3.1242 \(\int \cos ^6(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=201 \[ -\frac {\left (10 a^2+11 b^2\right ) \sin (c+d x) \cos ^7(c+d x)}{80 d}+\frac {\left (10 a^2+3 b^2\right ) \sin (c+d x) \cos ^5(c+d x)}{480 d}+\frac {\left (10 a^2+3 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{384 d}+\frac {\left (10 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{256 d}+\frac {1}{256} x \left (10 a^2+3 b^2\right )+\frac {2 a b \cos ^9(c+d x)}{9 d}-\frac {2 a b \cos ^7(c+d x)}{7 d}+\frac {b^2 \sin (c+d x) \cos ^9(c+d x)}{10 d} \]

[Out]

1/256*(10*a^2+3*b^2)*x-2/7*a*b*cos(d*x+c)^7/d+2/9*a*b*cos(d*x+c)^9/d+1/256*(10*a^2+3*b^2)*cos(d*x+c)*sin(d*x+c
)/d+1/384*(10*a^2+3*b^2)*cos(d*x+c)^3*sin(d*x+c)/d+1/480*(10*a^2+3*b^2)*cos(d*x+c)^5*sin(d*x+c)/d-1/80*(10*a^2
+11*b^2)*cos(d*x+c)^7*sin(d*x+c)/d+1/10*b^2*cos(d*x+c)^9*sin(d*x+c)/d

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Rubi [A]  time = 0.27, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2911, 2565, 14, 3200, 455, 385, 199, 203} \[ -\frac {\left (10 a^2+11 b^2\right ) \sin (c+d x) \cos ^7(c+d x)}{80 d}+\frac {\left (10 a^2+3 b^2\right ) \sin (c+d x) \cos ^5(c+d x)}{480 d}+\frac {\left (10 a^2+3 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{384 d}+\frac {\left (10 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{256 d}+\frac {1}{256} x \left (10 a^2+3 b^2\right )+\frac {2 a b \cos ^9(c+d x)}{9 d}-\frac {2 a b \cos ^7(c+d x)}{7 d}+\frac {b^2 \sin (c+d x) \cos ^9(c+d x)}{10 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((10*a^2 + 3*b^2)*x)/256 - (2*a*b*Cos[c + d*x]^7)/(7*d) + (2*a*b*Cos[c + d*x]^9)/(9*d) + ((10*a^2 + 3*b^2)*Cos
[c + d*x]*Sin[c + d*x])/(256*d) + ((10*a^2 + 3*b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(384*d) + ((10*a^2 + 3*b^2)*C
os[c + d*x]^5*Sin[c + d*x])/(480*d) - ((10*a^2 + 11*b^2)*Cos[c + d*x]^7*Sin[c + d*x])/(80*d) + (b^2*Cos[c + d*
x]^9*Sin[c + d*x])/(10*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 2565

