Integrand size = 29, antiderivative size = 111 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {2 (a+a \sin (c+d x))^6}{3 a^3 d}-\frac {12 (a+a \sin (c+d x))^7}{7 a^4 d}+\frac {13 (a+a \sin (c+d x))^8}{8 a^5 d}-\frac {2 (a+a \sin (c+d x))^9}{3 a^6 d}+\frac {(a+a \sin (c+d x))^{10}}{10 a^7 d} \]
2/3*(a+a*sin(d*x+c))^6/a^3/d-12/7*(a+a*sin(d*x+c))^7/a^4/d+13/8*(a+a*sin(d *x+c))^8/a^5/d-2/3*(a+a*sin(d*x+c))^9/a^6/d+1/10*(a+a*sin(d*x+c))^10/a^7/d
Time = 0.47 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.99 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 (-2835+34440 \cos (2 (c+d x))+5040 \cos (4 (c+d x))-4060 \cos (6 (c+d x))-1260 \cos (8 (c+d x))+84 \cos (10 (c+d x))-63840 \sin (c+d x)+8960 \sin (3 (c+d x))+8064 \sin (5 (c+d x))+240 \sin (7 (c+d x))-560 \sin (9 (c+d x)))}{430080 d} \]
-1/430080*(a^3*(-2835 + 34440*Cos[2*(c + d*x)] + 5040*Cos[4*(c + d*x)] - 4 060*Cos[6*(c + d*x)] - 1260*Cos[8*(c + d*x)] + 84*Cos[10*(c + d*x)] - 6384 0*Sin[c + d*x] + 8960*Sin[3*(c + d*x)] + 8064*Sin[5*(c + d*x)] + 240*Sin[7 *(c + d*x)] - 560*Sin[9*(c + d*x)]))/d
Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 27, 99, 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.
\(\displaystyle \int \sin ^2(c+d x) \cos ^5(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^2 \cos (c+d x)^5 (a \sin (c+d x)+a)^3dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin ^2(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^2 \sin ^2(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left ((\sin (c+d x) a+a)^9-6 a (\sin (c+d x) a+a)^8+13 a^2 (\sin (c+d x) a+a)^7-12 a^3 (\sin (c+d x) a+a)^6+4 a^4 (\sin (c+d x) a+a)^5\right )d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2}{3} a^4 (a \sin (c+d x)+a)^6-\frac {12}{7} a^3 (a \sin (c+d x)+a)^7+\frac {13}{8} a^2 (a \sin (c+d x)+a)^8+\frac {1}{10} (a \sin (c+d x)+a)^{10}-\frac {2}{3} a (a \sin (c+d x)+a)^9}{a^7 d}\) |
((2*a^4*(a + a*Sin[c + d*x])^6)/3 - (12*a^3*(a + a*Sin[c + d*x])^7)/7 + (1 3*a^2*(a + a*Sin[c + d*x])^8)/8 - (2*a*(a + a*Sin[c + d*x])^9)/3 + (a + a* Sin[c + d*x])^10/10)/(a^7*d)
3.6.21.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 0.63 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(\frac {a^{3} \left (\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}+\frac {\left (\sin ^{9}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {5 \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\) | \(89\) |
default | \(\frac {a^{3} \left (\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}+\frac {\left (\sin ^{9}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {5 \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\) | \(89\) |
parallelrisch | \(-\frac {a^{3} \left (\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (1356 \cos \left (2 d x +2 c \right )-21 \sin \left (7 d x +7 c \right )+252 \sin \left (5 d x +5 c \right )-140 \cos \left (6 d x +6 c \right )+3465 \sin \left (d x +c \right )+1834 \sin \left (3 d x +3 c \right )-360 \cos \left (4 d x +4 c \right )+3624\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{26880 d}\) | \(127\) |
risch | \(\frac {19 a^{3} \sin \left (d x +c \right )}{128 d}-\frac {a^{3} \cos \left (10 d x +10 c \right )}{5120 d}+\frac {a^{3} \sin \left (9 d x +9 c \right )}{768 d}+\frac {3 a^{3} \cos \left (8 d x +8 c \right )}{1024 d}-\frac {a^{3} \sin \left (7 d x +7 c \right )}{1792 d}+\frac {29 a^{3} \cos \left (6 d x +6 c \right )}{3072 d}-\frac {3 a^{3} \sin \left (5 d x +5 c \right )}{160 d}-\frac {3 a^{3} \cos \left (4 d x +4 c \right )}{256 d}-\frac {a^{3} \sin \left (3 d x +3 c \right )}{48 d}-\frac {41 a^{3} \cos \left (2 d x +2 c \right )}{512 d}\) | \(169\) |
a^3/d*(1/10*sin(d*x+c)^10+1/3*sin(d*x+c)^9+1/8*sin(d*x+c)^8-5/7*sin(d*x+c) ^7-5/6*sin(d*x+c)^6+1/5*sin(d*x+c)^5+3/4*sin(d*x+c)^4+1/3*sin(d*x+c)^3)
Time = 0.31 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {84 \, a^{3} \cos \left (d x + c\right )^{10} - 525 \, a^{3} \cos \left (d x + c\right )^{8} + 560 \, a^{3} \cos \left (d x + c\right )^{6} - 8 \, {\left (35 \, a^{3} \cos \left (d x + c\right )^{8} - 65 \, a^{3} \cos \left (d x + c\right )^{6} + 6 \, a^{3} \cos \left (d x + c\right )^{4} + 8 \, a^{3} \cos \left (d x + c\right )^{2} + 16 \, a^{3}\right )} \sin \left (d x + c\right )}{840 \, d} \]
-1/840*(84*a^3*cos(d*x + c)^10 - 525*a^3*cos(d*x + c)^8 + 560*a^3*cos(d*x + c)^6 - 8*(35*a^3*cos(d*x + c)^8 - 65*a^3*cos(d*x + c)^6 + 6*a^3*cos(d*x + c)^4 + 8*a^3*cos(d*x + c)^2 + 16*a^3)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (99) = 198\).
