Integrand size = 27, antiderivative size = 114 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \sin ^{1+n}(c+d x)}{d (1+n)}+\frac {4 a^4 \sin ^{2+n}(c+d x)}{d (2+n)}+\frac {6 a^4 \sin ^{3+n}(c+d x)}{d (3+n)}+\frac {4 a^4 \sin ^{4+n}(c+d x)}{d (4+n)}+\frac {a^4 \sin ^{5+n}(c+d x)}{d (5+n)} \]
a^4*sin(d*x+c)^(1+n)/d/(1+n)+4*a^4*sin(d*x+c)^(2+n)/d/(2+n)+6*a^4*sin(d*x+ c)^(3+n)/d/(3+n)+4*a^4*sin(d*x+c)^(4+n)/d/(4+n)+a^4*sin(d*x+c)^(5+n)/d/(5+ n)
Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.70 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \sin ^{1+n}(c+d x) \left (\frac {1}{1+n}+\frac {4 \sin (c+d x)}{2+n}+\frac {6 \sin ^2(c+d x)}{3+n}+\frac {4 \sin ^3(c+d x)}{4+n}+\frac {\sin ^4(c+d x)}{5+n}\right )}{d} \]
(a^4*Sin[c + d*x]^(1 + n)*((1 + n)^(-1) + (4*Sin[c + d*x])/(2 + n) + (6*Si n[c + d*x]^2)/(3 + n) + (4*Sin[c + d*x]^3)/(4 + n) + Sin[c + d*x]^4/(5 + n )))/d
Time = 0.32 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3042, 3312, 53, 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.
\(\displaystyle \int \cos (c+d x) (a \sin (c+d x)+a)^4 \sin ^n(c+d x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x) (a \sin (c+d x)+a)^4 \sin (c+d x)^ndx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \sin ^n(c+d x) (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\int \left (a^4 \sin ^n(c+d x)+4 a^4 \sin ^{n+1}(c+d x)+6 a^4 \sin ^{n+2}(c+d x)+4 a^4 \sin ^{n+3}(c+d x)+a^4 \sin ^{n+4}(c+d x)\right )d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^5 \sin ^{n+1}(c+d x)}{n+1}+\frac {4 a^5 \sin ^{n+2}(c+d x)}{n+2}+\frac {6 a^5 \sin ^{n+3}(c+d x)}{n+3}+\frac {4 a^5 \sin ^{n+4}(c+d x)}{n+4}+\frac {a^5 \sin ^{n+5}(c+d x)}{n+5}}{a d}\) |
((a^5*Sin[c + d*x]^(1 + n))/(1 + n) + (4*a^5*Sin[c + d*x]^(2 + n))/(2 + n) + (6*a^5*Sin[c + d*x]^(3 + n))/(3 + n) + (4*a^5*Sin[c + d*x]^(4 + n))/(4 + n) + (a^5*Sin[c + d*x]^(5 + n))/(5 + n))/(a*d)
3.3.58.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 3.78 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.34
method | result | size |
derivativedivides | \(\frac {a^{4} \sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (1+n \right )}+\frac {a^{4} \left (\sin ^{5}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (5+n \right )}+\frac {4 a^{4} \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (2+n \right )}+\frac {6 a^{4} \left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (3+n \right )}+\frac {4 a^{4} \left (\sin ^{4}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (4+n \right )}\) | \(153\) |
default | \(\frac {a^{4} \sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (1+n \right )}+\frac {a^{4} \left (\sin ^{5}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (5+n \right )}+\frac {4 a^{4} \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (2+n \right )}+\frac {6 a^{4} \left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (3+n \right )}+\frac {4 a^{4} \left (\sin ^{4}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (4+n \right )}\) | \(153\) |
parallelrisch | \(\frac {\left (2640-64 \left (5+n \right ) \left (1+n \right ) \left (3+n \right )^{2} \cos \left (2 d x +2 c \right )+8 \left (n^{4}+11 n^{3}+41 n^{2}+61 n +30\right ) \cos \left (4 d x +4 c \right )+\left (-29 n^{4}-338 n^{3}-1351 n^{2}-2122 n -1080\right ) \sin \left (3 d x +3 c \right )+\left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right ) \sin \left (5 d x +5 c \right )+2 \left (49 n^{4}+594 n^{3}+2507 n^{2}+4290 n +2520\right ) \sin \left (d x +c \right )+56 n^{4}+680 n^{3}+2872 n^{2}+4888 n \right ) \left (\sin ^{n}\left (d x +c \right )\right ) a^{4}}{16 \left (5+n \right ) \left (3+n \right ) \left (1+n \right ) \left (2+n \right ) d \left (4+n \right )}\) | \(199\) |
a^4/d/(1+n)*sin(d*x+c)*exp(n*ln(sin(d*x+c)))+a^4/d/(5+n)*sin(d*x+c)^5*exp( n*ln(sin(d*x+c)))+4*a^4/d/(2+n)*sin(d*x+c)^2*exp(n*ln(sin(d*x+c)))+6*a^4/d /(3+n)*sin(d*x+c)^3*exp(n*ln(sin(d*x+c)))+4*a^4/d/(4+n)*sin(d*x+c)^4*exp(n *ln(sin(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (114) = 228\).
