\(\int \frac {\cos ^7(c+d x) \sin ^5(c+d x)}{a+a \sin (c+d x)} \, dx\) [678]

Optimal result

Integrand size = 29, antiderivative size = 109 \[ \int \frac {\cos ^7(c+d x) \sin ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^6(c+d x)}{6 a d}-\frac {\sin ^7(c+d x)}{7 a d}-\frac {\sin ^8(c+d x)}{4 a d}+\frac {2 \sin ^9(c+d x)}{9 a d}+\frac {\sin ^{10}(c+d x)}{10 a d}-\frac {\sin ^{11}(c+d x)}{11 a d} \]

[Out]

1/6*sin(d*x+c)^6/a/d-1/7*sin(d*x+c)^7/a/d-1/4*sin(d*x+c)^8/a/d+2/9*sin(d*x+c)^9/a/d+1/10*sin(d*x+c)^10/a/d-1/1
1*sin(d*x+c)^11/a/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 12, 90} \[ \int \frac {\cos ^7(c+d x) \sin ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sin ^{11}(c+d x)}{11 a d}+\frac {\sin ^{10}(c+d x)}{10 a d}+\frac {2 \sin ^9(c+d x)}{9 a d}-\frac {\sin ^8(c+d x)}{4 a d}-\frac {\sin ^7(c+d x)}{7 a d}+\frac {\sin ^6(c+d x)}{6 a d} \]

[In]

Int[(Cos[c + d*x]^7*Sin[c + d*x]^5)/(a + a*Sin[c + d*x]),x]

[Out]

Sin[c + d*x]^6/(6*a*d) - Sin[c + d*x]^7/(7*a*d) - Sin[c + d*x]^8/(4*a*d) + (2*Sin[c + d*x]^9)/(9*a*d) + Sin[c
+ d*x]^10/(10*a*d) - Sin[c + d*x]^11/(11*a*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(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]))

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[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] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x)^3 x^5 (a+x)^2}{a^5} \, dx,x,a \sin (c+d x)\right )}{a^7 d} \\ & = \frac {\text {Subst}\left (\int (a-x)^3 x^5 (a+x)^2 \, dx,x,a \sin (c+d x)\right )}{a^{12} d} \\ & = \frac {\text {Subst}\left (\int \left (a^5 x^5-a^4 x^6-2 a^3 x^7+2 a^2 x^8+a x^9-x^{10}\right ) \, dx,x,a \sin (c+d x)\right )}{a^{12} d} \\ & = \frac {\sin ^6(c+d x)}{6 a d}-\frac {\sin ^7(c+d x)}{7 a d}-\frac {\sin ^8(c+d x)}{4 a d}+\frac {2 \sin ^9(c+d x)}{9 a d}+\frac {\sin ^{10}(c+d x)}{10 a d}-\frac {\sin ^{11}(c+d x)}{11 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.62 \[ \int \frac {\cos ^7(c+d x) \sin ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^6(c+d x) \left (2310-1980 \sin (c+d x)-3465 \sin ^2(c+d x)+3080 \sin ^3(c+d x)+1386 \sin ^4(c+d x)-1260 \sin ^5(c+d x)\right )}{13860 a d} \]

[In]

Integrate[(Cos[c + d*x]^7*Sin[c + d*x]^5)/(a + a*Sin[c + d*x]),x]

[Out]

(Sin[c + d*x]^6*(2310 - 1980*Sin[c + d*x] - 3465*Sin[c + d*x]^2 + 3080*Sin[c + d*x]^3 + 1386*Sin[c + d*x]^4 -
1260*Sin[c + d*x]^5))/(13860*a*d)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.64

method	result	size
derivativedivides	\(-\frac {\frac {\left (\sin ^{11}\left (d x +c \right )\right )}{11}-\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}-\frac {2 \left (\sin ^{9}\left (d x +c \right )\right )}{9}+\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{4}+\frac {\left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}}{a d}\)	\(70\)
default	\(-\frac {\frac {\left (\sin ^{11}\left (d x +c \right )\right )}{11}-\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}-\frac {2 \left (\sin ^{9}\left (d x +c \right )\right )}{9}+\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{4}+\frac {\left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}}{a d}\)	\(70\)
parallelrisch	\(\frac {\left (-10+\cos \left (3 d x +3 c \right )-6 \cos \left (2 d x +2 c \right )+15 \cos \left (d x +c \right )\right ) \left (-4158 \cos \left (2 d x +2 c \right )+315 \sin \left (5 d x +5 c \right )+1830 \sin \left (d x +c \right )+1505 \sin \left (3 d x +3 c \right )-693 \cos \left (4 d x +4 c \right )-4389\right ) \left (\cos \left (3 d x +3 c \right )+6 \cos \left (2 d x +2 c \right )+15 \cos \left (d x +c \right )+10\right )}{887040 a d}\)	\(123\)
risch	\(-\frac {5 \sin \left (d x +c \right )}{512 a d}+\frac {\sin \left (11 d x +11 c \right )}{11264 d a}-\frac {\cos \left (10 d x +10 c \right )}{5120 a d}-\frac {\sin \left (9 d x +9 c \right )}{9216 d a}-\frac {5 \sin \left (7 d x +7 c \right )}{7168 d a}+\frac {5 \cos \left (6 d x +6 c \right )}{3072 a d}+\frac {\sin \left (5 d x +5 c \right )}{1024 d a}+\frac {5 \sin \left (3 d x +3 c \right )}{1536 d a}-\frac {5 \cos \left (2 d x +2 c \right )}{512 a d}\)	\(152\)

