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

Optimal. Leaf size=134 \[ \frac{3 \left (a^2 \sin (d+e x)+a c \cos (d+e x)\right )}{16 c^4 e (a (-\cos (d+e x))+a+c \sin (d+e x))}-\frac{\left (3 a^2+c^2\right ) \log \left (a+c \cot \left (\frac{1}{2} (d+e x)\right )\right )}{16 c^5 e}-\frac{a \sin (d+e x)+c \cos (d+e x)}{16 c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))^2} \]

[Out]

-((3*a^2 + c^2)*Log[a + c*Cot[(d + e*x)/2]])/(16*c^5*e) - (c*Cos[d + e*x] + a*Sin[d + e*x])/(16*c^2*e*(a - a*C
os[d + e*x] + c*Sin[d + e*x])^2) + (3*(a*c*Cos[d + e*x] + a^2*Sin[d + e*x]))/(16*c^4*e*(a - a*Cos[d + e*x] + c
*Sin[d + e*x]))

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Rubi [A]  time = 0.112631, antiderivative size = 134, 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, 3121, 31} \[ \frac{3 \left (a^2 \sin (d+e x)+a c \cos (d+e x)\right )}{16 c^4 e (a (-\cos (d+e x))+a+c \sin (d+e x))}-\frac{\left (3 a^2+c^2\right ) \log \left (a+c \cot \left (\frac{1}{2} (d+e x)\right )\right )}{16 c^5 e}-\frac{a \sin (d+e x)+c \cos (d+e x)}{16 c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a^2 + c^2)*Log[a + c*Cot[(d + e*x)/2]])/(16*c^5*e) - (c*Cos[d + e*x] + a*Sin[d + e*x])/(16*c^2*e*(a - a*C
os[d + e*x] + c*Sin[d + e*x])^2) + (3*(a*c*Cos[d + e*x] + a^2*Sin[d + e*x]))/(16*c^4*e*(a - a*Cos[d + e*x] + c
*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 3121

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

Mathematica [C]  time = 0.610723, size = 350, normalized size = 2.61 \[ \frac{\sin \left (\frac{1}{2} (d+e x)\right ) \left (a \sin \left (\frac{1}{2} (d+e x)\right )+c \cos \left (\frac{1}{2} (d+e x)\right )\right ) \left (-6 a \left (a^2+c^2\right ) \sin ^3\left (\frac{1}{2} (d+e x)\right ) \left (a \sin \left (\frac{1}{2} (d+e x)\right )+c \cos \left (\frac{1}{2} (d+e x)\right )\right )+4 \left (3 a^2+c^2\right ) \sin ^2\left (\frac{1}{2} (d+e x)\right ) \log \left (\sin \left (\frac{1}{2} (d+e x)\right )\right ) \left (a \sin \left (\frac{1}{2} (d+e x)\right )+c \cos \left (\frac{1}{2} (d+e x)\right )\right )^2-4 \left (3 a^2+c^2\right ) \sin ^2\left (\frac{1}{2} (d+e x)\right ) \left (a \sin \left (\frac{1}{2} (d+e x)\right )+c \cos \left (\frac{1}{2} (d+e x)\right )\right )^2 \log \left (a \sin \left (\frac{1}{2} (d+e x)\right )+c \cos \left (\frac{1}{2} (d+e x)\right )\right )+c^2 (c-i a) (c+i a) \sin ^2\left (\frac{1}{2} (d+e x)\right )-c^2 \left (a \sin \left (\frac{1}{2} (d+e x)\right )+c \cos \left (\frac{1}{2} (d+e x)\right )\right )^2+3 a c \sin (d+e x) \left (a \sin \left (\frac{1}{2} (d+e x)\right )+c \cos \left (\frac{1}{2} (d+e x)\right )\right )^2\right )}{8 c^5 e (a (-\cos (d+e x))+a+c \sin (d+e x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[(d + e*x)/2]*(c*Cos[(d + e*x)/2] + a*Sin[(d + e*x)/2])*(c^2*((-I)*a + c)*(I*a + c)*Sin[(d + e*x)/2]^2 - 6
*a*(a^2 + c^2)*Sin[(d + e*x)/2]^3*(c*Cos[(d + e*x)/2] + a*Sin[(d + e*x)/2]) - c^2*(c*Cos[(d + e*x)/2] + a*Sin[
(d + e*x)/2])^2 + 4*(3*a^2 + c^2)*Log[Sin[(d + e*x)/2]]*Sin[(d + e*x)/2]^2*(c*Cos[(d + e*x)/2] + a*Sin[(d + e*
x)/2])^2 - 4*(3*a^2 + c^2)*Log[c*Cos[(d + e*x)/2] + a*Sin[(d + e*x)/2]]*Sin[(d + e*x)/2]^2*(c*Cos[(d + e*x)/2]
+ a*Sin[(d + e*x)/2])^2 + 3*a*c*(c*Cos[(d + e*x)/2] + a*Sin[(d + e*x)/2])^2*Sin[d + e*x]))/(8*c^5*e*(a - a*Co
s[d + e*x] + c*Sin[d + e*x])^3)

