\(\int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx\) [227]

Optimal result

Integrand size = 24, antiderivative size = 82 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {5}{8} a^3 A x-\frac {5 a^3 A \cos ^3(c+d x)}{12 d}+\frac {5 a^3 A \cos (c+d x) \sin (c+d x)}{8 d}-\frac {A \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{4 d} \]

[Out]

5/8*a^3*A*x-5/12*a^3*A*cos(d*x+c)^3/d+5/8*a^3*A*cos(d*x+c)*sin(d*x+c)/d-1/4*A*cos(d*x+c)^3*(a^3+a^3*sin(d*x+c)
)/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2815, 2757, 2748, 2715, 8} \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {5 a^3 A \cos ^3(c+d x)}{12 d}-\frac {A \cos ^3(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{4 d}+\frac {5 a^3 A \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5}{8} a^3 A x \]

[In]

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

[Out]

(5*a^3*A*x)/8 - (5*a^3*A*Cos[c + d*x]^3)/(12*d) + (5*a^3*A*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (A*Cos[c + d*x]^
3*(a^3 + a^3*Sin[c + d*x]))/(4*d)

Rule 8

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

Rule 2715

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] && Integ
erQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[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 2757

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 2815

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rubi steps \begin{align*} \text {integral}& = (a A) \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx \\ & = -\frac {A \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{4 d}+\frac {1}{4} \left (5 a^2 A\right ) \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx \\ & = -\frac {5 a^3 A \cos ^3(c+d x)}{12 d}-\frac {A \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{4 d}+\frac {1}{4} \left (5 a^3 A\right ) \int \cos ^2(c+d x) \, dx \\ & = -\frac {5 a^3 A \cos ^3(c+d x)}{12 d}+\frac {5 a^3 A \cos (c+d x) \sin (c+d x)}{8 d}-\frac {A \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{4 d}+\frac {1}{8} \left (5 a^3 A\right ) \int 1 \, dx \\ & = \frac {5}{8} a^3 A x-\frac {5 a^3 A \cos ^3(c+d x)}{12 d}+\frac {5 a^3 A \cos (c+d x) \sin (c+d x)}{8 d}-\frac {A \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.66 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^3 A (60 d x-48 \cos (c+d x)-16 \cos (3 (c+d x))+24 \sin (2 (c+d x))-3 \sin (4 (c+d x)))}{96 d} \]

[In]

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

[Out]

(a^3*A*(60*d*x - 48*Cos[c + d*x] - 16*Cos[3*(c + d*x)] + 24*Sin[2*(c + d*x)] - 3*Sin[4*(c + d*x)]))/(96*d)

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(-\frac {A \,a^{3} \left (-60 d x +48 \cos \left (d x +c \right )+3 \sin \left (4 d x +4 c \right )+16 \cos \left (3 d x +3 c \right )-24 \sin \left (2 d x +2 c \right )+64\right )}{96 d}\)	\(57\)
risch	\(\frac {5 a^{3} A x}{8}-\frac {a^{3} A \cos \left (d x +c \right )}{2 d}-\frac {A \,a^{3} \sin \left (4 d x +4 c \right )}{32 d}-\frac {A \,a^{3} \cos \left (3 d x +3 c \right )}{6 d}+\frac {A \,a^{3} \sin \left (2 d x +2 c \right )}{4 d}\)	\(78\)
derivativedivides	\(\frac {-A \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 A \,a^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}-2 A \,a^{3} \cos \left (d x +c \right )+A \,a^{3} \left (d x +c \right )}{d}\)	\(89\)
default	\(\frac {-A \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 A \,a^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}-2 A \,a^{3} \cos \left (d x +c \right )+A \,a^{3} \left (d x +c \right )}{d}\)	\(89\)
parts	\(a^{3} A x -\frac {2 a^{3} A \cos \left (d x +c \right )}{d}+\frac {2 A \,a^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3 d}-\frac {A \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(90\)
norman	\(\frac {-\frac {4 A \,a^{3}}{3 d}+\frac {5 a^{3} A x}{8}-\frac {4 A \,a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 A \,a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 A \,a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {11 A \,a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {11 A \,a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {3 A \,a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 a^{3} A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a^{3} A x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a^{3} A x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{3} A x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(244\)

