\(\int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx\) [392]

Optimal result

Integrand size = 24, antiderivative size = 83 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=-\frac {a \log \left (a+b \tan \left (\frac {d}{2}+\frac {\pi }{4}+\frac {e x}{2}\right )\right )}{4 b^3 e}+\frac {a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a+b \cos (d+e x)-a \sin (d+e x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 83, 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, 3201, 31} \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=\frac {a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a (-\sin (d+e x))+a+b \cos (d+e x))}-\frac {a \log \left (a+b \tan \left (\frac {d}{2}+\frac {e x}{2}+\frac {\pi }{4}\right )\right )}{4 b^3 e} \]

[In]

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

[Out]

-1/4*(a*Log[a + b*Tan[d/2 + Pi/4 + (e*x)/2]])/(b^3*e) + (a*Cos[d + e*x] + b*Sin[d + e*x])/(4*b^2*e*(a + b*Cos[
d + e*x] - a*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 3201

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Tan[(d + e*x)/2 + Pi/4], x]}, Dist[f/e, Subst[Int[1/(a + b*f*x), x], x, Tan[(d + e*x)/2 + Pi/4]/f], x]
] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a + c, 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)+b \sin (d+e x)}{4 b^2 e (a+b \cos (d+e x)-a \sin (d+e x))}+\frac {\int -\frac {2 a}{2 a+2 b \cos (d+e x)-2 a \sin (d+e x)} \, dx}{4 b^2} \\ & = \frac {a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a+b \cos (d+e x)-a \sin (d+e x))}-\frac {a \int \frac {1}{2 a+2 b \cos (d+e x)-2 a \sin (d+e x)} \, dx}{2 b^2} \\ & = \frac {a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a+b \cos (d+e x)-a \sin (d+e x))}-\frac {a \text {Subst}\left (\int \frac {1}{2 a+2 b x} \, dx,x,\tan \left (\frac {\pi }{4}+\frac {1}{2} (d+e x)\right )\right )}{2 b^2 e} \\ & = -\frac {a \log \left (a+b \tan \left (\frac {d}{2}+\frac {\pi }{4}+\frac {e x}{2}\right )\right )}{4 b^3 e}+\frac {a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a+b \cos (d+e x)-a \sin (d+e x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 166, normalized size of antiderivative = 2.00 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=\frac {a \log \left (\cos \left (\frac {1}{2} (d+e x)\right )-\sin \left (\frac {1}{2} (d+e x)\right )\right )-a \log \left ((a+b) \cos \left (\frac {1}{2} (d+e x)\right )+(-a+b) \sin \left (\frac {1}{2} (d+e x)\right )\right )+\frac {b \sin \left (\frac {1}{2} (d+e x)\right )}{\cos \left (\frac {1}{2} (d+e x)\right )-\sin \left (\frac {1}{2} (d+e x)\right )}+\frac {b \left (a^2+b^2\right ) \sin \left (\frac {1}{2} (d+e x)\right )}{(a+b) \left ((a+b) \cos \left (\frac {1}{2} (d+e x)\right )+(-a+b) \sin \left (\frac {1}{2} (d+e x)\right )\right )}}{4 b^3 e} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.17 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.57

method	result	size
derivativedivides	\(\frac {-\frac {a^{2}+b^{2}}{b^{2} \left (a -b \right ) \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}-\frac {a \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}{b^{3}}-\frac {1}{b^{2} \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}+\frac {a \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}{b^{3}}}{4 e}\)	\(130\)
default	\(\frac {-\frac {a^{2}+b^{2}}{b^{2} \left (a -b \right ) \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}-\frac {a \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}{b^{3}}-\frac {1}{b^{2} \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}+\frac {a \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}{b^{3}}}{4 e}\)	\(130\)
risch	\(\frac {i \left (-i a +b +a \,{\mathrm e}^{i \left (e x +d \right )}\right )}{2 b^{2} e \left (i a \,{\mathrm e}^{2 i \left (e x +d \right )}+b \,{\mathrm e}^{2 i \left (e x +d \right )}-i a +2 a \,{\mathrm e}^{i \left (e x +d \right )}+b \right )}+\frac {a \ln \left ({\mathrm e}^{i \left (e x +d \right )}-i\right )}{4 b^{3} e}-\frac {a \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a -b}{i b -a}\right )}{4 b^{3} e}\)	\(133\)
parallelrisch	\(\frac {\left (\cos \left (e x +d \right ) a^{2} b -a^{3} \left (\sin \left (e x +d \right )-1\right )\right ) \ln \left (-a -b +\left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+a^{2} \left (a \left (\sin \left (e x +d \right )-1\right )-b \cos \left (e x +d \right )\right ) \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )+\left (-a^{2} b -b^{3}\right ) \cos \left (e x +d \right )-a \,b^{2}}{4 b^{3} e a \left (a \left (\sin \left (e x +d \right )-1\right )-b \cos \left (e x +d \right )\right )}\)	\(145\)
norman	\(\frac {\frac {a^{2}+a b +b^{2}}{4 a \,b^{2} e}-\frac {\left (a^{2}-a b +b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{4 a \,b^{2} e}}{\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right ) \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}+\frac {a \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}{4 b^{3} e}-\frac {a \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-a -b \right )}{4 b^{3} e}\)	\(164\)

