Integrand size = 19, antiderivative size = 76 \[ \int (d \cos (a+b x))^n \sin ^5(a+b x) \, dx=-\frac {(d \cos (a+b x))^{1+n}}{b d (1+n)}+\frac {2 (d \cos (a+b x))^{3+n}}{b d^3 (3+n)}-\frac {(d \cos (a+b x))^{5+n}}{b d^5 (5+n)} \] Output:
-(d*cos(b*x+a))^(1+n)/b/d/(1+n)+2*(d*cos(b*x+a))^(3+n)/b/d^3/(3+n)-(d*cos( b*x+a))^(5+n)/b/d^5/(5+n)
Time = 0.70 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.09 \[ \int (d \cos (a+b x))^n \sin ^5(a+b x) \, dx=-\frac {\cos (a+b x) (d \cos (a+b x))^n \left (89+28 n+3 n^2-4 \left (7+8 n+n^2\right ) \cos (2 (a+b x))+\left (3+4 n+n^2\right ) \cos (4 (a+b x))\right )}{8 b (1+n) (3+n) (5+n)} \] Input:
Integrate[(d*Cos[a + b*x])^n*Sin[a + b*x]^5,x]
Output:
-1/8*(Cos[a + b*x]*(d*Cos[a + b*x])^n*(89 + 28*n + 3*n^2 - 4*(7 + 8*n + n^ 2)*Cos[2*(a + b*x)] + (3 + 4*n + n^2)*Cos[4*(a + b*x)]))/(b*(1 + n)*(3 + n )*(5 + n))
Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 3045, 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.
\(\displaystyle \int \sin ^5(a+b x) (d \cos (a+b x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^5 (d \cos (a+b x))^ndx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\int \frac {(d \cos (a+b x))^n \left (d^2-d^2 \cos ^2(a+b x)\right )^2}{d^4}d(d \cos (a+b x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int (d \cos (a+b x))^n \left (d^2-d^2 \cos ^2(a+b x)\right )^2d(d \cos (a+b x))}{b d^5}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {\int \left (d^4 (d \cos (a+b x))^n-2 d^2 (d \cos (a+b x))^{n+2}+(d \cos (a+b x))^{n+4}\right )d(d \cos (a+b x))}{b d^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {d^4 (d \cos (a+b x))^{n+1}}{n+1}-\frac {2 d^2 (d \cos (a+b x))^{n+3}}{n+3}+\frac {(d \cos (a+b x))^{n+5}}{n+5}}{b d^5}\) |
Input:
Int[(d*Cos[a + b*x])^n*Sin[a + b*x]^5,x]
Output:
-(((d^4*(d*Cos[a + b*x])^(1 + n))/(1 + n) - (2*d^2*(d*Cos[a + b*x])^(3 + n ))/(3 + n) + (d*Cos[a + b*x])^(5 + n)/(5 + n))/(b*d^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_)]*(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 = 6.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.14
method | result | size |
parallelrisch | \(-\frac {\left (\left (-\frac {3}{2} n^{2}-14 n -\frac {25}{2}\right ) \cos \left (3 b x +3 a \right )+\left (\frac {1}{2} n^{2}+2 n +\frac {3}{2}\right ) \cos \left (5 b x +5 a \right )+\cos \left (b x +a \right ) \left (n^{2}+12 n +75\right )\right ) \left (d \cos \left (b x +a \right )\right )^{n}}{8 b \left (n^{3}+9 n^{2}+23 n +15\right )}\) | \(87\) |
derivativedivides | \(-\frac {\cos \left (b x +a \right ) {\mathrm e}^{n \ln \left (d \cos \left (b x +a \right )\right )}}{b \left (1+n \right )}+\frac {2 \cos \left (b x +a \right )^{3} {\mathrm e}^{n \ln \left (d \cos \left (b x +a \right )\right )}}{b \left (3+n \right )}-\frac {\cos \left (b x +a \right )^{5} {\mathrm e}^{n \ln \left (d \cos \left (b x +a \right )\right )}}{b \left (5+n \right )}\) | \(90\) |
default | \(-\frac {\cos \left (b x +a \right ) {\mathrm e}^{n \ln \left (d \cos \left (b x +a \right )\right )}}{b \left (1+n \right )}+\frac {2 \cos \left (b x +a \right )^{3} {\mathrm e}^{n \ln \left (d \cos \left (b x +a \right )\right )}}{b \left (3+n \right )}-\frac {\cos \left (b x +a \right )^{5} {\mathrm e}^{n \ln \left (d \cos \left (b x +a \right )\right )}}{b \left (5+n \right )}\) | \(90\) |
Input:
int((d*cos(b*x+a))^n*sin(b*x+a)^5,x,method=_RETURNVERBOSE)
Output:
-1/8*((-3/2*n^2-14*n-25/2)*cos(3*b*x+3*a)+(1/2*n^2+2*n+3/2)*cos(5*b*x+5*a) +cos(b*x+a)*(n^2+12*n+75))*(d*cos(b*x+a))^n/b/(n^3+9*n^2+23*n+15)
Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11 \[ \int (d \cos (a+b x))^n \sin ^5(a+b x) \, dx=-\frac {{\left ({\left (n^{2} + 4 \, n + 3\right )} \cos \left (b x + a\right )^{5} - 2 \, {\left (n^{2} + 6 \, n + 5\right )} \cos \left (b x + a\right )^{3} + {\left (n^{2} + 8 \, n + 15\right )} \cos \left (b x + a\right )\right )} \left (d \cos \left (b x + a\right )\right )^{n}}{b n^{3} + 9 \, b n^{2} + 23 \, b n + 15 \, b} \] Input:
integrate((d*cos(b*x+a))^n*sin(b*x+a)^5,x, algorithm="fricas")
Output:
-((n^2 + 4*n + 3)*cos(b*x + a)^5 - 2*(n^2 + 6*n + 5)*cos(b*x + a)^3 + (n^2 + 8*n + 15)*cos(b*x + a))*(d*cos(b*x + a))^n/(b*n^3 + 9*b*n^2 + 23*b*n + 15*b)
Leaf count of result is larger than twice the leaf count of optimal. 2451 vs. \(2 (60) = 120\).
Time = 4.45 (sec) , antiderivative size = 2451, normalized size of antiderivative = 32.25 \[ \int (d \cos (a+b x))^n \sin ^5(a+b x) \, dx=\text {Too large to display} \] Input:
integrate((d*cos(b*x+a))**n*sin(b*x+a)**5,x)
Output:
Piecewise((x*(d*cos(a))**n*sin(a)**5, Eq(b, 0)), ((-log(cos(a + b*x))/b + sin(a + b*x)**4/(4*b*cos(a + b*x)**4) - sin(a + b*x)**2/(2*b*cos(a + b*x)* *2))/d**5, Eq(n, -5)), ((2*log(tan(a/2 + b*x/2) - 1)*tan(a/2 + b*x/2)**8/( b*tan(a/2 + b*x/2)**8 - 2*b*tan(a/2 + b*x/2)**4 + b) - 4*log(tan(a/2 + b*x /2) - 1)*tan(a/2 + b*x/2)**4/(b*tan(a/2 + b*x/2)**8 - 2*b*tan(a/2 + b*x/2) **4 + b) + 2*log(tan(a/2 + b*x/2) - 1)/(b*tan(a/2 + b*x/2)**8 - 2*b*tan(a/ 2 + b*x/2)**4 + b) + 2*log(tan(a/2 + b*x/2) + 1)*tan(a/2 + b*x/2)**8/(b*ta n(a/2 + b*x/2)**8 - 2*b*tan(a/2 + b*x/2)**4 + b) - 4*log(tan(a/2 + b*x/2) + 1)*tan(a/2 + b*x/2)**4/(b*tan(a/2 + b*x/2)**8 - 2*b*tan(a/2 + b*x/2)**4 + b) + 2*log(tan(a/2 + b*x/2) + 1)/(b*tan(a/2 + b*x/2)**8 - 2*b*tan(a/2 + b*x/2)**4 + b) - 2*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**8/(b*tan (a/2 + b*x/2)**8 - 2*b*tan(a/2 + b*x/2)**4 + b) + 4*log(tan(a/2 + b*x/2)** 2 + 1)*tan(a/2 + b*x/2)**4/(b*tan(a/2 + b*x/2)**8 - 2*b*tan(a/2 + b*x/2)** 4 + b) - 2*log(tan(a/2 + b*x/2)**2 + 1)/(b*tan(a/2 + b*x/2)**8 - 2*b*tan(a /2 + b*x/2)**4 + b) + 4*tan(a/2 + b*x/2)**6/(b*tan(a/2 + b*x/2)**8 - 2*b*t an(a/2 + b*x/2)**4 + b) + 4*tan(a/2 + b*x/2)**2/(b*tan(a/2 + b*x/2)**8 - 2 *b*tan(a/2 + b*x/2)**4 + b))/d**3, Eq(n, -3)), ((-log(tan(a/2 + b*x/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) - 1)*tan(a/2 + b*x/2)**6/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*...
Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.03 \[ \int (d \cos (a+b x))^n \sin ^5(a+b x) \, dx=-\frac {\frac {d^{n} \cos \left (b x + a\right )^{n} \cos \left (b x + a\right )^{5}}{n + 5} - \frac {2 \, d^{n} \cos \left (b x + a\right )^{n} \cos \left (b x + a\right )^{3}}{n + 3} + \frac {\left (d \cos \left (b x + a\right )\right )^{n + 1}}{d {\left (n + 1\right )}}}{b} \] Input:
integrate((d*cos(b*x+a))^n*sin(b*x+a)^5,x, algorithm="maxima")
Output:
-(d^n*cos(b*x + a)^n*cos(b*x + a)^5/(n + 5) - 2*d^n*cos(b*x + a)^n*cos(b*x + a)^3/(n + 3) + (d*cos(b*x + a))^(n + 1)/(d*(n + 1)))/b
Leaf count of result is larger than twice the leaf count of optimal. 249 vs. \(2 (76) = 152\).
Time = 0.13 (sec) , antiderivative size = 249, normalized size of antiderivative = 3.28 \[ \int (d \cos (a+b x))^n \sin ^5(a+b x) \, dx=-\frac {\left (d \cos \left (b x + a\right )\right )^{n} d^{5} n^{2} \cos \left (b x + a\right )^{5} + 4 \, \left (d \cos \left (b x + a\right )\right )^{n} d^{5} n \cos \left (b x + a\right )^{5} - 2 \, \left (d \cos \left (b x + a\right )\right )^{n} d^{5} n^{2} \cos \left (b x + a\right )^{3} + 3 \, \left (d \cos \left (b x + a\right )\right )^{n} d^{5} \cos \left (b x + a\right )^{5} - 12 \, \left (d \cos \left (b x + a\right )\right )^{n} d^{5} n \cos \left (b x + a\right )^{3} + \left (d \cos \left (b x + a\right )\right )^{n} d^{5} n^{2} \cos \left (b x + a\right ) - 10 \, \left (d \cos \left (b x + a\right )\right )^{n} d^{5} \cos \left (b x + a\right )^{3} + 8 \, \left (d \cos \left (b x + a\right )\right )^{n} d^{5} n \cos \left (b x + a\right ) + 15 \, \left (d \cos \left (b x + a\right )\right )^{n} d^{5} \cos \left (b x + a\right )}{{\left (d^{4} n^{3} + 9 \, d^{4} n^{2} + 23 \, d^{4} n + 15 \, d^{4}\right )} b d} \] Input:
integrate((d*cos(b*x+a))^n*sin(b*x+a)^5,x, algorithm="giac")
Output:
-((d*cos(b*x + a))^n*d^5*n^2*cos(b*x + a)^5 + 4*(d*cos(b*x + a))^n*d^5*n*c os(b*x + a)^5 - 2*(d*cos(b*x + a))^n*d^5*n^2*cos(b*x + a)^3 + 3*(d*cos(b*x + a))^n*d^5*cos(b*x + a)^5 - 12*(d*cos(b*x + a))^n*d^5*n*cos(b*x + a)^3 + (d*cos(b*x + a))^n*d^5*n^2*cos(b*x + a) - 10*(d*cos(b*x + a))^n*d^5*cos(b *x + a)^3 + 8*(d*cos(b*x + a))^n*d^5*n*cos(b*x + a) + 15*(d*cos(b*x + a))^ n*d^5*cos(b*x + a))/((d^4*n^3 + 9*d^4*n^2 + 23*d^4*n + 15*d^4)*b*d)
Time = 26.89 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.74 \[ \int (d \cos (a+b x))^n \sin ^5(a+b x) \, dx=-\frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^n\,\left (150\,\cos \left (a+b\,x\right )-25\,\cos \left (3\,a+3\,b\,x\right )+3\,\cos \left (5\,a+5\,b\,x\right )+24\,n\,\cos \left (a+b\,x\right )-28\,n\,\cos \left (3\,a+3\,b\,x\right )+4\,n\,\cos \left (5\,a+5\,b\,x\right )+2\,n^2\,\cos \left (a+b\,x\right )-3\,n^2\,\cos \left (3\,a+3\,b\,x\right )+n^2\,\cos \left (5\,a+5\,b\,x\right )\right )}{16\,b\,\left (n^3+9\,n^2+23\,n+15\right )} \] Input:
int(sin(a + b*x)^5*(d*cos(a + b*x))^n,x)
Output:
-((d*cos(a + b*x))^n*(150*cos(a + b*x) - 25*cos(3*a + 3*b*x) + 3*cos(5*a + 5*b*x) + 24*n*cos(a + b*x) - 28*n*cos(3*a + 3*b*x) + 4*n*cos(5*a + 5*b*x) + 2*n^2*cos(a + b*x) - 3*n^2*cos(3*a + 3*b*x) + n^2*cos(5*a + 5*b*x)))/(1 6*b*(23*n + 9*n^2 + n^3 + 15))
\[ \int (d \cos (a+b x))^n \sin ^5(a+b x) \, dx=d^{n} \left (\int \cos \left (b x +a \right )^{n} \sin \left (b x +a \right )^{5}d x \right ) \] Input:
int((d*cos(b*x+a))^n*sin(b*x+a)^5,x)
Output:
d**n*int(cos(a + b*x)**n*sin(a + b*x)**5,x)