Integrand size = 29, antiderivative size = 102 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {(A-B) (a+a \sin (c+d x))^4}{a^3 d}-\frac {4 (A-2 B) (a+a \sin (c+d x))^5}{5 a^4 d}+\frac {(A-5 B) (a+a \sin (c+d x))^6}{6 a^5 d}+\frac {B (a+a \sin (c+d x))^7}{7 a^6 d} \] Output:
(A-B)*(a+a*sin(d*x+c))^4/a^3/d-4/5*(A-2*B)*(a+a*sin(d*x+c))^5/a^4/d+1/6*(A -5*B)*(a+a*sin(d*x+c))^6/a^5/d+1/7*B*(a+a*sin(d*x+c))^7/a^6/d
Time = 0.69 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \left ((A-B) (1+\sin (c+d x))^4-\frac {4}{5} (A-2 B) (1+\sin (c+d x))^5+\frac {1}{6} (A-5 B) (1+\sin (c+d x))^6+\frac {1}{7} B (1+\sin (c+d x))^7\right )}{d} \] Input:
Integrate[Cos[c + d*x]^5*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(a*((A - B)*(1 + Sin[c + d*x])^4 - (4*(A - 2*B)*(1 + Sin[c + d*x])^5)/5 + ((A - 5*B)*(1 + Sin[c + d*x])^6)/6 + (B*(1 + Sin[c + d*x])^7)/7))/d
Time = 0.32 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 27, 86, 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.
\(\displaystyle \int \cos ^5(c+d x) (a \sin (c+d x)+a) (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^5 (a \sin (c+d x)+a) (A+B \sin (c+d x))dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3 (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (B (\sin (c+d x) a+a)^6+a (A-5 B) (\sin (c+d x) a+a)^5-4 a^2 (A-2 B) (\sin (c+d x) a+a)^4+4 a^3 (A-B) (\sin (c+d x) a+a)^3\right )d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 (A-B) (a \sin (c+d x)+a)^4-\frac {4}{5} a^2 (A-2 B) (a \sin (c+d x)+a)^5+\frac {1}{6} a (A-5 B) (a \sin (c+d x)+a)^6+\frac {1}{7} B (a \sin (c+d x)+a)^7}{a^6 d}\) |
Input:
Int[Cos[c + d*x]^5*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(a^3*(A - B)*(a + a*Sin[c + d*x])^4 - (4*a^2*(A - 2*B)*(a + a*Sin[c + d*x] )^5)/5 + (a*(A - 5*B)*(a + a*Sin[c + d*x])^6)/6 + (B*(a + a*Sin[c + d*x])^ 7)/7)/(a^6*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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 34.36 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(\frac {a \left (\frac {B \sin \left (d x +c \right )^{7}}{7}+\frac {\left (A +B \right ) \sin \left (d x +c \right )^{6}}{6}+\frac {\left (A -2 B \right ) \sin \left (d x +c \right )^{5}}{5}+\frac {\left (-2 B -2 A \right ) \sin \left (d x +c \right )^{4}}{4}+\frac {\left (B -2 A \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (A +B \right ) \sin \left (d x +c \right )^{2}}{2}+A \sin \left (d x +c \right )\right )}{d}\) | \(99\) |
default | \(\frac {a \left (\frac {B \sin \left (d x +c \right )^{7}}{7}+\frac {\left (A +B \right ) \sin \left (d x +c \right )^{6}}{6}+\frac {\left (A -2 B \right ) \sin \left (d x +c \right )^{5}}{5}+\frac {\left (-2 B -2 A \right ) \sin \left (d x +c \right )^{4}}{4}+\frac {\left (B -2 A \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (A +B \right ) \sin \left (d x +c \right )^{2}}{2}+A \sin \left (d x +c \right )\right )}{d}\) | \(99\) |
parallelrisch | \(\frac {5 \left (\frac {\left (-A -B \right ) \cos \left (2 d x +2 c \right )}{8}+\frac {\left (-A -B \right ) \cos \left (4 d x +4 c \right )}{20}+\frac {\left (-A -B \right ) \cos \left (6 d x +6 c \right )}{120}+\frac {\left (A -\frac {B}{20}\right ) \sin \left (3 d x +3 c \right )}{6}+\frac {\left (A -\frac {3 B}{4}\right ) \sin \left (5 d x +5 c \right )}{50}-\frac {B \sin \left (7 d x +7 c \right )}{280}+\left (A +\frac {B}{8}\right ) \sin \left (d x +c \right )+\frac {11 A}{60}+\frac {11 B}{60}\right ) a}{8 d}\) | \(124\) |
