3.393 \(\int \frac{1}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^3} \, dx\)

Optimal. Leaf size=142 \[ \frac{\left (3 a^2+b^2\right ) \log \left (a+b \tan \left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right )\right )}{16 b^5 e}-\frac{3 \left (a^2 \cos (d+e x)+a b \sin (d+e x)\right )}{16 b^4 e (a (-\sin (d+e x))+a+b \cos (d+e x))}+\frac{a \cos (d+e x)+b \sin (d+e x)}{16 b^2 e (a (-\sin (d+e x))+a+b \cos (d+e x))^2} \]

[Out]

((3*a^2 + b^2)*Log[a + b*Tan[d/2 + Pi/4 + (e*x)/2]])/(16*b^5*e) + (a*Cos[d + e*x] + b*Sin[d + e*x])/(16*b^2*e*
(a + b*Cos[d + e*x] - a*Sin[d + e*x])^2) - (3*(a^2*Cos[d + e*x] + a*b*Sin[d + e*x]))/(16*b^4*e*(a + b*Cos[d +
e*x] - a*Sin[d + e*x]))

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Rubi [A]  time = 0.107658, antiderivative size = 142, 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, 3153, 3122, 31} \[ \frac{\left (3 a^2+b^2\right ) \log \left (a+b \tan \left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right )\right )}{16 b^5 e}-\frac{3 \left (a^2 \cos (d+e x)+a b \sin (d+e x)\right )}{16 b^4 e (a (-\sin (d+e x))+a+b \cos (d+e x))}+\frac{a \cos (d+e x)+b \sin (d+e x)}{16 b^2 e (a (-\sin (d+e x))+a+b \cos (d+e x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(2*a + 2*b*Cos[d + e*x] - 2*a*Sin[d + e*x])^(-3),x]

[Out]

((3*a^2 + b^2)*Log[a + b*Tan[d/2 + Pi/4 + (e*x)/2]])/(16*b^5*e) + (a*Cos[d + e*x] + b*Sin[d + e*x])/(16*b^2*e*
(a + b*Cos[d + e*x] - a*Sin[d + e*x])^2) - (3*(a^2*Cos[d + e*x] + a*b*Sin[d + e*x]))/(16*b^4*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 3153

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

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))^3} \, dx &=\frac{a \cos (d+e x)+b \sin (d+e x)}{16 b^2 e (a+b \cos (d+e x)-a \sin (d+e x))^2}+\frac{\int \frac{-4 a+2 b \cos (d+e x)-2 a \sin (d+e x)}{(2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2} \, dx}{8 b^2}\\ &=\frac{a \cos (d+e x)+b \sin (d+e x)}{16 b^2 e (a+b \cos (d+e x)-a \sin (d+e x))^2}-\frac{3 \left (a^2 \cos (d+e x)+a b \sin (d+e x)\right )}{16 b^4 e (a+b \cos (d+e x)-a \sin (d+e x))}+\frac{\left (3 a^2+b^2\right ) \int \frac{1}{2 a+2 b \cos (d+e x)-2 a \sin (d+e x)} \, dx}{8 b^4}\\ &=\frac{a \cos (d+e x)+b \sin (d+e x)}{16 b^2 e (a+b \cos (d+e x)-a \sin (d+e x))^2}-\frac{3 \left (a^2 \cos (d+e x)+a b \sin (d+e x)\right )}{16 b^4 e (a+b \cos (d+e x)-a \sin (d+e x))}+\frac{\left (3 a^2+b^2\right ) \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 )}{8 b^4 e}\\ &=\frac{\left (3 a^2+b^2\right ) \log \left (a+b \tan \left (\frac{d}{2}+\frac{\pi }{4}+\frac{e x}{2}\right )\right )}{16 b^5 e}+\frac{a \cos (d+e x)+b \sin (d+e x)}{16 b^2 e (a+b \cos (d+e x)-a \sin (d+e x))^2}-\frac{3 \left (a^2 \cos (d+e x)+a b \sin (d+e x)\right )}{16 b^4 e (a+b \cos (d+e x)-a \sin (d+e x))}\\ \end{align*}

