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

Optimal. Leaf size=134 \[ -\frac {a^2 (5 A+2 B) \cos ^3(c+d x)}{12 d}-\frac {(5 A+2 B) \cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{20 d}+\frac {a^2 (5 A+2 B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a^2 x (5 A+2 B)-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d} \]

[Out]

1/8*a^2*(5*A+2*B)*x-1/12*a^2*(5*A+2*B)*cos(d*x+c)^3/d+1/8*a^2*(5*A+2*B)*cos(d*x+c)*sin(d*x+c)/d-1/5*B*cos(d*x+
c)^3*(a+a*sin(d*x+c))^2/d-1/20*(5*A+2*B)*cos(d*x+c)^3*(a^2+a^2*sin(d*x+c))/d

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Rubi [A]  time = 0.17, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ -\frac {a^2 (5 A+2 B) \cos ^3(c+d x)}{12 d}-\frac {(5 A+2 B) \cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{20 d}+\frac {a^2 (5 A+2 B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a^2 x (5 A+2 B)-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(5*A + 2*B)*x)/8 - (a^2*(5*A + 2*B)*Cos[c + d*x]^3)/(12*d) + (a^2*(5*A + 2*B)*Cos[c + d*x]*Sin[c + d*x])/
(8*d) - (B*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^2)/(5*d) - ((5*A + 2*B)*Cos[c + d*x]^3*(a^2 + a^2*Sin[c + d*x])
)/(20*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))^2 (A+B \sin (c+d x)) \, dx &=-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}+\frac {1}{5} (5 A+2 B) \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {(5 A+2 B) \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{20 d}+\frac {1}{4} (a (5 A+2 B)) \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac {a^2 (5 A+2 B) \cos ^3(c+d x)}{12 d}-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {(5 A+2 B) \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{20 d}+\frac {1}{4} \left (a^2 (5 A+2 B)\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {a^2 (5 A+2 B) \cos ^3(c+d x)}{12 d}+\frac {a^2 (5 A+2 B) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {(5 A+2 B) \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{20 d}+\frac {1}{8} \left (a^2 (5 A+2 B)\right ) \int 1 \, dx\\ &=\frac {1}{8} a^2 (5 A+2 B) x-\frac {a^2 (5 A+2 B) \cos ^3(c+d x)}{12 d}+\frac {a^2 (5 A+2 B) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {(5 A+2 B) \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{20 d}\\ \end {align*}

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Mathematica [A]  time = 0.98, size = 133, normalized size = 0.99 \[ -\frac {a^2 \cos (c+d x) \left (8 (10 A+7 B) \cos (2 (c+d x))+\frac {60 (5 A+2 B) \sin ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )}{\sqrt {\cos ^2(c+d x)}}-135 A \sin (c+d x)+15 A \sin (3 (c+d x))+80 A-30 B \sin (c+d x)+30 B \sin (3 (c+d x))-6 B \cos (4 (c+d x))+62 B\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/240*(a^2*Cos[c + d*x]*(80*A + 62*B + (60*(5*A + 2*B)*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]])/Sqrt[Cos[c + d
*x]^2] + 8*(10*A + 7*B)*Cos[2*(c + d*x)] - 6*B*Cos[4*(c + d*x)] - 135*A*Sin[c + d*x] - 30*B*Sin[c + d*x] + 15*
A*Sin[3*(c + d*x)] + 30*B*Sin[3*(c + d*x)]))/d

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fricas [A]  time = 0.68, size = 95, normalized size = 0.71 \[ \frac {24 \, B a^{2} \cos \left (d x + c\right )^{5} - 80 \, {\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{3} + 15 \, {\left (5 \, A + 2 \, B\right )} a^{2} d x - 15 \, {\left (2 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} - {\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(24*B*a^2*cos(d*x + c)^5 - 80*(A + B)*a^2*cos(d*x + c)^3 + 15*(5*A + 2*B)*a^2*d*x - 15*(2*(A + 2*B)*a^2*
cos(d*x + c)^3 - (5*A + 2*B)*a^2*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.23, size = 130, normalized size = 0.97 \[ \frac {B a^{2} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {A a^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {1}{8} \, {\left (5 \, A a^{2} + 2 \, B a^{2}\right )} x - \frac {{\left (8 \, A a^{2} + 5 \, B a^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {{\left (4 \, A a^{2} + 3 \, B a^{2}\right )} \cos \left (d x + c\right )}{8 \, d} - \frac {{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*B*a^2*cos(5*d*x + 5*c)/d + 1/4*A*a^2*sin(2*d*x + 2*c)/d + 1/8*(5*A*a^2 + 2*B*a^2)*x - 1/48*(8*A*a^2 + 5*B
*a^2)*cos(3*d*x + 3*c)/d - 1/8*(4*A*a^2 + 3*B*a^2)*cos(d*x + c)/d - 1/32*(A*a^2 + 2*B*a^2)*sin(4*d*x + 4*c)/d

