3.997 \(\int \cos ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=159 \[ -\frac {7 a^3 (2 A+B) \cos ^3(c+d x)}{24 d}-\frac {7 (2 A+B) \cos ^3(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{40 d}+\frac {7 a^3 (2 A+B) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {7}{16} a^3 x (2 A+B)-\frac {a (2 A+B) \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{10 d}-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 d} \]

[Out]

7/16*a^3*(2*A+B)*x-7/24*a^3*(2*A+B)*cos(d*x+c)^3/d+7/16*a^3*(2*A+B)*cos(d*x+c)*sin(d*x+c)/d-1/10*a*(2*A+B)*cos
(d*x+c)^3*(a+a*sin(d*x+c))^2/d-1/6*B*cos(d*x+c)^3*(a+a*sin(d*x+c))^3/d-7/40*(2*A+B)*cos(d*x+c)^3*(a^3+a^3*sin(
d*x+c))/d

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Rubi [A]  time = 0.22, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2860, 2678, 2669, 2635, 8} \[ -\frac {7 a^3 (2 A+B) \cos ^3(c+d x)}{24 d}-\frac {7 (2 A+B) \cos ^3(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{40 d}+\frac {7 a^3 (2 A+B) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {7}{16} a^3 x (2 A+B)-\frac {a (2 A+B) \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{10 d}-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(7*a^3*(2*A + B)*x)/16 - (7*a^3*(2*A + B)*Cos[c + d*x]^3)/(24*d) + (7*a^3*(2*A + B)*Cos[c + d*x]*Sin[c + d*x])
/(16*d) - (a*(2*A + B)*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^2)/(10*d) - (B*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^
3)/(6*d) - (7*(2*A + B)*Cos[c + d*x]^3*(a^3 + a^3*Sin[c + d*x]))/(40*d)

Rule 8

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

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 2669

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

Rule 2678

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

Rule 2860

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

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}+\frac {1}{2} (2 A+B) \int \cos ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\\ &=-\frac {a (2 A+B) \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}+\frac {1}{10} (7 a (2 A+B)) \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac {a (2 A+B) \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac {7 (2 A+B) \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{40 d}+\frac {1}{8} \left (7 a^2 (2 A+B)\right ) \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac {7 a^3 (2 A+B) \cos ^3(c+d x)}{24 d}-\frac {a (2 A+B) \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac {7 (2 A+B) \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{40 d}+\frac {1}{8} \left (7 a^3 (2 A+B)\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {7 a^3 (2 A+B) \cos ^3(c+d x)}{24 d}+\frac {7 a^3 (2 A+B) \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a (2 A+B) \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac {7 (2 A+B) \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{40 d}+\frac {1}{16} \left (7 a^3 (2 A+B)\right ) \int 1 \, dx\\ &=\frac {7}{16} a^3 (2 A+B) x-\frac {7 a^3 (2 A+B) \cos ^3(c+d x)}{24 d}+\frac {7 a^3 (2 A+B) \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a (2 A+B) \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac {7 (2 A+B) \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{40 d}\\ \end {align*}

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Mathematica [A]  time = 1.42, size = 146, normalized size = 0.92 \[ -\frac {a^3 \cos (c+d x) \left (16 (17 A+11 B) \cos (2 (c+d x))-12 (A+3 B) \cos (4 (c+d x))+\frac {420 (2 A+B) \sin ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )}{\sqrt {\cos ^2(c+d x)}}-330 A \sin (c+d x)+90 A \sin (3 (c+d x))+284 A-95 B \sin (c+d x)+110 B \sin (3 (c+d x))-5 B \sin (5 (c+d x))+212 B\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

-1/480*(a^3*Cos[c + d*x]*(284*A + 212*B + (420*(2*A + B)*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]])/Sqrt[Cos[c +
d*x]^2] + 16*(17*A + 11*B)*Cos[2*(c + d*x)] - 12*(A + 3*B)*Cos[4*(c + d*x)] - 330*A*Sin[c + d*x] - 95*B*Sin[c
+ d*x] + 90*A*Sin[3*(c + d*x)] + 110*B*Sin[3*(c + d*x)] - 5*B*Sin[5*(c + d*x)]))/d

