3.1239 \(\int \cos ^6(c+d x) \sin ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=238 \[ -\frac{\left (a^2+3 b^2\right ) \cos ^{11}(c+d x)}{11 d}+\frac{\left (2 a^2+3 b^2\right ) \cos ^9(c+d x)}{9 d}-\frac{\left (a^2+b^2\right ) \cos ^7(c+d x)}{7 d}-\frac{a b \sin ^5(c+d x) \cos ^7(c+d x)}{6 d}-\frac{a b \sin ^3(c+d x) \cos ^7(c+d x)}{12 d}-\frac{a b \sin (c+d x) \cos ^7(c+d x)}{32 d}+\frac{a b \sin (c+d x) \cos ^5(c+d x)}{192 d}+\frac{5 a b \sin (c+d x) \cos ^3(c+d x)}{768 d}+\frac{5 a b \sin (c+d x) \cos (c+d x)}{512 d}+\frac{5 a b x}{512}+\frac{b^2 \cos ^{13}(c+d x)}{13 d} \]

[Out]

(5*a*b*x)/512 - ((a^2 + b^2)*Cos[c + d*x]^7)/(7*d) + ((2*a^2 + 3*b^2)*Cos[c + d*x]^9)/(9*d) - ((a^2 + 3*b^2)*C
os[c + d*x]^11)/(11*d) + (b^2*Cos[c + d*x]^13)/(13*d) + (5*a*b*Cos[c + d*x]*Sin[c + d*x])/(512*d) + (5*a*b*Cos
[c + d*x]^3*Sin[c + d*x])/(768*d) + (a*b*Cos[c + d*x]^5*Sin[c + d*x])/(192*d) - (a*b*Cos[c + d*x]^7*Sin[c + d*
x])/(32*d) - (a*b*Cos[c + d*x]^7*Sin[c + d*x]^3)/(12*d) - (a*b*Cos[c + d*x]^7*Sin[c + d*x]^5)/(6*d)

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Rubi [A]  time = 0.371907, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2911, 2568, 2635, 8, 3201, 446, 77} \[ -\frac{\left (a^2+3 b^2\right ) \cos ^{11}(c+d x)}{11 d}+\frac{\left (2 a^2+3 b^2\right ) \cos ^9(c+d x)}{9 d}-\frac{\left (a^2+b^2\right ) \cos ^7(c+d x)}{7 d}-\frac{a b \sin ^5(c+d x) \cos ^7(c+d x)}{6 d}-\frac{a b \sin ^3(c+d x) \cos ^7(c+d x)}{12 d}-\frac{a b \sin (c+d x) \cos ^7(c+d x)}{32 d}+\frac{a b \sin (c+d x) \cos ^5(c+d x)}{192 d}+\frac{5 a b \sin (c+d x) \cos ^3(c+d x)}{768 d}+\frac{5 a b \sin (c+d x) \cos (c+d x)}{512 d}+\frac{5 a b x}{512}+\frac{b^2 \cos ^{13}(c+d x)}{13 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*Sin[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]

[Out]

(5*a*b*x)/512 - ((a^2 + b^2)*Cos[c + d*x]^7)/(7*d) + ((2*a^2 + 3*b^2)*Cos[c + d*x]^9)/(9*d) - ((a^2 + 3*b^2)*C
os[c + d*x]^11)/(11*d) + (b^2*Cos[c + d*x]^13)/(13*d) + (5*a*b*Cos[c + d*x]*Sin[c + d*x])/(512*d) + (5*a*b*Cos
[c + d*x]^3*Sin[c + d*x])/(768*d) + (a*b*Cos[c + d*x]^5*Sin[c + d*x])/(192*d) - (a*b*Cos[c + d*x]^7*Sin[c + d*
x])/(32*d) - (a*b*Cos[c + d*x]^7*Sin[c + d*x]^3)/(12*d) - (a*b*Cos[c + d*x]^7*Sin[c + d*x]^5)/(6*d)

Rule 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3201

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2
)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[(ff*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]
), Subst[Int[(d*ff*x)^n*(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{
a, b, d, e, f, n, p}, x] && IntegerQ[m/2]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((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]))))