Mathematica [A]  time = 2.62173, size = 261, normalized size = 1.84 \[ -\frac{\frac{b^2 \left (a^2+b^2\right )}{\left ((b-a) \sin \left (\frac{1}{2} (d+e x)\right )+(a+b) \cos \left (\frac{1}{2} (d+e x)\right )\right )^2}+\frac{6 a b \left (a^2+b^2\right ) \sin \left (\frac{1}{2} (d+e x)\right )}{(a+b) \left ((b-a) \sin \left (\frac{1}{2} (d+e x)\right )+(a+b) \cos \left (\frac{1}{2} (d+e x)\right )\right )}+2 \left (3 a^2+b^2\right ) \log \left (\cos \left (\frac{1}{2} (d+e x)\right )-\sin \left (\frac{1}{2} (d+e x)\right )\right )-2 \left (3 a^2+b^2\right ) \log \left ((b-a) \sin \left (\frac{1}{2} (d+e x)\right )+(a+b) \cos \left (\frac{1}{2} (d+e x)\right )\right )+\frac{6 a b \sin \left (\frac{1}{2} (d+e x)\right )}{\cos \left (\frac{1}{2} (d+e x)\right )-\sin \left (\frac{1}{2} (d+e x)\right )}-\frac{b^2}{\left (\cos \left (\frac{1}{2} (d+e x)\right )-\sin \left (\frac{1}{2} (d+e x)\right )\right )^2}}{32 b^5 e} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(2*a + 2*b*Cos[d + e*x] - 2*a*Sin[d + e*x])^(-3),x]

[Out]

-(2*(3*a^2 + b^2)*Log[Cos[(d + e*x)/2] - Sin[(d + e*x)/2]] - 2*(3*a^2 + b^2)*Log[(a + b)*Cos[(d + e*x)/2] + (-
a + b)*Sin[(d + e*x)/2]] - b^2/(Cos[(d + e*x)/2] - Sin[(d + e*x)/2])^2 + (6*a*b*Sin[(d + e*x)/2])/(Cos[(d + e*
x)/2] - Sin[(d + e*x)/2]) + (b^2*(a^2 + b^2))/((a + b)*Cos[(d + e*x)/2] + (-a + b)*Sin[(d + e*x)/2])^2 + (6*a*
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])))/(32*b^5*e)

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Maple [B]  time = 0.17, size = 687, normalized size = 4.8 \begin{align*} \text{result too large to display} \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))^3,x)

[Out]

1/16/e/b^3/(tan(1/2*d+1/2*e*x)-1)^2+3/16/e/b^4/(tan(1/2*d+1/2*e*x)-1)*a+1/16/e/b^3/(tan(1/2*d+1/2*e*x)-1)-3/16
/e/b^5*ln(tan(1/2*d+1/2*e*x)-1)*a^2-1/16/e/b^3*ln(tan(1/2*d+1/2*e*x)-1)+3/16/e/b^5/(a-b)*ln(a*tan(1/2*d+1/2*e*
x)-b*tan(1/2*d+1/2*e*x)-a-b)*a^3-3/16/e/b^4/(a-b)*ln(a*tan(1/2*d+1/2*e*x)-b*tan(1/2*d+1/2*e*x)-a-b)*a^2+1/16/e
/b^3/(a-b)*ln(a*tan(1/2*d+1/2*e*x)-b*tan(1/2*d+1/2*e*x)-a-b)*a-1/16/e/b^2/(a-b)*ln(a*tan(1/2*d+1/2*e*x)-b*tan(
1/2*d+1/2*e*x)-a-b)-1/16/e/b^3/(a-b)^2/(a*tan(1/2*d+1/2*e*x)-b*tan(1/2*d+1/2*e*x)-a-b)^2*a^4-1/8/e/b/(a-b)^2/(
a*tan(1/2*d+1/2*e*x)-b*tan(1/2*d+1/2*e*x)-a-b)^2*a^2-1/16/e*b/(a-b)^2/(a*tan(1/2*d+1/2*e*x)-b*tan(1/2*d+1/2*e*
x)-a-b)^2+3/16/e/b^4/(a-b)^2/(a*tan(1/2*d+1/2*e*x)-b*tan(1/2*d+1/2*e*x)-a-b)*a^4-1/4/e/b^3/(a-b)^2/(a*tan(1/2*
d+1/2*e*x)-b*tan(1/2*d+1/2*e*x)-a-b)*a^3+1/8/e/b^2/(a-b)^2/(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/b/(a-b)^2/(a*tan(1/2*d+1/2*e*x)-b*tan(1/2*d+1/2*e*x)-a-b)*a-1/16/e/(a-b)^2/(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 = 1.10031, size = 663, normalized size = 4.67 \begin{align*} -\frac{\frac{2 \,{\left (3 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4} - \frac{{\left (9 \, a^{5} - 9 \, a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 5 \, a b^{4} + b^{5}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac{{\left (9 \, a^{5} - 18 \, a^{4} b + 12 \, a^{3} b^{2} - 6 \, a^{2} b^{3} + a b^{4}\right )} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - \frac{{\left (3 \, a^{5} - 9 \, a^{4} b + 10 \, a^{3} b^{2} - 6 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}\right )}}{a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8} - \frac{4 \,{\left (a^{4} b^{4} - a^{3} b^{5} - a^{2} b^{6} + a b^{7}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac{2 \,{\left (3 \, a^{4} b^{4} - 6 \, a^{3} b^{5} + 2 \, a^{2} b^{6} + 2 \, a b^{7} - b^{8}\right )} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - \frac{4 \,{\left (a^{4} b^{4} - 3 \, a^{3} b^{5} + 3 \, a^{2} b^{6} - a b^{7}\right )} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}} + \frac{{\left (a^{4} b^{4} - 4 \, a^{3} b^{5} + 6 \, a^{2} b^{6} - 4 \, a b^{7} + b^{8}\right )} \sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}}} - \frac{{\left (3 \, a^{2} + b^{2}\right )} \log \left (a + b - \frac{{\left (a - b\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{b^{5}} + \frac{{\left (3 \, a^{2} + b^{2}\right )} \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - 1\right )}{b^{5}}}{16 \, 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))^3,x, algorithm="maxima")

