\(\int \frac {1}{(\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x))^2} \, dx\) [287]

Optimal result

Integrand size = 30, antiderivative size = 129 \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2} \, dx=\frac {-c \cos (d+e x)+b \sin (d+e x)}{3 \sqrt {b^2+c^2} e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}-\frac {c-\sqrt {b^2+c^2} \sin (d+e x)}{3 c \sqrt {b^2+c^2} e (c \cos (d+e x)-b \sin (d+e x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2} \, dx=\frac {-2 c \sqrt {b^2+c^2}+2 b c \cos ^3(d+e x)+2 c^2 \sin (d+e x)+c^2 \cos ^2(d+e x) \sin (d+e x)+b^2 \sin ^3(d+e x)}{3 c e (c \cos (d+e x)-b \sin (d+e x))^3} \] Input:

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

Output:

(-2*c*Sqrt[b^2 + c^2] + 2*b*c*Cos[d + e*x]^3 + 2*c^2*Sin[d + e*x] + c^2*Co 
s[d + e*x]^2*Sin[d + e*x] + b^2*Sin[d + e*x]^3)/(3*c*e*(c*Cos[d + e*x] - b 
*Sin[d + e*x])^3)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3042, 3595, 3042, 3593}

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.

\(\displaystyle \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}dx\)

\(\Big \downarrow \) 3595

\(\displaystyle \frac {\int \frac {1}{b \cos (d+e x)+c \sin (d+e x)+\sqrt {b^2+c^2}}dx}{3 \sqrt {b^2+c^2}}-\frac {c \cos (d+e x)-b \sin (d+e x)}{3 e \sqrt {b^2+c^2} \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {1}{b \cos (d+e x)+c \sin (d+e x)+\sqrt {b^2+c^2}}dx}{3 \sqrt {b^2+c^2}}-\frac {c \cos (d+e x)-b \sin (d+e x)}{3 e \sqrt {b^2+c^2} \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}\)

\(\Big \downarrow \) 3593

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3593

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ 
(-1), x_Symbol] :> Simp[-(c - a*Sin[d + e*x])/(c*e*(c*Cos[d + e*x] - b*Sin[ 
d + e*x])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0]
 

rule 3595

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/(a*e*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) 
 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] && LtQ[n, -1]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.71 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {2 \left (i \sqrt {b^{2}+c^{2}}\, c +3 b^{2} {\mathrm e}^{i \left (e x +d \right )}+3 c^{2} {\mathrm e}^{i \left (e x +d \right )}+\sqrt {b^{2}+c^{2}}\, b \right ) \left (i b^{2}-i c^{2}-2 c b \right )}{3 \left (i \sqrt {b^{2}+c^{2}}\, c +b^{2} {\mathrm e}^{i \left (e x +d \right )}+c^{2} {\mathrm e}^{i \left (e x +d \right )}+\sqrt {b^{2}+c^{2}}\, b \right )^{3} e}\)	\(129\)
derivativedivides	\(\frac {2 \left (\sqrt {b^{2}+c^{2}}+b \right ) \left (-\frac {\left (\sqrt {b^{2}+c^{2}}+b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{c^{2}}-\frac {\left (2 b^{2}+c^{2}+2 \sqrt {b^{2}+c^{2}}\, b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{c^{3}}-\frac {2 \left (2 \sqrt {b^{2}+c^{2}}\, b^{2}+\sqrt {b^{2}+c^{2}}\, c^{2}+2 b^{3}+2 b \,c^{2}\right )}{3 c^{4}}\right )}{e \,c^{2} \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+\frac {2 \sqrt {b^{2}+c^{2}}\, \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{c}+\frac {2 b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{c}+\frac {2 b \sqrt {b^{2}+c^{2}}}{c^{2}}+\frac {2 b^{2}}{c^{2}}+1\right ) \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+\frac {\sqrt {b^{2}+c^{2}}}{c}+\frac {b}{c}\right )}\)	\(233\)
default	\(\frac {2 \left (\sqrt {b^{2}+c^{2}}+b \right ) \left (-\frac {\left (\sqrt {b^{2}+c^{2}}+b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{c^{2}}-\frac {\left (2 b^{2}+c^{2}+2 \sqrt {b^{2}+c^{2}}\, b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{c^{3}}-\frac {2 \left (2 \sqrt {b^{2}+c^{2}}\, b^{2}+\sqrt {b^{2}+c^{2}}\, c^{2}+2 b^{3}+2 b \,c^{2}\right )}{3 c^{4}}\right )}{e \,c^{2} \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+\frac {2 \sqrt {b^{2}+c^{2}}\, \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{c}+\frac {2 b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{c}+\frac {2 b \sqrt {b^{2}+c^{2}}}{c^{2}}+\frac {2 b^{2}}{c^{2}}+1\right ) \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+\frac {\sqrt {b^{2}+c^{2}}}{c}+\frac {b}{c}\right )}\)	\(233\)

Input:

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

Output:

