Integrand size = 30, antiderivative size = 116 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=\frac {3}{2} \left (b^2+c^2\right ) x-\frac {3 c \sqrt {b^2+c^2} \cos (d+e x)}{2 e}+\frac {3 b \sqrt {b^2+c^2} \sin (d+e x)}{2 e}-\frac {(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{2 e} \]
3/2*(b^2+c^2)*x-3/2*c*cos(e*x+d)*(b^2+c^2)^(1/2)/e+3/2*b*sin(e*x+d)*(b^2+c ^2)^(1/2)/e-1/2*(c*cos(e*x+d)-b*sin(e*x+d))*(b*cos(e*x+d)+c*sin(e*x+d)+(b^ 2+c^2)^(1/2))/e
Time = 1.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.96 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=\frac {6 b^2 d+6 c^2 d+6 b^2 e x+6 c^2 e x-8 c \sqrt {b^2+c^2} \cos (d+e x)-2 b c \cos (2 (d+e x))+8 b \sqrt {b^2+c^2} \sin (d+e x)+b^2 \sin (2 (d+e x))-c^2 \sin (2 (d+e x))}{4 e} \]
(6*b^2*d + 6*c^2*d + 6*b^2*e*x + 6*c^2*e*x - 8*c*Sqrt[b^2 + c^2]*Cos[d + e *x] - 2*b*c*Cos[2*(d + e*x)] + 8*b*Sqrt[b^2 + c^2]*Sin[d + e*x] + b^2*Sin[ 2*(d + e*x)] - c^2*Sin[2*(d + e*x)])/(4*e)
Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3042, 3592, 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.
\(\displaystyle \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2dx\) |
\(\Big \downarrow \) 3592 |
\(\displaystyle \frac {3}{2} \sqrt {b^2+c^2} \int \left (b \cos (d+e x)+c \sin (d+e x)+\sqrt {b^2+c^2}\right )dx-\frac {(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{2 e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{2} \sqrt {b^2+c^2} \left (x \sqrt {b^2+c^2}+\frac {b \sin (d+e x)}{e}-\frac {c \cos (d+e x)}{e}\right )-\frac {(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{2 e}\) |
-1/2*((c*Cos[d + e*x] - b*Sin[d + e*x])*(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x]))/e + (3*Sqrt[b^2 + c^2]*(Sqrt[b^2 + c^2]*x - (c*Cos[d + e*x])/e + (b*Sin[d + e*x])/e))/2
3.4.57.3.1 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[a*((2*n - 1)/n) Int[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e }, x] && EqQ[a^2 - b^2 - c^2, 0] && GtQ[n, 0]
Time = 1.02 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\frac {3 x \,b^{2}}{2}+\frac {3 x \,c^{2}}{2}-\frac {2 c \cos \left (e x +d \right ) \sqrt {b^{2}+c^{2}}}{e}+\frac {2 b \sin \left (e x +d \right ) \sqrt {b^{2}+c^{2}}}{e}-\frac {c b \cos \left (2 e x +2 d \right )}{2 e}+\frac {\sin \left (2 e x +2 d \right ) b^{2}}{4 e}-\frac {\sin \left (2 e x +2 d \right ) c^{2}}{4 e}\) | \(106\) |
derivativedivides | \(\frac {b^{2} \left (\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-c b \cos \left (e x +d \right )^{2}+c^{2} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+2 \sqrt {b^{2}+c^{2}}\, b \sin \left (e x +d \right )-2 \sqrt {b^{2}+c^{2}}\, c \cos \left (e x +d \right )+b^{2} \left (e x +d \right )+c^{2} \left (e x +d \right )}{e}\) | \(124\) |
default | \(\frac {b^{2} \left (\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-c b \cos \left (e x +d \right )^{2}+c^{2} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+2 \sqrt {b^{2}+c^{2}}\, b \sin \left (e x +d \right )-2 \sqrt {b^{2}+c^{2}}\, c \cos \left (e x +d \right )+b^{2} \left (e x +d \right )+c^{2} \left (e x +d \right )}{e}\) | \(124\) |
parts | \(x \,b^{2}+x \,c^{2}+\frac {2 b \left (\frac {\sin \left (e x +d \right )^{2} c}{2}+\sin \left (e x +d \right ) \sqrt {b^{2}+c^{2}}\right )}{e}+\frac {b^{2} \left (\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )}{e}+\frac {c^{2} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )}{e}-\frac {2 c \cos \left (e x +d \right ) \sqrt {b^{2}+c^{2}}}{e}\) | \(125\) |
norman | \(\frac {\left (\frac {3 b^{2}}{2}+\frac {3 c^{2}}{2}\right ) x -\frac {4 c \sqrt {b^{2}+c^{2}}}{e}+\left (\frac {3 b^{2}}{2}+\frac {3 c^{2}}{2}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}+\frac {\left (4 \sqrt {b^{2}+c^{2}}\, b +b^{2}-c^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e}+\left (3 b^{2}+3 c^{2}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}-\frac {\left (b^{2}-c^{2}-4 \sqrt {b^{2}+c^{2}}\, b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{e}+\frac {2 \left (-2 \sqrt {b^{2}+c^{2}}\, c +2 c b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{e}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}\) | \(198\) |