Time = 0.29 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.65 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {{\left (8 \, a^{4} n^{4} + 96 \, a^{4} n^{3} + 400 \, a^{4} n^{2} + 672 \, a^{4} n + 4 \, {\left (a^{4} n^{4} + 11 \, a^{4} n^{3} + 41 \, a^{4} n^{2} + 61 \, a^{4} n + 30 \, a^{4}\right )} \cos \left (d x + c\right )^{4} + 360 \, a^{4} - 4 \, {\left (3 \, a^{4} n^{4} + 35 \, a^{4} n^{3} + 141 \, a^{4} n^{2} + 229 \, a^{4} n + 120 \, a^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (8 \, a^{4} n^{4} + 96 \, a^{4} n^{3} + 400 \, a^{4} n^{2} + 672 \, a^{4} n + {\left (a^{4} n^{4} + 10 \, a^{4} n^{3} + 35 \, a^{4} n^{2} + 50 \, a^{4} n + 24 \, a^{4}\right )} \cos \left (d x + c\right )^{4} + 384 \, a^{4} - 4 \, {\left (2 \, a^{4} n^{4} + 23 \, a^{4} n^{3} + 91 \, a^{4} n^{2} + 142 \, a^{4} n + 72 \, a^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{d n^{5} + 15 \, d n^{4} + 85 \, d n^{3} + 225 \, d n^{2} + 274 \, d n + 120 \, d} \]
(8*a^4*n^4 + 96*a^4*n^3 + 400*a^4*n^2 + 672*a^4*n + 4*(a^4*n^4 + 11*a^4*n^ 3 + 41*a^4*n^2 + 61*a^4*n + 30*a^4)*cos(d*x + c)^4 + 360*a^4 - 4*(3*a^4*n^ 4 + 35*a^4*n^3 + 141*a^4*n^2 + 229*a^4*n + 120*a^4)*cos(d*x + c)^2 + (8*a^ 4*n^4 + 96*a^4*n^3 + 400*a^4*n^2 + 672*a^4*n + (a^4*n^4 + 10*a^4*n^3 + 35* a^4*n^2 + 50*a^4*n + 24*a^4)*cos(d*x + c)^4 + 384*a^4 - 4*(2*a^4*n^4 + 23* a^4*n^3 + 91*a^4*n^2 + 142*a^4*n + 72*a^4)*cos(d*x + c)^2)*sin(d*x + c))*s in(d*x + c)^n/(d*n^5 + 15*d*n^4 + 85*d*n^3 + 225*d*n^2 + 274*d*n + 120*d)
Leaf count of result is larger than twice the leaf count of optimal. 1833 vs. \(2 (97) = 194\).
