\(\int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^3} \, dx\) [386]

Optimal result

Integrand size = 24, antiderivative size = 142 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=-\frac {\left (3 a^2+b^2\right ) \log \left (a+b \cot \left (\frac {d}{2}+\frac {\pi }{4}+\frac {e x}{2}\right )\right )}{16 b^5 e}-\frac {a \cos (d+e x)-b \sin (d+e x)}{16 b^2 e (a+b \cos (d+e x)+a \sin (d+e x))^2}+\frac {3 \left (a^2 \cos (d+e x)-a b \sin (d+e x)\right )}{16 b^4 e (a+b \cos (d+e x)+a \sin (d+e x))} \]

[Out]

-1/16*(3*a^2+b^2)*ln(a+b*cot(1/2*d+1/4*Pi+1/2*e*x))/b^5/e+1/16*(-a*cos(e*x+d)+b*sin(e*x+d))/b^2/e/(a+b*cos(e*x
+d)+a*sin(e*x+d))^2+3/16*(a^2*cos(e*x+d)-a*b*sin(e*x+d))/b^4/e/(a+b*cos(e*x+d)+a*sin(e*x+d))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 142, 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, 3232, 3202, 31} \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=\frac {3 \left (a^2 \cos (d+e x)-a b \sin (d+e x)\right )}{16 b^4 e (a \sin (d+e x)+a+b \cos (d+e x))}-\frac {\left (3 a^2+b^2\right ) \log \left (a+b \cot \left (\frac {d}{2}+\frac {e x}{2}+\frac {\pi }{4}\right )\right )}{16 b^5 e}-\frac {a \cos (d+e x)-b \sin (d+e x)}{16 b^2 e (a \sin (d+e x)+a+b \cos (d+e x))^2} \]

[In]

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

[Out]

-1/16*((3*a^2 + b^2)*Log[a + b*Cot[d/2 + Pi/4 + (e*x)/2]])/(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 31

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

Rule 3202

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Cot[(d + e*x)/2 + Pi/4], x]}, Dist[-f/e, Subst[Int[1/(a + b*f*x), x], x, Cot[(d + e*x)/2 + Pi/4]/f], x
]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a - c, 0] && NeQ[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]

Rule 3232

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]

Rubi steps \begin{align*} \text {integral}& = -\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 ) \text {Subst}\left (\int \frac {1}{2 a+2 b x} \, dx,x,\cot \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 \cot \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] (verified)

Time = 1.92 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.80 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=-\frac {-2 \left (3 a^2+b^2\right ) \log \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )+2 \left (3 a^2+b^2\right ) \log \left ((a+b) \cos \left (\frac {1}{2} (d+e x)\right )+(a-b) \sin \left (\frac {1}{2} (d+e x)\right )\right )+\frac {b^2}{\left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )^2}+\frac {6 a b \sin \left (\frac {1}{2} (d+e x)\right )}{\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )}-\frac {b^2 \left (a^2+b^2\right )}{\left ((a+b) \cos \left (\frac {1}{2} (d+e x)\right )+(a-b) \sin \left (\frac {1}{2} (d+e x)\right )\right )^2}+\frac {6 a b \left (a^2+b^2\right ) \sin \left (\frac {1}{2} (d+e x)\right )}{(a+b) \left ((a+b) \cos \left (\frac {1}{2} (d+e x)\right )+(a-b) \sin \left (\frac {1}{2} (d+e x)\right )\right )}}{32 b^5 e} \]

[In]

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

[Out]

-1/32*(-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])))/(b^5*e)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(263\) vs. \(2(130)=260\).

