\(\int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx\) [371]

Optimal result

Integrand size = 24, antiderivative size = 75 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {\log \left (1+\tan \left (\frac {1}{2} (d+e x)\right )\right )}{4 a^2 e}-\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )} \]

[Out]

-1/4*ln(1+tan(1/2*e*x+1/2*d))/a^2/e+1/4*(-a*cos(e*x+d)+a*sin(e*x+d))/e/(a^3+a^3*cos(e*x+d)+a^3*sin(e*x+d))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3208, 12, 3203, 31} \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}-\frac {\log \left (\tan \left (\frac {1}{2} (d+e x)\right )+1\right )}{4 a^2 e} \]

[In]

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

[Out]

-1/4*Log[1 + Tan[(d + e*x)/2]]/(a^2*e) - (a*Cos[d + e*x] - a*Sin[d + e*x])/(4*e*(a^3 + a^3*Cos[d + e*x] + a^3*
Sin[d + e*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3203

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Tan[(d + e*x)/2], x]}, Dist[2*(f/e), Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d +
e*x)/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]

Rule 3208

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 + 1)*(a^2 - b^2 - c^2))), x] +
Dist[1/((n + 1)*(a^2 - b^2 - c^2)), Int[(a*(n + 1) - b*(n + 2)*Cos[d + e*x] - c*(n + 2)*Sin[d + e*x])*(a + b*C
os[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && LtQ[n
, -1] && NeQ[n, -3/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )}+\frac {\int -\frac {2 a}{2 a+2 a \cos (d+e x)+2 a \sin (d+e x)} \, dx}{4 a^2} \\ & = -\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )}-\frac {\int \frac {1}{2 a+2 a \cos (d+e x)+2 a \sin (d+e x)} \, dx}{2 a} \\ & = -\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )}-\frac {\text {Subst}\left (\int \frac {1}{4 a+4 a x} \, dx,x,\tan \left (\frac {1}{2} (d+e x)\right )\right )}{a e} \\ & = -\frac {\log \left (1+\tan \left (\frac {1}{2} (d+e x)\right )\right )}{4 a^2 e}-\frac {a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\frac {2 \log \left (\cos \left (\frac {1}{2} (d+e x)\right )\right )-2 \log \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )+\frac {2 \sin \left (\frac {1}{2} (d+e x)\right )}{\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )}+\tan \left (\frac {1}{2} (d+e x)\right )}{8 a^2 e} \]

[In]

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

[Out]

(2*Log[Cos[(d + e*x)/2]] - 2*Log[Cos[(d + e*x)/2] + Sin[(d + e*x)/2]] + (2*Sin[(d + e*x)/2])/(Cos[(d + e*x)/2]
+ Sin[(d + e*x)/2]) + Tan[(d + e*x)/2])/(8*a^2*e)

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.64

method	result	size
derivativedivides	\(\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-2 \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )-\frac {2}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )}}{8 e \,a^{2}}\)	\(48\)
default	\(\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-2 \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )-\frac {2}{1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )}}{8 e \,a^{2}}\)	\(48\)
norman	\(\frac {\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{8 a e}-\frac {3}{8 a e}}{a \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}-\frac {\ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{4 a^{2} e}\)	\(67\)
parallelrisch	\(\frac {\left (-2 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-2\right ) \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+3 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{8 e \,a^{2} \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}\)	\(71\)
risch	\(\frac {\left (-\frac {1}{4}+\frac {i}{4}\right ) \left ({\mathrm e}^{i \left (e x +d \right )}+1+i\right )}{a^{2} e \left (i {\mathrm e}^{i \left (e x +d \right )}+{\mathrm e}^{2 i \left (e x +d \right )}+i+{\mathrm e}^{i \left (e x +d \right )}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+i\right )}{4 a^{2} e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+1\right )}{4 a^{2} e}\)	\(101\)

[In]

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

[Out]