risch | \(\frac {5 a A \sin \left (d x +c \right )}{8 d}+\frac {5 a B \sin \left (d x +c \right )}{64 d}-\frac {\sin \left (7 d x +7 c \right ) a B}{448 d}-\frac {a \cos \left (6 d x +6 c \right ) A}{192 d}-\frac {a \cos \left (6 d x +6 c \right ) B}{192 d}+\frac {\sin \left (5 d x +5 c \right ) a A}{80 d}-\frac {3 \sin \left (5 d x +5 c \right ) a B}{320 d}-\frac {a \cos \left (4 d x +4 c \right ) A}{32 d}-\frac {a \cos \left (4 d x +4 c \right ) B}{32 d}+\frac {5 a A \sin \left (3 d x +3 c \right )}{48 d}-\frac {\sin \left (3 d x +3 c \right ) a B}{192 d}-\frac {5 a \cos \left (2 d x +2 c \right ) A}{64 d}-\frac {5 a \cos \left (2 d x +2 c \right ) B}{64 d}\) | \(204\) |
norman | \(\frac {\frac {\left (2 a A +2 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {\left (2 a A +2 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {\left (2 a A +2 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {\left (2 a A +2 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {5 \left (4 a A +4 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}+\frac {5 \left (4 a A +4 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d}+\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {4 a \left (5 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {4 a \left (5 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {8 a \left (91 A +38 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{35 d}+\frac {2 a \left (113 A -16 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {2 a \left (113 A -16 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{15 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}\) | \(318\) |
orering | \(\text {Expression too large to display}\) | \(2330\) |
Input:
int(cos(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
a/d*(1/7*B*sin(d*x+c)^7+1/6*(A+B)*sin(d*x+c)^6+1/5*(A-2*B)*sin(d*x+c)^5+1/ 4*(-2*B-2*A)*sin(d*x+c)^4+1/3*(B-2*A)*sin(d*x+c)^3+1/2*(A+B)*sin(d*x+c)^2+ A*sin(d*x+c))
Time = 0.08 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {35 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{6} + 2 \, {\left (15 \, B a \cos \left (d x + c\right )^{6} - 3 \, {\left (7 \, A + B\right )} a \cos \left (d x + c\right )^{4} - 4 \, {\left (7 \, A + B\right )} a \cos \left (d x + c\right )^{2} - 8 \, {\left (7 \, A + B\right )} a\right )} \sin \left (d x + c\right )}{210 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fri cas")
Output:
-1/210*(35*(A + B)*a*cos(d*x + c)^6 + 2*(15*B*a*cos(d*x + c)^6 - 3*(7*A + B)*a*cos(d*x + c)^4 - 4*(7*A + B)*a*cos(d*x + c)^2 - 8*(7*A + B)*a)*sin(d* x + c))/d
Time = 0.47 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.75 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {8 A a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 A a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {A a \cos ^{6}{\left (c + d x \right )}}{6 d} + \frac {8 B a \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 B a \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {B a \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac {B a \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right ) \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
Piecewise((8*A*a*sin(c + d*x)**5/(15*d) + 4*A*a*sin(c + d*x)**3*cos(c + d* x)**2/(3*d) + A*a*sin(c + d*x)*cos(c + d*x)**4/d - A*a*cos(c + d*x)**6/(6* d) + 8*B*a*sin(c + d*x)**7/(105*d) + 4*B*a*sin(c + d*x)**5*cos(c + d*x)**2 /(15*d) + B*a*sin(c + d*x)**3*cos(c + d*x)**4/(3*d) - B*a*cos(c + d*x)**6/ (6*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)*cos(c)**5, True))