3/2*x*b^2+3/2*x*c^2-2*c*cos(e*x+d)*(b^2+c^2)^(1/2)/e+2*b*sin(e*x+d)*(b^2+c ^2)^(1/2)/e-1/2*c*b/e*cos(2*e*x+2*d)+1/4/e*sin(2*e*x+2*d)*b^2-1/4/e*sin(2* e*x+2*d)*c^2
Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=-\frac {2 \, b c \cos \left (e x + d\right )^{2} - 3 \, {\left (b^{2} + c^{2}\right )} e x - {\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right ) + 4 \, \sqrt {b^{2} + c^{2}} {\left (c \cos \left (e x + d\right ) - b \sin \left (e x + d\right )\right )}}{2 \, e} \]
-1/2*(2*b*c*cos(e*x + d)^2 - 3*(b^2 + c^2)*e*x - (b^2 - c^2)*cos(e*x + d)* sin(e*x + d) + 4*sqrt(b^2 + c^2)*(c*cos(e*x + d) - b*sin(e*x + d)))/e
Time = 0.16 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.66 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=\begin {cases} \frac {b^{2} x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {b^{2} x \cos ^{2}{\left (d + e x \right )}}{2} + b^{2} x + \frac {b^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} + \frac {b c \sin ^{2}{\left (d + e x \right )}}{e} + \frac {2 b \sqrt {b^{2} + c^{2}} \sin {\left (d + e x \right )}}{e} + \frac {c^{2} x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {c^{2} x \cos ^{2}{\left (d + e x \right )}}{2} + c^{2} x - \frac {c^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} - \frac {2 c \sqrt {b^{2} + c^{2}} \cos {\left (d + e x \right )}}{e} & \text {for}\: e \neq 0 \\x \left (b \cos {\left (d \right )} + c \sin {\left (d \right )} + \sqrt {b^{2} + c^{2}}\right )^{2} & \text {otherwise} \end {cases} \]
Piecewise((b**2*x*sin(d + e*x)**2/2 + b**2*x*cos(d + e*x)**2/2 + b**2*x + b**2*sin(d + e*x)*cos(d + e*x)/(2*e) + b*c*sin(d + e*x)**2/e + 2*b*sqrt(b* *2 + c**2)*sin(d + e*x)/e + c**2*x*sin(d + e*x)**2/2 + c**2*x*cos(d + e*x) **2/2 + c**2*x - c**2*sin(d + e*x)*cos(d + e*x)/(2*e) - 2*c*sqrt(b**2 + c* *2)*cos(d + e*x)/e, Ne(e, 0)), (x*(b*cos(d) + c*sin(d) + sqrt(b**2 + c**2) )**2, True))
Time = 0.24 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.97 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=b^{2} x + c^{2} x - \frac {b c \cos \left (e x + d\right )^{2}}{e} + \frac {{\left (2 \, e x + 2 \, d + \sin \left (2 \, e x + 2 \, d\right )\right )} b^{2}}{4 \, e} + \frac {{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} c^{2}}{4 \, e} - 2 \, \sqrt {b^{2} + c^{2}} {\left (\frac {c \cos \left (e x + d\right )}{e} - \frac {b \sin \left (e x + d\right )}{e}\right )} \]
b^2*x + c^2*x - b*c*cos(e*x + d)^2/e + 1/4*(2*e*x + 2*d + sin(2*e*x + 2*d) )*b^2/e + 1/4*(2*e*x + 2*d - sin(2*e*x + 2*d))*c^2/e - 2*sqrt(b^2 + c^2)*( c*cos(e*x + d)/e - b*sin(e*x + d)/e)
Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.79 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=\frac {3}{2} \, {\left (b^{2} + c^{2}\right )} x - \frac {b c \cos \left (2 \, e x + 2 \, d\right )}{2 \, e} - \frac {2 \, \sqrt {b^{2} + c^{2}} c \cos \left (e x + d\right )}{e} + \frac {2 \, \sqrt {b^{2} + c^{2}} b \sin \left (e x + d\right )}{e} + \frac {{\left (b^{2} - c^{2}\right )} \sin \left (2 \, e x + 2 \, d\right )}{4 \, e} \]
3/2*(b^2 + c^2)*x - 1/2*b*c*cos(2*e*x + 2*d)/e - 2*sqrt(b^2 + c^2)*c*cos(e *x + d)/e + 2*sqrt(b^2 + c^2)*b*sin(e*x + d)/e + 1/4*(b^2 - c^2)*sin(2*e*x + 2*d)/e
Time = 28.23 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.86 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=\frac {b^2\,\sin \left (2\,d+2\,e\,x\right )-c^2\,\sin \left (2\,d+2\,e\,x\right )+16\,c\,{\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\sqrt {b^2+c^2}+8\,b\,\sin \left (d+e\,x\right )\,\sqrt {b^2+c^2}+4\,b\,c\,{\sin \left (d+e\,x\right )}^2+6\,b^2\,e\,x+6\,c^2\,e\,x}{4\,e} \]