Rubi steps

\begin{align*} \int \cos ^6(c+d x) \sin ^5(c+d x) (a+b \sin (c+d x))^2 \, dx &=(2 a b) \int \cos ^6(c+d x) \sin ^6(c+d x) \, dx+\int \cos ^6(c+d x) \sin ^5(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=-\frac{a b \cos ^7(c+d x) \sin ^5(c+d x)}{6 d}+\frac{1}{6} (5 a b) \int \cos ^6(c+d x) \sin ^4(c+d x) \, dx+\frac{\left (\sqrt{\cos ^2(c+d x)} \sec (c+d x)\right ) \operatorname{Subst}\left (\int x^5 \left (1-x^2\right )^{5/2} \left (a^2+b^2 x^2\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{a b \cos ^7(c+d x) \sin ^3(c+d x)}{12 d}-\frac{a b \cos ^7(c+d x) \sin ^5(c+d x)}{6 d}+\frac{1}{4} (a b) \int \cos ^6(c+d x) \sin ^2(c+d x) \, dx+\frac{\left (\sqrt{\cos ^2(c+d x)} \sec (c+d x)\right ) \operatorname{Subst}\left (\int (1-x)^{5/2} x^2 \left (a^2+b^2 x\right ) \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=-\frac{a b \cos ^7(c+d x) \sin (c+d x)}{32 d}-\frac{a b \cos ^7(c+d x) \sin ^3(c+d x)}{12 d}-\frac{a b \cos ^7(c+d x) \sin ^5(c+d x)}{6 d}+\frac{1}{32} (a b) \int \cos ^6(c+d x) \, dx+\frac{\left (\sqrt{\cos ^2(c+d x)} \sec (c+d x)\right ) \operatorname{Subst}\left (\int \left (\left (a^2+b^2\right ) (1-x)^{5/2}+\left (-2 a^2-3 b^2\right ) (1-x)^{7/2}+\left (a^2+3 b^2\right ) (1-x)^{9/2}-b^2 (1-x)^{11/2}\right ) \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=-\frac{\left (a^2+b^2\right ) \cos ^7(c+d x)}{7 d}+\frac{\left (2 a^2+3 b^2\right ) \cos ^9(c+d x)}{9 d}-\frac{\left (a^2+3 b^2\right ) \cos ^{11}(c+d x)}{11 d}+\frac{b^2 \cos ^{13}(c+d x)}{13 d}+\frac{a b \cos ^5(c+d x) \sin (c+d x)}{192 d}-\frac{a b \cos ^7(c+d x) \sin (c+d x)}{32 d}-\frac{a b \cos ^7(c+d x) \sin ^3(c+d x)}{12 d}-\frac{a b \cos ^7(c+d x) \sin ^5(c+d x)}{6 d}+\frac{1}{192} (5 a b) \int \cos ^4(c+d x) \, dx\\ &=-\frac{\left (a^2+b^2\right ) \cos ^7(c+d x)}{7 d}+\frac{\left (2 a^2+3 b^2\right ) \cos ^9(c+d x)}{9 d}-\frac{\left (a^2+3 b^2\right ) \cos ^{11}(c+d x)}{11 d}+\frac{b^2 \cos ^{13}(c+d x)}{13 d}+\frac{5 a b \cos ^3(c+d x) \sin (c+d x)}{768 d}+\frac{a b \cos ^5(c+d x) \sin (c+d x)}{192 d}-\frac{a b \cos ^7(c+d x) \sin (c+d x)}{32 d}-\frac{a b \cos ^7(c+d x) \sin ^3(c+d x)}{12 d}-\frac{a b \cos ^7(c+d x) \sin ^5(c+d x)}{6 d}+\frac{1}{256} (5 a b) \int \cos ^2(c+d x) \, dx\\ &=-\frac{\left (a^2+b^2\right ) \cos ^7(c+d x)}{7 d}+\frac{\left (2 a^2+3 b^2\right ) \cos ^9(c+d x)}{9 d}-\frac{\left (a^2+3 b^2\right ) \cos ^{11}(c+d x)}{11 d}+\frac{b^2 \cos ^{13}(c+d x)}{13 d}+\frac{5 a b \cos (c+d x) \sin (c+d x)}{512 d}+\frac{5 a b \cos ^3(c+d x) \sin (c+d x)}{768 d}+\frac{a b \cos ^5(c+d x) \sin (c+d x)}{192 d}-\frac{a b \cos ^7(c+d x) \sin (c+d x)}{32 d}-\frac{a b \cos ^7(c+d x) \sin ^3(c+d x)}{12 d}-\frac{a b \cos ^7(c+d x) \sin ^5(c+d x)}{6 d}+\frac{1}{512} (5 a b) \int 1 \, dx\\ &=\frac{5 a b x}{512}-\frac{\left (a^2+b^2\right ) \cos ^7(c+d x)}{7 d}+\frac{\left (2 a^2+3 b^2\right ) \cos ^9(c+d x)}{9 d}-\frac{\left (a^2+3 b^2\right ) \cos ^{11}(c+d x)}{11 d}+\frac{b^2 \cos ^{13}(c+d x)}{13 d}+\frac{5 a b \cos (c+d x) \sin (c+d x)}{512 d}+\frac{5 a b \cos ^3(c+d x) \sin (c+d x)}{768 d}+\frac{a b \cos ^5(c+d x) \sin (c+d x)}{192 d}-\frac{a b \cos ^7(c+d x) \sin (c+d x)}{32 d}-\frac{a b \cos ^7(c+d x) \sin ^3(c+d x)}{12 d}-\frac{a b \cos ^7(c+d x) \sin ^5(c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 2.15656, size = 210, normalized size = 0.88 \[ \frac{-180180 \left (2 a^2+b^2\right ) \cos (c+d x)-15015 \left (8 a^2+3 b^2\right ) \cos (3 (c+d x))+36036 a^2 \cos (5 (c+d x))+25740 a^2 \cos (7 (c+d x))-4004 a^2 \cos (9 (c+d x))-3276 a^2 \cos (11 (c+d x))-135135 a b \sin (4 (c+d x))+27027 a b \sin (8 (c+d x))-3003 a b \sin (12 (c+d x))+360360 a b c+360360 a b d x+27027 b^2 \cos (5 (c+d x))+7722 b^2 \cos (7 (c+d x))-6006 b^2 \cos (9 (c+d x))-819 b^2 \cos (11 (c+d x))+693 b^2 \cos (13 (c+d x))}{36900864 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*Sin[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]