1/8/e/a^2*(tan(1/2*e*x+1/2*d)-2*ln(1+tan(1/2*e*x+1/2*d))-2/(1+tan(1/2*e*x+1/2*d)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.33 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\frac {{\left (\cos \left (e x + d\right ) + \sin \left (e x + d\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (e x + d\right ) + \frac {1}{2}\right ) - {\left (\cos \left (e x + d\right ) + \sin \left (e x + d\right ) + 1\right )} \log \left (\sin \left (e x + d\right ) + 1\right ) - 2 \, \cos \left (e x + d\right ) + 2 \, \sin \left (e x + d\right )}{8 \, {\left (a^{2} e \cos \left (e x + d\right ) + a^{2} e \sin \left (e x + d\right ) + a^{2} e\right )}} \]

[In]

integrate(1/(2*a+2*a*cos(e*x+d)+2*a*sin(e*x+d))^2,x, algorithm="fricas")

[Out]

1/8*((cos(e*x + d) + sin(e*x + d) + 1)*log(1/2*cos(e*x + d) + 1/2) - (cos(e*x + d) + sin(e*x + d) + 1)*log(sin
(e*x + d) + 1) - 2*cos(e*x + d) + 2*sin(e*x + d))/(a^2*e*cos(e*x + d) + a^2*e*sin(e*x + d) + a^2*e)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (65) = 130\).

Time = 0.83 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.24 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\begin {cases} - \frac {2 \log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )} \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )}}{8 a^{2} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 8 a^{2} e} - \frac {2 \log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )}}{8 a^{2} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 8 a^{2} e} + \frac {\tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )}}{8 a^{2} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 8 a^{2} e} - \frac {3}{8 a^{2} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 8 a^{2} e} & \text {for}\: e \neq 0 \\\frac {x}{\left (2 a \sin {\left (d \right )} + 2 a \cos {\left (d \right )} + 2 a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-2*log(tan(d/2 + e*x/2) + 1)*tan(d/2 + e*x/2)/(8*a**2*e*tan(d/2 + e*x/2) + 8*a**2*e) - 2*log(tan(d/
2 + e*x/2) + 1)/(8*a**2*e*tan(d/2 + e*x/2) + 8*a**2*e) + tan(d/2 + e*x/2)**2/(8*a**2*e*tan(d/2 + e*x/2) + 8*a*
*2*e) - 3/(8*a**2*e*tan(d/2 + e*x/2) + 8*a**2*e), Ne(e, 0)), (x/(2*a*sin(d) + 2*a*cos(d) + 2*a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {\frac {2}{a^{2} + \frac {a^{2} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}} + \frac {2 \, \log \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right )}{a^{2}} - \frac {\sin \left (e x + d\right )}{a^{2} {\left (\cos \left (e x + d\right ) + 1\right )}}}{8 \, e} \]

[In]

integrate(1/(2*a+2*a*cos(e*x+d)+2*a*sin(e*x+d))^2,x, algorithm="maxima")

[Out]

-1/8*(2/(a^2 + a^2*sin(e*x + d)/(cos(e*x + d) + 1)) + 2*log(sin(e*x + d)/(cos(e*x + d) + 1) + 1)/a^2 - sin(e*x
+ d)/(a^2*(cos(e*x + d) + 1)))/e

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=-\frac {\frac {2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 1 \right |}\right )}{a^{2}} - \frac {\tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{a^{2}} - \frac {2 \, \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{a^{2} {\left (\tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 1\right )}}}{8 \, e} \]

[In]

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

[Out]

-1/8*(2*log(abs(tan(1/2*e*x + 1/2*d) + 1))/a^2 - tan(1/2*e*x + 1/2*d)/a^2 - 2*tan(1/2*e*x + 1/2*d)/(a^2*(tan(1
/2*e*x + 1/2*d) + 1)))/e

Mupad [B] (verification not implemented)

Time = 27.88 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{8\,a^2\,e}-\frac {\ln \left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1\right )}{4\,a^2\,e}-\frac {1}{4\,a^2\,e\,\left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1\right )} \]

[In]

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

[Out]

tan(d/2 + (e*x)/2)/(8*a^2*e) - log(tan(d/2 + (e*x)/2) + 1)/(4*a^2*e) - 1/(4*a^2*e*(tan(d/2 + (e*x)/2) + 1))