\(\int (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x))^2 \, dx\) [284]

Optimal result

Integrand size = 30, antiderivative size = 116 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=\frac {3}{2} \left (b^2+c^2\right ) x-\frac {3 c \sqrt {b^2+c^2} \cos (d+e x)}{2 e}+\frac {3 b \sqrt {b^2+c^2} \sin (d+e x)}{2 e}-\frac {(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{2 e} \] Output:

3/2*(b^2+c^2)*x-3/2*c*(b^2+c^2)^(1/2)*cos(e*x+d)/e+3/2*b*(b^2+c^2)^(1/2)*s 
in(e*x+d)/e-1/2*(c*cos(e*x+d)-b*sin(e*x+d))*((b^2+c^2)^(1/2)+b*cos(e*x+d)+ 
c*sin(e*x+d))/e
 

Mathematica [A] (verified)

Time = 1.22 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.96 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=\frac {6 b^2 d+6 c^2 d+6 b^2 e x+6 c^2 e x-8 c \sqrt {b^2+c^2} \cos (d+e x)-2 b c \cos (2 (d+e x))+8 b \sqrt {b^2+c^2} \sin (d+e x)+b^2 \sin (2 (d+e x))-c^2 \sin (2 (d+e x))}{4 e} \] Input:

Integrate[(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^2,x]
 

Output:

(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)
 

Rubi [A] (verified)

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

\(\displaystyle \frac {3}{2} \sqrt {b^2+c^2} \left (x \sqrt {b^2+c^2}+\frac {b \sin (d+e x)}{e}-\frac {c \cos (d+e x)}{e}\right )-\frac {(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{2 e}\)

Input:

Int[(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^2,x]
 

Output:

-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
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3592

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]
 

