\(\int (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x))^3 \, dx\) [283]

Optimal result

Integrand size = 30, antiderivative size = 178 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^3 \, dx=\frac {5}{2} \left (b^2+c^2\right )^{3/2} x-\frac {5 c \left (b^2+c^2\right ) \cos (d+e x)}{2 e}+\frac {5 b \left (b^2+c^2\right ) \sin (d+e x)}{2 e}-\frac {5 \sqrt {b^2+c^2} (c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{6 e}-\frac {(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}{3 e} \] Output:

5/2*(b^2+c^2)^(3/2)*x-5/2*c*(b^2+c^2)*cos(e*x+d)/e+5/2*b*(b^2+c^2)*sin(e*x 
+d)/e-5/6*(b^2+c^2)^(1/2)*(c*cos(e*x+d)-b*sin(e*x+d))*((b^2+c^2)^(1/2)+b*c 
os(e*x+d)+c*sin(e*x+d))/e-1/3*(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))^2/e
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.53 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.92 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^3 \, dx=\frac {30 (b-i c) (b+i c) \sqrt {b^2+c^2} (d+e x)-45 c \left (b^2+c^2\right ) \cos (d+e x)-18 b c \sqrt {b^2+c^2} \cos (2 (d+e x))+c \left (-3 b^2+c^2\right ) \cos (3 (d+e x))+45 b \left (b^2+c^2\right ) \sin (d+e x)+9 \left (b^2-c^2\right ) \sqrt {b^2+c^2} \sin (2 (d+e x))+b \left (b^2-3 c^2\right ) \sin (3 (d+e x))}{12 e} \] Input:

Integrate[(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^3,x]
 

Output:

(30*(b - I*c)*(b + I*c)*Sqrt[b^2 + c^2]*(d + e*x) - 45*c*(b^2 + c^2)*Cos[d 
 + e*x] - 18*b*c*Sqrt[b^2 + c^2]*Cos[2*(d + e*x)] + c*(-3*b^2 + c^2)*Cos[3 
*(d + e*x)] + 45*b*(b^2 + c^2)*Sin[d + e*x] + 9*(b^2 - c^2)*Sqrt[b^2 + c^2 
]*Sin[2*(d + e*x)] + b*(b^2 - 3*c^2)*Sin[3*(d + e*x)])/(12*e)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3592, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3592

\(\displaystyle \frac {5}{3} \sqrt {b^2+c^2} \int \left (b \cos (d+e x)+c \sin (d+e x)+\sqrt {b^2+c^2}\right )^2dx-\frac {(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}{3 e}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{3} \sqrt {b^2+c^2} \int \left (b \cos (d+e x)+c \sin (d+e x)+\sqrt {b^2+c^2}\right )^2dx-\frac {(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}{3 e}\)

\(\Big \downarrow \) 3592

\(\displaystyle \frac {5}{3} \sqrt {b^2+c^2} \left (\frac {3}{2} \sqrt {b^2+c^2} \int \left (b \cos (d+e x)+c \sin (d+e x)+\sqrt {b^2+c^2}\right )dx-\frac {(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{2 e}\right )-\frac {(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}{3 e}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {5}{3} \sqrt {b^2+c^2} \left (\frac {3}{2} \sqrt {b^2+c^2} \left (x \sqrt {b^2+c^2}+\frac {b \sin (d+e x)}{e}-\frac {c \cos (d+e x)}{e}\right )-\frac {(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{2 e}\right )-\frac {(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}{3 e}\)

Input:

Int[(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^3,x]
 

Output:

-1/3*((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])^2)/e + (5*Sqrt[b^2 + c^2]*(-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*Sqr 
t[b^2 + c^2]*(Sqrt[b^2 + c^2]*x - (c*Cos[d + e*x])/e + (b*Sin[d + e*x])/e) 
)/2))/3
 

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 = 0.80 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.40

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

Input:

int(((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^3,x)
                                                                                    
                                                                                    
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.81 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^3 \, dx=-\frac {2 \, {\left (3 \, b^{2} c - c^{3}\right )} \cos \left (e x + d\right )^{3} + 6 \, {\left (3 \, b^{2} c + 4 \, c^{3}\right )} \cos \left (e x + d\right ) - 2 \, {\left (11 \, b^{3} + 12 \, b c^{2} + {\left (b^{3} - 3 \, b c^{2}\right )} \cos \left (e x + d\right )^{2}\right )} \sin \left (e x + d\right ) + 3 \, {\left (6 \, b c \cos \left (e x + d\right )^{2} - 5 \, {\left (b^{2} + c^{2}\right )} e x - 3 \, {\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right )\right )} \sqrt {b^{2} + c^{2}}}{6 \, e} \] Input:

integrate(((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^3,x, algorithm="fric 
as")
 

Output:

-1/6*(2*(3*b^2*c - c^3)*cos(e*x + d)^3 + 6*(3*b^2*c + 4*c^3)*cos(e*x + d) 
- 2*(11*b^3 + 12*b*c^2 + (b^3 - 3*b*c^2)*cos(e*x + d)^2)*sin(e*x + d) + 3* 
(6*b*c*cos(e*x + d)^2 - 5*(b^2 + c^2)*e*x - 3*(b^2 - c^2)*cos(e*x + d)*sin 
(e*x + d))*sqrt(b^2 + c^2))/e
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 415 vs. \(2 (165) = 330\).

