Integrand size = 24, antiderivative size = 81 \[ \int (2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2 \, dx=2 \left (3 a^2+b^2\right ) x-\frac {6 a^2 \cos (d+e x)}{e}+\frac {6 a b \sin (d+e x)}{e}-\frac {2 (a+b \cos (d+e x)+a \sin (d+e x)) (a \cos (d+e x)-b \sin (d+e x))}{e} \]
2*(3*a^2+b^2)*x-6*a^2*cos(e*x+d)/e+6*a*b*sin(e*x+d)/e-2*(a+b*cos(e*x+d)+a* sin(e*x+d))*(a*cos(e*x+d)-b*sin(e*x+d))/e
Time = 0.48 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.14 \[ \int (2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2 \, dx=4 \left (\frac {\left (3 a^2+b^2\right ) (d+e x)}{2 e}-\frac {2 a^2 \cos (d+e x)}{e}-\frac {a b \cos (2 (d+e x))}{2 e}+\frac {2 a b \sin (d+e x)}{e}-\frac {\left (a^2-b^2\right ) \sin (2 (d+e x))}{4 e}\right ) \]
4*(((3*a^2 + b^2)*(d + e*x))/(2*e) - (2*a^2*Cos[d + e*x])/e - (a*b*Cos[2*( d + e*x)])/(2*e) + (2*a*b*Sin[d + e*x])/e - ((a^2 - b^2)*Sin[2*(d + e*x)]) /(4*e))
Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3042, 3599, 2009}
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.
\(\displaystyle \int (2 a \sin (d+e x)+2 a+2 b \cos (d+e x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (2 a \sin (d+e x)+2 a+2 b \cos (d+e x))^2dx\) |
\(\Big \downarrow \) 3599 |
\(\displaystyle \frac {1}{2} \int \left (12 \sin (d+e x) a^2+12 b \cos (d+e x) a+4 \left (3 a^2+b^2\right )\right )dx-\frac {2 (a \sin (d+e x)+a+b \cos (d+e x)) (a \cos (d+e x)-b \sin (d+e x))}{e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (4 x \left (3 a^2+b^2\right )-\frac {12 a^2 \cos (d+e x)}{e}+\frac {12 a b \sin (d+e x)}{e}\right )-\frac {2 (a \sin (d+e x)+a+b \cos (d+e x)) (a \cos (d+e x)-b \sin (d+e x))}{e}\) |
(-2*(a + b*Cos[d + e*x] + a*Sin[d + e*x])*(a*Cos[d + e*x] - b*Sin[d + e*x] ))/e + (4*(3*a^2 + b^2)*x - (12*a^2*Cos[d + e*x])/e + (12*a*b*Sin[d + e*x] )/e)/2
3.4.82.3.1 Defintions of rubi rules used
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ (n_), x_Symbol] :> Simp[(-(c*Cos[d + e*x] - b*Sin[d + e*x]))*((a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1)/(e*n)), x] + Simp[1/n Int[Simp[n*a^2 + ( n - 1)*(b^2 + c^2) + a*b*(2*n - 1)*Cos[d + e*x] + a*c*(2*n - 1)*Sin[d + e*x ], x]*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && GtQ[n, 1]
Time = 0.82 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99
method | result | size |
parallelrisch | \(\frac {\left (-a^{2}+b^{2}\right ) \sin \left (2 e x +2 d \right )-2 a b \cos \left (2 e x +2 d \right )-8 a^{2} \cos \left (e x +d \right )+8 a b \sin \left (e x +d \right )+\left (6 e x +8\right ) a^{2}+2 a b +2 b^{2} e x}{e}\) | \(80\) |
risch | \(6 a^{2} x +2 x \,b^{2}-\frac {8 a^{2} \cos \left (e x +d \right )}{e}+\frac {8 a b \sin \left (e x +d \right )}{e}-\frac {2 a b \cos \left (2 e x +2 d \right )}{e}-\frac {\sin \left (2 e x +2 d \right ) a^{2}}{e}+\frac {\sin \left (2 e x +2 d \right ) b^{2}}{e}\) | \(90\) |
derivativedivides | \(\frac {4 b^{2} \left (\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-4 a b \cos \left (e x +d \right )^{2}+8 a b \sin \left (e x +d \right )+4 a^{2} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-8 a^{2} \cos \left (e x +d \right )+4 a^{2} \left (e x +d \right )}{e}\) | \(101\) |
default | \(\frac {4 b^{2} \left (\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-4 a b \cos \left (e x +d \right )^{2}+8 a b \sin \left (e x +d \right )+4 a^{2} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-8 a^{2} \cos \left (e x +d \right )+4 a^{2} \left (e x +d \right )}{e}\) | \(101\) |
parts | \(\frac {8 a b \left (\frac {\sin \left (e x +d \right )^{2}}{2}+\sin \left (e x +d \right )\right )}{e}+4 a^{2} x -\frac {8 a^{2} \cos \left (e x +d \right )}{e}+\frac {4 a^{2} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )}{e}+\frac {4 b^{2} \left (\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )}{e}\) | \(106\) |
norman | \(\frac {\left (6 a^{2}+2 b^{2}\right ) x +\left (6 a^{2}+2 b^{2}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}+\left (12 a^{2}+4 b^{2}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+\frac {16 a^{2} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{e}-\frac {4 \left (a^{2}-4 a b -b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e}+\frac {4 \left (a^{2}+4 a b -b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{e}+\frac {2 \left (8 a^{2}+8 a b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{e}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}\) | \(180\) |
((-a^2+b^2)*sin(2*e*x+2*d)-2*a*b*cos(2*e*x+2*d)-8*a^2*cos(e*x+d)+8*a*b*sin (e*x+d)+(6*e*x+8)*a^2+2*a*b+2*b^2*e*x)/e
Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89 \[ \int (2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2 \, dx=-\frac {2 \, {\left (2 \, a b \cos \left (e x + d\right )^{2} - {\left (3 \, a^{2} + b^{2}\right )} e x + 4 \, a^{2} \cos \left (e x + d\right ) - {\left (4 \, a b - {\left (a^{2} - b^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )}}{e} \]
-2*(2*a*b*cos(e*x + d)^2 - (3*a^2 + b^2)*e*x + 4*a^2*cos(e*x + d) - (4*a*b - (a^2 - b^2)*cos(e*x + d))*sin(e*x + d))/e
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (80) = 160\).