Time = 1.95 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.86

method	result	size
derivativedivides	\(\frac {-\frac {1}{2 b^{3} \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}-\frac {-3 a -b}{2 b^{4} \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}+\frac {\left (3 a^{2}+b^{2}\right ) \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{2 b^{5}}+\frac {\left (-3 a^{3}+3 a^{2} b -a \,b^{2}+b^{3}\right ) \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )}{2 b^{5} \left (a -b \right )}-\frac {-a^{4}-2 a^{2} b^{2}-b^{4}}{2 b^{3} \left (a -b \right )^{2} \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )^{2}}-\frac {-3 a^{4}+4 a^{3} b -2 a^{2} b^{2}+4 a \,b^{3}+b^{4}}{2 b^{4} \left (a -b \right )^{2} \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )}}{8 e}\)	\(264\)
default	\(\frac {-\frac {1}{2 b^{3} \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}-\frac {-3 a -b}{2 b^{4} \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}+\frac {\left (3 a^{2}+b^{2}\right ) \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{2 b^{5}}+\frac {\left (-3 a^{3}+3 a^{2} b -a \,b^{2}+b^{3}\right ) \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )}{2 b^{5} \left (a -b \right )}-\frac {-a^{4}-2 a^{2} b^{2}-b^{4}}{2 b^{3} \left (a -b \right )^{2} \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )^{2}}-\frac {-3 a^{4}+4 a^{3} b -2 a^{2} b^{2}+4 a \,b^{3}+b^{4}}{2 b^{4} \left (a -b \right )^{2} \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )}}{8 e}\)	\(264\)
risch	\(-\frac {i \left (3 a^{2} b \,{\mathrm e}^{3 i \left (e x +d \right )}+b^{3} {\mathrm e}^{3 i \left (e x +d \right )}+9 a^{3} {\mathrm e}^{2 i \left (e x +d \right )}+3 a \,b^{2} {\mathrm e}^{2 i \left (e x +d \right )}-3 i a^{3} {\mathrm e}^{3 i \left (e x +d \right )}-i a \,b^{2} {\mathrm e}^{3 i \left (e x +d \right )}+9 a^{2} b \,{\mathrm e}^{i \left (e x +d \right )}-b^{3} {\mathrm e}^{i \left (e x +d \right )}-3 a^{3}+3 a \,b^{2}+9 i a^{3} {\mathrm e}^{i \left (e x +d \right )}-i {\mathrm e}^{i \left (e x +d \right )} a \,b^{2}+6 i b \,a^{2}\right )}{8 \left (-i a \,{\mathrm e}^{2 i \left (e x +d \right )}+b \,{\mathrm e}^{2 i \left (e x +d \right )}+i a +2 a \,{\mathrm e}^{i \left (e x +d \right )}+b \right )^{2} b^{4} e}-\frac {3 \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a +b}{i b +a}\right ) a^{2}}{16 b^{5} e}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a +b}{i b +a}\right )}{16 b^{3} e}+\frac {3 \ln \left ({\mathrm e}^{i \left (e x +d \right )}+i\right ) a^{2}}{16 b^{5} e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+i\right )}{16 b^{3} e}\)	\(338\)
norman	\(\frac {\frac {9 a^{5}+18 a^{4} b +12 a^{3} b^{2}+6 a^{2} b^{3}+a \,b^{4}}{16 b^{4} e \left (3 a^{2}-b^{2}\right )}-\frac {\left (9 a^{5}-9 a^{4} b +6 a^{3} b^{2}-6 a^{2} b^{3}+3 a \,b^{4}+b^{5}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{8 b^{4} e \left (3 a^{2}-b^{2}\right )}+\frac {\left (9 a^{5}+9 a^{4} b +6 a^{3} b^{2}+6 a^{2} b^{3}+3 a \,b^{4}-b^{5}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{8 b^{4} e \left (3 a^{2}-b^{2}\right )}-\frac {\left (9 a^{5}-18 a^{4} b +12 a^{3} b^{2}-6 a^{2} b^{3}+a \,b^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{16 b^{4} e \left (3 a^{2}-b^{2}\right )}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2} \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )^{2}}+\frac {\left (3 a^{2}+b^{2}\right ) \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{16 b^{5} e}-\frac {\left (3 a^{2}+b^{2}\right ) \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )}{16 b^{5} e}\)	\(375\)
parallelrisch	\(\frac {-9 \left (a^{2}+\frac {b^{2}}{3}\right ) \left (\left (a^{2}-b^{2}\right ) \cos \left (2 e x +2 d \right )-4 a^{2} \sin \left (e x +d \right )-4 a b \cos \left (e x +d \right )-2 a b \sin \left (2 e x +2 d \right )-3 a^{2}-b^{2}\right ) \left (a^{2}-\frac {b^{2}}{3}\right ) \ln \left (a +b +\left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+9 \left (a^{2}+\frac {b^{2}}{3}\right ) \left (\left (a^{2}-b^{2}\right ) \cos \left (2 e x +2 d \right )-4 a^{2} \sin \left (e x +d \right )-4 a b \cos \left (e x +d \right )-2 a b \sin \left (2 e x +2 d \right )-3 a^{2}-b^{2}\right ) \left (a^{2}-\frac {b^{2}}{3}\right ) \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\left (-9 a^{4} b^{2}-3 a^{2} b^{4}\right ) \cos \left (2 e x +2 d \right )+\left (-9 a^{5} b -6 a^{3} b^{3}-3 a \,b^{5}\right ) \sin \left (2 e x +2 d \right )+\left (-18 a^{5} b -24 a^{3} b^{3}-2 a \,b^{5}\right ) \cos \left (e x +d \right )+\left (-18 a^{4} b^{2}-12 a^{2} b^{4}+2 b^{6}\right ) \sin \left (e x +d \right )-27 a^{4} b^{2}-9 a^{2} b^{4}}{48 b^{5} \left (\left (a^{2}-b^{2}\right ) \cos \left (2 e x +2 d \right )-4 a^{2} \sin \left (e x +d \right )-4 a b \cos \left (e x +d \right )-2 a b \sin \left (2 e x +2 d \right )-3 a^{2}-b^{2}\right ) e \left (a^{2}-\frac {b^{2}}{3}\right )}\)	\(417\)

