3.494 \(\int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=97 \[ \frac {a \sin ^{11}(c+d x)}{11 d}+\frac {a \sin ^{10}(c+d x)}{10 d}-\frac {2 a \sin ^9(c+d x)}{9 d}-\frac {a \sin ^8(c+d x)}{4 d}+\frac {a \sin ^7(c+d x)}{7 d}+\frac {a \sin ^6(c+d x)}{6 d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2836, 12, 88} \[ \frac {a \sin ^{11}(c+d x)}{11 d}+\frac {a \sin ^{10}(c+d x)}{10 d}-\frac {2 a \sin ^9(c+d x)}{9 d}-\frac {a \sin ^8(c+d x)}{4 d}+\frac {a \sin ^7(c+d x)}{7 d}+\frac {a \sin ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sin[c + d*x]^6)/(6*d) + (a*Sin[c + d*x]^7)/(7*d) - (a*Sin[c + d*x]^8)/(4*d) - (2*a*Sin[c + d*x]^9)/(9*d) +
(a*Sin[c + d*x]^10)/(10*d) + (a*Sin[c + d*x]^11)/(11*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 88

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 2836

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*x)/b
)^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*} \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(a-x)^2 x^5 (a+x)^3}{a^5} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {\operatorname {Subst}\left (\int (a-x)^2 x^5 (a+x)^3 \, dx,x,a \sin (c+d x)\right )}{a^{10} d}\\ &=\frac {\operatorname {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^{10} d}\\ &=\frac {a \sin ^6(c+d x)}{6 d}+\frac {a \sin ^7(c+d x)}{7 d}-\frac {a \sin ^8(c+d x)}{4 d}-\frac {2 a \sin ^9(c+d x)}{9 d}+\frac {a \sin ^{10}(c+d x)}{10 d}+\frac {a \sin ^{11}(c+d x)}{11 d}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 97, normalized size = 1.00 \[ -\frac {a (-34650 \sin (c+d x)+11550 \sin (3 (c+d x))+3465 \sin (5 (c+d x))-2475 \sin (7 (c+d x))-385 \sin (9 (c+d x))+315 \sin (11 (c+d x))+34650 \cos (2 (c+d x))-5775 \cos (6 (c+d x))+693 \cos (10 (c+d x)))}{3548160 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3548160*(a*(34650*Cos[2*(c + d*x)] - 5775*Cos[6*(c + d*x)] + 693*Cos[10*(c + d*x)] - 34650*Sin[c + d*x] + 1
1550*Sin[3*(c + d*x)] + 3465*Sin[5*(c + d*x)] - 2475*Sin[7*(c + d*x)] - 385*Sin[9*(c + d*x)] + 315*Sin[11*(c +
 d*x)]))/d

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fricas [A]  time = 0.46, size = 106, normalized size = 1.09 \[ -\frac {1386 \, a \cos \left (d x + c\right )^{10} - 3465 \, a \cos \left (d x + c\right )^{8} + 2310 \, a \cos \left (d x + c\right )^{6} + 20 \, {\left (63 \, a \cos \left (d x + c\right )^{10} - 161 \, a \cos \left (d x + c\right )^{8} + 113 \, a \cos \left (d x + c\right )^{6} - 3 \, a \cos \left (d x + c\right )^{4} - 4 \, a \cos \left (d x + c\right )^{2} - 8 \, a\right )} \sin \left (d x + c\right )}{13860 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.31, size = 133, normalized size = 1.37 \[ -\frac {a \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac {5 \, a \cos \left (6 \, d x + 6 \, c\right )}{3072 \, d} - \frac {5 \, a \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} - \frac {a \sin \left (11 \, d x + 11 \, c\right )}{11264 \, d} + \frac {a \sin \left (9 \, d x + 9 \, c\right )}{9216 \, d} + \frac {5 \, a \sin \left (7 \, d x + 7 \, c\right )}{7168 \, d} - \frac {a \sin \left (5 \, d x + 5 \, c\right )}{1024 \, d} - \frac {5 \, a \sin \left (3 \, d x + 3 \, c\right )}{1536 \, d} + \frac {5 \, a \sin \left (d x + c\right )}{512 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5120*a*cos(10*d*x + 10*c)/d + 5/3072*a*cos(6*d*x + 6*c)/d - 5/512*a*cos(2*d*x + 2*c)/d - 1/11264*a*sin(11*d
*x + 11*c)/d + 1/9216*a*sin(9*d*x + 9*c)/d + 5/7168*a*sin(7*d*x + 7*c)/d - 1/1024*a*sin(5*d*x + 5*c)/d - 5/153
6*a*sin(3*d*x + 3*c)/d + 5/512*a*sin(d*x + c)/d

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maple [A]  time = 0.23, size = 138, normalized size = 1.42 \[ \frac {a \left (-\frac {\left (\sin ^{5}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{11}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{99}-\frac {5 \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{231}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{231}\right )+a \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{10}-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{20}-\frac {\left (\cos ^{6}\left (d x +c \right )\right )}{60}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(-1/11*sin(d*x+c)^5*cos(d*x+c)^6-5/99*sin(d*x+c)^3*cos(d*x+c)^6-5/231*sin(d*x+c)*cos(d*x+c)^6+1/231*(8/
3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))+a*(-1/10*sin(d*x+c)^4*cos(d*x+c)^6-1/20*sin(d*x+c)^2*cos(d*x+c)^6
-1/60*cos(d*x+c)^6))

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maxima [A]  time = 0.35, size = 72, normalized size = 0.74 \[ \frac {1260 \, a \sin \left (d x + c\right )^{11} + 1386 \, a \sin \left (d x + c\right )^{10} - 3080 \, a \sin \left (d x + c\right )^{9} - 3465 \, a \sin \left (d x + c\right )^{8} + 1980 \, a \sin \left (d x + c\right )^{7} + 2310 \, a \sin \left (d x + c\right )^{6}}{13860 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13860*(1260*a*sin(d*x + c)^11 + 1386*a*sin(d*x + c)^10 - 3080*a*sin(d*x + c)^9 - 3465*a*sin(d*x + c)^8 + 198
0*a*sin(d*x + c)^7 + 2310*a*sin(d*x + c)^6)/d

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mupad [B]  time = 8.72, size = 71, normalized size = 0.73 \[ \frac {\frac {a\,{\sin \left (c+d\,x\right )}^{11}}{11}+\frac {a\,{\sin \left (c+d\,x\right )}^{10}}{10}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^9}{9}-\frac {a\,{\sin \left (c+d\,x\right )}^8}{4}+\frac {a\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {a\,{\sin \left (c+d\,x\right )}^6}{6}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 42.24, size = 136, normalized size = 1.40 \[ \begin {cases} \frac {8 a \sin ^{11}{\left (c + d x \right )}}{693 d} + \frac {4 a \sin ^{9}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{63 d} + \frac {a \sin ^{7}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{7 d} - \frac {a \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{12 d} - \frac {a \cos ^{10}{\left (c + d x \right )}}{60 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right ) \sin ^{5}{\relax (c )} \cos ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*a*sin(c + d*x)**11/(693*d) + 4*a*sin(c + d*x)**9*cos(c + d*x)**2/(63*d) + a*sin(c + d*x)**7*cos(c
 + d*x)**4/(7*d) - a*sin(c + d*x)**4*cos(c + d*x)**6/(6*d) - a*sin(c + d*x)**2*cos(c + d*x)**8/(12*d) - a*cos(
c + d*x)**10/(60*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)**5*cos(c)**5, True))

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