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

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 3200

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

Rubi steps

\begin {align*} \int \cos ^6(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx &=(2 a b) \int \cos ^6(c+d x) \sin ^3(c+d x) \, dx+\int \cos ^6(c+d x) \sin ^2(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (a^2+\left (a^2+b^2\right ) x^2\right )}{\left (1+x^2\right )^6} \, dx,x,\tan (c+d x)\right )}{d}-\frac {(2 a b) \operatorname {Subst}\left (\int x^6 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {b^2 \cos ^9(c+d x) \sin (c+d x)}{10 d}-\frac {\operatorname {Subst}\left (\int \frac {b^2-10 \left (a^2+b^2\right ) x^2}{\left (1+x^2\right )^5} \, dx,x,\tan (c+d x)\right )}{10 d}-\frac {(2 a b) \operatorname {Subst}\left (\int \left (x^6-x^8\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {2 a b \cos ^7(c+d x)}{7 d}+\frac {2 a b \cos ^9(c+d x)}{9 d}-\frac {\left (10 a^2+11 b^2\right ) \cos ^7(c+d x) \sin (c+d x)}{80 d}+\frac {b^2 \cos ^9(c+d x) \sin (c+d x)}{10 d}+\frac {\left (10 a^2+3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^4} \, dx,x,\tan (c+d x)\right )}{80 d}\\ &=-\frac {2 a b \cos ^7(c+d x)}{7 d}+\frac {2 a b \cos ^9(c+d x)}{9 d}+\frac {\left (10 a^2+3 b^2\right ) \cos ^5(c+d x) \sin (c+d x)}{480 d}-\frac {\left (10 a^2+11 b^2\right ) \cos ^7(c+d x) \sin (c+d x)}{80 d}+\frac {b^2 \cos ^9(c+d x) \sin (c+d x)}{10 d}+\frac {\left (10 a^2+3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{96 d}\\ &=-\frac {2 a b \cos ^7(c+d x)}{7 d}+\frac {2 a b \cos ^9(c+d x)}{9 d}+\frac {\left (10 a^2+3 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{384 d}+\frac {\left (10 a^2+3 b^2\right ) \cos ^5(c+d x) \sin (c+d x)}{480 d}-\frac {\left (10 a^2+11 b^2\right ) \cos ^7(c+d x) \sin (c+d x)}{80 d}+\frac {b^2 \cos ^9(c+d x) \sin (c+d x)}{10 d}+\frac {\left (10 a^2+3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{128 d}\\ &=-\frac {2 a b \cos ^7(c+d x)}{7 d}+\frac {2 a b \cos ^9(c+d x)}{9 d}+\frac {\left (10 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{256 d}+\frac {\left (10 a^2+3 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{384 d}+\frac {\left (10 a^2+3 b^2\right ) \cos ^5(c+d x) \sin (c+d x)}{480 d}-\frac {\left (10 a^2+11 b^2\right ) \cos ^7(c+d x) \sin (c+d x)}{80 d}+\frac {b^2 \cos ^9(c+d x) \sin (c+d x)}{10 d}+\frac {\left (10 a^2+3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{256 d}\\ &=\frac {1}{256} \left (10 a^2+3 b^2\right ) x-\frac {2 a b \cos ^7(c+d x)}{7 d}+\frac {2 a b \cos ^9(c+d x)}{9 d}+\frac {\left (10 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{256 d}+\frac {\left (10 a^2+3 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{384 d}+\frac {\left (10 a^2+3 b^2\right ) \cos ^5(c+d x) \sin (c+d x)}{480 d}-\frac {\left (10 a^2+11 b^2\right ) \cos ^7(c+d x) \sin (c+d x)}{80 d}+\frac {b^2 \cos ^9(c+d x) \sin (c+d x)}{10 d}\\ \end {align*}

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Mathematica [A]  time = 0.90, size = 193, normalized size = 0.96 \[ \frac {5040 a^2 \sin (2 (c+d x))-2520 a^2 \sin (4 (c+d x))-1680 a^2 \sin (6 (c+d x))-315 a^2 \sin (8 (c+d x))+12600 a^2 d x-15120 a b \cos (c+d x)-6720 a b \cos (3 (c+d x))+1080 a b \cos (7 (c+d x))+280 a b \cos (9 (c+d x))+630 b^2 \sin (2 (c+d x))-1260 b^2 \sin (4 (c+d x))-315 b^2 \sin (6 (c+d x))+\frac {315}{2} b^2 \sin (8 (c+d x))+63 b^2 \sin (10 (c+d x))+6300 b^2 c+3780 b^2 d x}{322560 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6300*b^2*c + 12600*a^2*d*x + 3780*b^2*d*x - 15120*a*b*Cos[c + d*x] - 6720*a*b*Cos[3*(c + d*x)] + 1080*a*b*Cos
[7*(c + d*x)] + 280*a*b*Cos[9*(c + d*x)] + 5040*a^2*Sin[2*(c + d*x)] + 630*b^2*Sin[2*(c + d*x)] - 2520*a^2*Sin
[4*(c + d*x)] - 1260*b^2*Sin[4*(c + d*x)] - 1680*a^2*Sin[6*(c + d*x)] - 315*b^2*Sin[6*(c + d*x)] - 315*a^2*Sin
[8*(c + d*x)] + (315*b^2*Sin[8*(c + d*x)])/2 + 63*b^2*Sin[10*(c + d*x)])/(322560*d)