Time = 1.26 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.30 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\begin {cases} \frac {8 a^{3} \sin ^{9}{\left (c + d x \right )}}{105 d} + \frac {12 a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {8 a^{3} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {3 a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {4 a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} - \frac {a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac {a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{12 d} - \frac {a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{2 d} - \frac {a^{3} \cos ^{10}{\left (c + d x \right )}}{60 d} - \frac {a^{3} \cos ^{8}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{3} \sin ^{2}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((8*a**3*sin(c + d*x)**9/(105*d) + 12*a**3*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 8*a**3*sin(c + d*x)**7/(105*d) + 3*a**3*sin(c + d*x)**5 *cos(c + d*x)**4/(5*d) + 4*a**3*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) - a **3*sin(c + d*x)**4*cos(c + d*x)**6/(6*d) + a**3*sin(c + d*x)**3*cos(c + d *x)**4/(3*d) - a**3*sin(c + d*x)**2*cos(c + d*x)**8/(12*d) - a**3*sin(c + d*x)**2*cos(c + d*x)**6/(2*d) - a**3*cos(c + d*x)**10/(60*d) - a**3*cos(c + d*x)**8/(8*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*sin(c)**2*cos(c)**5, True ))
Time = 0.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.99 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {84 \, a^{3} \sin \left (d x + c\right )^{10} + 280 \, a^{3} \sin \left (d x + c\right )^{9} + 105 \, a^{3} \sin \left (d x + c\right )^{8} - 600 \, a^{3} \sin \left (d x + c\right )^{7} - 700 \, a^{3} \sin \left (d x + c\right )^{6} + 168 \, a^{3} \sin \left (d x + c\right )^{5} + 630 \, a^{3} \sin \left (d x + c\right )^{4} + 280 \, a^{3} \sin \left (d x + c\right )^{3}}{840 \, d} \]
1/840*(84*a^3*sin(d*x + c)^10 + 280*a^3*sin(d*x + c)^9 + 105*a^3*sin(d*x + c)^8 - 600*a^3*sin(d*x + c)^7 - 700*a^3*sin(d*x + c)^6 + 168*a^3*sin(d*x + c)^5 + 630*a^3*sin(d*x + c)^4 + 280*a^3*sin(d*x + c)^3)/d
Time = 0.63 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.51 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^{3} \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac {3 \, a^{3} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} + \frac {29 \, a^{3} \cos \left (6 \, d x + 6 \, c\right )}{3072 \, d} - \frac {3 \, a^{3} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {41 \, a^{3} \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} + \frac {a^{3} \sin \left (9 \, d x + 9 \, c\right )}{768 \, d} - \frac {a^{3} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac {3 \, a^{3} \sin \left (5 \, d x + 5 \, c\right )}{160 \, d} - \frac {a^{3} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {19 \, a^{3} \sin \left (d x + c\right )}{128 \, d} \]
-1/5120*a^3*cos(10*d*x + 10*c)/d + 3/1024*a^3*cos(8*d*x + 8*c)/d + 29/3072 *a^3*cos(6*d*x + 6*c)/d - 3/256*a^3*cos(4*d*x + 4*c)/d - 41/512*a^3*cos(2* d*x + 2*c)/d + 1/768*a^3*sin(9*d*x + 9*c)/d - 1/1792*a^3*sin(7*d*x + 7*c)/ d - 3/160*a^3*sin(5*d*x + 5*c)/d - 1/48*a^3*sin(3*d*x + 3*c)/d + 19/128*a^ 3*sin(d*x + c)/d
Time = 9.21 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.98 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^{10}}{10}+\frac {a^3\,{\sin \left (c+d\,x\right )}^9}{3}+\frac {a^3\,{\sin \left (c+d\,x\right )}^8}{8}-\frac {5\,a^3\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {5\,a^3\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {a^3\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {3\,a^3\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a^3\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \]