Integrand size = 27, antiderivative size = 67 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {(a+a \sin (c+d x))^5}{5 a d}-\frac {(a+a \sin (c+d x))^6}{3 a^2 d}+\frac {(a+a \sin (c+d x))^7}{7 a^3 d} \] Output:
1/5*(a+a*sin(d*x+c))^5/a/d-1/3*(a+a*sin(d*x+c))^6/a^2/d+1/7*(a+a*sin(d*x+c ))^7/a^3/d
Time = 0.17 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.19 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {a^4 (-630+5460 \cos (2 (c+d x))-1680 \cos (4 (c+d x))+140 \cos (6 (c+d x))-7245 \sin (c+d x)+3395 \sin (3 (c+d x))-609 \sin (5 (c+d x))+15 \sin (7 (c+d x)))}{6720 d} \] Input:
Integrate[Cos[c + d*x]*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^4,x]
Output:
-1/6720*(a^4*(-630 + 5460*Cos[2*(c + d*x)] - 1680*Cos[4*(c + d*x)] + 140*C os[6*(c + d*x)] - 7245*Sin[c + d*x] + 3395*Sin[3*(c + d*x)] - 609*Sin[5*(c + d*x)] + 15*Sin[7*(c + d*x)]))/d
Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3312, 27, 49, 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.
\(\displaystyle \int \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^4 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^2 \cos (c+d x) (a \sin (c+d x)+a)^4dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \sin ^2(c+d x) (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^2 \sin ^2(c+d x) (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left ((\sin (c+d x) a+a)^6-2 a (\sin (c+d x) a+a)^5+a^2 (\sin (c+d x) a+a)^4\right )d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{5} a^2 (a \sin (c+d x)+a)^5+\frac {1}{7} (a \sin (c+d x)+a)^7-\frac {1}{3} a (a \sin (c+d x)+a)^6}{a^3 d}\) |
Input:
Int[Cos[c + d*x]*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^4,x]
Output:
((a^2*(a + a*Sin[c + d*x])^5)/5 - (a*(a + a*Sin[c + d*x])^6)/3 + (a + a*Si n[c + d*x])^7/7)/(a^3*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_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[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 = 292.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.04
method | result | size |
derivativedivides | \(\frac {\frac {a^{4} \sin \left (d x +c \right )^{7}}{7}+\frac {2 a^{4} \sin \left (d x +c \right )^{6}}{3}+\frac {6 a^{4} \sin \left (d x +c \right )^{5}}{5}+a^{4} \sin \left (d x +c \right )^{4}+\frac {a^{4} \sin \left (d x +c \right )^{3}}{3}}{d}\) | \(70\) |
default | \(\frac {\frac {a^{4} \sin \left (d x +c \right )^{7}}{7}+\frac {2 a^{4} \sin \left (d x +c \right )^{6}}{3}+\frac {6 a^{4} \sin \left (d x +c \right )^{5}}{5}+a^{4} \sin \left (d x +c \right )^{4}+\frac {a^{4} \sin \left (d x +c \right )^{3}}{3}}{d}\) | \(70\) |
parallelrisch | \(-\frac {a^{4} \left (\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (829-564 \cos \left (2 d x +2 c \right )+1260 \sin \left (d x +c \right )-140 \sin \left (3 d x +3 c \right )+15 \cos \left (4 d x +4 c \right )\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{1680 d}\) | \(94\) |
risch | \(\frac {69 a^{4} \sin \left (d x +c \right )}{64 d}-\frac {a^{4} \sin \left (7 d x +7 c \right )}{448 d}-\frac {a^{4} \cos \left (6 d x +6 c \right )}{48 d}+\frac {29 a^{4} \sin \left (5 d x +5 c \right )}{320 d}+\frac {a^{4} \cos \left (4 d x +4 c \right )}{4 d}-\frac {97 a^{4} \sin \left (3 d x +3 c \right )}{192 d}-\frac {13 a^{4} \cos \left (2 d x +2 c \right )}{16 d}\) | \(118\) |
norman | \(\frac {\frac {16 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {16 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {8 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {736 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {3888 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{35 d}+\frac {736 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{15 d}+\frac {8 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {272 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}+\frac {272 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}\) | \(189\) |
orering | \(\text {Expression too large to display}\) | \(2522\) |
Input:
int(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/d*(1/7*a^4*sin(d*x+c)^7+2/3*a^4*sin(d*x+c)^6+6/5*a^4*sin(d*x+c)^5+a^4*si n(d*x+c)^4+1/3*a^4*sin(d*x+c)^3)
Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.45 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {70 \, a^{4} \cos \left (d x + c\right )^{6} - 315 \, a^{4} \cos \left (d x + c\right )^{4} + 420 \, a^{4} \cos \left (d x + c\right )^{2} + {\left (15 \, a^{4} \cos \left (d x + c\right )^{6} - 171 \, a^{4} \cos \left (d x + c\right )^{4} + 332 \, a^{4} \cos \left (d x + c\right )^{2} - 176 \, a^{4}\right )} \sin \left (d x + c\right )}{105 \, d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^4,x, algorithm="fricas" )
Output:
-1/105*(70*a^4*cos(d*x + c)^6 - 315*a^4*cos(d*x + c)^4 + 420*a^4*cos(d*x + c)^2 + (15*a^4*cos(d*x + c)^6 - 171*a^4*cos(d*x + c)^4 + 332*a^4*cos(d*x + c)^2 - 176*a^4)*sin(d*x + c))/d
Time = 0.47 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.42 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\begin {cases} \frac {a^{4} \sin ^{7}{\left (c + d x \right )}}{7 d} + \frac {2 a^{4} \sin ^{6}{\left (c + d x \right )}}{3 d} + \frac {6 a^{4} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {a^{4} \sin ^{4}{\left (c + d x \right )}}{d} + \frac {a^{4} \sin ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{4} \sin ^{2}{\left (c \right )} \cos {\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)**2*(a+a*sin(d*x+c))**4,x)
Output:
Piecewise((a**4*sin(c + d*x)**7/(7*d) + 2*a**4*sin(c + d*x)**6/(3*d) + 6*a **4*sin(c + d*x)**5/(5*d) + a**4*sin(c + d*x)**4/d + a**4*sin(c + d*x)**3/ (3*d), Ne(d, 0)), (x*(a*sin(c) + a)**4*sin(c)**2*cos(c), True))
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {15 \, a^{4} \sin \left (d x + c\right )^{7} + 70 \, a^{4} \sin \left (d x + c\right )^{6} + 126 \, a^{4} \sin \left (d x + c\right )^{5} + 105 \, a^{4} \sin \left (d x + c\right )^{4} + 35 \, a^{4} \sin \left (d x + c\right )^{3}}{105 \, d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^4,x, algorithm="maxima" )
Output:
1/105*(15*a^4*sin(d*x + c)^7 + 70*a^4*sin(d*x + c)^6 + 126*a^4*sin(d*x + c )^5 + 105*a^4*sin(d*x + c)^4 + 35*a^4*sin(d*x + c)^3)/d
Time = 0.17 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {15 \, a^{4} \sin \left (d x + c\right )^{7} + 70 \, a^{4} \sin \left (d x + c\right )^{6} + 126 \, a^{4} \sin \left (d x + c\right )^{5} + 105 \, a^{4} \sin \left (d x + c\right )^{4} + 35 \, a^{4} \sin \left (d x + c\right )^{3}}{105 \, d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
1/105*(15*a^4*sin(d*x + c)^7 + 70*a^4*sin(d*x + c)^6 + 126*a^4*sin(d*x + c )^5 + 105*a^4*sin(d*x + c)^4 + 35*a^4*sin(d*x + c)^3)/d
Time = 18.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.03 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {\frac {a^4\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {2\,a^4\,{\sin \left (c+d\,x\right )}^6}{3}+\frac {6\,a^4\,{\sin \left (c+d\,x\right )}^5}{5}+a^4\,{\sin \left (c+d\,x\right )}^4+\frac {a^4\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \] Input:
int(cos(c + d*x)*sin(c + d*x)^2*(a + a*sin(c + d*x))^4,x)
Output:
((a^4*sin(c + d*x)^3)/3 + a^4*sin(c + d*x)^4 + (6*a^4*sin(c + d*x)^5)/5 + (2*a^4*sin(c + d*x)^6)/3 + (a^4*sin(c + d*x)^7)/7)/d
Time = 0.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.84 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {\sin \left (d x +c \right )^{3} a^{4} \left (15 \sin \left (d x +c \right )^{4}+70 \sin \left (d x +c \right )^{3}+126 \sin \left (d x +c \right )^{2}+105 \sin \left (d x +c \right )+35\right )}{105 d} \] Input:
int(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^4,x)
Output:
(sin(c + d*x)**3*a**4*(15*sin(c + d*x)**4 + 70*sin(c + d*x)**3 + 126*sin(c + d*x)**2 + 105*sin(c + d*x) + 35))/(105*d)