Integrand size = 19, antiderivative size = 74 \[ \int \cos ^5(a+b x) (c \sin (a+b x))^m \, dx=\frac {(c \sin (a+b x))^{1+m}}{b c (1+m)}-\frac {2 (c \sin (a+b x))^{3+m}}{b c^3 (3+m)}+\frac {(c \sin (a+b x))^{5+m}}{b c^5 (5+m)} \] Output:
(c*sin(b*x+a))^(1+m)/b/c/(1+m)-2*(c*sin(b*x+a))^(3+m)/b/c^3/(3+m)+(c*sin(b *x+a))^(5+m)/b/c^5/(5+m)
Time = 0.36 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.74 \[ \int \cos ^5(a+b x) (c \sin (a+b x))^m \, dx=\frac {\sin (a+b x) (c \sin (a+b x))^m \left (\frac {1}{1+m}-\frac {2 \sin ^2(a+b x)}{3+m}+\frac {\sin ^4(a+b x)}{5+m}\right )}{b} \] Input:
Integrate[Cos[a + b*x]^5*(c*Sin[a + b*x])^m,x]
Output:
(Sin[a + b*x]*(c*Sin[a + b*x])^m*((1 + m)^(-1) - (2*Sin[a + b*x]^2)/(3 + m ) + Sin[a + b*x]^4/(5 + m)))/b
Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 3044, 27, 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.
\(\displaystyle \int \cos ^5(a+b x) (c \sin (a+b x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (a+b x)^5 (c \sin (a+b x))^mdx\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {\int \frac {(c \sin (a+b x))^m \left (c^2-c^2 \sin ^2(a+b x)\right )^2}{c^4}d(c \sin (a+b x))}{b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (c \sin (a+b x))^m \left (c^2-c^2 \sin ^2(a+b x)\right )^2d(c \sin (a+b x))}{b c^5}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (c^4 (c \sin (a+b x))^m-2 c^2 (c \sin (a+b x))^{m+2}+(c \sin (a+b x))^{m+4}\right )d(c \sin (a+b x))}{b c^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {c^4 (c \sin (a+b x))^{m+1}}{m+1}-\frac {2 c^2 (c \sin (a+b x))^{m+3}}{m+3}+\frac {(c \sin (a+b x))^{m+5}}{m+5}}{b c^5}\) |
Input:
Int[Cos[a + b*x]^5*(c*Sin[a + b*x])^m,x]
Output:
((c^4*(c*Sin[a + b*x])^(1 + m))/(1 + m) - (2*c^2*(c*Sin[a + b*x])^(3 + m)) /(3 + m) + (c*Sin[a + b*x])^(5 + m)/(5 + m))/(b*c^5)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 = 5.84 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.18
method | result | size |
parallelrisch | \(\frac {\left (\left (\frac {3}{2} m^{2}+14 m +\frac {25}{2}\right ) \sin \left (3 b x +3 a \right )+\left (\frac {1}{2} m^{2}+2 m +\frac {3}{2}\right ) \sin \left (5 b x +5 a \right )+\sin \left (b x +a \right ) \left (m^{2}+12 m +75\right )\right ) \left (c \sin \left (b x +a \right )\right )^{m}}{8 b \left (m^{3}+9 m^{2}+23 m +15\right )}\) | \(87\) |
derivativedivides | \(\frac {\sin \left (b x +a \right ) {\mathrm e}^{m \ln \left (c \sin \left (b x +a \right )\right )}}{b \left (1+m \right )}+\frac {\sin \left (b x +a \right )^{5} {\mathrm e}^{m \ln \left (c \sin \left (b x +a \right )\right )}}{b \left (5+m \right )}-\frac {2 \sin \left (b x +a \right )^{3} {\mathrm e}^{m \ln \left (c \sin \left (b x +a \right )\right )}}{b \left (3+m \right )}\) | \(88\) |
default | \(\frac {\sin \left (b x +a \right ) {\mathrm e}^{m \ln \left (c \sin \left (b x +a \right )\right )}}{b \left (1+m \right )}+\frac {\sin \left (b x +a \right )^{5} {\mathrm e}^{m \ln \left (c \sin \left (b x +a \right )\right )}}{b \left (5+m \right )}-\frac {2 \sin \left (b x +a \right )^{3} {\mathrm e}^{m \ln \left (c \sin \left (b x +a \right )\right )}}{b \left (3+m \right )}\) | \(88\) |
Input:
int(cos(b*x+a)^5*(c*sin(b*x+a))^m,x,method=_RETURNVERBOSE)
Output:
1/8*((3/2*m^2+14*m+25/2)*sin(3*b*x+3*a)+(1/2*m^2+2*m+3/2)*sin(5*b*x+5*a)+s in(b*x+a)*(m^2+12*m+75))*(c*sin(b*x+a))^m/b/(m^3+9*m^2+23*m+15)
Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.95 \[ \int \cos ^5(a+b x) (c \sin (a+b x))^m \, dx=\frac {{\left ({\left (m^{2} + 4 \, m + 3\right )} \cos \left (b x + a\right )^{4} + 4 \, {\left (m + 1\right )} \cos \left (b x + a\right )^{2} + 8\right )} \left (c \sin \left (b x + a\right )\right )^{m} \sin \left (b x + a\right )}{b m^{3} + 9 \, b m^{2} + 23 \, b m + 15 \, b} \] Input:
integrate(cos(b*x+a)^5*(c*sin(b*x+a))^m,x, algorithm="fricas")
Output:
((m^2 + 4*m + 3)*cos(b*x + a)^4 + 4*(m + 1)*cos(b*x + a)^2 + 8)*(c*sin(b*x + a))^m*sin(b*x + a)/(b*m^3 + 9*b*m^2 + 23*b*m + 15*b)
Leaf count of result is larger than twice the leaf count of optimal. 2040 vs. \(2 (60) = 120\).
Time = 4.32 (sec) , antiderivative size = 2040, normalized size of antiderivative = 27.57 \[ \int \cos ^5(a+b x) (c \sin (a+b x))^m \, dx=\text {Too large to display} \] Input:
integrate(cos(b*x+a)**5*(c*sin(b*x+a))**m,x)
Output:
Piecewise((x*(c*sin(a))**m*cos(a)**5, Eq(b, 0)), ((log(sin(a + b*x))/b + c os(a + b*x)**2/(2*b*sin(a + b*x)**2) - cos(a + b*x)**4/(4*b*sin(a + b*x)** 4))/c**5, Eq(m, -5)), ((16*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)** 6/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/ 2)**2) + 32*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 16*log (tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**2/(8*b*tan(a/2 + b*x/2)**6 + 1 6*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 16*log(tan(a/2 + b*x/ 2))*tan(a/2 + b*x/2)**6/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)** 4 + 8*b*tan(a/2 + b*x/2)**2) - 32*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)** 4/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/ 2)**2) - 16*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**2/(8*b*tan(a/2 + b*x/2 )**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - tan(a/2 + b*x /2)**8/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 18*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a /2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 1/(8*b*tan(a/2 + b*x/2)**6 + 1 6*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2))/c**3, Eq(m, -3)), ((-l og(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**8/(b*tan(a/2 + b*x/2)**8 + 4 *b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) - 4*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**6/(b*tan(a/2 +...
Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.04 \[ \int \cos ^5(a+b x) (c \sin (a+b x))^m \, dx=\frac {\frac {c^{m} \sin \left (b x + a\right )^{m} \sin \left (b x + a\right )^{5}}{m + 5} - \frac {2 \, c^{m} \sin \left (b x + a\right )^{m} \sin \left (b x + a\right )^{3}}{m + 3} + \frac {\left (c \sin \left (b x + a\right )\right )^{m + 1}}{c {\left (m + 1\right )}}}{b} \] Input:
integrate(cos(b*x+a)^5*(c*sin(b*x+a))^m,x, algorithm="maxima")
Output:
(c^m*sin(b*x + a)^m*sin(b*x + a)^5/(m + 5) - 2*c^m*sin(b*x + a)^m*sin(b*x + a)^3/(m + 3) + (c*sin(b*x + a))^(m + 1)/(c*(m + 1)))/b
Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (74) = 148\).
