Integrand size = 21, antiderivative size = 78 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {5 a^2 x}{8}-\frac {5 a^2 \cos ^3(c+d x)}{12 d}+\frac {5 a^2 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{4 d} \] Output:
5/8*a^2*x-5/12*a^2*cos(d*x+c)^3/d+5/8*a^2*cos(d*x+c)*sin(d*x+c)/d-1/4*cos( d*x+c)^3*(a^2+a^2*sin(d*x+c))/d
Time = 0.32 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.68 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \cos ^3(c+d x) \left (30 \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}+\sqrt {1+\sin (c+d x)} \left (16-25 \sin (c+d x)-7 \sin ^2(c+d x)+10 \sin ^3(c+d x)+6 \sin ^4(c+d x)\right )\right )}{24 d (-1+\sin (c+d x))^2 (1+\sin (c+d x))^{3/2}} \] Input:
Integrate[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
Output:
-1/24*(a^2*Cos[c + d*x]^3*(30*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[ 1 - Sin[c + d*x]] + Sqrt[1 + Sin[c + d*x]]*(16 - 25*Sin[c + d*x] - 7*Sin[c + d*x]^2 + 10*Sin[c + d*x]^3 + 6*Sin[c + d*x]^4)))/(d*(-1 + Sin[c + d*x]) ^2*(1 + Sin[c + d*x])^(3/2))
Time = 0.37 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {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.
\(\displaystyle \int \cos ^2(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^2 (a \sin (c+d x)+a)^2dx\) |
\(\Big \downarrow \) 3157 |
\(\displaystyle \frac {5}{4} a \int \cos ^2(c+d x) (\sin (c+d x) a+a)dx-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{4} a \int \cos (c+d x)^2 (\sin (c+d x) a+a)dx-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\) |
\(\Big \downarrow \) 3148 |
\(\displaystyle \frac {5}{4} a \left (a \int \cos ^2(c+d x)dx-\frac {a \cos ^3(c+d x)}{3 d}\right )-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{4} a \left (a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a \cos ^3(c+d x)}{3 d}\right )-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {5}{4} a \left (a \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {a \cos ^3(c+d x)}{3 d}\right )-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {5}{4} a \left (a \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {a \cos ^3(c+d x)}{3 d}\right )-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\) |
Input:
Int[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
Output:
-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
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]
Time = 2.54 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.72
method | result | size |
parallelrisch | \(-\frac {a^{2} \left (-60 d x +48 \cos \left (d x +c \right )+3 \sin \left (4 d x +4 c \right )+16 \cos \left (3 d x +3 c \right )-24 \sin \left (2 d x +2 c \right )+64\right )}{96 d}\) | \(56\) |
risch | \(\frac {5 a^{2} x}{8}-\frac {a^{2} \cos \left (d x +c \right )}{2 d}-\frac {a^{2} \sin \left (4 d x +4 c \right )}{32 d}-\frac {a^{2} \cos \left (3 d x +3 c \right )}{6 d}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{4 d}\) | \(73\) |
derivativedivides | \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {2 a^{2} \cos \left (d x +c \right )^{3}}{3}+a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(87\) |
default | \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {2 a^{2} \cos \left (d x +c \right )^{3}}{3}+a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(87\) |
norman | \(\frac {\frac {5 a^{2} x}{8}-\frac {4 a^{2}}{3 d}+\frac {3 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {11 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}-\frac {11 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}-\frac {3 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {5 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {15 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {5 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}+\frac {5 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\) | \(231\) |
orering | \(\text {Expression too large to display}\) | \(1044\) |
Input:
int(cos(d*x+c)^2*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-1/96*a^2*(-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
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.76 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {16 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, a^{2} d x + 3 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{3} - 5 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="fricas")
Output:
-1/24*(16*a^2*cos(d*x + c)^3 - 15*a^2*d*x + 3*(2*a^2*cos(d*x + c)^3 - 5*a^ 2*cos(d*x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (71) = 142\).
Time = 0.19 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.31 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {2 a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**2*(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((a**2*x*sin(c + d*x)**4/8 + a**2*x*sin(c + d*x)**2*cos(c + d*x)* *2/4 + a**2*x*sin(c + d*x)**2/2 + a**2*x*cos(c + d*x)**4/8 + a**2*x*cos(c + d*x)**2/2 + a**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) - a**2*sin(c + d*x)* cos(c + d*x)**3/(8*d) + a**2*sin(c + d*x)*cos(c + d*x)/(2*d) - 2*a**2*cos( c + d*x)**3/(3*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*cos(c)**2, True))
Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.83 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {64 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{96 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxima")
Output:
-1/96*(64*a^2*cos(d*x + c)^3 - 3*(4*d*x + 4*c - sin(4*d*x + 4*c))*a^2 - 24 *(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2)/d
Time = 0.14 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.92 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {5}{8} \, a^{2} x - \frac {a^{2} \cos \left (3 \, d x + 3 \, c\right )}{6 \, d} - \frac {a^{2} \cos \left (d x + c\right )}{2 \, d} - \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
5/8*a^2*x - 1/6*a^2*cos(3*d*x + 3*c)/d - 1/2*a^2*cos(d*x + c)/d - 1/32*a^2 *sin(4*d*x + 4*c)/d + 1/4*a^2*sin(2*d*x + 2*c)/d
Time = 28.07 (sec) , antiderivative size = 237, normalized size of antiderivative = 3.04 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {5\,a^2\,x}{8}-\frac {\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {a^2\,\left (15\,c+15\,d\,x\right )}{24}-\frac {a^2\,\left (15\,c+15\,d\,x-32\right )}{24}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (15\,c+15\,d\,x\right )}{6}-\frac {a^2\,\left (60\,c+60\,d\,x-32\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^2\,\left (15\,c+15\,d\,x\right )}{6}-\frac {a^2\,\left (60\,c+60\,d\,x-96\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2\,\left (15\,c+15\,d\,x\right )}{4}-\frac {a^2\,\left (90\,c+90\,d\,x-96\right )}{24}\right )-\frac {3\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \] Input:
int(cos(c + d*x)^2*(a + a*sin(c + d*x))^2,x)
Output:
(5*a^2*x)/8 - ((11*a^2*tan(c/2 + (d*x)/2)^5)/4 - (11*a^2*tan(c/2 + (d*x)/2 )^3)/4 + (3*a^2*tan(c/2 + (d*x)/2)^7)/4 + (a^2*(15*c + 15*d*x))/24 - (a^2* (15*c + 15*d*x - 32))/24 + tan(c/2 + (d*x)/2)^2*((a^2*(15*c + 15*d*x))/6 - (a^2*(60*c + 60*d*x - 32))/24) + tan(c/2 + (d*x)/2)^6*((a^2*(15*c + 15*d* x))/6 - (a^2*(60*c + 60*d*x - 96))/24) + tan(c/2 + (d*x)/2)^4*((a^2*(15*c + 15*d*x))/4 - (a^2*(90*c + 90*d*x - 96))/24) - (3*a^2*tan(c/2 + (d*x)/2)) /4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^4)
Time = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.87 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+9 \cos \left (d x +c \right ) \sin \left (d x +c \right )-16 \cos \left (d x +c \right )+15 d x +16\right )}{24 d} \] Input:
int(cos(d*x+c)^2*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*(6*cos(c + d*x)*sin(c + d*x)**3 + 16*cos(c + d*x)*sin(c + d*x)**2 + 9*cos(c + d*x)*sin(c + d*x) - 16*cos(c + d*x) + 15*d*x + 16))/(24*d)