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

Optimal. Leaf size=83 \[ \frac{a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a (-\sin (d+e x))+a+b \cos (d+e x))}-\frac{a \log \left (a+b \tan \left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right )\right )}{4 b^3 e} \]

[Out]

-(a*Log[a + b*Tan[d/2 + Pi/4 + (e*x)/2]])/(4*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]))

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Rubi [A]  time = 0.0529481, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3129, 12, 3122, 31} \[ \frac{a \cos (d+e x)+b \sin (d+e x)}{4 b^2 e (a (-\sin (d+e x))+a+b \cos (d+e x))}-\frac{a \log \left (a+b \tan \left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right )\right )}{4 b^3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Log[a + b*Tan[d/2 + Pi/4 + (e*x)/2]])/(4*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 3129

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]

Rule 12

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

Rule 3122

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 31

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

Rubi steps

\begin{align*} \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+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 \operatorname{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]  time = 0.556618, size = 166, normalized size = 2. \[ \frac{\frac{b \left (a^2+b^2\right ) \sin \left (\frac{1}{2} (d+e x)\right )}{(a+b) \left ((b-a) \sin \left (\frac{1}{2} (d+e x)\right )+(a+b) \cos \left (\frac{1}{2} (d+e x)\right )\right )}-a \log \left ((b-a) \sin \left (\frac{1}{2} (d+e x)\right )+(a+b) \cos \left (\frac{1}{2} (d+e x)\right )\right )+a \log \left (\cos \left (\frac{1}{2} (d+e x)\right )-\sin \left (\frac{1}{2} (d+e x)\right )\right )+\frac{b \sin \left (\frac{1}{2} (d+e x)\right )}{\cos \left (\frac{1}{2} (d+e x)\right )-\sin \left (\frac{1}{2} (d+e x)\right )}}{4 b^3 e} \]

Antiderivative was successfully verified.

[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)

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Maple [B]  time = 0.148, size = 178, normalized size = 2.1 \begin{align*} -{\frac{1}{4\,{b}^{2}e} \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) -1 \right ) ^{-1}}+{\frac{a}{4\,{b}^{3}e}\ln \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) -1 \right ) }-{\frac{{a}^{2}}{4\,{b}^{2}e \left ( a-b \right ) } \left ( a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) -b\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) -a-b \right ) ^{-1}}-{\frac{1}{4\,e \left ( a-b \right ) } \left ( a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) -b\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) -a-b \right ) ^{-1}}-{\frac{a}{4\,{b}^{3}e}\ln \left ( a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) -b\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) -a-b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.00558, size = 246, normalized size = 2.96 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [B]  time = 2.27783, size = 377, normalized size = 4.54 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.16489, size = 258, normalized size = 3.11 \begin{align*} -\frac{1}{4} \,{\left (\frac{{\left (a^{2} - a b\right )} \log \left ({\left | a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - a - b \right |}\right )}{a b^{3} - b^{4}} + \frac{2 \,{\left (a^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - a b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + b^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - a^{2}\right )}}{{\left (a b^{2} - b^{3}\right )}{\left (a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} - b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} - 2 \, a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + a + b\right )}} - \frac{a \log \left ({\left | \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - 1 \right |}\right )}{b^{3}}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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