[In]

int(1/(2*a+2*b*cos(e*x+d)+2*a*sin(e*x+d))^3,x,method=_RETURNVERBOSE)

[Out]

1/8/e*(-1/2/b^3/(1+tan(1/2*e*x+1/2*d))^2-1/2*(-3*a-b)/b^4/(1+tan(1/2*e*x+1/2*d))+1/2*(3*a^2+b^2)/b^5*ln(1+tan(
1/2*e*x+1/2*d))+1/2*(-3*a^3+3*a^2*b-a*b^2+b^3)/b^5/(a-b)*ln(a*tan(1/2*e*x+1/2*d)-b*tan(1/2*e*x+1/2*d)+a+b)-1/2
*(-a^4-2*a^2*b^2-b^4)/b^3/(a-b)^2/(a*tan(1/2*e*x+1/2*d)-b*tan(1/2*e*x+1/2*d)+a+b)^2-1/2*(-3*a^4+4*a^3*b-2*a^2*
b^2+4*a*b^3+b^4)/b^4/(a-b)^2/(a*tan(1/2*e*x+1/2*d)-b*tan(1/2*e*x+1/2*d)+a+b))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 420 vs. \(2 (130) = 260\).

Time = 0.26 (sec) , antiderivative size = 420, normalized size of antiderivative = 2.96 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=\frac {12 \, a^{2} b^{2} \cos \left (e x + d\right )^{2} - 6 \, a^{2} b^{2} + 2 \, {\left (3 \, a^{3} b - a b^{3}\right )} \cos \left (e x + d\right ) - {\left (6 \, a^{4} + 2 \, a^{2} b^{2} - {\left (3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (e x + d\right )^{2} + 2 \, {\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (e x + d\right ) + 2 \, {\left (3 \, a^{4} + a^{2} b^{2} + {\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )} \log \left (2 \, a b \cos \left (e x + d\right ) + a^{2} + b^{2} + {\left (a^{2} - b^{2}\right )} \sin \left (e x + d\right )\right ) + {\left (6 \, a^{4} + 2 \, a^{2} b^{2} - {\left (3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (e x + d\right )^{2} + 2 \, {\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (e x + d\right ) + 2 \, {\left (3 \, a^{4} + a^{2} b^{2} + {\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )} \log \left (\sin \left (e x + d\right ) + 1\right ) - 2 \, {\left (3 \, a^{2} b^{2} - b^{4} - 3 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{32 \, {\left (2 \, a b^{6} e \cos \left (e x + d\right ) + 2 \, a^{2} b^{5} e - {\left (a^{2} b^{5} - b^{7}\right )} e \cos \left (e x + d\right )^{2} + 2 \, {\left (a b^{6} e \cos \left (e x + d\right ) + a^{2} b^{5} e\right )} \sin \left (e x + d\right )\right )}} \]