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91

method	result	size
risch	\(\frac {3 b^{2} x}{2}+\frac {3 x \,c^{2}}{2}-\frac {2 c \sqrt {b^{2}+c^{2}}\, \cos \left (e x +d \right )}{e}+\frac {2 b \sqrt {b^{2}+c^{2}}\, \sin \left (e x +d \right )}{e}-\frac {b c \cos \left (2 e x +2 d \right )}{2 e}+\frac {\sin \left (2 e x +2 d \right ) b^{2}}{4 e}-\frac {\sin \left (2 e x +2 d \right ) c^{2}}{4 e}\)	\(106\)
derivativedivides	\(\frac {b^{2} \left (\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-c b \cos \left (e x +d \right )^{2}+c^{2} \left (-\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+2 \sqrt {b^{2}+c^{2}}\, b \sin \left (e x +d \right )-2 \sqrt {b^{2}+c^{2}}\, c \cos \left (e x +d \right )+b^{2} \left (e x +d \right )+c^{2} \left (e x +d \right )}{e}\)	\(124\)
default	\(\frac {b^{2} \left (\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-c b \cos \left (e x +d \right )^{2}+c^{2} \left (-\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+2 \sqrt {b^{2}+c^{2}}\, b \sin \left (e x +d \right )-2 \sqrt {b^{2}+c^{2}}\, c \cos \left (e x +d \right )+b^{2} \left (e x +d \right )+c^{2} \left (e x +d \right )}{e}\)	\(124\)
parts	\(b^{2} x +x \,c^{2}+\frac {2 b \left (\frac {c \sin \left (e x +d \right )^{2}}{2}+\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d \right )\right )}{e}+\frac {b^{2} \left (\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )}{e}+\frac {c^{2} \left (-\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )}{e}-\frac {2 c \sqrt {b^{2}+c^{2}}\, \cos \left (e x +d \right )}{e}\)	\(125\)
norman	\(\frac {\left (\frac {3 b^{2}}{2}+\frac {3 c^{2}}{2}\right ) x -\frac {4 c \sqrt {b^{2}+c^{2}}}{e}+\left (\frac {3 b^{2}}{2}+\frac {3 c^{2}}{2}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}+\frac {\left (4 \sqrt {b^{2}+c^{2}}\, b +b^{2}-c^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e}+\left (3 b^{2}+3 c^{2}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}-\frac {\left (b^{2}-c^{2}-4 \sqrt {b^{2}+c^{2}}\, b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{e}+\frac {2 \left (-2 \sqrt {b^{2}+c^{2}}\, c +2 c b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{e}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}\)	\(198\)
orering	\(x \left (\sqrt {b^{2}+c^{2}}+b \cos \left (e x +d \right )+c \sin \left (e x +d \right )\right )^{2}-\frac {5 \left (\sqrt {b^{2}+c^{2}}+b \cos \left (e x +d \right )+c \sin \left (e x +d \right )\right ) \left (-b e \sin \left (e x +d \right )+c e \cos \left (e x +d \right )\right )}{2 e^{2}}+\frac {5 x \left (2 \left (-b e \sin \left (e x +d \right )+c e \cos \left (e x +d \right )\right )^{2}+2 \left (\sqrt {b^{2}+c^{2}}+b \cos \left (e x +d \right )+c \sin \left (e x +d \right )\right ) \left (-b \,e^{2} \cos \left (e x +d \right )-c \,e^{2} \sin \left (e x +d \right )\right )\right )}{4 e^{2}}-\frac {6 \left (-b e \sin \left (e x +d \right )+c e \cos \left (e x +d \right )\right ) \left (-b \,e^{2} \cos \left (e x +d \right )-c \,e^{2} \sin \left (e x +d \right )\right )+2 \left (\sqrt {b^{2}+c^{2}}+b \cos \left (e x +d \right )+c \sin \left (e x +d \right )\right ) \left (b \,e^{3} \sin \left (e x +d \right )-c \,e^{3} \cos \left (e x +d \right )\right )}{4 e^{4}}+\frac {x \left (6 \left (-b \,e^{2} \cos \left (e x +d \right )-c \,e^{2} \sin \left (e x +d \right )\right )^{2}+8 \left (-b e \sin \left (e x +d \right )+c e \cos \left (e x +d \right )\right ) \left (b \,e^{3} \sin \left (e x +d \right )-c \,e^{3} \cos \left (e x +d \right )\right )+2 \left (\sqrt {b^{2}+c^{2}}+b \cos \left (e x +d \right )+c \sin \left (e x +d \right )\right ) \left (b \,e^{4} \cos \left (e x +d \right )+c \,e^{4} \sin \left (e x +d \right )\right )\right )}{4 e^{4}}\)	\(405\)

Input:

int(((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^2,x,method=_RETURNVERBOSE)
 

Output:

3/2*b^2*x+3/2*x*c^2-2*c*(b^2+c^2)^(1/2)*cos(e*x+d)/e+2*b*(b^2+c^2)^(1/2)*s 
in(e*x+d)/e-1/2*b*c/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
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=-\frac {2 \, b c \cos \left (e x + d\right )^{2} - 3 \, {\left (b^{2} + c^{2}\right )} e x - {\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right ) + 4 \, \sqrt {b^{2} + c^{2}} {\left (c \cos \left (e x + d\right ) - b \sin \left (e x + d\right )\right )}}{2 \, e} \] Input:

integrate(((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^2,x, algorithm="fric 
as")
 

Output:

-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
 

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.66 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=\begin {cases} \frac {b^{2} x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {b^{2} x \cos ^{2}{\left (d + e x \right )}}{2} + b^{2} x + \frac {b^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} - \frac {b c \cos ^{2}{\left (d + e x \right )}}{e} + \frac {2 b \sqrt {b^{2} + c^{2}} \sin {\left (d + e x \right )}}{e} + \frac {c^{2} x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {c^{2} x \cos ^{2}{\left (d + e x \right )}}{2} + c^{2} x - \frac {c^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} - \frac {2 c \sqrt {b^{2} + c^{2}} \cos {\left (d + e x \right )}}{e} & \text {for}\: e \neq 0 \\x \left (b \cos {\left (d \right )} + c \sin {\left (d \right )} + \sqrt {b^{2} + c^{2}}\right )^{2} & \text {otherwise} \end {cases} \] Input:

integrate(((b**2+c**2)**(1/2)+b*cos(e*x+d)+c*sin(e*x+d))**2,x)
 

