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

Optimal. Leaf size=78 \[ -\frac {B (a \sin (c+d x)+a)^5}{5 a^4 d}-\frac {(A-3 B) (a \sin (c+d x)+a)^4}{4 a^3 d}+\frac {2 (A-B) (a \sin (c+d x)+a)^3}{3 a^2 d} \]

[Out]

2/3*(A-B)*(a+a*sin(d*x+c))^3/a^2/d-1/4*(A-3*B)*(a+a*sin(d*x+c))^4/a^3/d-1/5*B*(a+a*sin(d*x+c))^5/a^4/d

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2836, 77} \[ -\frac {(A-3 B) (a \sin (c+d x)+a)^4}{4 a^3 d}+\frac {2 (A-B) (a \sin (c+d x)+a)^3}{3 a^2 d}-\frac {B (a \sin (c+d x)+a)^5}{5 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A - B)*(a + a*Sin[c + d*x])^3)/(3*a^2*d) - ((A - 3*B)*(a + a*Sin[c + d*x])^4)/(4*a^3*d) - (B*(a + a*Sin[c
+ d*x])^5)/(5*a^4*d)

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]))))

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 ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx &=\frac {\operatorname {Subst}\left (\int (a-x) (a+x)^2 \left (A+\frac {B x}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (2 a (A-B) (a+x)^2+(-A+3 B) (a+x)^3-\frac {B (a+x)^4}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {2 (A-B) (a+a \sin (c+d x))^3}{3 a^2 d}-\frac {(A-3 B) (a+a \sin (c+d x))^4}{4 a^3 d}-\frac {B (a+a \sin (c+d x))^5}{5 a^4 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.81, size = 78, normalized size = 1.00 \[ -\frac {a (-4 (100 A+11 B) \sin (c+d x)+3 \cos (4 (c+d x)) (5 (A+B)+4 B \sin (c+d x))+\cos (2 (c+d x)) ((32 B-80 A) \sin (c+d x)+60 (A+B)))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/480*(a*(-4*(100*A + 11*B)*Sin[c + d*x] + 3*Cos[4*(c + d*x)]*(5*(A + B) + 4*B*Sin[c + d*x]) + Cos[2*(c + d*x
)]*(60*(A + B) + (-80*A + 32*B)*Sin[c + d*x])))/d

________________________________________________________________________________________

fricas [A]  time = 0.79, size = 65, normalized size = 0.83 \[ -\frac {15 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, B a \cos \left (d x + c\right )^{4} - {\left (5 \, A + B\right )} a \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, A + B\right )} a\right )} \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*(A + B)*a*cos(d*x + c)^4 + 4*(3*B*a*cos(d*x + c)^4 - (5*A + B)*a*cos(d*x + c)^2 - 2*(5*A + B)*a)*sin
(d*x + c))/d

________________________________________________________________________________________

giac [A]  time = 0.20, size = 100, normalized size = 1.28 \[ -\frac {12 \, B a \sin \left (d x + c\right )^{5} + 15 \, A a \sin \left (d x + c\right )^{4} + 15 \, B a \sin \left (d x + c\right )^{4} + 20 \, A a \sin \left (d x + c\right )^{3} - 20 \, B a \sin \left (d x + c\right )^{3} - 30 \, A a \sin \left (d x + c\right )^{2} - 30 \, B a \sin \left (d x + c\right )^{2} - 60 \, A a \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(12*B*a*sin(d*x + c)^5 + 15*A*a*sin(d*x + c)^4 + 15*B*a*sin(d*x + c)^4 + 20*A*a*sin(d*x + c)^3 - 20*B*a*
sin(d*x + c)^3 - 30*A*a*sin(d*x + c)^2 - 30*B*a*sin(d*x + c)^2 - 60*A*a*sin(d*x + c))/d

________________________________________________________________________________________

maple [A]  time = 0.46, size = 88, normalized size = 1.13 \[ \frac {a B \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-\frac {a A \left (\cos ^{4}\left (d x +c \right )\right )}{4}-\frac {a B \left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {a A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 72, normalized size = 0.92 \[ -\frac {12 \, B a \sin \left (d x + c\right )^{5} + 15 \, {\left (A + B\right )} a \sin \left (d x + c\right )^{4} + 20 \, {\left (A - B\right )} a \sin \left (d x + c\right )^{3} - 30 \, {\left (A + B\right )} a \sin \left (d x + c\right )^{2} - 60 \, A a \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(12*B*a*sin(d*x + c)^5 + 15*(A + B)*a*sin(d*x + c)^4 + 20*(A - B)*a*sin(d*x + c)^3 - 30*(A + B)*a*sin(d*
x + c)^2 - 60*A*a*sin(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 9.01, size = 72, normalized size = 0.92 \[ -\frac {\frac {B\,a\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a\,\left (A+B\right )\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a\,\left (A-B\right )\,{\sin \left (c+d\,x\right )}^3}{3}-\frac {a\,\left (A+B\right )\,{\sin \left (c+d\,x\right )}^2}{2}-A\,a\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a*sin(c + d*x)^4*(A + B))/4 - (a*sin(c + d*x)^2*(A + B))/2 - A*a*sin(c + d*x) + (a*sin(c + d*x)^3*(A - B))/
3 + (B*a*sin(c + d*x)^5)/5)/d

________________________________________________________________________________________

sympy [A]  time = 1.77, size = 128, normalized size = 1.64 \[ \begin {cases} \frac {2 A a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {A a \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 B a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {B a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {B a \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a \sin {\relax (c )} + a\right ) \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________