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Maple [B]  time = 0.196, size = 272, normalized size = 2. \begin{align*}{\frac{3\,{a}^{2}}{32\,{c}^{4}e} \left ( c+a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-1}}+{\frac{1}{16\,{c}^{2}e} \left ( c+a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-1}}-{\frac{1}{32\,{a}^{2}e} \left ( c+a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-1}}+{\frac{{a}^{2}}{64\,e{c}^{3}} \left ( c+a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-2}}+{\frac{1}{32\,ce} \left ( c+a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-2}}+{\frac{c}{64\,{a}^{2}e} \left ( c+a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-2}}-{\frac{3\,{a}^{2}}{16\,e{c}^{5}}\ln \left ( c+a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) }-{\frac{1}{16\,e{c}^{3}}\ln \left ( c+a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) }-{\frac{1}{64\,e{c}^{3}} \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-2}}+{\frac{3\,{a}^{2}}{16\,e{c}^{5}}\ln \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) }+{\frac{1}{16\,e{c}^{3}}\ln \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) }+{\frac{3\,a}{32\,{c}^{4}e} \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*a-2*a*cos(e*x+d)+2*c*sin(e*x+d))^3,x)

[Out]

3/32/e*a^2/c^4/(c+a*tan(1/2*d+1/2*e*x))+1/16/e/c^2/(c+a*tan(1/2*d+1/2*e*x))-1/32/e/a^2/(c+a*tan(1/2*d+1/2*e*x)
)+1/64/e*a^2/c^3/(c+a*tan(1/2*d+1/2*e*x))^2+1/32/e/c/(c+a*tan(1/2*d+1/2*e*x))^2+1/64/e/a^2*c/(c+a*tan(1/2*d+1/
2*e*x))^2-3/16/e/c^5*ln(c+a*tan(1/2*d+1/2*e*x))*a^2-1/16/e/c^3*ln(c+a*tan(1/2*d+1/2*e*x))-1/64/e/c^3/tan(1/2*d
+1/2*e*x)^2+3/16/e/c^5*ln(tan(1/2*d+1/2*e*x))*a^2+1/16/e/c^3*ln(tan(1/2*d+1/2*e*x))+3/32/e/c^4*a/tan(1/2*d+1/2
*e*x)

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Maxima [B]  time = 1.06342, size = 358, normalized size = 2.67 \begin{align*} -\frac{\frac{a^{2} c^{3} - \frac{4 \, a^{3} c^{2} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - \frac{{\left (18 \, a^{4} c + 6 \, a^{2} c^{3} - c^{5}\right )} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - \frac{2 \,{\left (6 \, a^{5} + 2 \, a^{3} c^{2} - a c^{4}\right )} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}}{\frac{a^{2} c^{6} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac{2 \, a^{3} c^{5} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}} + \frac{a^{4} c^{4} \sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}}} + \frac{4 \,{\left (3 \, a^{2} + c^{2}\right )} \log \left (c + \frac{a \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{c^{5}} - \frac{4 \,{\left (3 \, a^{2} + c^{2}\right )} \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{c^{5}}}{64 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*a-2*a*cos(e*x+d)+2*c*sin(e*x+d))^3,x, algorithm="maxima")

[Out]

