Integrand size = 17, antiderivative size = 46 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=-\frac {\cos ^5(a+b x)}{5 b}+\frac {2 \cos ^7(a+b x)}{7 b}-\frac {\cos ^9(a+b x)}{9 b} \] Output:
-1/5*cos(b*x+a)^5/b+2/7*cos(b*x+a)^7/b-1/9*cos(b*x+a)^9/b
Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=\frac {\cos ^5(a+b x) (-249+220 \cos (2 (a+b x))-35 \cos (4 (a+b x)))}{2520 b} \] Input:
Integrate[Cos[a + b*x]^4*Sin[a + b*x]^5,x]
Output:
(Cos[a + b*x]^5*(-249 + 220*Cos[2*(a + b*x)] - 35*Cos[4*(a + b*x)]))/(2520 *b)
Time = 0.37 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 3045, 244, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sin ^5(a+b x) \cos ^4(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^5 \cos (a+b x)^4dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\int \cos ^4(a+b x) \left (1-\cos ^2(a+b x)\right )^2d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {\int \left (\cos ^8(a+b x)-2 \cos ^6(a+b x)+\cos ^4(a+b x)\right )d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {1}{9} \cos ^9(a+b x)-\frac {2}{7} \cos ^7(a+b x)+\frac {1}{5} \cos ^5(a+b x)}{b}\) |
Input:
Int[Cos[a + b*x]^4*Sin[a + b*x]^5,x]
Output:
-((Cos[a + b*x]^5/5 - (2*Cos[a + b*x]^7)/7 + Cos[a + b*x]^9/9)/b)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(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])
Time = 12.68 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(-\frac {\frac {\cos \left (b x +a \right )^{9}}{9}-\frac {2 \cos \left (b x +a \right )^{7}}{7}+\frac {\cos \left (b x +a \right )^{5}}{5}}{b}\) | \(37\) |
default | \(-\frac {\frac {\cos \left (b x +a \right )^{9}}{9}-\frac {2 \cos \left (b x +a \right )^{7}}{7}+\frac {\cos \left (b x +a \right )^{5}}{5}}{b}\) | \(37\) |
parallelrisch | \(\frac {-2048-420 \cos \left (3 b x +3 a \right )-1890 \cos \left (b x +a \right )+252 \cos \left (5 b x +5 a \right )+45 \cos \left (7 b x +7 a \right )-35 \cos \left (9 b x +9 a \right )}{80640 b}\) | \(60\) |
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 | \(\frac {-\frac {16}{315 b}-\frac {112 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{8}}{5 b}-\frac {32 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{12}}{3 b}+\frac {16 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{10}}{b}+\frac {32 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}}{5 b}-\frac {16 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{35 b}-\frac {64 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}{35 b}}{\left (1+\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )^{9}}\) | \(119\) |
orering | \(-\frac {117469 \left (-4 \cos \left (b x +a \right )^{3} \sin \left (b x +a \right )^{6} b +5 \cos \left (b x +a \right )^{5} \sin \left (b x +a \right )^{4} b \right )}{99225 b^{2}}-\frac {34562 \left (-24 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{8} b^{3}+280 \cos \left (b x +a \right )^{3} \sin \left (b x +a \right )^{6} b^{3}-365 \cos \left (b x +a \right )^{5} \sin \left (b x +a \right )^{4} b^{3}+60 \cos \left (b x +a \right )^{7} \sin \left (b x +a \right )^{2} b^{3}\right )}{178605 b^{4}}-\frac {418 \left (2280 b^{5} \sin \left (b x +a \right )^{8} \cos \left (b x +a \right )-21244 \cos \left (b x +a \right )^{3} \sin \left (b x +a \right )^{6} b^{5}+28805 \cos \left (b x +a \right )^{5} \sin \left (b x +a \right )^{4} b^{5}-6600 \cos \left (b x +a \right )^{7} \sin \left (b x +a \right )^{2} b^{5}+120 \cos \left (b x +a \right )^{9} b^{5}\right )}{42525 