Time = 0.14 (sec) , antiderivative size = 248, normalized size of antiderivative = 3.35 \[ \int \cos ^5(a+b x) (c \sin (a+b x))^m \, dx=\frac {\left (c \sin \left (b x + a\right )\right )^{m} c^{5} m^{2} \sin \left (b x + a\right )^{5} + 4 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} m \sin \left (b x + a\right )^{5} - 2 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} m^{2} \sin \left (b x + a\right )^{3} + 3 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} \sin \left (b x + a\right )^{5} - 12 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} m \sin \left (b x + a\right )^{3} + \left (c \sin \left (b x + a\right )\right )^{m} c^{5} m^{2} \sin \left (b x + a\right ) - 10 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} \sin \left (b x + a\right )^{3} + 8 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} m \sin \left (b x + a\right ) + 15 \, \left (c \sin \left (b x + a\right )\right )^{m} c^{5} \sin \left (b x + a\right )}{{\left (c^{4} m^{3} + 9 \, c^{4} m^{2} + 23 \, c^{4} m + 15 \, c^{4}\right )} b c} \] Input:
integrate(cos(b*x+a)^5*(c*sin(b*x+a))^m,x, algorithm="giac")
Output:
((c*sin(b*x + a))^m*c^5*m^2*sin(b*x + a)^5 + 4*(c*sin(b*x + a))^m*c^5*m*si n(b*x + a)^5 - 2*(c*sin(b*x + a))^m*c^5*m^2*sin(b*x + a)^3 + 3*(c*sin(b*x + a))^m*c^5*sin(b*x + a)^5 - 12*(c*sin(b*x + a))^m*c^5*m*sin(b*x + a)^3 + (c*sin(b*x + a))^m*c^5*m^2*sin(b*x + a) - 10*(c*sin(b*x + a))^m*c^5*sin(b* x + a)^3 + 8*(c*sin(b*x + a))^m*c^5*m*sin(b*x + a) + 15*(c*sin(b*x + a))^m *c^5*sin(b*x + a))/((c^4*m^3 + 9*c^4*m^2 + 23*c^4*m + 15*c^4)*b*c)
Time = 26.98 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.78 \[ \int \cos ^5(a+b x) (c \sin (a+b x))^m \, dx=\frac {{\left (c\,\sin \left (a+b\,x\right )\right )}^m\,\left (150\,\sin \left (a+b\,x\right )+25\,\sin \left (3\,a+3\,b\,x\right )+3\,\sin \left (5\,a+5\,b\,x\right )+24\,m\,\sin \left (a+b\,x\right )+28\,m\,\sin \left (3\,a+3\,b\,x\right )+4\,m\,\sin \left (5\,a+5\,b\,x\right )+2\,m^2\,\sin \left (a+b\,x\right )+3\,m^2\,\sin \left (3\,a+3\,b\,x\right )+m^2\,\sin \left (5\,a+5\,b\,x\right )\right )}{16\,b\,\left (m^3+9\,m^2+23\,m+15\right )} \] Input:
int(cos(a + b*x)^5*(c*sin(a + b*x))^m,x)
Output:
((c*sin(a + b*x))^m*(150*sin(a + b*x) + 25*sin(3*a + 3*b*x) + 3*sin(5*a + 5*b*x) + 24*m*sin(a + b*x) + 28*m*sin(3*a + 3*b*x) + 4*m*sin(5*a + 5*b*x) + 2*m^2*sin(a + b*x) + 3*m^2*sin(3*a + 3*b*x) + m^2*sin(5*a + 5*b*x)))/(16 *b*(23*m + 9*m^2 + m^3 + 15))
\[ \int \cos ^5(a+b x) (c \sin (a+b x))^m \, dx=c^{m} \left (\int \sin \left (b x +a \right )^{m} \cos \left (b x +a \right )^{5}d x \right ) \] Input:
int(cos(b*x+a)^5*(c*sin(b*x+a))^m,x)
Output:
c**m*int(sin(a + b*x)**m*cos(a + b*x)**5,x)