-1/64*((a^2*c^3 - 4*a^3*c^2*sin(e*x + d)/(cos(e*x + d) + 1) - (18*a^4*c + 6*a^2*c^3 - c^5)*sin(e*x + d)^2/(cos
(e*x + d) + 1)^2 - 2*(6*a^5 + 2*a^3*c^2 - a*c^4)*sin(e*x + d)^3/(cos(e*x + d) + 1)^3)/(a^2*c^6*sin(e*x + d)^2/
(cos(e*x + d) + 1)^2 + 2*a^3*c^5*sin(e*x + d)^3/(cos(e*x + d) + 1)^3 + a^4*c^4*sin(e*x + d)^4/(cos(e*x + d) +
1)^4) + 4*(3*a^2 + c^2)*log(c + a*sin(e*x + d)/(cos(e*x + d) + 1))/c^5 - 4*(3*a^2 + c^2)*log(sin(e*x + d)/(cos
(e*x + d) + 1))/c^5)/e

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Fricas [B]  time = 2.46851, size = 987, normalized size = 7.37 \begin{align*} \frac{12 \, a^{2} c^{2} \cos \left (e x + d\right )^{2} - 6 \, a^{2} c^{2} - 2 \,{\left (3 \, a^{2} c^{2} - c^{4}\right )} \cos \left (e x + d\right ) +{\left (3 \, a^{4} + 4 \, a^{2} c^{2} + c^{4} +{\left (3 \, a^{4} - 2 \, a^{2} c^{2} - c^{4}\right )} \cos \left (e x + d\right )^{2} - 2 \,{\left (3 \, a^{4} + a^{2} c^{2}\right )} \cos \left (e x + d\right ) + 2 \,{\left (3 \, a^{3} c + a c^{3} -{\left (3 \, a^{3} c + a c^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\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 (3 \, a^{4} + 4 \, a^{2} c^{2} + c^{4} +{\left (3 \, a^{4} - 2 \, a^{2} c^{2} - c^{4}\right )} \cos \left (e x + d\right )^{2} - 2 \,{\left (3 \, a^{4} + a^{2} c^{2}\right )} \cos \left (e x + d\right ) + 2 \,{\left (3 \, a^{3} c + a c^{3} -{\left (3 \, a^{3} c + a c^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (e x + d\right ) + \frac{1}{2}\right ) - 2 \,{\left (3 \, a^{3} c - a c^{3} - 3 \,{\left (a^{3} c - a c^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{32 \,{\left (2 \, a^{2} c^{5} e \cos \left (e x + d\right ) -{\left (a^{2} c^{5} - c^{7}\right )} e \cos \left (e x + d\right )^{2} -{\left (a^{2} c^{5} + c^{7}\right )} e + 2 \,{\left (a c^{6} e \cos \left (e x + d\right ) - a c^{6} 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*a*cos(e*x+d)+2*c*sin(e*x+d))^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.29775, size = 323, normalized size = 2.41 \begin{align*} \frac{1}{64} \,{\left (\frac{4 \,{\left (3 \, a^{2} + c^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) \right |}\right )}{c^{5}} - \frac{4 \,{\left (3 \, a^{3} + a c^{2}\right )} \log \left ({\left | a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + c \right |}\right )}{a c^{5}} + \frac{12 \, a^{5} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} + 4 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} - 2 \, a c^{4} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} + 18 \, a^{4} c \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 6 \, a^{2} c^{3} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} - c^{5} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 4 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - a^{2} c^{3}}{{\left (a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + c \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )\right )}^{2} a^{2} c^{4}}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*a-2*a*cos(e*x+d)+2*c*sin(e*x+d))^3,x, algorithm="giac")

[Out]

1/64*(4*(3*a^2 + c^2)*log(abs(tan(1/2*x*e + 1/2*d)))/c^5 - 4*(3*a^3 + a*c^2)*log(abs(a*tan(1/2*x*e + 1/2*d) +
c))/(a*c^5) + (12*a^5*tan(1/2*x*e + 1/2*d)^3 + 4*a^3*c^2*tan(1/2*x*e + 1/2*d)^3 - 2*a*c^4*tan(1/2*x*e + 1/2*d)
^3 + 18*a^4*c*tan(1/2*x*e + 1/2*d)^2 + 6*a^2*c^3*tan(1/2*x*e + 1/2*d)^2 - c^5*tan(1/2*x*e + 1/2*d)^2 + 4*a^3*c
^2*tan(1/2*x*e + 1/2*d) - a^2*c^3)/((a*tan(1/2*x*e + 1/2*d)^2 + c*tan(1/2*x*e + 1/2*d))^2*a^2*c^4))*e^(-1)