Integrand size = 24, antiderivative size = 82 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {5}{8} a^3 A x-\frac {5 a^3 A \cos ^3(c+d x)}{12 d}+\frac {5 a^3 A \cos (c+d x) \sin (c+d x)}{8 d}-\frac {A \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{4 d} \]
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
Time = 0.56 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.66 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^3 A (60 d x-48 \cos (c+d x)-16 \cos (3 (c+d x))+24 \sin (2 (c+d x))-3 \sin (4 (c+d x)))}{96 d} \]
(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)
Time = 0.44 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3215, 3042, 3157, 3042, 3148, 3042, 3115, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (a \sin (c+d x)+a)^3 (A-A \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (c+d x)+a)^3 (A-A \sin (c+d x))dx\) |
\(\Big \downarrow \) 3215 |
\(\displaystyle a A \int \cos ^2(c+d x) (\sin (c+d x) a+a)^2dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a A \int \cos (c+d x)^2 (\sin (c+d x) a+a)^2dx\) |
\(\Big \downarrow \) 3157 |
\(\displaystyle a A \left (\frac {5}{4} a \int \cos ^2(c+d x) (\sin (c+d x) a+a)dx-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a A \left (\frac {5}{4} a \int \cos (c+d x)^2 (\sin (c+d x) a+a)dx-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\right )\) |
\(\Big \downarrow \) 3148 |
\(\displaystyle a A \left (\frac {5}{4} a \left (a \int \cos ^2(c+d x)dx-\frac {a \cos ^3(c+d x)}{3 d}\right )-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a A \left (\frac {5}{4} a \left (a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a \cos ^3(c+d x)}{3 d}\right )-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\right )\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle a A \left (\frac {5}{4} a \left (a \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {a \cos ^3(c+d x)}{3 d}\right )-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle a A \left (\frac {5}{4} a \left (a \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {a \cos ^3(c+d x)}{3 d}\right )-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\right )\) |
a*A*(-1/4*(Cos[c + d*x]^3*(a^2 + a^2*Sin[c + d*x]))/d + (5*a*(-1/3*(a*Cos[ c + d*x]^3)/d + a*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d))))/4)
3.3.27.3.1 Defintions of rubi rules used
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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] + Simp[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])
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] + Simp[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] && Integers Q[2*m, 2*p]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[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] && ((Lt Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
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 | \(\frac {-\frac {4 A \,a^{3}}{3 d}+\frac {5 a^{3} A x}{8}-\frac {4 A \,a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 A \,a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 A \,a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {11 A \,a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {11 A \,a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {3 A \,a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 a^{3} A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a^{3} A x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a^{3} A x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{3} A x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\) | \(244\) |
-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*s in(2*d*x+2*c)+64)/d
Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.77 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {16 \, A a^{3} \cos \left (d x + c\right )^{3} - 15 \, A a^{3} d x + 3 \, {\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} - 5 \, A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]
-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
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 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\begin {cases} - \frac {3 A a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} - \frac {3 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} - \frac {3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + A a^{3} x + \frac {5 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 A a^{3} \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {3 A a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {4 A a^{3} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 A a^{3} \cos {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (- A \sin {\left (c \right )} + A\right ) \left (a \sin {\left (c \right )} + a\right )^{3} & \text {otherwise} \end {cases} \]
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))
Time = 0.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {64 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} A a^{3} + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 96 \, {\left (d x + c\right )} A a^{3} + 192 \, A a^{3} \cos \left (d x + c\right )}{96 \, d} \]
-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
Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.94 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {5}{8} \, A a^{3} x - \frac {A a^{3} \cos \left (3 \, d x + 3 \, c\right )}{6 \, d} - \frac {A a^{3} \cos \left (d x + c\right )}{2 \, d} - \frac {A a^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {A a^{3} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]
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
Time = 14.51 (sec) , antiderivative size = 250, normalized size of antiderivative = 3.05 \[ \int (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {5\,A\,a^3\,x}{8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{6}-\frac {A\,a^3\,\left (60\,c+60\,d\,x-32\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{6}-\frac {A\,a^3\,\left (60\,c+60\,d\,x-96\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{4}-\frac {A\,a^3\,\left (90\,c+90\,d\,x-96\right )}{24}\right )-\frac {3\,A\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {11\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {11\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {3\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{24}-\frac {A\,a^3\,\left (15\,c+15\,d\,x-32\right )}{24}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]
(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)