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fricas [A]  time = 0.61, size = 111, normalized size = 0.70 \[ \frac {48 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} - 320 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{3} + 105 \, {\left (2 \, A + B\right )} a^{3} d x + 5 \, {\left (8 \, B a^{3} \cos \left (d x + c\right )^{5} - 2 \, {\left (18 \, A + 25 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 21 \, {\left (2 \, A + B\right )} a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/240*(48*(A + 3*B)*a^3*cos(d*x + c)^5 - 320*(A + B)*a^3*cos(d*x + c)^3 + 105*(2*A + B)*a^3*d*x + 5*(8*B*a^3*c
os(d*x + c)^5 - 2*(18*A + 25*B)*a^3*cos(d*x + c)^3 + 21*(2*A + B)*a^3*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.29, size = 165, normalized size = 1.04 \[ \frac {B a^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {7}{16} \, {\left (2 \, A a^{3} + B a^{3}\right )} x + \frac {{\left (A a^{3} + 3 \, B a^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {{\left (13 \, A a^{3} + 7 \, B a^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {{\left (7 \, A a^{3} + 5 \, B a^{3}\right )} \cos \left (d x + c\right )}{8 \, d} - \frac {{\left (6 \, A a^{3} + 7 \, B a^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (16 \, A a^{3} - B a^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/192*B*a^3*sin(6*d*x + 6*c)/d + 7/16*(2*A*a^3 + B*a^3)*x + 1/80*(A*a^3 + 3*B*a^3)*cos(5*d*x + 5*c)/d - 1/48*(
13*A*a^3 + 7*B*a^3)*cos(3*d*x + 3*c)/d - 1/8*(7*A*a^3 + 5*B*a^3)*cos(d*x + c)/d - 1/64*(6*A*a^3 + 7*B*a^3)*sin
(4*d*x + 4*c)/d + 1/64*(16*A*a^3 - B*a^3)*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.35, size = 279, normalized size = 1.75 \[ \frac {a^{3} A \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+B \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+3 a^{3} A \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+3 B \,a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )-a^{3} A \left (\cos ^{3}\left (d x +c \right )\right )+3 B \,a^{3} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+a^{3} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {B \,a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^3*A*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)+B*a^3*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*sin(d*
x+c)*cos(d*x+c)^3+1/16*cos(d*x+c)*sin(d*x+c)+1/16*d*x+1/16*c)+3*a^3*A*(-1/4*sin(d*x+c)*cos(d*x+c)^3+1/8*cos(d*
x+c)*sin(d*x+c)+1/8*d*x+1/8*c)+3*B*a^3*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)-a^3*A*cos(d*x+c)^3+3
*B*a^3*(-1/4*sin(d*x+c)*cos(d*x+c)^3+1/8*cos(d*x+c)*sin(d*x+c)+1/8*d*x+1/8*c)+a^3*A*(1/2*cos(d*x+c)*sin(d*x+c)
+1/2*d*x+1/2*c)-1/3*B*a^3*cos(d*x+c)^3)

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maxima [A]  time = 0.32, size = 199, normalized size = 1.25 \[ -\frac {960 \, A a^{3} \cos \left (d x + c\right )^{3} + 320 \, B a^{3} \cos \left (d x + c\right )^{3} - 64 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} A a^{3} - 90 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} A a^{3} - 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 192 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} B a^{3} + 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} B a^{3} - 90 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} B a^{3}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/960*(960*A*a^3*cos(d*x + c)^3 + 320*B*a^3*cos(d*x + c)^3 - 64*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*A*a^3 -
90*(4*d*x + 4*c - sin(4*d*x + 4*c))*A*a^3 - 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 - 192*(3*cos(d*x + c)^
5 - 5*cos(d*x + c)^3)*B*a^3 + 5*(4*sin(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*c))*B*a^3 - 90*(4*d*x
+ 4*c - sin(4*d*x + 4*c))*B*a^3)/d

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mupad [B]  time = 10.73, size = 451, normalized size = 2.84 \[ \frac {7\,a^3\,\mathrm {atan}\left (\frac {7\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A+B\right )}{8\,\left (\frac {7\,A\,a^3}{4}+\frac {7\,B\,a^3}{8}\right )}\right )\,\left (2\,A+B\right )}{8\,d}-\frac {\frac {34\,A\,a^3}{15}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A\,a^3}{4}-\frac {7\,B\,a^3}{8}\right )+\frac {22\,B\,a^3}{15}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (6\,A\,a^3+2\,B\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (12\,A\,a^3+4\,B\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {A\,a^3}{4}-\frac {7\,B\,a^3}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (22\,A\,a^3+18\,B\,a^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {13\,A\,a^3}{2}+\frac {37\,B\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {13\,A\,a^3}{2}+\frac {37\,B\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {38\,A\,a^3}{5}+\frac {34\,B\,a^3}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {68\,A\,a^3}{3}+\frac {44\,B\,a^3}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {27\,A\,a^3}{4}+\frac {73\,B\,a^3}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {27\,A\,a^3}{4}+\frac {73\,B\,a^3}{24}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {7\,a^3\,\left (2\,A+B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3,x)