[In]

int((a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/96*A*a^3*(-60*d*x+48*cos(d*x+c)+3*sin(4*d*x+4*c)+16*cos(3*d*x+3*c)-24*sin(2*d*x+2*c)+64)/d

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.77 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {16 \, A a^{3} \cos \left (d x + c\right )^{3} - 15 \, A a^{3} d x + 3 \, {\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} - 5 \, A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

[In]

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

[Out]

-1/24*(16*A*a^3*cos(d*x + c)^3 - 15*A*a^3*d*x + 3*(2*A*a^3*cos(d*x + c)^3 - 5*A*a^3*cos(d*x + c))*sin(d*x + c)
)/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (78) = 156\).

Time = 0.17 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.39 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\begin {cases} - \frac {3 A a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} - \frac {3 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} - \frac {3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + A a^{3} x + \frac {5 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 A a^{3} \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {3 A a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {4 A a^{3} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 A a^{3} \cos {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (- A \sin {\left (c \right )} + A\right ) \left (a \sin {\left (c \right )} + a\right )^{3} & \text {otherwise} \end {cases} \]

[In]

integrate((a+a*sin(d*x+c))**3*(A-A*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 - 3*A*a**3*x*cos(c + d
*x)**4/8 + A*a**3*x + 5*A*a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*A*a**3*sin(c + d*x)**2*cos(c + d*x)/d +
3*A*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 4*A*a**3*cos(c + d*x)**3/(3*d) - 2*A*a**3*cos(c + d*x)/d, Ne(d,
0)), (x*(-A*sin(c) + A)*(a*sin(c) + a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {64 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} A a^{3} + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 96 \, {\left (d x + c\right )} A a^{3} + 192 \, A a^{3} \cos \left (d x + c\right )}{96 \, d} \]

[In]

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

[Out]

-1/96*(64*(cos(d*x + c)^3 - 3*cos(d*x + c))*A*a^3 + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*
A*a^3 - 96*(d*x + c)*A*a^3 + 192*A*a^3*cos(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.94 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {5}{8} \, A a^{3} x - \frac {A a^{3} \cos \left (3 \, d x + 3 \, c\right )}{6 \, d} - \frac {A a^{3} \cos \left (d x + c\right )}{2 \, d} - \frac {A a^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {A a^{3} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

[In]

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

[Out]

5/8*A*a^3*x - 1/6*A*a^3*cos(3*d*x + 3*c)/d - 1/2*A*a^3*cos(d*x + c)/d - 1/32*A*a^3*sin(4*d*x + 4*c)/d + 1/4*A*
a^3*sin(2*d*x + 2*c)/d

Mupad [B] (verification not implemented)

Time = 14.51 (sec) , antiderivative size = 250, normalized size of antiderivative = 3.05 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {5\,A\,a^3\,x}{8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{6}-\frac {A\,a^3\,\left (60\,c+60\,d\,x-32\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{6}-\frac {A\,a^3\,\left (60\,c+60\,d\,x-96\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{4}-\frac {A\,a^3\,\left (90\,c+90\,d\,x-96\right )}{24}\right )-\frac {3\,A\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {11\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {11\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {3\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{24}-\frac {A\,a^3\,\left (15\,c+15\,d\,x-32\right )}{24}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]

[In]

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

[Out]

(5*A*a^3*x)/8 - (tan(c/2 + (d*x)/2)^2*((A*a^3*(15*c + 15*d*x))/6 - (A*a^3*(60*c + 60*d*x - 32))/24) + tan(c/2
+ (d*x)/2)^6*((A*a^3*(15*c + 15*d*x))/6 - (A*a^3*(60*c + 60*d*x - 96))/24) + tan(c/2 + (d*x)/2)^4*((A*a^3*(15*
c + 15*d*x))/4 - (A*a^3*(90*c + 90*d*x - 96))/24) - (3*A*a^3*tan(c/2 + (d*x)/2))/4 - (11*A*a^3*tan(c/2 + (d*x)
/2)^3)/4 + (11*A*a^3*tan(c/2 + (d*x)/2)^5)/4 + (3*A*a^3*tan(c/2 + (d*x)/2)^7)/4 + (A*a^3*(15*c + 15*d*x))/24 -
(A*a^3*(15*c + 15*d*x - 32))/24)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^4)