\(\int \cos ^2(a+b x) \sin ^3(a+b x) \, dx\) [69]

Optimal result

Integrand size = 17, antiderivative size = 31 \[ \int \cos ^2(a+b x) \sin ^3(a+b x) \, dx=-\frac {\cos ^3(a+b x)}{3 b}+\frac {\cos ^5(a+b x)}{5 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2645, 14} \[ \int \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\frac {\cos ^5(a+b x)}{5 b}-\frac {\cos ^3(a+b x)}{3 b} \]

[In]

Int[Cos[a + b*x]^2*Sin[a + b*x]^3,x]

[Out]

-1/3*Cos[a + b*x]^3/b + Cos[a + b*x]^5/(5*b)

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]]

Rule 2645

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\frac {\cos ^3(a+b x) (-7+3 \cos (2 (a+b x)))}{30 b} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {\frac {\left (\cos ^{5}\left (b x +a \right )\right )}{5}-\frac {\left (\cos ^{3}\left (b x +a \right )\right )}{3}}{b}\)	\(26\)
default	\(\frac {\frac {\left (\cos ^{5}\left (b x +a \right )\right )}{5}-\frac {\left (\cos ^{3}\left (b x +a \right )\right )}{3}}{b}\)	\(26\)
parallelrisch	\(\frac {-32-30 \cos \left (b x +a \right )-5 \cos \left (3 b x +3 a \right )+3 \cos \left (5 b x +5 a \right )}{240 b}\)	\(38\)
risch	\(-\frac {\cos \left (b x +a \right )}{8 b}+\frac {\cos \left (5 b x +5 a \right )}{80 b}-\frac {\cos \left (3 b x +3 a \right )}{48 b}\)	\(41\)
norman	\(\frac {-\frac {4}{15 b}-\frac {4 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {4 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}+\frac {4 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{5}}\)	\(71\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\frac {3 \, \cos \left (b x + a\right )^{5} - 5 \, \cos \left (b x + a\right )^{3}}{15 \, b} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).

Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\begin {cases} - \frac {\sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac {2 \cos ^{5}{\left (a + b x \right )}}{15 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} \cos ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\frac {3 \, \cos \left (b x + a\right )^{5} - 5 \, \cos \left (b x + a\right )^{3}}{15 \, b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\frac {\cos \left (b x + a\right )^{5}}{5 \, b} - \frac {\cos \left (b x + a\right )^{3}}{3 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \cos ^2(a+b x) \sin ^3(a+b x) \, dx=-\frac {5\,{\cos \left (a+b\,x\right )}^3-3\,{\cos \left (a+b\,x\right )}^5}{15\,b} \]

[In]

int(cos(a + b*x)^2*sin(a + b*x)^3,x)

[Out]

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