\(\int \cos ^4(a+b x) \sin ^5(a+b x) \, dx\) [101]

Optimal result

Integrand size = 17, antiderivative size = 46 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=-\frac {\cos ^5(a+b x)}{5 b}+\frac {2 \cos ^7(a+b x)}{7 b}-\frac {\cos ^9(a+b x)}{9 b} \]

[Out]

-1/5*cos(b*x+a)^5/b+2/7*cos(b*x+a)^7/b-1/9*cos(b*x+a)^9/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, 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, 276} \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=-\frac {\cos ^9(a+b x)}{9 b}+\frac {2 \cos ^7(a+b x)}{7 b}-\frac {\cos ^5(a+b x)}{5 b} \]

[In]

Int[Cos[a + b*x]^4*Sin[a + b*x]^5,x]

[Out]

-1/5*Cos[a + b*x]^5/b + (2*Cos[a + b*x]^7)/(7*b) - Cos[a + b*x]^9/(9*b)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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^4 \left (1-x^2\right )^2 \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {\text {Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {\cos ^5(a+b x)}{5 b}+\frac {2 \cos ^7(a+b x)}{7 b}-\frac {\cos ^9(a+b x)}{9 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=\frac {\cos ^5(a+b x) (-249+220 \cos (2 (a+b x))-35 \cos (4 (a+b x)))}{2520 b} \]

[In]

Integrate[Cos[a + b*x]^4*Sin[a + b*x]^5,x]

[Out]

(Cos[a + b*x]^5*(-249 + 220*Cos[2*(a + b*x)] - 35*Cos[4*(a + b*x)]))/(2520*b)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(-\frac {\frac {\left (\cos ^{9}\left (b x +a \right )\right )}{9}-\frac {2 \left (\cos ^{7}\left (b x +a \right )\right )}{7}+\frac {\left (\cos ^{5}\left (b x +a \right )\right )}{5}}{b}\)	\(37\)
default	\(-\frac {\frac {\left (\cos ^{9}\left (b x +a \right )\right )}{9}-\frac {2 \left (\cos ^{7}\left (b x +a \right )\right )}{7}+\frac {\left (\cos ^{5}\left (b x +a \right )\right )}{5}}{b}\)	\(37\)
parallelrisch	\(\frac {-2048-35 \cos \left (9 b x +9 a \right )-1890 \cos \left (b x +a \right )+252 \cos \left (5 b x +5 a \right )-420 \cos \left (3 b x +3 a \right )+45 \cos \left (7 b x +7 a \right )}{80640 b}\)	\(60\)
risch	\(-\frac {3 \cos \left (b x +a \right )}{128 b}-\frac {\cos \left (9 b x +9 a \right )}{2304 b}+\frac {\cos \left (7 b x +7 a \right )}{1792 b}+\frac {\cos \left (5 b x +5 a \right )}{320 b}-\frac {\cos \left (3 b x +3 a \right )}{192 b}\)	\(69\)
norman	\(\frac {-\frac {16}{315 b}-\frac {112 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 b}-\frac {32 \left (\tan ^{12}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}+\frac {16 \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {32 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 b}-\frac {16 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{35 b}-\frac {64 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{35 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{9}}\)	\(119\)

[In]

int(cos(b*x+a)^4*sin(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

-1/b*(1/9*cos(b*x+a)^9-2/7*cos(b*x+a)^7+1/5*cos(b*x+a)^5)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=-\frac {35 \, \cos \left (b x + a\right )^{9} - 90 \, \cos \left (b x + a\right )^{7} + 63 \, \cos \left (b x + a\right )^{5}}{315 \, b} \]

[In]

integrate(cos(b*x+a)^4*sin(b*x+a)^5,x, algorithm="fricas")

[Out]

-1/315*(35*cos(b*x + a)^9 - 90*cos(b*x + a)^7 + 63*cos(b*x + a)^5)/b

Sympy [A] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.48 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=\begin {cases} - \frac {\sin ^{4}{\left (a + b x \right )} \cos ^{5}{\left (a + b x \right )}}{5 b} - \frac {4 \sin ^{2}{\left (a + b x \right )} \cos ^{7}{\left (a + b x \right )}}{35 b} - \frac {8 \cos ^{9}{\left (a + b x \right )}}{315 b} & \text {for}\: b \neq 0 \\x \sin ^{5}{\left (a \right )} \cos ^{4}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)**4*sin(b*x+a)**5,x)

[Out]

Piecewise((-sin(a + b*x)**4*cos(a + b*x)**5/(5*b) - 4*sin(a + b*x)**2*cos(a + b*x)**7/(35*b) - 8*cos(a + b*x)*
*9/(315*b), Ne(b, 0)), (x*sin(a)**5*cos(a)**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=-\frac {35 \, \cos \left (b x + a\right )^{9} - 90 \, \cos \left (b x + a\right )^{7} + 63 \, \cos \left (b x + a\right )^{5}}{315 \, b} \]

[In]

integrate(cos(b*x+a)^4*sin(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/315*(35*cos(b*x + a)^9 - 90*cos(b*x + a)^7 + 63*cos(b*x + a)^5)/b

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.48 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=-\frac {\cos \left (9 \, b x + 9 \, a\right )}{2304 \, b} + \frac {\cos \left (7 \, b x + 7 \, a\right )}{1792 \, b} + \frac {\cos \left (5 \, b x + 5 \, a\right )}{320 \, b} - \frac {\cos \left (3 \, b x + 3 \, a\right )}{192 \, b} - \frac {3 \, \cos \left (b x + a\right )}{128 \, b} \]

[In]

integrate(cos(b*x+a)^4*sin(b*x+a)^5,x, algorithm="giac")

[Out]

-1/2304*cos(9*b*x + 9*a)/b + 1/1792*cos(7*b*x + 7*a)/b + 1/320*cos(5*b*x + 5*a)/b - 1/192*cos(3*b*x + 3*a)/b -
3/128*cos(b*x + a)/b

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=-\frac {35\,{\cos \left (a+b\,x\right )}^9-90\,{\cos \left (a+b\,x\right )}^7+63\,{\cos \left (a+b\,x\right )}^5}{315\,b} \]

[In]

int(cos(a + b*x)^4*sin(a + b*x)^5,x)

[Out]

-(63*cos(a + b*x)^5 - 90*cos(a + b*x)^7 + 35*cos(a + b*x)^9)/(315*b)