[ \frac{1}{2} a \sin ^2(x)+\frac{b x}{2}-\frac{1}{2} b \sin (x) \cos (x) \]

[Out]

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

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Rubi [A]  time = 0.026968, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {3089, 2564, 30, 2635, 8} \[ \frac{1}{2} a \sin ^2(x)+\frac{b x}{2}-\frac{1}{2} b \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]*(a*Cos[x] + b*Sin[x]),x]

[Out]

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

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

Mathematica [A]  time = 0.0040089, size = 25, normalized size = 1. \[ -\frac{1}{2} a \cos ^2(x)+\frac{b x}{2}-\frac{1}{4} b \sin (2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x)/2 - (a*Cos[x]^2)/2 - (b*Sin[2*x])/4

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Maple [A]  time = 0.018, size = 21, normalized size = 0.8 \begin{align*} b \left ( -{\frac{\cos \left ( x \right ) \sin \left ( x \right ) }{2}}+{\frac{x}{2}} \right ) -{\frac{ \left ( \cos \left ( x \right ) \right ) ^{2}a}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.09505, size = 28, normalized size = 1.12 \begin{align*} -\frac{1}{2} \, a \cos \left (x\right )^{2} + \frac{1}{4} \, b{\left (2 \, x - \sin \left (2 \, x\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.472956, size = 66, normalized size = 2.64 \begin{align*} -\frac{1}{2} \, a \cos \left (x\right )^{2} - \frac{1}{2} \, b \cos \left (x\right ) \sin \left (x\right ) + \frac{1}{2} \, b x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.204378, size = 37, normalized size = 1.48 \begin{align*} - \frac{a \cos ^{2}{\left (x \right )}}{2} + \frac{b x \sin ^{2}{\left (x \right )}}{2} + \frac{b x \cos ^{2}{\left (x \right )}}{2} - \frac{b \sin{\left (x \right )} \cos{\left (x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11448, size = 26, normalized size = 1.04 \begin{align*} \frac{1}{2} \, b x - \frac{1}{4} \, a \cos \left (2 \, 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)*(a*cos(x)+b*sin(x)),x, algorithm="giac")

[Out]

1/2*b*x - 1/4*a*cos(2*x) - 1/4*b*sin(2*x)