Time = 0.11 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.10 \[ \int (2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2 \, dx=\begin {cases} 2 a^{2} x \sin ^{2}{\left (d + e x \right )} + 2 a^{2} x \cos ^{2}{\left (d + e x \right )} + 4 a^{2} x - \frac {2 a^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} - \frac {8 a^{2} \cos {\left (d + e x \right )}}{e} + \frac {4 a b \sin ^{2}{\left (d + e x \right )}}{e} + \frac {8 a b \sin {\left (d + e x \right )}}{e} + 2 b^{2} x \sin ^{2}{\left (d + e x \right )} + 2 b^{2} x \cos ^{2}{\left (d + e x \right )} + \frac {2 b^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} & \text {for}\: e \neq 0 \\x \left (2 a \sin {\left (d \right )} + 2 a + 2 b \cos {\left (d \right )}\right )^{2} & \text {otherwise} \end {cases} \]
Piecewise((2*a**2*x*sin(d + e*x)**2 + 2*a**2*x*cos(d + e*x)**2 + 4*a**2*x - 2*a**2*sin(d + e*x)*cos(d + e*x)/e - 8*a**2*cos(d + e*x)/e + 4*a*b*sin(d + e*x)**2/e + 8*a*b*sin(d + e*x)/e + 2*b**2*x*sin(d + e*x)**2 + 2*b**2*x* cos(d + e*x)**2 + 2*b**2*sin(d + e*x)*cos(d + e*x)/e, Ne(e, 0)), (x*(2*a*s in(d) + 2*a + 2*b*cos(d))**2, True))
Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.22 \[ \int (2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2 \, dx=4 \, a^{2} x - \frac {4 \, a b \cos \left (e x + d\right )^{2}}{e} + \frac {{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} a^{2}}{e} + \frac {{\left (2 \, e x + 2 \, d + \sin \left (2 \, e x + 2 \, d\right )\right )} b^{2}}{e} - 8 \, a {\left (\frac {a \cos \left (e x + d\right )}{e} - \frac {b \sin \left (e x + d\right )}{e}\right )} \]
4*a^2*x - 4*a*b*cos(e*x + d)^2/e + (2*e*x + 2*d - sin(2*e*x + 2*d))*a^2/e + (2*e*x + 2*d + sin(2*e*x + 2*d))*b^2/e - 8*a*(a*cos(e*x + d)/e - b*sin(e *x + d)/e)
Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98 \[ \int (2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2 \, dx=2 \, {\left (3 \, a^{2} + b^{2}\right )} x - \frac {2 \, a b \cos \left (2 \, e x + 2 \, d\right )}{e} - \frac {8 \, a^{2} \cos \left (e x + d\right )}{e} + \frac {8 \, a b \sin \left (e x + d\right )}{e} - \frac {{\left (a^{2} - b^{2}\right )} \sin \left (2 \, e x + 2 \, d\right )}{e} \]
2*(3*a^2 + b^2)*x - 2*a*b*cos(2*e*x + 2*d)/e - 8*a^2*cos(e*x + d)/e + 8*a* b*sin(e*x + d)/e - (a^2 - b^2)*sin(2*e*x + 2*d)/e
Time = 27.88 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.57 \[ \int (2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^2 \, dx=\frac {x\,\left (12\,a^2+4\,b^2\right )}{2}+\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (16\,a\,b-16\,a^2\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (4\,a^2+16\,a\,b-4\,b^2\right )-16\,a^2+\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (-4\,a^2+16\,a\,b+4\,b^2\right )}{e\,\left ({\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+1\right )} \]