[Out]

-1/16*(2*(3*a^5 - 4*a^3*b^2 - a*b^4 - (9*a^5 - 9*a^4*b - 2*a^3*b^2 + 2*a^2*b^3 - 5*a*b^4 + b^5)*sin(e*x + d)/(
cos(e*x + d) + 1) + (9*a^5 - 18*a^4*b + 12*a^3*b^2 - 6*a^2*b^3 + a*b^4)*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 -
(3*a^5 - 9*a^4*b + 10*a^3*b^2 - 6*a^2*b^3 + a*b^4 + b^5)*sin(e*x + d)^3/(cos(e*x + d) + 1)^3)/(a^4*b^4 - 2*a^2
*b^6 + b^8 - 4*(a^4*b^4 - a^3*b^5 - a^2*b^6 + a*b^7)*sin(e*x + d)/(cos(e*x + d) + 1) + 2*(3*a^4*b^4 - 6*a^3*b^
5 + 2*a^2*b^6 + 2*a*b^7 - b^8)*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 - 4*(a^4*b^4 - 3*a^3*b^5 + 3*a^2*b^6 - a*b^
7)*sin(e*x + d)^3/(cos(e*x + d) + 1)^3 + (a^4*b^4 - 4*a^3*b^5 + 6*a^2*b^6 - 4*a*b^7 + b^8)*sin(e*x + d)^4/(cos
(e*x + d) + 1)^4) - (3*a^2 + b^2)*log(a + b - (a - b)*sin(e*x + d)/(cos(e*x + d) + 1))/b^5 + (3*a^2 + b^2)*log
(sin(e*x + d)/(cos(e*x + d) + 1) - 1)/b^5)/e