2/3*(I*(b^2+c^2)^(1/2)*c+3*b^2*exp(I*(e*x+d))+3*c^2*exp(I*(e*x+d))+(b^2+c^ 
2)^(1/2)*b)*(I*b^2-I*c^2-2*c*b)/(I*(b^2+c^2)^(1/2)*c+b^2*exp(I*(e*x+d))+c^ 
2*exp(I*(e*x+d))+(b^2+c^2)^(1/2)*b)^3/e
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.49 \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2} \, dx=-\frac {3 \, b^{3} \cos \left (e x + d\right ) - {\left (b^{3} - 3 \, b c^{2}\right )} \cos \left (e x + d\right )^{3} + {\left (3 \, b^{2} c + 2 \, c^{3} - {\left (3 \, b^{2} c - c^{3}\right )} \cos \left (e x + d\right )^{2}\right )} \sin \left (e x + d\right ) - 2 \, {\left (b^{2} + c^{2}\right )}^{\frac {3}{2}}}{3 \, {\left ({\left (3 \, b^{4} c + 2 \, b^{2} c^{3} - c^{5}\right )} e \cos \left (e x + d\right )^{3} - 3 \, {\left (b^{4} c + b^{2} c^{3}\right )} e \cos \left (e x + d\right ) - {\left ({\left (b^{5} - 2 \, b^{3} c^{2} - 3 \, b c^{4}\right )} e \cos \left (e x + d\right )^{2} - {\left (b^{5} + b^{3} c^{2}\right )} e\right )} \sin \left (e x + d\right )\right )}} \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

\[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2} \, dx=\int { \frac {1}{{\left (b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) + \sqrt {b^{2} + c^{2}}\right )}^{2}} \,d x } \] Input:

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

Output:

sage0*x
 

Mupad [B] (verification not implemented)

Time = 16.25 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.12 \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2} \, dx=-\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (\frac {4\,b^2+2\,c^2}{c^4}+\frac {4\,b\,\sqrt {b^2+c^2}}{c^4}\right )+\frac {\frac {16\,b^4}{3}+\frac {20\,b^2\,c^2}{3}+\frac {4\,c^4}{3}}{c^6}+\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (\frac {8\,b^3+6\,b\,c^2}{c^5}+\frac {\left (8\,b^2+2\,c^2\right )\,\sqrt {b^2+c^2}}{c^5}\right )+\frac {\left (\frac {16\,b^3}{3}+4\,b\,c^2\right )\,\sqrt {b^2+c^2}}{c^6}}{e\,\left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (\frac {6\,b^2+3\,c^2}{c^2}+\frac {6\,b\,\sqrt {b^2+c^2}}{c^2}\right )+\frac {4\,b^3+3\,b\,c^2}{c^3}+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (\frac {3\,\sqrt {b^2+c^2}}{c}+\frac {3\,b}{c}\right )+\frac {\left (4\,b^2+c^2\right )\,\sqrt {b^2+c^2}}{c^3}\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 1090, normalized size of antiderivative = 8.45 \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(3*cos(d + e*x)**3*sin(d + e*x)**2*b**6*c + cos(d + e*x)**3*sin(d + e*x)** 
2*b**4*c**3 - 2*cos(d + e*x)**3*sin(d + e*x)**2*b**2*c**5 + 2*cos(d + e*x) 
**3*b**4*c**3 + 2*cos(d + e*x)**3*b**2*c**5 - 4*sqrt(b**2 + c**2)*cos(d + 
e*x)**2*b**3*c**3 - cos(d + e*x)**2*sin(d + e*x)**3*b**7 + 9*cos(d + e*x)* 
*2*sin(d + e*x)**3*b**5*c**2 + 6*cos(d + e*x)**2*sin(d + e*x)**3*b**3*c**4 
 - 4*cos(d + e*x)**2*sin(d + e*x)**3*b*c**6 + 4*cos(d + e*x)**2*sin(d + e* 
x)*b**3*c**4 + 4*cos(d + e*x)**2*sin(d + e*x)*b*c**6 - 8*sqrt(b**2 + c**2) 
*cos(d + e*x)*sin(d + e*x)*b**2*c**4 - 2*cos(d + e*x)*sin(d + e*x)**4*b**6 
*c + 9*cos(d + e*x)*sin(d + e*x)**4*b**4*c**3 + 9*cos(d + e*x)*sin(d + e*x 
)**4*b**2*c**5 - 2*cos(d + e*x)*sin(d + e*x)**4*c**7 - 3*cos(d + e*x)*sin( 
d + e*x)**2*b**6*c - 13*cos(d + e*x)*sin(d + e*x)**2*b**4*c**3 + 6*cos(d + 
 e*x)*sin(d + e*x)**2*b**2*c**5 + 4*cos(d + e*x)*sin(d + e*x)**2*c**7 - 2* 
cos(d + e*x)*b**4*c**3 - 7*cos(d + e*x)*b**2*c**5 - 2*cos(d + e*x)*c**7 - 
4*sqrt(b**2 + c**2)*sin(d + e*x)**2*b*c**5 + 4*sqrt(b**2 + c**2)*b**3*c**3 
 + 4*sqrt(b**2 + c**2)*b*c**5 - sin(d + e*x)**5*b**5*c**2 + 3*sin(d + e*x) 
**5*b**3*c**4 + 4*sin(d + e*x)**5*b*c**6 + sin(d + e*x)**3*b**7 + sin(d + 
e*x)**3*b**5*c**2 - 16*sin(d + e*x)**3*b**3*c**4 - 4*sin(d + e*x)**3*b*c** 
6 + 9*sin(d + e*x)*b**3*c**4)/(6*b*c**3*e*(3*cos(d + e*x)**3*sin(d + e*x)* 
*2*b**4*c - cos(d + e*x)**3*sin(d + e*x)**2*b**2*c**3 + cos(d + e*x)**3*b* 
*2*c**3 - cos(d + e*x)**2*sin(d + e*x)**3*b**5 + 9*cos(d + e*x)**2*sin(...