[In]

int(cos(d*x+c)^7*sin(d*x+c)^5/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/a/d*(1/11*sin(d*x+c)^11-1/10*sin(d*x+c)^10-2/9*sin(d*x+c)^9+1/4*sin(d*x+c)^8+1/7*sin(d*x+c)^7-1/6*sin(d*x+c
)^6)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^7(c+d x) \sin ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {1386 \, \cos \left (d x + c\right )^{10} - 3465 \, \cos \left (d x + c\right )^{8} + 2310 \, \cos \left (d x + c\right )^{6} - 20 \, {\left (63 \, \cos \left (d x + c\right )^{10} - 161 \, \cos \left (d x + c\right )^{8} + 113 \, \cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right )}{13860 \, a d} \]

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/13860*(1386*cos(d*x + c)^10 - 3465*cos(d*x + c)^8 + 2310*cos(d*x + c)^6 - 20*(63*cos(d*x + c)^10 - 161*cos(
d*x + c)^8 + 113*cos(d*x + c)^6 - 3*cos(d*x + c)^4 - 4*cos(d*x + c)^2 - 8)*sin(d*x + c))/(a*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^7(c+d x) \sin ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**7*sin(d*x+c)**5/(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.63 \[ \int \frac {\cos ^7(c+d x) \sin ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {1260 \, \sin \left (d x + c\right )^{11} - 1386 \, \sin \left (d x + c\right )^{10} - 3080 \, \sin \left (d x + c\right )^{9} + 3465 \, \sin \left (d x + c\right )^{8} + 1980 \, \sin \left (d x + c\right )^{7} - 2310 \, \sin \left (d x + c\right )^{6}}{13860 \, a d} \]

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/13860*(1260*sin(d*x + c)^11 - 1386*sin(d*x + c)^10 - 3080*sin(d*x + c)^9 + 3465*sin(d*x + c)^8 + 1980*sin(d
*x + c)^7 - 2310*sin(d*x + c)^6)/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.63 \[ \int \frac {\cos ^7(c+d x) \sin ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {1260 \, \sin \left (d x + c\right )^{11} - 1386 \, \sin \left (d x + c\right )^{10} - 3080 \, \sin \left (d x + c\right )^{9} + 3465 \, \sin \left (d x + c\right )^{8} + 1980 \, \sin \left (d x + c\right )^{7} - 2310 \, \sin \left (d x + c\right )^{6}}{13860 \, a d} \]

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/13860*(1260*sin(d*x + c)^11 - 1386*sin(d*x + c)^10 - 3080*sin(d*x + c)^9 + 3465*sin(d*x + c)^8 + 1980*sin(d
*x + c)^7 - 2310*sin(d*x + c)^6)/(a*d)

Mupad [B] (verification not implemented)

Time = 9.85 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^7(c+d x) \sin ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^6}{6\,a}-\frac {{\sin \left (c+d\,x\right )}^7}{7\,a}-\frac {{\sin \left (c+d\,x\right )}^8}{4\,a}+\frac {2\,{\sin \left (c+d\,x\right )}^9}{9\,a}+\frac {{\sin \left (c+d\,x\right )}^{10}}{10\,a}-\frac {{\sin \left (c+d\,x\right )}^{11}}{11\,a}}{d} \]

[In]

int((cos(c + d*x)^7*sin(c + d*x)^5)/(a + a*sin(c + d*x)),x)

[Out]

(sin(c + d*x)^6/(6*a) - sin(c + d*x)^7/(7*a) - sin(c + d*x)^8/(4*a) + (2*sin(c + d*x)^9)/(9*a) + sin(c + d*x)^
10/(10*a) - sin(c + d*x)^11/(11*a))/d