Integrand size = 29, antiderivative size = 111 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {2 (a+a \sin (c+d x))^6}{3 a^3 d}-\frac {12 (a+a \sin (c+d x))^7}{7 a^4 d}+\frac {13 (a+a \sin (c+d x))^8}{8 a^5 d}-\frac {2 (a+a \sin (c+d x))^9}{3 a^6 d}+\frac {(a+a \sin (c+d x))^{10}}{10 a^7 d} \] Output:
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 = 0.72 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.99 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 (-2835+34440 \cos (2 (c+d x))+5040 \cos (4 (c+d x))-4060 \cos (6 (c+d x))-1260 \cos (8 (c+d x))+84 \cos (10 (c+d x))-63840 \sin (c+d x)+8960 \sin (3 (c+d x))+8064 \sin (5 (c+d x))+240 \sin (7 (c+d x))-560 \sin (9 (c+d x)))}{430080 d} \] Input:
Integrate[Cos[c + d*x]^5*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]
Output:
-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 = 0.34 (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 |
\(\displaystyle \frac {\frac {2}{3} a^4 (a \sin (c+d x)+a)^6-\frac {12}{7} a^3 (a \sin (c+d x)+a)^7+\frac {13}{8} a^2 (a \sin (c+d x)+a)^8+\frac {1}{10} (a \sin (c+d x)+a)^{10}-\frac {2}{3} a (a \sin (c+d x)+a)^9}{a^7 d}\) |
Input:
Int[Cos[c + d*x]^5*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]
Output:
((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)
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 = 0.06 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.80
\[\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{10}}{10}+\frac {\sin \left (d x +c \right )^{9}}{3}+\frac {\sin \left (d x +c \right )^{8}}{8}-\frac {5 \sin \left (d x +c \right )^{7}}{7}-\frac {5 \sin \left (d x +c \right )^{6}}{6}+\frac {\sin \left (d x +c \right )^{5}}{5}+\frac {3 \sin \left (d x +c \right )^{4}}{4}+\frac {\sin \left (d x +c \right )^{3}}{3}\right )}{d}\]
Input:
int(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x)
Output:
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 = 0.13 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {84 \, a^{3} \cos \left (d x + c\right )^{10} - 525 \, a^{3} \cos \left (d x + c\right )^{8} + 560 \, a^{3} \cos \left (d x + c\right )^{6} - 8 \, {\left (35 \, a^{3} \cos \left (d x + c\right )^{8} - 65 \, a^{3} \cos \left (d x + c\right )^{6} + 6 \, a^{3} \cos \left (d x + c\right )^{4} + 8 \, a^{3} \cos \left (d x + c\right )^{2} + 16 \, a^{3}\right )} \sin \left (d x + c\right )}{840 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
-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. 280 vs. \(2 (99) = 198\).
Time = 1.36 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.52 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\begin {cases} \frac {a^{3} \sin ^{10}{\left (c + d x \right )}}{60 d} + \frac {8 a^{3} \sin ^{9}{\left (c + d x \right )}}{105 d} + \frac {a^{3} \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{12 d} + \frac {a^{3} \sin ^{8}{\left (c + d x \right )}}{8 d} + \frac {12 a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {8 a^{3} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {a^{3} \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{6 d} + \frac {a^{3} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac {3 a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {4 a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {3 a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{3} \sin ^{2}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**2*(a+a*sin(d*x+c))**3,x)
Output:
Piecewise((a**3*sin(c + d*x)**10/(60*d) + 8*a**3*sin(c + d*x)**9/(105*d) + a**3*sin(c + d*x)**8*cos(c + d*x)**2/(12*d) + a**3*sin(c + d*x)**8/(8*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) + a**3*sin(c + d*x)**6*cos(c + d*x)**4/(6*d) + a**3*sin(c + d*x)** 6*cos(c + d*x)**2/(2*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) + 3*a**3*sin(c + d*x)**4*cos( c + d*x)**4/(4*d) + a**3*sin(c + d*x)**3*cos(c + d*x)**4/(3*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*sin(c)**2*cos(c)**5, True))
Time = 0.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.99 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {84 \, a^{3} \sin \left (d x + c\right )^{10} + 280 \, a^{3} \sin \left (d x + c\right )^{9} + 105 \, a^{3} \sin \left (d x + c\right )^{8} - 600 \, a^{3} \sin \left (d x + c\right )^{7} - 700 \, a^{3} \sin \left (d x + c\right )^{6} + 168 \, a^{3} \sin \left (d x + c\right )^{5} + 630 \, a^{3} \sin \left (d x + c\right )^{4} + 280 \, a^{3} \sin \left (d x + c\right )^{3}}{840 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
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 = 0.21 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.99 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {84 \, a^{3} \sin \left (d x + c\right )^{10} + 280 \, a^{3} \sin \left (d x + c\right )^{9} + 105 \, a^{3} \sin \left (d x + c\right )^{8} - 600 \, a^{3} \sin \left (d x + c\right )^{7} - 700 \, a^{3} \sin \left (d x + c\right )^{6} + 168 \, a^{3} \sin \left (d x + c\right )^{5} + 630 \, a^{3} \sin \left (d x + c\right )^{4} + 280 \, a^{3} \sin \left (d x + c\right )^{3}}{840 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
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 = 31.86 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.98 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^{10}}{10}+\frac {a^3\,{\sin \left (c+d\,x\right )}^9}{3}+\frac {a^3\,{\sin \left (c+d\,x\right )}^8}{8}-\frac {5\,a^3\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {5\,a^3\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {a^3\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {3\,a^3\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a^3\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \] Input:
int(cos(c + d*x)^5*sin(c + d*x)^2*(a + a*sin(c + d*x))^3,x)
Output:
((a^3*sin(c + d*x)^3)/3 + (3*a^3*sin(c + d*x)^4)/4 + (a^3*sin(c + d*x)^5)/ 5 - (5*a^3*sin(c + d*x)^6)/6 - (5*a^3*sin(c + d*x)^7)/7 + (a^3*sin(c + d*x )^8)/8 + (a^3*sin(c + d*x)^9)/3 + (a^3*sin(c + d*x)^10)/10)/d
Time = 0.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.77 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {\sin \left (d x +c \right )^{3} a^{3} \left (84 \sin \left (d x +c \right )^{7}+280 \sin \left (d x +c \right )^{6}+105 \sin \left (d x +c \right )^{5}-600 \sin \left (d x +c \right )^{4}-700 \sin \left (d x +c \right )^{3}+168 \sin \left (d x +c \right )^{2}+630 \sin \left (d x +c \right )+280\right )}{840 d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x)
Output:
(sin(c + d*x)**3*a**3*(84*sin(c + d*x)**7 + 280*sin(c + d*x)**6 + 105*sin( c + d*x)**5 - 600*sin(c + d*x)**4 - 700*sin(c + d*x)**3 + 168*sin(c + d*x) **2 + 630*sin(c + d*x) + 280))/(840*d)