[In]

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

[Out]

1/4/e*(-(a^2+b^2)/b^2/(a-b)/(a*tan(1/2*e*x+1/2*d)-b*tan(1/2*e*x+1/2*d)-a-b)-a/b^3*ln(a*tan(1/2*e*x+1/2*d)-b*ta
n(1/2*e*x+1/2*d)-a-b)-1/b^2/(tan(1/2*e*x+1/2*d)-1)+a/b^3*ln(tan(1/2*e*x+1/2*d)-1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (73) = 146\).

Time = 0.26 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.86 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=\frac {2 \, a b \cos \left (e x + d\right ) + 2 \, b^{2} \sin \left (e x + d\right ) - {\left (a b \cos \left (e x + d\right ) - a^{2} \sin \left (e x + d\right ) + a^{2}\right )} \log \left (2 \, a b \cos \left (e x + d\right ) + a^{2} + b^{2} - {\left (a^{2} - b^{2}\right )} \sin \left (e x + d\right )\right ) + {\left (a b \cos \left (e x + d\right ) - a^{2} \sin \left (e x + d\right ) + a^{2}\right )} \log \left (-\sin \left (e x + d\right ) + 1\right )}{8 \, {\left (b^{4} e \cos \left (e x + d\right ) - a b^{3} e \sin \left (e x + d\right ) + a b^{3} e\right )}} \]

[In]

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

[Out]

1/8*(2*a*b*cos(e*x + d) + 2*b^2*sin(e*x + d) - (a*b*cos(e*x + d) - a^2*sin(e*x + d) + a^2)*log(2*a*b*cos(e*x +
d) + a^2 + b^2 - (a^2 - b^2)*sin(e*x + d)) + (a*b*cos(e*x + d) - a^2*sin(e*x + d) + a^2)*log(-sin(e*x + d) +
1))/(b^4*e*cos(e*x + d) - a*b^3*e*sin(e*x + d) + a*b^3*e)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (73) = 146\).

Time = 0.24 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.19 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=\frac {\frac {2 \, {\left (a^{2} - \frac {{\left (a^{2} - a b + b^{2}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}}{a^{2} b^{2} - b^{4} - \frac {2 \, {\left (a^{2} b^{2} - a b^{3}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {{\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}} - \frac {a \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{b^{3}} + \frac {a \log \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - 1\right )}{b^{3}}}{4 \, e} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (73) = 146\).

Time = 0.30 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.28 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=-\frac {\frac {2 \, {\left (a^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - a b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - a^{2}\right )}}{{\left (a b^{2} - b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - 2 \, a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a + b\right )}} + \frac {a \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, a - 2 \, {\left | b \right |} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, a + 2 \, {\left | b \right |} \right |}}\right )}{b^{2} {\left | b \right |}}}{4 \, e} \]

[In]

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

[Out]

-1/4*(2*(a^2*tan(1/2*e*x + 1/2*d) - a*b*tan(1/2*e*x + 1/2*d) + b^2*tan(1/2*e*x + 1/2*d) - a^2)/((a*b^2 - b^3)*
(a*tan(1/2*e*x + 1/2*d)^2 - b*tan(1/2*e*x + 1/2*d)^2 - 2*a*tan(1/2*e*x + 1/2*d) + a + b)) + a*log(abs(2*a*tan(
1/2*e*x + 1/2*d) - 2*b*tan(1/2*e*x + 1/2*d) - 2*a - 2*abs(b))/abs(2*a*tan(1/2*e*x + 1/2*d) - 2*b*tan(1/2*e*x +
1/2*d) - 2*a + 2*abs(b)))/(b^2*abs(b)))/e

Mupad [B] (verification not implemented)

Time = 27.15 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.52 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx=\frac {\frac {a^2}{b^2\,\left (a-b\right )}-\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (a^2-a\,b+b^2\right )}{b^2\,\left (a-b\right )}}{e\,\left (\left (2\,a-2\,b\right )\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2-4\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+2\,a+2\,b\right )}-\frac {a\,\mathrm {atanh}\left (\frac {a-\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,a-2\,b\right )}{2}}{b}\right )}{2\,b^3\,e} \]

[In]

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

[Out]

(a^2/(b^2*(a - b)) - (tan(d/2 + (e*x)/2)*(a^2 - a*b + b^2))/(b^2*(a - b)))/(e*(2*a + 2*b + tan(d/2 + (e*x)/2)^
2*(2*a - 2*b) - 4*a*tan(d/2 + (e*x)/2))) - (a*atanh((a - (tan(d/2 + (e*x)/2)*(2*a - 2*b))/2)/b))/(2*b^3*e)