[Out]

(360360*a*b*c + 360360*a*b*d*x - 180180*(2*a^2 + b^2)*Cos[c + d*x] - 15015*(8*a^2 + 3*b^2)*Cos[3*(c + d*x)] +
36036*a^2*Cos[5*(c + d*x)] + 27027*b^2*Cos[5*(c + d*x)] + 25740*a^2*Cos[7*(c + d*x)] + 7722*b^2*Cos[7*(c + d*x
)] - 4004*a^2*Cos[9*(c + d*x)] - 6006*b^2*Cos[9*(c + d*x)] - 3276*a^2*Cos[11*(c + d*x)] - 819*b^2*Cos[11*(c +
d*x)] + 693*b^2*Cos[13*(c + d*x)] - 135135*a*b*Sin[4*(c + d*x)] + 27027*a*b*Sin[8*(c + d*x)] - 3003*a*b*Sin[12
*(c + d*x)])/(36900864*d)

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Maple [A]  time = 0.049, size = 225, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{11}}-{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{99}}-{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{693}} \right ) +2\,ab \left ( -1/12\, \left ( \sin \left ( dx+c \right ) \right ) ^{5} \left ( \cos \left ( dx+c \right ) \right ) ^{7}-1/24\, \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{7}-{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{64}}+{\frac{\sin \left ( dx+c \right ) }{384} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+5/4\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{1024}}+{\frac{5\,c}{1024}} \right ) +{b}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{13}}-{\frac{6\, \left ( \sin \left ( dx+c \right ) \right ) ^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{143}}-{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{429}}-{\frac{16\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{3003}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*sin(d*x+c)^5*(a+b*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(-1/11*sin(d*x+c)^4*cos(d*x+c)^7-4/99*sin(d*x+c)^2*cos(d*x+c)^7-8/693*cos(d*x+c)^7)+2*a*b*(-1/12*sin(
d*x+c)^5*cos(d*x+c)^7-1/24*sin(d*x+c)^3*cos(d*x+c)^7-1/64*sin(d*x+c)*cos(d*x+c)^7+1/384*(cos(d*x+c)^5+5/4*cos(
d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/1024*d*x+5/1024*c)+b^2*(-1/13*sin(d*x+c)^6*cos(d*x+c)^7-6/143*sin(d*x+c
)^4*cos(d*x+c)^7-8/429*sin(d*x+c)^2*cos(d*x+c)^7-16/3003*cos(d*x+c)^7))

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Maxima [A]  time = 0.995607, size = 182, normalized size = 0.76 \begin{align*} -\frac{53248 \,{\left (63 \, \cos \left (d x + c\right )^{11} - 154 \, \cos \left (d x + c\right )^{9} + 99 \, \cos \left (d x + c\right )^{7}\right )} a^{2} - 3003 \,{\left (4 \, \sin \left (4 \, d x + 4 \, c\right )^{3} + 120 \, d x + 120 \, c + 9 \, \sin \left (8 \, d x + 8 \, c\right ) - 48 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b - 12288 \,{\left (231 \, \cos \left (d x + c\right )^{13} - 819 \, \cos \left (d x + c\right )^{11} + 1001 \, \cos \left (d x + c\right )^{9} - 429 \, \cos \left (d x + c\right )^{7}\right )} b^{2}}{36900864 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/36900864*(53248*(63*cos(d*x + c)^11 - 154*cos(d*x + c)^9 + 99*cos(d*x + c)^7)*a^2 - 3003*(4*sin(4*d*x + 4*c
)^3 + 120*d*x + 120*c + 9*sin(8*d*x + 8*c) - 48*sin(4*d*x + 4*c))*a*b - 12288*(231*cos(d*x + c)^13 - 819*cos(d
*x + c)^11 + 1001*cos(d*x + c)^9 - 429*cos(d*x + c)^7)*b^2)/d

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Fricas [A]  time = 1.99853, size = 456, normalized size = 1.92 \begin{align*} \frac{354816 \, b^{2} \cos \left (d x + c\right )^{13} - 419328 \,{\left (a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{11} + 512512 \,{\left (2 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{9} - 658944 \,{\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{7} + 45045 \, a b d x - 3003 \,{\left (256 \, a b \cos \left (d x + c\right )^{11} - 640 \, a b \cos \left (d x + c\right )^{9} + 432 \, a b \cos \left (d x + c\right )^{7} - 8 \, a b \cos \left (d x + c\right )^{5} - 10 \, a b \cos \left (d x + c\right )^{3} - 15 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4612608 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/4612608*(354816*b^2*cos(d*x + c)^13 - 419328*(a^2 + 3*b^2)*cos(d*x + c)^11 + 512512*(2*a^2 + 3*b^2)*cos(d*x
+ c)^9 - 658944*(a^2 + b^2)*cos(d*x + c)^7 + 45045*a*b*d*x - 3003*(256*a*b*cos(d*x + c)^11 - 640*a*b*cos(d*x +
c)^9 + 432*a*b*cos(d*x + c)^7 - 8*a*b*cos(d*x + c)^5 - 10*a*b*cos(d*x + c)^3 - 15*a*b*cos(d*x + c))*sin(d*x +
c))/d

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Sympy [A]  time = 130.965, size = 488, normalized size = 2.05 \begin{align*} \begin{cases} - \frac{a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{7 d} - \frac{4 a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{9}{\left (c + d x \right )}}{63 d} - \frac{8 a^{2} \cos ^{11}{\left (c + d x \right )}}{693 d} + \frac{5 a b x \sin ^{12}{\left (c + d x \right )}}{512} + \frac{15 a b x \sin ^{10}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{256} + \frac{75 a b x \sin ^{8}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{512} + \frac{25 a b x \sin ^{6}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{128} + \frac{75 a b x \sin ^{4}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{512} + \frac{15 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{10}{\left (c + d x \right )}}{256} + \frac{5 a b x \cos ^{12}{\left (c + d x \right )}}{512} + \frac{5 a b \sin ^{11}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{512 d} + \frac{85 a b \sin ^{9}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{1536 d} + \frac{33 a b \sin ^{7}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{256 d} - \frac{33 a b \sin ^{5}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{256 d} - \frac{85 a b \sin ^{3}{\left (c + d x \right )} \cos ^{9}{\left (c + d x \right )}}{1536 d} - \frac{5 a b \sin{\left (c + d x \right )} \cos ^{11}{\left (c + d x \right )}}{512 d} - \frac{b^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{7 d} - \frac{2 b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{9}{\left (c + d x \right )}}{21 d} - \frac{8 b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{11}{\left (c + d x \right )}}{231 d} - \frac{16 b^{2} \cos ^{13}{\left (c + d x \right )}}{3003 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{2} \sin ^{5}{\left (c \right )} \cos ^{6}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*sin(d*x+c)**5*(a+b*sin(d*x+c))**2,x)