[Out]

(7*a^3*atan((7*a^3*tan(c/2 + (d*x)/2)*(2*A + B))/(8*((7*A*a^3)/4 + (7*B*a^3)/8)))*(2*A + B))/(8*d) - ((34*A*a^
3)/15 - tan(c/2 + (d*x)/2)*((A*a^3)/4 - (7*B*a^3)/8) + (22*B*a^3)/15 + tan(c/2 + (d*x)/2)^10*(6*A*a^3 + 2*B*a^
3) + tan(c/2 + (d*x)/2)^4*(12*A*a^3 + 4*B*a^3) + tan(c/2 + (d*x)/2)^11*((A*a^3)/4 - (7*B*a^3)/8) + tan(c/2 + (
d*x)/2)^8*(22*A*a^3 + 18*B*a^3) - tan(c/2 + (d*x)/2)^5*((13*A*a^3)/2 + (37*B*a^3)/4) + tan(c/2 + (d*x)/2)^7*((
13*A*a^3)/2 + (37*B*a^3)/4) + tan(c/2 + (d*x)/2)^2*((38*A*a^3)/5 + (34*B*a^3)/5) + tan(c/2 + (d*x)/2)^6*((68*A
*a^3)/3 + (44*B*a^3)/3) - tan(c/2 + (d*x)/2)^3*((27*A*a^3)/4 + (73*B*a^3)/24) + tan(c/2 + (d*x)/2)^9*((27*A*a^
3)/4 + (73*B*a^3)/24))/(d*(6*tan(c/2 + (d*x)/2)^2 + 15*tan(c/2 + (d*x)/2)^4 + 20*tan(c/2 + (d*x)/2)^6 + 15*tan
(c/2 + (d*x)/2)^8 + 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) - (7*a^3*(2*A + B)*(atan(tan(c/2 + (
d*x)/2)) - (d*x)/2))/(8*d)

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sympy [A]  time = 4.38, size = 588, normalized size = 3.70 \[ \begin {cases} \frac {3 A a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {A a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {A a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {3 A a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {A a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {2 A a^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {A a^{3} \cos ^{3}{\left (c + d x \right )}}{d} + \frac {B a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 B a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 B a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 B a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {B a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 B a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {B a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {3 B a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {B a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} - \frac {B a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {3 B a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {2 B a^{3} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {B a^{3} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a \sin {\relax (c )} + a\right )^{3} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((3*A*a**3*x*sin(c + d*x)**4/8 + 3*A*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + A*a**3*x*sin(c + d*x)
**2/2 + 3*A*a**3*x*cos(c + d*x)**4/8 + A*a**3*x*cos(c + d*x)**2/2 + 3*A*a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d
) - A*a**3*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - 3*A*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + A*a**3*sin(c
+ d*x)*cos(c + d*x)/(2*d) - 2*A*a**3*cos(c + d*x)**5/(15*d) - A*a**3*cos(c + d*x)**3/d + B*a**3*x*sin(c + d*x)
**6/16 + 3*B*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*B*a**3*x*sin(c + d*x)**4/8 + 3*B*a**3*x*sin(c + d*x
)**2*cos(c + d*x)**4/16 + 3*B*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + B*a**3*x*cos(c + d*x)**6/16 + 3*B*a**
3*x*cos(c + d*x)**4/8 + B*a**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) - B*a**3*sin(c + d*x)**3*cos(c + d*x)**3/(6
*d) + 3*B*a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d) - B*a**3*sin(c + d*x)**2*cos(c + d*x)**3/d - B*a**3*sin(c +
d*x)*cos(c + d*x)**5/(16*d) - 3*B*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) - 2*B*a**3*cos(c + d*x)**5/(5*d) - B
*a**3*cos(c + d*x)**3/(3*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**3*cos(c)**2, True))

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