b^{6}}-\frac {11 \left (1659760 b^{7} \sin \left (b x +a \right )^{6} \cos \left (b x +a \right )^{3}-184464 b^{7} \sin \left (b x +a \right )^{8} \cos \left (b x +a \right )-2325965 \cos \left (b x +a \right )^{5} \sin \left (b x +a \right )^{4} b^{7}+598500 \cos \left (b x +a \right )^{7} \sin \left (b x +a \right )^{2} b^{7}-14280 \cos \left (b x +a \right )^{9} b^{7}\right )}{59535 b^{8}}-\frac {188902085 b^{9} \sin \left (b x +a \right )^{4} \cos \left (b x +a \right )^{5}-131538484 b^{9} \sin \left (b x +a \right )^{6} \cos \left (b x +a \right )^{3}+14570160 b^{9} \sin \left (b x +a \right )^{8} \cos \left (b x +a \right )-51084240 \cos \left (b x +a \right )^{7} \sin \left (b x +a \right )^{2} b^{9}+1325520 \cos \left (b x +a \right )^{9} b^{9}}{893025 b^{10}}\) | \(437\) |
Input:
int(cos(b*x+a)^4*sin(b*x+a)^5,x,method=_RETURNVERBOSE)
Output:
-1/b*(1/9*cos(b*x+a)^9-2/7*cos(b*x+a)^7+1/5*cos(b*x+a)^5)
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=-\frac {35 \, \cos \left (b x + a\right )^{9} - 90 \, \cos \left (b x + a\right )^{7} + 63 \, \cos \left (b x + a\right )^{5}}{315 \, b} \] Input:
integrate(cos(b*x+a)^4*sin(b*x+a)^5,x, algorithm="fricas")
Output:
-1/315*(35*cos(b*x + a)^9 - 90*cos(b*x + a)^7 + 63*cos(b*x + a)^5)/b
Time = 0.91 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.48 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=\begin {cases} - \frac {\sin ^{4}{\left (a + b x \right )} \cos ^{5}{\left (a + b x \right )}}{5 b} - \frac {4 \sin ^{2}{\left (a + b x \right )} \cos ^{7}{\left (a + b x \right )}}{35 b} - \frac {8 \cos ^{9}{\left (a + b x \right )}}{315 b} & \text {for}\: b \neq 0 \\x \sin ^{5}{\left (a \right )} \cos ^{4}{\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(b*x+a)**4*sin(b*x+a)**5,x)
Output:
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*co s(a)**4, True))
Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=-\frac {35 \, \cos \left (b x + a\right )^{9} - 90 \, \cos \left (b x + a\right )^{7} + 63 \, \cos \left (b x + a\right )^{5}}{315 \, b} \] Input:
integrate(cos(b*x+a)^4*sin(b*x+a)^5,x, algorithm="maxima")
Output:
-1/315*(35*cos(b*x + a)^9 - 90*cos(b*x + a)^7 + 63*cos(b*x + a)^5)/b
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=-\frac {35 \, \cos \left (b x + a\right )^{9} - 90 \, \cos \left (b x + a\right )^{7} + 63 \, \cos \left (b x + a\right )^{5}}{315 \, b} \] Input:
integrate(cos(b*x+a)^4*sin(b*x+a)^5,x, algorithm="giac")
Output:
-1/315*(35*cos(b*x + a)^9 - 90*cos(b*x + a)^7 + 63*cos(b*x + a)^5)/b
Time = 25.80 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=-\frac {35\,{\cos \left (a+b\,x\right )}^9-90\,{\cos \left (a+b\,x\right )}^7+63\,{\cos \left (a+b\,x\right )}^5}{315\,b} \] Input:
int(cos(a + b*x)^4*sin(a + b*x)^5,x)
Output:
-(63*cos(a + b*x)^5 - 90*cos(a + b*x)^7 + 35*cos(a + b*x)^9)/(315*b)
Time = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.72 \[ \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx=\frac {-35 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{8}+50 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{6}-3 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{4}-4 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}-8 \cos \left (b x +a \right )+8}{315 b} \] Input:
int(cos(b*x+a)^4*sin(b*x+a)^5,x)
Output:
( - 35*cos(a + b*x)*sin(a + b*x)**8 + 50*cos(a + b*x)*sin(a + b*x)**6 - 3* cos(a + b*x)*sin(a + b*x)**4 - 4*cos(a + b*x)*sin(a + b*x)**2 - 8*cos(a + b*x) + 8)/(315*b)