3.51 \(\int \cos ^3(a+b x) \sin ^2(a+b x) \, dx\)

Optimal. Leaf size=31 \[ \frac{\sin ^3(a+b x)}{3 b}-\frac{\sin ^5(a+b x)}{5 b} \]

[Out]

Sin[a + b*x]^3/(3*b) - Sin[a + b*x]^5/(5*b)

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Rubi [A]  time = 0.0348122, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2564, 14} \[ \frac{\sin ^3(a+b x)}{3 b}-\frac{\sin ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*Sin[a + b*x]^2,x]

[Out]

Sin[a + b*x]^3/(3*b) - Sin[a + b*x]^5/(5*b)

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \cos ^3(a+b x) \sin ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{\sin ^3(a+b x)}{3 b}-\frac{\sin ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0585984, size = 27, normalized size = 0.87 \[ \frac{\sin ^3(a+b x) (3 \cos (2 (a+b x))+7)}{30 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3*Sin[a + b*x]^2,x]

[Out]

((7 + 3*Cos[2*(a + b*x)])*Sin[a + b*x]^3)/(30*b)

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Maple [A]  time = 0.036, size = 40, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ( -{\frac{\sin \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{5}}+{\frac{ \left ( 2+ \left ( \cos \left ( bx+a \right ) \right ) ^{2} \right ) \sin \left ( bx+a \right ) }{15}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3*sin(b*x+a)^2,x)

[Out]

1/b*(-1/5*sin(b*x+a)*cos(b*x+a)^4+1/15*(2+cos(b*x+a)^2)*sin(b*x+a))

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Maxima [A]  time = 0.98687, size = 35, normalized size = 1.13 \begin{align*} -\frac{3 \, \sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3}}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/15*(3*sin(b*x + a)^5 - 5*sin(b*x + a)^3)/b

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Fricas [A]  time = 1.97271, size = 84, normalized size = 2.71 \begin{align*} -\frac{{\left (3 \, \cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2} - 2\right )} \sin \left (b x + a\right )}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/15*(3*cos(b*x + a)^4 - cos(b*x + a)^2 - 2)*sin(b*x + a)/b

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Sympy [A]  time = 2.1003, size = 44, normalized size = 1.42 \begin{align*} \begin{cases} \frac{2 \sin ^{5}{\left (a + b x \right )}}{15 b} + \frac{\sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b} & \text{for}\: b \neq 0 \\x \sin ^{2}{\left (a \right )} \cos ^{3}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**3*sin(b*x+a)**2,x)

[Out]

Piecewise((2*sin(a + b*x)**5/(15*b) + sin(a + b*x)**3*cos(a + b*x)**2/(3*b), Ne(b, 0)), (x*sin(a)**2*cos(a)**3
, True))

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Giac [A]  time = 1.2339, size = 35, normalized size = 1.13 \begin{align*} -\frac{3 \, \sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3}}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(b*x+a)^2,x, algorithm="giac")

[Out]

-1/15*(3*sin(b*x + a)^5 - 5*sin(b*x + a)^3)/b