Time = 3.61 (sec) , antiderivative size = 1833, normalized size of antiderivative = 16.08 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^4 \, dx=\text {Too large to display} \]
Piecewise((x*(a*sin(c) + a)**4*sin(c)**n*cos(c), Eq(d, 0)), (a**4*log(sin( c + d*x))/d - 4*a**4/(d*sin(c + d*x)) - 3*a**4/(d*sin(c + d*x)**2) - 4*a** 4/(3*d*sin(c + d*x)**3) - a**4/(4*d*sin(c + d*x)**4), Eq(n, -5)), (4*a**4* log(sin(c + d*x))/d + a**4*sin(c + d*x)/d - 6*a**4/(d*sin(c + d*x)) - 2*a* *4/(d*sin(c + d*x)**2) - a**4/(3*d*sin(c + d*x)**3), Eq(n, -4)), (6*a**4*l og(sin(c + d*x))/d + a**4*sin(c + d*x)**2/(2*d) + 4*a**4*sin(c + d*x)/d - 4*a**4/(d*sin(c + d*x)) - a**4/(2*d*sin(c + d*x)**2), Eq(n, -3)), (4*a**4* log(sin(c + d*x))/d + a**4*sin(c + d*x)**3/(3*d) + 2*a**4*sin(c + d*x)**2/ d + 6*a**4*sin(c + d*x)/d - a**4/(d*sin(c + d*x)), Eq(n, -2)), (a**4*log(s in(c + d*x))/d + a**4*sin(c + d*x)**4/(4*d) + 4*a**4*sin(c + d*x)**3/(3*d) + 3*a**4*sin(c + d*x)**2/d + 4*a**4*sin(c + d*x)/d, Eq(n, -1)), (a**4*n** 4*sin(c + d*x)**5*sin(c + d*x)**n/(d*n**5 + 15*d*n**4 + 85*d*n**3 + 225*d* n**2 + 274*d*n + 120*d) + 4*a**4*n**4*sin(c + d*x)**4*sin(c + d*x)**n/(d*n **5 + 15*d*n**4 + 85*d*n**3 + 225*d*n**2 + 274*d*n + 120*d) + 6*a**4*n**4* sin(c + d*x)**3*sin(c + d*x)**n/(d*n**5 + 15*d*n**4 + 85*d*n**3 + 225*d*n* *2 + 274*d*n + 120*d) + 4*a**4*n**4*sin(c + d*x)**2*sin(c + d*x)**n/(d*n** 5 + 15*d*n**4 + 85*d*n**3 + 225*d*n**2 + 274*d*n + 120*d) + a**4*n**4*sin( c + d*x)*sin(c + d*x)**n/(d*n**5 + 15*d*n**4 + 85*d*n**3 + 225*d*n**2 + 27 4*d*n + 120*d) + 10*a**4*n**3*sin(c + d*x)**5*sin(c + d*x)**n/(d*n**5 + 15 *d*n**4 + 85*d*n**3 + 225*d*n**2 + 274*d*n + 120*d) + 44*a**4*n**3*sin(...
Time = 0.20 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.90 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {\frac {a^{4} \sin \left (d x + c\right )^{n + 5}}{n + 5} + \frac {4 \, a^{4} \sin \left (d x + c\right )^{n + 4}}{n + 4} + \frac {6 \, a^{4} \sin \left (d x + c\right )^{n + 3}}{n + 3} + \frac {4 \, a^{4} \sin \left (d x + c\right )^{n + 2}}{n + 2} + \frac {a^{4} \sin \left (d x + c\right )^{n + 1}}{n + 1}}{d} \]
(a^4*sin(d*x + c)^(n + 5)/(n + 5) + 4*a^4*sin(d*x + c)^(n + 4)/(n + 4) + 6 *a^4*sin(d*x + c)^(n + 3)/(n + 3) + 4*a^4*sin(d*x + c)^(n + 2)/(n + 2) + a ^4*sin(d*x + c)^(n + 1)/(n + 1))/d
