\(\int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx\) [378]

Optimal result

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

[Out]

1/4*a*ln(a+c*cot(1/2*e*x+1/2*d))/c^3/e+1/4*(-c*cos(e*x+d)-a*sin(e*x+d))/c^2/e/(a-a*cos(e*x+d)+c*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, 3200, 31} \[ \int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx=\frac {a \log \left (a+c \cot \left (\frac {1}{2} (d+e x)\right )\right )}{4 c^3 e}-\frac {a \sin (d+e x)+c \cos (d+e x)}{4 c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))} \]

[In]

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

[Out]

(a*Log[a + c*Cot[(d + e*x)/2]])/(4*c^3*e) - (c*Cos[d + e*x] + a*Sin[d + e*x])/(4*c^2*e*(a - a*Cos[d + e*x] + c
*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 3200

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(229\) vs. \(2(75)=150\).

Time = 5.41 (sec) , antiderivative size = 229, normalized size of antiderivative = 3.05 \[ \int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx=-\frac {\sin \left (\frac {1}{2} (d+e x)\right ) \left (c \cos \left (\frac {1}{2} (d+e x)\right )+a \sin \left (\frac {1}{2} (d+e x)\right )\right ) \left (\cos (d+e x) \left (a^2+2 c^2-2 a^2 \log \left (\sin \left (\frac {1}{2} (d+e x)\right )\right )+2 a^2 \log \left (c \cos \left (\frac {1}{2} (d+e x)\right )+a \sin \left (\frac {1}{2} (d+e x)\right )\right )\right )+a \left (a \left (-1+2 \log \left (\sin \left (\frac {1}{2} (d+e x)\right )\right )-2 \log \left (c \cos \left (\frac {1}{2} (d+e x)\right )+a \sin \left (\frac {1}{2} (d+e x)\right )\right )\right )+c \left (1+2 \log \left (\sin \left (\frac {1}{2} (d+e x)\right )\right )-2 \log \left (c \cos \left (\frac {1}{2} (d+e x)\right )+a \sin \left (\frac {1}{2} (d+e x)\right )\right )\right ) \sin (d+e x)\right )\right )}{4 c^3 e (a-a \cos (d+e x)+c \sin (d+e x))^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {-\frac {a^{2}+c^{2}}{2 c^{2} a \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}+\frac {a \ln \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{c^{3}}-\frac {1}{2 c^{2} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}-\frac {a \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{c^{3}}}{4 e}\)	\(88\)
default	\(\frac {-\frac {a^{2}+c^{2}}{2 c^{2} a \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}+\frac {a \ln \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{c^{3}}-\frac {1}{2 c^{2} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}-\frac {a \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{c^{3}}}{4 e}\)	\(88\)
norman	\(\frac {-\frac {1}{8 c e}-\frac {\left (2 a^{2}+c^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{8 a \,c^{2} e}}{\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}-\frac {a \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{4 c^{3} e}+\frac {a \ln \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{4 c^{3} e}\)	\(109\)
risch	\(-\frac {i \left (-i a \,{\mathrm e}^{i \left (e x +d \right )}+i a +c \right )}{2 c^{2} e \left (c \,{\mathrm e}^{2 i \left (e x +d \right )}-i a \,{\mathrm e}^{2 i \left (e x +d \right )}-c +2 i a \,{\mathrm e}^{i \left (e x +d \right )}-i a \right )}+\frac {a \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i c -a}{i c +a}\right )}{4 c^{3} e}-\frac {a \ln \left ({\mathrm e}^{i \left (e x +d \right )}-1\right )}{4 c^{3} e}\)	\(135\)
parallelrisch	\(\frac {-2 \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right ) a^{2}+2 \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) \ln \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right ) a^{2}-2 \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right ) a c +2 \ln \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right ) a c -\cot \left (\frac {e x}{2}+\frac {d}{2}\right ) c^{2}+2 \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a^{2}+\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) c^{2}}{8 \left (c +a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right ) c^{3} e}\)	\(150\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (72) = 144\).

Time = 0.26 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.16 \[ \int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx=\frac {2 \, c^{2} \cos \left (e x + d\right ) + 2 \, a c \sin \left (e x + d\right ) + {\left (a^{2} \cos \left (e x + d\right ) - a c \sin \left (e x + d\right ) - a^{2}\right )} \log \left (a c \sin \left (e x + d\right ) + \frac {1}{2} \, a^{2} + \frac {1}{2} \, c^{2} - \frac {1}{2} \, {\left (a^{2} - c^{2}\right )} \cos \left (e x + d\right )\right ) - {\left (a^{2} \cos \left (e x + d\right ) - a c \sin \left (e x + d\right ) - a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (e x + d\right ) + \frac {1}{2}\right )}{8 \, {\left (a c^{3} e \cos \left (e x + d\right ) - c^{4} e \sin \left (e x + d\right ) - a c^{3} e\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.83 \[ \int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx=-\frac {\frac {a c + \frac {{\left (2 \, a^{2} + c^{2}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}}{\frac {a c^{3} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {a^{2} c^{2} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}} - \frac {2 \, a \log \left (c + \frac {a \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{c^{3}} + \frac {2 \, a \log \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{c^{3}}}{8 \, e} \]

[In]

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

[Out]

-1/8*((a*c + (2*a^2 + c^2)*sin(e*x + d)/(cos(e*x + d) + 1))/(a*c^3*sin(e*x + d)/(cos(e*x + d) + 1) + a^2*c^2*s
in(e*x + d)^2/(cos(e*x + d) + 1)^2) - 2*a*log(c + a*sin(e*x + d)/(cos(e*x + d) + 1))/c^3 + 2*a*log(sin(e*x + d
)/(cos(e*x + d) + 1))/c^3)/e

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.47 \[ \int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx=\frac {\frac {2 \, a \log \left ({\left | a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + c \right |}\right )}{c^{3}} - \frac {2 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) \right |}\right )}{c^{3}} - \frac {2 \, a^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + c^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a c}{{\left (a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )\right )} a c^{2}}}{8 \, e} \]

[In]

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

[Out]

1/8*(2*a*log(abs(a*tan(1/2*e*x + 1/2*d) + c))/c^3 - 2*a*log(abs(tan(1/2*e*x + 1/2*d)))/c^3 - (2*a^2*tan(1/2*e*
x + 1/2*d) + c^2*tan(1/2*e*x + 1/2*d) + a*c)/((a*tan(1/2*e*x + 1/2*d)^2 + c*tan(1/2*e*x + 1/2*d))*a*c^2))/e

Mupad [B] (verification not implemented)

Time = 27.32 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2} \, dx=\frac {a\,\mathrm {atanh}\left (\frac {2\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{c}+1\right )}{2\,c^3\,e}-\frac {\frac {1}{c}+\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,a^2+c^2\right )}{a\,c^2}}{e\,\left (8\,a\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+8\,c\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\right )} \]

[In]

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

[Out]

(a*atanh((2*a*tan(d/2 + (e*x)/2))/c + 1))/(2*c^3*e) - (1/c + (tan(d/2 + (e*x)/2)*(2*a^2 + c^2))/(a*c^2))/(e*(8
*c*tan(d/2 + (e*x)/2) + 8*a*tan(d/2 + (e*x)/2)^2))