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maple [A]  time = 0.39, size = 182, normalized size = 1.36 \[ \frac {a^{2} A \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+B \,a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )-\frac {2 a^{2} A \left (\cos ^{3}\left (d x +c \right )\right )}{3}+2 B \,a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {B \,a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 134, normalized size = 1.00 \[ -\frac {320 \, A a^{2} \cos \left (d x + c\right )^{3} + 160 \, B a^{2} \cos \left (d x + c\right )^{3} - 15 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} A a^{2} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 32 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} B a^{2} - 30 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} B a^{2}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480*(320*A*a^2*cos(d*x + c)^3 + 160*B*a^2*cos(d*x + c)^3 - 15*(4*d*x + 4*c - sin(4*d*x + 4*c))*A*a^2 - 120*
(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 - 32*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*B*a^2 - 30*(4*d*x + 4*c - si
n(4*d*x + 4*c))*B*a^2)/d

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mupad [B]  time = 10.47, size = 367, normalized size = 2.74 \[ \frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,A+2\,B\right )}{4\,\left (\frac {5\,A\,a^2}{4}+\frac {B\,a^2}{2}\right )}\right )\,\left (5\,A+2\,B\right )}{4\,d}-\frac {a^2\,\left (5\,A+2\,B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d}-\frac {\frac {4\,A\,a^2}{3}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A\,a^2}{4}-\frac {B\,a^2}{2}\right )+\frac {14\,B\,a^2}{15}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (4\,A\,a^2+2\,B\,a^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {7\,A\,a^2}{2}+3\,B\,a^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {7\,A\,a^2}{2}+3\,B\,a^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {3\,A\,a^2}{4}-\frac {B\,a^2}{2}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (8\,A\,a^2+8\,B\,a^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {8\,A\,a^2}{3}+\frac {8\,B\,a^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {16\,A\,a^2}{3}+\frac {4\,B\,a^2}{3}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

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))^2,x)

[Out]

(a^2*atan((a^2*tan(c/2 + (d*x)/2)*(5*A + 2*B))/(4*((5*A*a^2)/4 + (B*a^2)/2)))*(5*A + 2*B))/(4*d) - (a^2*(5*A +
2*B)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(4*d) - ((4*A*a^2)/3 - tan(c/2 + (d*x)/2)*((3*A*a^2)/4 - (B*a^2)/2
) + (14*B*a^2)/15 + tan(c/2 + (d*x)/2)^8*(4*A*a^2 + 2*B*a^2) - tan(c/2 + (d*x)/2)^3*((7*A*a^2)/2 + 3*B*a^2) +
tan(c/2 + (d*x)/2)^7*((7*A*a^2)/2 + 3*B*a^2) + tan(c/2 + (d*x)/2)^9*((3*A*a^2)/4 - (B*a^2)/2) + tan(c/2 + (d*x
)/2)^6*(8*A*a^2 + 8*B*a^2) + tan(c/2 + (d*x)/2)^2*((8*A*a^2)/3 + (8*B*a^2)/3) + tan(c/2 + (d*x)/2)^4*((16*A*a^
2)/3 + (4*B*a^2)/3))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/
2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1))

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sympy [A]  time = 2.43, size = 371, normalized size = 2.77 \[ \begin {cases} \frac {A a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {A a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {A a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {A a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {2 A a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {B a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {B a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} - \frac {B a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} - \frac {2 B a^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {B a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a \sin {\relax (c )} + a\right )^{2} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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