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fricas [A]  time = 0.89, size = 149, normalized size = 0.74 \[ \frac {17920 \, a b \cos \left (d x + c\right )^{9} - 23040 \, a b \cos \left (d x + c\right )^{7} + 315 \, {\left (10 \, a^{2} + 3 \, b^{2}\right )} d x + 21 \, {\left (384 \, b^{2} \cos \left (d x + c\right )^{9} - 48 \, {\left (10 \, a^{2} + 11 \, b^{2}\right )} \cos \left (d x + c\right )^{7} + 8 \, {\left (10 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{5} + 10 \, {\left (10 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (10 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{80640 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80640*(17920*a*b*cos(d*x + c)^9 - 23040*a*b*cos(d*x + c)^7 + 315*(10*a^2 + 3*b^2)*d*x + 21*(384*b^2*cos(d*x
+ c)^9 - 48*(10*a^2 + 11*b^2)*cos(d*x + c)^7 + 8*(10*a^2 + 3*b^2)*cos(d*x + c)^5 + 10*(10*a^2 + 3*b^2)*cos(d*x
+ c)^3 + 15*(10*a^2 + 3*b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.36, size = 189, normalized size = 0.94 \[ \frac {1}{256} \, {\left (10 \, a^{2} + 3 \, b^{2}\right )} x + \frac {a b \cos \left (9 \, d x + 9 \, c\right )}{1152 \, d} + \frac {3 \, a b \cos \left (7 \, d x + 7 \, c\right )}{896 \, d} - \frac {a b \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {3 \, a b \cos \left (d x + c\right )}{64 \, d} + \frac {b^{2} \sin \left (10 \, d x + 10 \, c\right )}{5120 \, d} - \frac {{\left (2 \, a^{2} - b^{2}\right )} \sin \left (8 \, d x + 8 \, c\right )}{2048 \, d} - \frac {{\left (16 \, a^{2} + 3 \, b^{2}\right )} \sin \left (6 \, d x + 6 \, c\right )}{3072 \, d} - \frac {{\left (2 \, a^{2} + b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{256 \, d} + \frac {{\left (8 \, a^{2} + b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{512 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(10*a^2 + 3*b^2)*x + 1/1152*a*b*cos(9*d*x + 9*c)/d + 3/896*a*b*cos(7*d*x + 7*c)/d - 1/48*a*b*cos(3*d*x +
3*c)/d - 3/64*a*b*cos(d*x + c)/d + 1/5120*b^2*sin(10*d*x + 10*c)/d - 1/2048*(2*a^2 - b^2)*sin(8*d*x + 8*c)/d
- 1/3072*(16*a^2 + 3*b^2)*sin(6*d*x + 6*c)/d - 1/256*(2*a^2 + b^2)*sin(4*d*x + 4*c)/d + 1/512*(8*a^2 + b^2)*si
n(2*d*x + 2*c)/d

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maple [A]  time = 0.33, size = 183, normalized size = 0.91 \[ \frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )}{8}+\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{48}+\frac {5 d x}{128}+\frac {5 c}{128}\right )+2 a b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{7}\left (d x +c \right )\right )}{9}-\frac {2 \left (\cos ^{7}\left (d x +c \right )\right )}{63}\right )+b^{2} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{7}\left (d x +c \right )\right )}{10}-\frac {3 \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )}{80}+\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{160}+\frac {3 d x}{256}+\frac {3 c}{256}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(-1/8*sin(d*x+c)*cos(d*x+c)^7+1/48*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/128*d
*x+5/128*c)+2*a*b*(-1/9*sin(d*x+c)^2*cos(d*x+c)^7-2/63*cos(d*x+c)^7)+b^2*(-1/10*sin(d*x+c)^3*cos(d*x+c)^7-3/80
*sin(d*x+c)*cos(d*x+c)^7+1/160*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+3/256*d*x+3/256*c))

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maxima [A]  time = 0.67, size = 127, normalized size = 0.63 \[ \frac {210 \, {\left (64 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 120 \, d x + 120 \, c - 3 \, \sin \left (8 \, d x + 8 \, c\right ) - 24 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} + 20480 \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} a b + 63 \, {\left (32 \, \sin \left (2 \, d x + 2 \, c\right )^{5} + 120 \, d x + 120 \, c + 5 \, \sin \left (8 \, d x + 8 \, c\right ) - 40 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b^{2}}{645120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/645120*(210*(64*sin(2*d*x + 2*c)^3 + 120*d*x + 120*c - 3*sin(8*d*x + 8*c) - 24*sin(4*d*x + 4*c))*a^2 + 20480
*(7*cos(d*x + c)^9 - 9*cos(d*x + c)^7)*a*b + 63*(32*sin(2*d*x + 2*c)^5 + 120*d*x + 120*c + 5*sin(8*d*x + 8*c)
- 40*sin(4*d*x + 4*c))*b^2)/d