Time = 0.34 (sec) , antiderivative size = 415, normalized size of antiderivative = 2.33 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^3 \, dx=\begin {cases} \frac {2 b^{3} \sin ^{3}{\left (d + e x \right )}}{3 e} + \frac {b^{3} \sin {\left (d + e x \right )} \cos ^{2}{\left (d + e x \right )}}{e} + \frac {3 b^{3} \sin {\left (d + e x \right )}}{e} - \frac {b^{2} c \cos ^{3}{\left (d + e x \right )}}{e} - \frac {3 b^{2} c \cos {\left (d + e x \right )}}{e} + \frac {3 b^{2} x \sqrt {b^{2} + c^{2}} \sin ^{2}{\left (d + e x \right )}}{2} + \frac {3 b^{2} x \sqrt {b^{2} + c^{2}} \cos ^{2}{\left (d + e x \right )}}{2} + b^{2} x \sqrt {b^{2} + c^{2}} + \frac {3 b^{2} \sqrt {b^{2} + c^{2}} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} + \frac {b c^{2} \sin ^{3}{\left (d + e x \right )}}{e} + \frac {3 b c^{2} \sin {\left (d + e x \right )}}{e} - \frac {3 b c \sqrt {b^{2} + c^{2}} \cos ^{2}{\left (d + e x \right )}}{e} - \frac {c^{3} \sin ^{2}{\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} - \frac {2 c^{3} \cos ^{3}{\left (d + e x \right )}}{3 e} - \frac {3 c^{3} \cos {\left (d + e x \right )}}{e} + \frac {3 c^{2} x \sqrt {b^{2} + c^{2}} \sin ^{2}{\left (d + e x \right )}}{2} + \frac {3 c^{2} x \sqrt {b^{2} + c^{2}} \cos ^{2}{\left (d + e x \right )}}{2} + c^{2} x \sqrt {b^{2} + c^{2}} - \frac {3 c^{2} \sqrt {b^{2} + c^{2}} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} & \text {for}\: e \neq 0 \\x \left (b \cos {\left (d \right )} + c \sin {\left (d \right )} + \sqrt {b^{2} + c^{2}}\right )^{3} & \text {otherwise} \end {cases} \] Input:

integrate(((b**2+c**2)**(1/2)+b*cos(e*x+d)+c*sin(e*x+d))**3,x)
 

Output:

Piecewise((2*b**3*sin(d + e*x)**3/(3*e) + b**3*sin(d + e*x)*cos(d + e*x)** 
2/e + 3*b**3*sin(d + e*x)/e - b**2*c*cos(d + e*x)**3/e - 3*b**2*c*cos(d + 
e*x)/e + 3*b**2*x*sqrt(b**2 + c**2)*sin(d + e*x)**2/2 + 3*b**2*x*sqrt(b**2 
 + c**2)*cos(d + e*x)**2/2 + b**2*x*sqrt(b**2 + c**2) + 3*b**2*sqrt(b**2 + 
 c**2)*sin(d + e*x)*cos(d + e*x)/(2*e) + b*c**2*sin(d + e*x)**3/e + 3*b*c* 
*2*sin(d + e*x)/e - 3*b*c*sqrt(b**2 + c**2)*cos(d + e*x)**2/e - c**3*sin(d 
 + e*x)**2*cos(d + e*x)/e - 2*c**3*cos(d + e*x)**3/(3*e) - 3*c**3*cos(d + 
e*x)/e + 3*c**2*x*sqrt(b**2 + c**2)*sin(d + e*x)**2/2 + 3*c**2*x*sqrt(b**2 
 + c**2)*cos(d + e*x)**2/2 + c**2*x*sqrt(b**2 + c**2) - 3*c**2*sqrt(b**2 + 
 c**2)*sin(d + e*x)*cos(d + e*x)/(2*e), Ne(e, 0)), (x*(b*cos(d) + c*sin(d) 
 + sqrt(b**2 + c**2))**3, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.16 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^3 \, dx=-\frac {b^{2} c \cos \left (e x + d\right )^{3}}{e} + \frac {b c^{2} \sin \left (e x + d\right )^{3}}{e} - \frac {{\left (\sin \left (e x + d\right )^{3} - 3 \, \sin \left (e x + d\right )\right )} b^{3}}{3 \, e} + \frac {{\left (\cos \left (e x + d\right )^{3} - 3 \, \cos \left (e x + d\right )\right )} c^{3}}{3 \, e} + {\left (b^{2} + c^{2}\right )}^{\frac {3}{2}} x - 3 \, {\left (b^{2} + c^{2}\right )} {\left (\frac {c \cos \left (e x + d\right )}{e} - \frac {b \sin \left (e x + d\right )}{e}\right )} - \frac {3}{4} \, {\left (\frac {4 \, b c \cos \left (e x + d\right )^{2}}{e} - \frac {{\left (2 \, e x + 2 \, d + \sin \left (2 \, e x + 2 \, d\right )\right )} b^{2}}{e} - \frac {{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} c^{2}}{e}\right )} \sqrt {b^{2} + c^{2}} \] Input:

integrate(((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^3,x, algorithm="maxi 
ma")
 

Output:

-b^2*c*cos(e*x + d)^3/e + b*c^2*sin(e*x + d)^3/e - 1/3*(sin(e*x + d)^3 - 3 
*sin(e*x + d))*b^3/e + 1/3*(cos(e*x + d)^3 - 3*cos(e*x + d))*c^3/e + (b^2 
+ c^2)^(3/2)*x - 3*(b^2 + c^2)*(c*cos(e*x + d)/e - b*sin(e*x + d)/e) - 3/4 
*(4*b*c*cos(e*x + d)^2/e - (2*e*x + 2*d + sin(2*e*x + 2*d))*b^2/e - (2*e*x 
 + 2*d - sin(2*e*x + 2*d))*c^2/e)*sqrt(b^2 + c^2)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.12 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^3 \, dx={\left (b^{2} + c^{2}\right )}^{\frac {3}{2}} x - \frac {3 \, \sqrt {b^{2} + c^{2}} b c \cos \left (2 \, e x + 2 \, d\right )}{2 \, e} + \frac {3}{2} \, {\left (\sqrt {b^{2} + c^{2}} b^{2} + \sqrt {b^{2} + c^{2}} c^{2}\right )} x - \frac {{\left (3 \, b^{2} c - c^{3}\right )} \cos \left (3 \, e x + 3 \, d\right )}{12 \, e} - \frac {15 \, {\left (b^{2} c + c^{3}\right )} \cos \left (e x + d\right )}{4 \, e} + \frac {{\left (b^{3} - 3 \, b c^{2}\right )} \sin \left (3 \, e x + 3 \, d\right )}{12 \, e} + \frac {3 \, {\left (\sqrt {b^{2} + c^{2}} b^{2} - \sqrt {b^{2} + c^{2}} c^{2}\right )} \sin \left (2 \, e x + 2 \, d\right )}{4 \, e} + \frac {15 \, {\left (b^{3} + b c^{2}\right )} \sin \left (e x + d\right )}{4 \, e} \] Input:

integrate(((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^3,x, algorithm="giac 
")
 

Output:

(b^2 + c^2)^(3/2)*x - 3/2*sqrt(b^2 + c^2)*b*c*cos(2*e*x + 2*d)/e + 3/2*(sq 
rt(b^2 + c^2)*b^2 + sqrt(b^2 + c^2)*c^2)*x - 1/12*(3*b^2*c - c^3)*cos(3*e* 
x + 3*d)/e - 15/4*(b^2*c + c^3)*cos(e*x + d)/e + 1/12*(b^3 - 3*b*c^2)*sin( 
3*e*x + 3*d)/e + 3/4*(sqrt(b^2 + c^2)*b^2 - sqrt(b^2 + c^2)*c^2)*sin(2*e*x 
 + 2*d)/e + 15/4*(b^3 + b*c^2)*sin(e*x + d)/e
 

Mupad [B] (verification not implemented)

Time = 20.04 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.47 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^3 \, dx=\frac {5\,x\,{\left (b^2+c^2\right )}^{3/2}}{2}-\frac {8\,b^2\,c-\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (\left (3\,b^2-3\,c^2\right )\,\sqrt {b^2+c^2}+6\,b\,c^2+8\,b^3\right )-{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (\frac {40\,b^3}{3}+20\,b\,c^2\right )+\frac {22\,c^3}{3}-{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^5\,\left (6\,b\,c^2-\left (3\,b^2-3\,c^2\right )\,\sqrt {b^2+c^2}+8\,b^3\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4\,\left (12\,b^2\,c+6\,c^3-12\,b\,c\,\sqrt {b^2+c^2}\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (12\,b^2\,c+16\,c^3-12\,b\,c\,\sqrt {b^2+c^2}\right )}{e\,\left ({\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+1\right )} \] Input:

int((b*cos(d + e*x) + c*sin(d + e*x) + (b^2 + c^2)^(1/2))^3,x)
 

