Integrand size = 17, antiderivative size = 46 \[ \int \cos ^5(a+b x) \sin ^4(a+b x) \, dx=\frac {\sin ^5(a+b x)}{5 b}-\frac {2 \sin ^7(a+b x)}{7 b}+\frac {\sin ^9(a+b x)}{9 b} \] Output:
1/5*sin(b*x+a)^5/b-2/7*sin(b*x+a)^7/b+1/9*sin(b*x+a)^9/b
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \cos ^5(a+b x) \sin ^4(a+b x) \, dx=\frac {(249+220 \cos (2 (a+b x))+35 \cos (4 (a+b x))) \sin ^5(a+b x)}{2520 b} \] Input:
Integrate[Cos[a + b*x]^5*Sin[a + b*x]^4,x]
Output:
((249 + 220*Cos[2*(a + b*x)] + 35*Cos[4*(a + b*x)])*Sin[a + b*x]^5)/(2520* b)
Time = 0.37 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 3044, 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.
\(\displaystyle \int \sin ^4(a+b x) \cos ^5(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^4 \cos (a+b x)^5dx\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {\int \sin ^4(a+b x) \left (1-\sin ^2(a+b x)\right )^2d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\sin ^8(a+b x)-2 \sin ^6(a+b x)+\sin ^4(a+b x)\right )d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{9} \sin ^9(a+b x)-\frac {2}{7} \sin ^7(a+b x)+\frac {1}{5} \sin ^5(a+b x)}{b}\) |
Input:
Int[Cos[a + b*x]^5*Sin[a + b*x]^4,x]
Output:
(Sin[a + b*x]^5/5 - (2*Sin[a + b*x]^7)/7 + Sin[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_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[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] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Time = 21.33 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (b x +a \right )^{9}}{9}-\frac {2 \sin \left (b x +a \right )^{7}}{7}+\frac {\sin \left (b x +a \right )^{5}}{5}}{b}\) | \(36\) |
default | \(\frac {\frac {\sin \left (b x +a \right )^{9}}{9}-\frac {2 \sin \left (b x +a \right )^{7}}{7}+\frac {\sin \left (b x +a \right )^{5}}{5}}{b}\) | \(36\) |
risch | \(\frac {3 \sin \left (b x +a \right )}{128 b}+\frac {\sin \left (9 b x +9 a \right )}{2304 b}+\frac {\sin \left (7 b x +7 a \right )}{1792 b}-\frac {\sin \left (5 b x +5 a \right )}{320 b}-\frac {\sin \left (3 b x +3 a \right )}{192 b}\) | \(69\) |
parallelrisch | \(\frac {\left (\sin \left (\frac {5 b x}{2}+\frac {5 a}{2}\right )-5 \sin \left (\frac {3 b x}{2}+\frac {3 a}{2}\right )+10 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (249+35 \cos \left (4 b x +4 a \right )+220 \cos \left (2 b x +2 a \right )\right ) \left (\cos \left (\frac {5 b x}{2}+\frac {5 a}{2}\right )+5 \cos \left (\frac {3 b x}{2}+\frac {3 a}{2}\right )+10 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{20160 b}\) | \(94\) |
norman | \(\frac {\frac {32 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{5}}{5 b}-\frac {384 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{7}}{35 b}+\frac {6976 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{9}}{315 b}-\frac {384 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{11}}{35 b}+\frac {32 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{13}}{5 b}}{\left (1+\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )^{9}}\) | \(98\) |
orering | \(-\frac {117469 \left (-5 \cos \left (b x +a \right )^{4} \sin \left (b x +a \right )^{5} b +4 \cos \left (b x +a \right )^{6} \sin \left (b x +a \right )^{3} b \right )}{99225 b^{2}}-\frac {34562 \left (-60 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{7} b^{3}+365 \cos \left (b x +a \right )^{4} \sin \left (b x +a \right )^{5} b^{3}-280 \cos \left (b x +a \right )^{6} \sin \left (b x +a \right )^{3} b^{3}+24 \cos \left (b x +a \right )^{8} \sin \left (b x +a \right ) b^{3}\right )}{178605 b^{4}}-\frac {418 \left (-120 b^{5} \sin \left (b x +a \right )^{9}+6600 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{7} b^{5}-28805 \cos \left (b x +a \right )^{4} \sin \left (b x +a \right )^{5} b^{5}+21244 \cos \left (b x +a \right )^{6} \sin \left (b x +a \right )^{3} b^{5}-2280 \cos \left (b x +a \right )^{8} \sin \left (b x +a \right ) b^{5}\right )}{42525 b^{6}}-\frac {11 \left (-598500 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{7} b^{7}+14280 b^{7} \sin \left (b x +a \right )^{9}+2325965 \cos \left (b x +a \right )^{4} \sin \left (b x +a \right )^{5} b^{7}-1659760 \cos \left (b x +a \right )^{6} \sin \left (b x +a \right )^{3} b^{7}+184464 \cos \left (b x +a \right )^{8} \sin \left (b x +a \right ) b^{7}\right )}{59535 b^{8}}-\frac {-1325520 b^{9} \sin \left (b x +a \right )^{9}+51084240 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{7} b^{9}-188902085 \cos \left (b x +a \right )^{4} \sin \left (b x +a \right )^{5} b^{9}+131538484 \cos \left (b x +a \right )^{6} \sin \left (b x +a \right )^{3} b^{9}-14570160 \cos \left (b x +a \right )^{8} \sin \left (b x +a \right ) b^{9}}{893025 b^{10}}\) | \(437\) |
Input:
int(cos(b*x+a)^5*sin(b*x+a)^4,x,method=_RETURNVERBOSE)
Output:
1/b*(1/9*sin(b*x+a)^9-2/7*sin(b*x+a)^7+1/5*sin(b*x+a)^5)
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.15 \[ \int \cos ^5(a+b x) \sin ^4(a+b x) \, dx=\frac {{\left (35 \, \cos \left (b x + a\right )^{8} - 50 \, \cos \left (b x + a\right )^{6} + 3 \, \cos \left (b x + a\right )^{4} + 4 \, \cos \left (b x + a\right )^{2} + 8\right )} \sin \left (b x + a\right )}{315 \, b} \] Input:
integrate(cos(b*x+a)^5*sin(b*x+a)^4,x, algorithm="fricas")
Output:
1/315*(35*cos(b*x + a)^8 - 50*cos(b*x + a)^6 + 3*cos(b*x + a)^4 + 4*cos(b* x + a)^2 + 8)*sin(b*x + a)/b
Time = 0.90 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.43 \[ \int \cos ^5(a+b x) \sin ^4(a+b x) \, dx=\begin {cases} \frac {8 \sin ^{9}{\left (a + b x \right )}}{315 b} + \frac {4 \sin ^{7}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{35 b} + \frac {\sin ^{5}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{5 b} & \text {for}\: b \neq 0 \\x \sin ^{4}{\left (a \right )} \cos ^{5}{\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(b*x+a)**5*sin(b*x+a)**4,x)
Output:
Piecewise((8*sin(a + b*x)**9/(315*b) + 4*sin(a + b*x)**7*cos(a + b*x)**2/( 35*b) + sin(a + b*x)**5*cos(a + b*x)**4/(5*b), Ne(b, 0)), (x*sin(a)**4*cos (a)**5, True))
Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos ^5(a+b x) \sin ^4(a+b x) \, dx=\frac {35 \, \sin \left (b x + a\right )^{9} - 90 \, \sin \left (b x + a\right )^{7} + 63 \, \sin \left (b x + a\right )^{5}}{315 \, b} \] Input:
integrate(cos(b*x+a)^5*sin(b*x+a)^4,x, algorithm="maxima")
Output:
1/315*(35*sin(b*x + a)^9 - 90*sin(b*x + a)^7 + 63*sin(b*x + a)^5)/b
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos ^5(a+b x) \sin ^4(a+b x) \, dx=\frac {35 \, \sin \left (b x + a\right )^{9} - 90 \, \sin \left (b x + a\right )^{7} + 63 \, \sin \left (b x + a\right )^{5}}{315 \, b} \] Input:
integrate(cos(b*x+a)^5*sin(b*x+a)^4,x, algorithm="giac")
Output:
1/315*(35*sin(b*x + a)^9 - 90*sin(b*x + a)^7 + 63*sin(b*x + a)^5)/b
Time = 25.44 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos ^5(a+b x) \sin ^4(a+b x) \, dx=\frac {35\,{\sin \left (a+b\,x\right )}^9-90\,{\sin \left (a+b\,x\right )}^7+63\,{\sin \left (a+b\,x\right )}^5}{315\,b} \] Input:
int(cos(a + b*x)^5*sin(a + b*x)^4,x)
Output:
(63*sin(a + b*x)^5 - 90*sin(a + b*x)^7 + 35*sin(a + b*x)^9)/(315*b)
Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76 \[ \int \cos ^5(a+b x) \sin ^4(a+b x) \, dx=\frac {\sin \left (b x +a \right )^{5} \left (35 \sin \left (b x +a \right )^{4}-90 \sin \left (b x +a \right )^{2}+63\right )}{315 b} \] Input:
int(cos(b*x+a)^5*sin(b*x+a)^4,x)
Output:
(sin(a + b*x)**5*(35*sin(a + b*x)**4 - 90*sin(a + b*x)**2 + 63))/(315*b)