[Out]

Piecewise((-a**2*sin(c + d*x)**4*cos(c + d*x)**7/(7*d) - 4*a**2*sin(c + d*x)**2*cos(c + d*x)**9/(63*d) - 8*a**
2*cos(c + d*x)**11/(693*d) + 5*a*b*x*sin(c + d*x)**12/512 + 15*a*b*x*sin(c + d*x)**10*cos(c + d*x)**2/256 + 75
*a*b*x*sin(c + d*x)**8*cos(c + d*x)**4/512 + 25*a*b*x*sin(c + d*x)**6*cos(c + d*x)**6/128 + 75*a*b*x*sin(c + d
*x)**4*cos(c + d*x)**8/512 + 15*a*b*x*sin(c + d*x)**2*cos(c + d*x)**10/256 + 5*a*b*x*cos(c + d*x)**12/512 + 5*
a*b*sin(c + d*x)**11*cos(c + d*x)/(512*d) + 85*a*b*sin(c + d*x)**9*cos(c + d*x)**3/(1536*d) + 33*a*b*sin(c + d
*x)**7*cos(c + d*x)**5/(256*d) - 33*a*b*sin(c + d*x)**5*cos(c + d*x)**7/(256*d) - 85*a*b*sin(c + d*x)**3*cos(c
+ d*x)**9/(1536*d) - 5*a*b*sin(c + d*x)*cos(c + d*x)**11/(512*d) - b**2*sin(c + d*x)**6*cos(c + d*x)**7/(7*d)
- 2*b**2*sin(c + d*x)**4*cos(c + d*x)**9/(21*d) - 8*b**2*sin(c + d*x)**2*cos(c + d*x)**11/(231*d) - 16*b**2*c
os(c + d*x)**13/(3003*d), Ne(d, 0)), (x*(a + b*sin(c))**2*sin(c)**5*cos(c)**6, True))

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Giac [A]  time = 1.30154, size = 289, normalized size = 1.21 \begin{align*} \frac{5}{512} \, a b x + \frac{b^{2} \cos \left (13 \, d x + 13 \, c\right )}{53248 \, d} - \frac{a b \sin \left (12 \, d x + 12 \, c\right )}{12288 \, d} + \frac{3 \, a b \sin \left (8 \, d x + 8 \, c\right )}{4096 \, d} - \frac{15 \, a b \sin \left (4 \, d x + 4 \, c\right )}{4096 \, d} - \frac{{\left (4 \, a^{2} + b^{2}\right )} \cos \left (11 \, d x + 11 \, c\right )}{45056 \, d} - \frac{{\left (2 \, a^{2} + 3 \, b^{2}\right )} \cos \left (9 \, d x + 9 \, c\right )}{18432 \, d} + \frac{{\left (10 \, a^{2} + 3 \, b^{2}\right )} \cos \left (7 \, d x + 7 \, c\right )}{14336 \, d} + \frac{{\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{4096 \, d} - \frac{5 \,{\left (8 \, a^{2} + 3 \, b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{12288 \, d} - \frac{5 \,{\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )}{1024 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

5/512*a*b*x + 1/53248*b^2*cos(13*d*x + 13*c)/d - 1/12288*a*b*sin(12*d*x + 12*c)/d + 3/4096*a*b*sin(8*d*x + 8*c
)/d - 15/4096*a*b*sin(4*d*x + 4*c)/d - 1/45056*(4*a^2 + b^2)*cos(11*d*x + 11*c)/d - 1/18432*(2*a^2 + 3*b^2)*co
s(9*d*x + 9*c)/d + 1/14336*(10*a^2 + 3*b^2)*cos(7*d*x + 7*c)/d + 1/4096*(4*a^2 + 3*b^2)*cos(5*d*x + 5*c)/d - 5
/12288*(8*a^2 + 3*b^2)*cos(3*d*x + 3*c)/d - 5/1024*(2*a^2 + b^2)*cos(d*x + c)/d