Time = 0.50 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.11 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {\frac {a^{4} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{5}}{n + 5} + \frac {4 \, a^{4} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{4}}{n + 4} + \frac {6 \, a^{4} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3}}{n + 3} + \frac {4 \, a^{4} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2}}{n + 2} + \frac {a^{4} \sin \left (d x + c\right )^{n + 1}}{n + 1}}{d} \]
(a^4*sin(d*x + c)^n*sin(d*x + c)^5/(n + 5) + 4*a^4*sin(d*x + c)^n*sin(d*x + c)^4/(n + 4) + 6*a^4*sin(d*x + c)^n*sin(d*x + c)^3/(n + 3) + 4*a^4*sin(d *x + c)^n*sin(d*x + c)^2/(n + 2) + a^4*sin(d*x + c)^(n + 1)/(n + 1))/d
Time = 13.88 (sec) , antiderivative size = 370, normalized size of antiderivative = 3.25 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4\,{\sin \left (c+d\,x\right )}^n\,\left (4888\,n+5040\,\sin \left (c+d\,x\right )-2880\,\cos \left (2\,c+2\,d\,x\right )+240\,\cos \left (4\,c+4\,d\,x\right )-1080\,\sin \left (3\,c+3\,d\,x\right )+24\,\sin \left (5\,c+5\,d\,x\right )+8580\,n\,\sin \left (c+d\,x\right )-5376\,n\,\cos \left (2\,c+2\,d\,x\right )+488\,n\,\cos \left (4\,c+4\,d\,x\right )-2122\,n\,\sin \left (3\,c+3\,d\,x\right )+50\,n\,\sin \left (5\,c+5\,d\,x\right )+5014\,n^2\,\sin \left (c+d\,x\right )+1188\,n^3\,\sin \left (c+d\,x\right )+98\,n^4\,\sin \left (c+d\,x\right )+2872\,n^2+680\,n^3+56\,n^4-3200\,n^2\,\cos \left (2\,c+2\,d\,x\right )-768\,n^3\,\cos \left (2\,c+2\,d\,x\right )-64\,n^4\,\cos \left (2\,c+2\,d\,x\right )+328\,n^2\,\cos \left (4\,c+4\,d\,x\right )+88\,n^3\,\cos \left (4\,c+4\,d\,x\right )+8\,n^4\,\cos \left (4\,c+4\,d\,x\right )-1351\,n^2\,\sin \left (3\,c+3\,d\,x\right )-338\,n^3\,\sin \left (3\,c+3\,d\,x\right )-29\,n^4\,\sin \left (3\,c+3\,d\,x\right )+35\,n^2\,\sin \left (5\,c+5\,d\,x\right )+10\,n^3\,\sin \left (5\,c+5\,d\,x\right )+n^4\,\sin \left (5\,c+5\,d\,x\right )+2640\right )}{16\,d\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )} \]
(a^4*sin(c + d*x)^n*(4888*n + 5040*sin(c + d*x) - 2880*cos(2*c + 2*d*x) + 240*cos(4*c + 4*d*x) - 1080*sin(3*c + 3*d*x) + 24*sin(5*c + 5*d*x) + 8580* n*sin(c + d*x) - 5376*n*cos(2*c + 2*d*x) + 488*n*cos(4*c + 4*d*x) - 2122*n *sin(3*c + 3*d*x) + 50*n*sin(5*c + 5*d*x) + 5014*n^2*sin(c + d*x) + 1188*n ^3*sin(c + d*x) + 98*n^4*sin(c + d*x) + 2872*n^2 + 680*n^3 + 56*n^4 - 3200 *n^2*cos(2*c + 2*d*x) - 768*n^3*cos(2*c + 2*d*x) - 64*n^4*cos(2*c + 2*d*x) + 328*n^2*cos(4*c + 4*d*x) + 88*n^3*cos(4*c + 4*d*x) + 8*n^4*cos(4*c + 4* d*x) - 1351*n^2*sin(3*c + 3*d*x) - 338*n^3*sin(3*c + 3*d*x) - 29*n^4*sin(3 *c + 3*d*x) + 35*n^2*sin(5*c + 5*d*x) + 10*n^3*sin(5*c + 5*d*x) + n^4*sin( 5*c + 5*d*x) + 2640))/(16*d*(274*n + 225*n^2 + 85*n^3 + 15*n^4 + n^5 + 120 ))