Integrand size = 21, antiderivative size = 86 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {1}{8} \left (4 a^2+b^2\right ) x-\frac {5 a b \cos ^3(c+d x)}{12 d}+\frac {\left (4 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d} \]
1/8*(4*a^2+b^2)*x-5/12*a*b*cos(d*x+c)^3/d+1/8*(4*a^2+b^2)*cos(d*x+c)*sin(d *x+c)/d-1/4*b*cos(d*x+c)^3*(a+b*sin(d*x+c))/d
Time = 0.40 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-48 a b \cos (c+d x)-16 a b \cos (3 (c+d x))+3 \left (16 a^2 c+4 b^2 c+16 a^2 d x+4 b^2 d x+8 a^2 \sin (2 (c+d x))-b^2 \sin (4 (c+d x))\right )}{96 d} \]
(-48*a*b*Cos[c + d*x] - 16*a*b*Cos[3*(c + d*x)] + 3*(16*a^2*c + 4*b^2*c + 16*a^2*d*x + 4*b^2*d*x + 8*a^2*Sin[2*(c + d*x)] - b^2*Sin[4*(c + d*x)]))/( 96*d)
Time = 0.40 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3171, 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.
\(\displaystyle \int \cos ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^2 (a+b \sin (c+d x))^2dx\) |
\(\Big \downarrow \) 3171 |
\(\displaystyle \frac {1}{4} \int \cos ^2(c+d x) \left (4 a^2+5 b \sin (c+d x) a+b^2\right )dx-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \int \cos (c+d x)^2 \left (4 a^2+5 b \sin (c+d x) a+b^2\right )dx-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\) |
\(\Big \downarrow \) 3148 |
\(\displaystyle \frac {1}{4} \left (\left (4 a^2+b^2\right ) \int \cos ^2(c+d x)dx-\frac {5 a b \cos ^3(c+d x)}{3 d}\right )-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (\left (4 a^2+b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {5 a b \cos ^3(c+d x)}{3 d}\right )-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{4} \left (\left (4 a^2+b^2\right ) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {5 a b \cos ^3(c+d x)}{3 d}\right )-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{4} \left (\left (4 a^2+b^2\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {5 a b \cos ^3(c+d x)}{3 d}\right )-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\) |
-1/4*(b*Cos[c + d*x]^3*(a + b*Sin[c + d*x]))/d + ((-5*a*b*Cos[c + d*x]^3)/ (3*d) + (4*a^2 + b^2)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4
3.4.95.3.1 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[1/(m + p) Int[(g*Cos[e + f*x])^p* (a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1) *Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ[m])
Time = 0.75 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {48 a^{2} d x +12 b^{2} d x -48 \cos \left (d x +c \right ) a b -3 \sin \left (4 d x +4 c \right ) b^{2}-16 a b \cos \left (3 d x +3 c \right )+24 \sin \left (2 d x +2 c \right ) a^{2}-64 a b}{96 d}\) | \(76\) |
risch | \(\frac {a^{2} x}{2}+\frac {b^{2} x}{8}-\frac {a b \cos \left (d x +c \right )}{2 d}-\frac {\sin \left (4 d x +4 c \right ) b^{2}}{32 d}-\frac {a b \cos \left (3 d x +3 c \right )}{6 d}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{4 d}\) | \(77\) |
derivativedivides | \(\frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {2 a b \left (\cos ^{3}\left (d x +c \right )\right )}{3}+b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )}{d}\) | \(86\) |
default | \(\frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {2 a b \left (\cos ^{3}\left (d x +c \right )\right )}{3}+b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )}{d}\) | \(86\) |
norman | \(\frac {\left (\frac {a^{2}}{2}+\frac {b^{2}}{8}\right ) x +\left (2 a^{2}+\frac {b^{2}}{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 a^{2}+\frac {b^{2}}{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a^{2}+\frac {3 b^{2}}{4}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {a^{2}}{2}+\frac {b^{2}}{8}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 a b}{3 d}+\frac {\left (4 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (4 a^{2}-b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (4 a^{2}+7 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {\left (4 a^{2}+7 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {4 a b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 a b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\) | \(294\) |
1/96*(48*a^2*d*x+12*b^2*d*x-48*cos(d*x+c)*a*b-3*sin(4*d*x+4*c)*b^2-16*a*b* cos(3*d*x+3*c)+24*sin(2*d*x+2*c)*a^2-64*a*b)/d
Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.81 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {16 \, a b \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, a^{2} + b^{2}\right )} d x + 3 \, {\left (2 \, b^{2} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]
-1/24*(16*a*b*cos(d*x + c)^3 - 3*(4*a^2 + b^2)*d*x + 3*(2*b^2*cos(d*x + c) ^3 - (4*a^2 + b^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. \(2 (76) = 152\).
Time = 0.17 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.09 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {2 a b \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((a**2*x*sin(c + d*x)**2/2 + a**2*x*cos(c + d*x)**2/2 + a**2*sin( c + d*x)*cos(c + d*x)/(2*d) - 2*a*b*cos(c + d*x)**3/(3*d) + b**2*x*sin(c + d*x)**4/8 + b**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + b**2*x*cos(c + d*x )**4/8 + b**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) - b**2*sin(c + d*x)*cos(c + d*x)**3/(8*d), Ne(d, 0)), (x*(a + b*sin(c))**2*cos(c)**2, True))
Time = 0.18 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {64 \, a b \cos \left (d x + c\right )^{3} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 3 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} b^{2}}{96 \, d} \]
-1/96*(64*a*b*cos(d*x + c)^3 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2 - 3 *(4*d*x + 4*c - sin(4*d*x + 4*c))*b^2)/d
Time = 0.31 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.88 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {1}{8} \, {\left (4 \, a^{2} + b^{2}\right )} x - \frac {a b \cos \left (3 \, d x + 3 \, c\right )}{6 \, d} - \frac {a b \cos \left (d x + c\right )}{2 \, d} - \frac {b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]
1/8*(4*a^2 + b^2)*x - 1/6*a*b*cos(3*d*x + 3*c)/d - 1/2*a*b*cos(d*x + c)/d - 1/32*b^2*sin(4*d*x + 4*c)/d + 1/4*a^2*sin(2*d*x + 2*c)/d
Time = 5.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {6\,a^2\,\sin \left (2\,c+2\,d\,x\right )-\frac {3\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{4}-12\,a\,b\,\cos \left (c+d\,x\right )-4\,a\,b\,\cos \left (3\,c+3\,d\,x\right )+12\,a^2\,d\,x+3\,b^2\,d\,x}{24\,d} \]