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mupad [B]  time = 11.93, size = 237, normalized size = 1.18 \[ \frac {5\,a^2\,x}{128}+\frac {3\,b^2\,x}{256}+\frac {5\,a^2\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{192\,d}+\frac {a^2\,{\cos \left (c+d\,x\right )}^5\,\sin \left (c+d\,x\right )}{48\,d}-\frac {a^2\,{\cos \left (c+d\,x\right )}^7\,\sin \left (c+d\,x\right )}{8\,d}+\frac {b^2\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{128\,d}+\frac {b^2\,{\cos \left (c+d\,x\right )}^5\,\sin \left (c+d\,x\right )}{160\,d}-\frac {11\,b^2\,{\cos \left (c+d\,x\right )}^7\,\sin \left (c+d\,x\right )}{80\,d}+\frac {b^2\,{\cos \left (c+d\,x\right )}^9\,\sin \left (c+d\,x\right )}{10\,d}-\frac {2\,a\,b\,{\cos \left (c+d\,x\right )}^7}{7\,d}+\frac {2\,a\,b\,{\cos \left (c+d\,x\right )}^9}{9\,d}+\frac {5\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{128\,d}+\frac {3\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{256\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a^2*x)/128 + (3*b^2*x)/256 + (5*a^2*cos(c + d*x)^3*sin(c + d*x))/(192*d) + (a^2*cos(c + d*x)^5*sin(c + d*x)
)/(48*d) - (a^2*cos(c + d*x)^7*sin(c + d*x))/(8*d) + (b^2*cos(c + d*x)^3*sin(c + d*x))/(128*d) + (b^2*cos(c +
d*x)^5*sin(c + d*x))/(160*d) - (11*b^2*cos(c + d*x)^7*sin(c + d*x))/(80*d) + (b^2*cos(c + d*x)^9*sin(c + d*x))
/(10*d) - (2*a*b*cos(c + d*x)^7)/(7*d) + (2*a*b*cos(c + d*x)^9)/(9*d) + (5*a^2*cos(c + d*x)*sin(c + d*x))/(128
*d) + (3*b^2*cos(c + d*x)*sin(c + d*x))/(256*d)

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sympy [A]  time = 23.77, size = 529, normalized size = 2.63 \[ \begin {cases} \frac {5 a^{2} x \sin ^{8}{\left (c + d x \right )}}{128} + \frac {5 a^{2} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac {15 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac {5 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac {5 a^{2} x \cos ^{8}{\left (c + d x \right )}}{128} + \frac {5 a^{2} \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {55 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{384 d} + \frac {73 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{384 d} - \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} - \frac {2 a b \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{7 d} - \frac {4 a b \cos ^{9}{\left (c + d x \right )}}{63 d} + \frac {3 b^{2} x \sin ^{10}{\left (c + d x \right )}}{256} + \frac {15 b^{2} x \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{256} + \frac {15 b^{2} x \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{128} + \frac {15 b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{128} + \frac {15 b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{256} + \frac {3 b^{2} x \cos ^{10}{\left (c + d x \right )}}{256} + \frac {3 b^{2} \sin ^{9}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{256 d} + \frac {7 b^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} + \frac {b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{10 d} - \frac {7 b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} - \frac {3 b^{2} \sin {\left (c + d x \right )} \cos ^{9}{\left (c + d x \right )}}{256 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{2} \sin ^{2}{\relax (c )} \cos ^{6}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((5*a**2*x*sin(c + d*x)**8/128 + 5*a**2*x*sin(c + d*x)**6*cos(c + d*x)**2/32 + 15*a**2*x*sin(c + d*x)
**4*cos(c + d*x)**4/64 + 5*a**2*x*sin(c + d*x)**2*cos(c + d*x)**6/32 + 5*a**2*x*cos(c + d*x)**8/128 + 5*a**2*s
in(c + d*x)**7*cos(c + d*x)/(128*d) + 55*a**2*sin(c + d*x)**5*cos(c + d*x)**3/(384*d) + 73*a**2*sin(c + d*x)**
3*cos(c + d*x)**5/(384*d) - 5*a**2*sin(c + d*x)*cos(c + d*x)**7/(128*d) - 2*a*b*sin(c + d*x)**2*cos(c + d*x)**
7/(7*d) - 4*a*b*cos(c + d*x)**9/(63*d) + 3*b**2*x*sin(c + d*x)**10/256 + 15*b**2*x*sin(c + d*x)**8*cos(c + d*x
)**2/256 + 15*b**2*x*sin(c + d*x)**6*cos(c + d*x)**4/128 + 15*b**2*x*sin(c + d*x)**4*cos(c + d*x)**6/128 + 15*
b**2*x*sin(c + d*x)**2*cos(c + d*x)**8/256 + 3*b**2*x*cos(c + d*x)**10/256 + 3*b**2*sin(c + d*x)**9*cos(c + d*
x)/(256*d) + 7*b**2*sin(c + d*x)**7*cos(c + d*x)**3/(128*d) + b**2*sin(c + d*x)**5*cos(c + d*x)**5/(10*d) - 7*
b**2*sin(c + d*x)**3*cos(c + d*x)**7/(128*d) - 3*b**2*sin(c + d*x)*cos(c + d*x)**9/(256*d), Ne(d, 0)), (x*(a +
b*sin(c))**2*sin(c)**2*cos(c)**6, True))

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