Output:

(5*x*(b^2 + c^2)^(3/2))/2 - (8*b^2*c - tan(d/2 + (e*x)/2)*((3*b^2 - 3*c^2) 
*(b^2 + c^2)^(1/2) + 6*b*c^2 + 8*b^3) - tan(d/2 + (e*x)/2)^3*(20*b*c^2 + ( 
40*b^3)/3) + (22*c^3)/3 - tan(d/2 + (e*x)/2)^5*(6*b*c^2 - (3*b^2 - 3*c^2)* 
(b^2 + c^2)^(1/2) + 8*b^3) + tan(d/2 + (e*x)/2)^4*(12*b^2*c + 6*c^3 - 12*b 
*c*(b^2 + c^2)^(1/2)) + tan(d/2 + (e*x)/2)^2*(12*b^2*c + 16*c^3 - 12*b*c*( 
b^2 + c^2)^(1/2)))/(e*(3*tan(d/2 + (e*x)/2)^2 + 3*tan(d/2 + (e*x)/2)^4 + t 
an(d/2 + (e*x)/2)^6 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.89 \[ \int \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^3 \, dx=\frac {-6 \cos \left (e x +d \right )^{3} b^{2} c -4 \cos \left (e x +d \right )^{3} c^{3}+9 \sqrt {b^{2}+c^{2}}\, \cos \left (e x +d \right )^{2} b^{2} e x -18 \sqrt {b^{2}+c^{2}}\, \cos \left (e x +d \right )^{2} b c +9 \sqrt {b^{2}+c^{2}}\, \cos \left (e x +d \right )^{2} c^{2} e x +6 \cos \left (e x +d \right )^{2} \sin \left (e x +d \right ) b^{3}+9 \sqrt {b^{2}+c^{2}}\, \cos \left (e x +d \right ) \sin \left (e x +d \right ) b^{2}-9 \sqrt {b^{2}+c^{2}}\, \cos \left (e x +d \right ) \sin \left (e x +d \right ) c^{2}-6 \cos \left (e x +d \right ) \sin \left (e x +d \right )^{2} c^{3}-18 \cos \left (e x +d \right ) b^{2} c -18 \cos \left (e x +d \right ) c^{3}+9 \sqrt {b^{2}+c^{2}}\, \sin \left (e x +d \right )^{2} b^{2} e x +9 \sqrt {b^{2}+c^{2}}\, \sin \left (e x +d \right )^{2} c^{2} e x +6 \sqrt {b^{2}+c^{2}}\, b^{2} e x +6 \sqrt {b^{2}+c^{2}}\, c^{2} e x +4 \sin \left (e x +d \right )^{3} b^{3}+6 \sin \left (e x +d \right )^{3} b \,c^{2}+18 \sin \left (e x +d \right ) b^{3}+18 \sin \left (e x +d \right ) b \,c^{2}}{6 e} \] Input:

int(((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^3,x)
 

Output:

( - 6*cos(d + e*x)**3*b**2*c - 4*cos(d + e*x)**3*c**3 + 9*sqrt(b**2 + c**2 
)*cos(d + e*x)**2*b**2*e*x - 18*sqrt(b**2 + c**2)*cos(d + e*x)**2*b*c + 9* 
sqrt(b**2 + c**2)*cos(d + e*x)**2*c**2*e*x + 6*cos(d + e*x)**2*sin(d + e*x 
)*b**3 + 9*sqrt(b**2 + c**2)*cos(d + e*x)*sin(d + e*x)*b**2 - 9*sqrt(b**2 
+ c**2)*cos(d + e*x)*sin(d + e*x)*c**2 - 6*cos(d + e*x)*sin(d + e*x)**2*c* 
*3 - 18*cos(d + e*x)*b**2*c - 18*cos(d + e*x)*c**3 + 9*sqrt(b**2 + c**2)*s 
in(d + e*x)**2*b**2*e*x + 9*sqrt(b**2 + c**2)*sin(d + e*x)**2*c**2*e*x + 6 
*sqrt(b**2 + c**2)*b**2*e*x + 6*sqrt(b**2 + c**2)*c**2*e*x + 4*sin(d + e*x 
)**3*b**3 + 6*sin(d + e*x)**3*b*c**2 + 18*sin(d + e*x)*b**3 + 18*sin(d + e 
*x)*b*c**2)/(6*e)