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Fricas [B]  time = 2.4585, size = 942, normalized size = 6.63 \begin{align*} -\frac{12 \, a^{2} b^{2} \cos \left (e x + d\right )^{2} - 6 \, a^{2} b^{2} + 2 \,{\left (3 \, a^{3} b - a b^{3}\right )} \cos \left (e x + d\right ) -{\left (6 \, a^{4} + 2 \, a^{2} b^{2} -{\left (3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (e x + d\right )^{2} + 2 \,{\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (e x + d\right ) - 2 \,{\left (3 \, a^{4} + a^{2} b^{2} +{\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\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 (6 \, a^{4} + 2 \, a^{2} b^{2} -{\left (3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (e x + d\right )^{2} + 2 \,{\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (e x + d\right ) - 2 \,{\left (3 \, a^{4} + a^{2} b^{2} +{\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )} \log \left (-\sin \left (e x + d\right ) + 1\right ) + 2 \,{\left (3 \, a^{2} b^{2} - b^{4} - 3 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{32 \,{\left (2 \, a b^{6} e \cos \left (e x + d\right ) + 2 \, a^{2} b^{5} e -{\left (a^{2} b^{5} - b^{7}\right )} e \cos \left (e x + d\right )^{2} - 2 \,{\left (a b^{6} e \cos \left (e x + d\right ) + a^{2} b^{5} e\right )} \sin \left (e x + d\right )\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))^3,x, algorithm="fricas")

[Out]

-1/32*(12*a^2*b^2*cos(e*x + d)^2 - 6*a^2*b^2 + 2*(3*a^3*b - a*b^3)*cos(e*x + d) - (6*a^4 + 2*a^2*b^2 - (3*a^4
- 2*a^2*b^2 - b^4)*cos(e*x + d)^2 + 2*(3*a^3*b + a*b^3)*cos(e*x + d) - 2*(3*a^4 + a^2*b^2 + (3*a^3*b + a*b^3)*
cos(e*x + d))*sin(e*x + d))*log(2*a*b*cos(e*x + d) + a^2 + b^2 - (a^2 - b^2)*sin(e*x + d)) + (6*a^4 + 2*a^2*b^
2 - (3*a^4 - 2*a^2*b^2 - b^4)*cos(e*x + d)^2 + 2*(3*a^3*b + a*b^3)*cos(e*x + d) - 2*(3*a^4 + a^2*b^2 + (3*a^3*
b + a*b^3)*cos(e*x + d))*sin(e*x + d))*log(-sin(e*x + d) + 1) + 2*(3*a^2*b^2 - b^4 - 3*(a^3*b - a*b^3)*cos(e*x
+ d))*sin(e*x + d))/(2*a*b^6*e*cos(e*x + d) + 2*a^2*b^5*e - (a^2*b^5 - b^7)*e*cos(e*x + d)^2 - 2*(a*b^6*e*cos
(e*x + d) + a^2*b^5*e)*sin(e*x + d))

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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))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.20773, size = 659, normalized size = 4.64 \begin{align*} \text{result too large to display} \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))^3,x, algorithm="giac")

[Out]

1/16*((3*a^3 - 3*a^2*b + a*b^2 - b^3)*log(abs(a*tan(1/2*x*e + 1/2*d) - b*tan(1/2*x*e + 1/2*d) - a - b))/(a*b^5
- b^6) + 2*(3*a^5*tan(1/2*x*e + 1/2*d)^3 - 9*a^4*b*tan(1/2*x*e + 1/2*d)^3 + 10*a^3*b^2*tan(1/2*x*e + 1/2*d)^3
- 6*a^2*b^3*tan(1/2*x*e + 1/2*d)^3 + a*b^4*tan(1/2*x*e + 1/2*d)^3 + b^5*tan(1/2*x*e + 1/2*d)^3 - 9*a^5*tan(1/
2*x*e + 1/2*d)^2 + 18*a^4*b*tan(1/2*x*e + 1/2*d)^2 - 12*a^3*b^2*tan(1/2*x*e + 1/2*d)^2 + 6*a^2*b^3*tan(1/2*x*e
+ 1/2*d)^2 - a*b^4*tan(1/2*x*e + 1/2*d)^2 + 9*a^5*tan(1/2*x*e + 1/2*d) - 9*a^4*b*tan(1/2*x*e + 1/2*d) - 2*a^3
*b^2*tan(1/2*x*e + 1/2*d) + 2*a^2*b^3*tan(1/2*x*e + 1/2*d) - 5*a*b^4*tan(1/2*x*e + 1/2*d) + b^5*tan(1/2*x*e +
1/2*d) - 3*a^5 + 4*a^3*b^2 + a*b^4)/((a^2*b^4 - 2*a*b^5 + b^6)*(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)^2) - (3*a^2 + b^2)*log(abs(tan(1/2*x*e + 1/2*d) - 1))/b^5)*e^(-1)