Output:

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*cos(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))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.97 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=b^{2} x + c^{2} x - \frac {b c \cos \left (e x + d\right )^{2}}{e} + \frac {{\left (2 \, e x + 2 \, d + \sin \left (2 \, e x + 2 \, d\right )\right )} b^{2}}{4 \, e} + \frac {{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} c^{2}}{4 \, e} - 2 \, \sqrt {b^{2} + c^{2}} {\left (\frac {c \cos \left (e x + d\right )}{e} - \frac {b \sin \left (e x + d\right )}{e}\right )} \] Input:

integrate(((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^2,x, algorithm="maxi 
ma")
 

Output:

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)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.79 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=\frac {3}{2} \, {\left (b^{2} + c^{2}\right )} x - \frac {b c \cos \left (2 \, e x + 2 \, d\right )}{2 \, e} - \frac {2 \, \sqrt {b^{2} + c^{2}} c \cos \left (e x + d\right )}{e} + \frac {2 \, \sqrt {b^{2} + c^{2}} b \sin \left (e x + d\right )}{e} + \frac {{\left (b^{2} - c^{2}\right )} \sin \left (2 \, e x + 2 \, d\right )}{4 \, e} \] Input:

integrate(((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^2,x, algorithm="giac 
")
 

Output:

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
 

Mupad [B] (verification not implemented)

Time = 15.78 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.86 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=\frac {b^2\,\sin \left (2\,d+2\,e\,x\right )-c^2\,\sin \left (2\,d+2\,e\,x\right )+16\,c\,{\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\sqrt {b^2+c^2}+8\,b\,\sin \left (d+e\,x\right )\,\sqrt {b^2+c^2}+4\,b\,c\,{\sin \left (d+e\,x\right )}^2+6\,b^2\,e\,x+6\,c^2\,e\,x}{4\,e} \] Input:

int((b*cos(d + e*x) + c*sin(d + e*x) + (b^2 + c^2)^(1/2))^2,x)
 

Output:

(b^2*sin(2*d + 2*e*x) - c^2*sin(2*d + 2*e*x) + 16*c*sin(d/2 + (e*x)/2)^2*( 
b^2 + c^2)^(1/2) + 8*b*sin(d + e*x)*(b^2 + c^2)^(1/2) + 4*b*c*sin(d + e*x) 
^2 + 6*b^2*e*x + 6*c^2*e*x)/(4*e)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.34 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx=\frac {\cos \left (e x +d \right )^{2} b^{2} e x -2 \cos \left (e x +d \right )^{2} b c +\cos \left (e x +d \right )^{2} c^{2} e x -4 \sqrt {b^{2}+c^{2}}\, \cos \left (e x +d \right ) c +\cos \left (e x +d \right ) \sin \left (e x +d \right ) b^{2}-\cos \left (e x +d \right ) \sin \left (e x +d \right ) c^{2}+4 \sqrt {b^{2}+c^{2}}\, \sin \left (e x +d \right ) b +\sin \left (e x +d \right )^{2} b^{2} e x +\sin \left (e x +d \right )^{2} c^{2} e x +2 b^{2} e x +2 c^{2} e x}{2 e} \] Input:

int(((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^2,x)
 

Output:

(cos(d + e*x)**2*b**2*e*x - 2*cos(d + e*x)**2*b*c + cos(d + e*x)**2*c**2*e 
*x - 4*sqrt(b**2 + c**2)*cos(d + e*x)*c + cos(d + e*x)*sin(d + e*x)*b**2 - 
 cos(d + e*x)*sin(d + e*x)*c**2 + 4*sqrt(b**2 + c**2)*sin(d + e*x)*b + sin 
(d + e*x)**2*b**2*e*x + sin(d + e*x)**2*c**2*e*x + 2*b**2*e*x + 2*c**2*e*x 
)/(2*e)