Time = 0.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {30 \, B a \sin \left (d x + c\right )^{7} + 35 \, {\left (A + B\right )} a \sin \left (d x + c\right )^{6} + 42 \, {\left (A - 2 \, B\right )} a \sin \left (d x + c\right )^{5} - 105 \, {\left (A + B\right )} a \sin \left (d x + c\right )^{4} - 70 \, {\left (2 \, A - B\right )} a \sin \left (d x + c\right )^{3} + 105 \, {\left (A + B\right )} a \sin \left (d x + c\right )^{2} + 210 \, A a \sin \left (d x + c\right )}{210 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="max ima")
Output:
1/210*(30*B*a*sin(d*x + c)^7 + 35*(A + B)*a*sin(d*x + c)^6 + 42*(A - 2*B)* a*sin(d*x + c)^5 - 105*(A + B)*a*sin(d*x + c)^4 - 70*(2*A - B)*a*sin(d*x + c)^3 + 105*(A + B)*a*sin(d*x + c)^2 + 210*A*a*sin(d*x + c))/d
Time = 0.15 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.45 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {30 \, B a \sin \left (d x + c\right )^{7} + 35 \, A a \sin \left (d x + c\right )^{6} + 35 \, B a \sin \left (d x + c\right )^{6} + 42 \, A a \sin \left (d x + c\right )^{5} - 84 \, B a \sin \left (d x + c\right )^{5} - 105 \, A a \sin \left (d x + c\right )^{4} - 105 \, B a \sin \left (d x + c\right )^{4} - 140 \, A a \sin \left (d x + c\right )^{3} + 70 \, B a \sin \left (d x + c\right )^{3} + 105 \, A a \sin \left (d x + c\right )^{2} + 105 \, B a \sin \left (d x + c\right )^{2} + 210 \, A a \sin \left (d x + c\right )}{210 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="gia c")
Output:
1/210*(30*B*a*sin(d*x + c)^7 + 35*A*a*sin(d*x + c)^6 + 35*B*a*sin(d*x + c) ^6 + 42*A*a*sin(d*x + c)^5 - 84*B*a*sin(d*x + c)^5 - 105*A*a*sin(d*x + c)^ 4 - 105*B*a*sin(d*x + c)^4 - 140*A*a*sin(d*x + c)^3 + 70*B*a*sin(d*x + c)^ 3 + 105*A*a*sin(d*x + c)^2 + 105*B*a*sin(d*x + c)^2 + 210*A*a*sin(d*x + c) )/d
Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\frac {B\,a\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {a\,\left (A+B\right )\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {a\,\left (A-2\,B\right )\,{\sin \left (c+d\,x\right )}^5}{5}-\frac {a\,\left (A+B\right )\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {a\,\left (2\,A-B\right )\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {a\,\left (A+B\right )\,{\sin \left (c+d\,x\right )}^2}{2}+A\,a\,\sin \left (c+d\,x\right )}{d} \] Input:
int(cos(c + d*x)^5*(A + B*sin(c + d*x))*(a + a*sin(c + d*x)),x)
Output:
(A*a*sin(c + d*x) - (a*sin(c + d*x)^3*(2*A - B))/3 + (a*sin(c + d*x)^2*(A + B))/2 - (a*sin(c + d*x)^4*(A + B))/2 + (a*sin(c + d*x)^6*(A + B))/6 + (a *sin(c + d*x)^5*(A - 2*B))/5 + (B*a*sin(c + d*x)^7)/7)/d
Time = 0.17 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.30 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right ) a \left (30 \sin \left (d x +c \right )^{6} b +35 \sin \left (d x +c \right )^{5} a +35 \sin \left (d x +c \right )^{5} b +42 \sin \left (d x +c \right )^{4} a -84 \sin \left (d x +c \right )^{4} b -105 \sin \left (d x +c \right )^{3} a -105 \sin \left (d x +c \right )^{3} b -140 \sin \left (d x +c \right )^{2} a +70 \sin \left (d x +c \right )^{2} b +105 \sin \left (d x +c \right ) a +105 \sin \left (d x +c \right ) b +210 a \right )}{210 d} \] Input:
int(cos(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
(sin(c + d*x)*a*(30*sin(c + d*x)**6*b + 35*sin(c + d*x)**5*a + 35*sin(c + d*x)**5*b + 42*sin(c + d*x)**4*a - 84*sin(c + d*x)**4*b - 105*sin(c + d*x) **3*a - 105*sin(c + d*x)**3*b - 140*sin(c + d*x)**2*a + 70*sin(c + d*x)**2 *b + 105*sin(c + d*x)*a + 105*sin(c + d*x)*b + 210*a))/(210*d)