[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)*c
os(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))

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=\text {Timed out} \]

[In]

integrate(1/(2*a+2*b*cos(e*x+d)+2*a*sin(e*x+d))**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 493 vs. \(2 (130) = 260\).

Time = 0.23 (sec) , antiderivative size = 493, normalized size of antiderivative = 3.47 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=\frac {\frac {2 \, {\left (3 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4} + \frac {{\left (9 \, a^{5} - 9 \, a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 5 \, a b^{4} + b^{5}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {{\left (9 \, a^{5} - 18 \, a^{4} b + 12 \, a^{3} b^{2} - 6 \, a^{2} b^{3} + a b^{4}\right )} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac {{\left (3 \, a^{5} - 9 \, a^{4} b + 10 \, a^{3} b^{2} - 6 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}\right )}}{a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8} + \frac {4 \, {\left (a^{4} b^{4} - a^{3} b^{5} - a^{2} b^{6} + a b^{7}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {2 \, {\left (3 \, a^{4} b^{4} - 6 \, a^{3} b^{5} + 2 \, a^{2} b^{6} + 2 \, a b^{7} - b^{8}\right )} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac {4 \, {\left (a^{4} b^{4} - 3 \, a^{3} b^{5} + 3 \, a^{2} b^{6} - a b^{7}\right )} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}} + \frac {{\left (a^{4} b^{4} - 4 \, a^{3} b^{5} + 6 \, a^{2} b^{6} - 4 \, a b^{7} + b^{8}\right )} \sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}}} - \frac {{\left (3 \, a^{2} + b^{2}\right )} \log \left (-a - b - \frac {{\left (a - b\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{b^{5}} + \frac {{\left (3 \, a^{2} + b^{2}\right )} \log \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right )}{b^{5}}}{16 \, e} \]

[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)/(c
os(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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 458 vs. \(2 (130) = 260\).

Time = 0.30 (sec) , antiderivative size = 458, normalized size of antiderivative = 3.23 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=\frac {\frac {2 \, {\left (3 \, a^{5} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} - 9 \, a^{4} b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + 10 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} - 6 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + a b^{4} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + b^{5} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + 9 \, a^{5} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - 18 \, a^{4} b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 12 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - 6 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + a b^{4} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 9 \, a^{5} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 9 \, a^{4} b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 2 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 5 \, a b^{4} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + b^{5} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 3 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4}\right )}}{{\left (a^{2} b^{4} - 2 \, a b^{5} + b^{6}\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 )}^{2}} + \frac {{\left (3 \, a^{2} + b^{2}\right )} \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^{4} {\left | b \right |}}}{16 \, e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 30.88 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.54 \[ \int \frac {1}{(2 a+2 b \cos (d+e x)+2 a \sin (d+e x))^3} \, dx=-\frac {\frac {-3\,a^5+4\,a^3\,b^2+a\,b^4}{2\,b^4\,{\left (a-b\right )}^2}+\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (-9\,a^5+9\,a^4\,b+2\,a^3\,b^2-2\,a^2\,b^3+5\,a\,b^4-b^5\right )}{2\,b^4\,{\left (a-b\right )}^2}+\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (-3\,a^4+6\,a^3\,b-4\,a^2\,b^2+2\,a\,b^3+b^4\right )}{2\,b^4\,\left (a-b\right )}-\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (9\,a^5-18\,a^4\,b+12\,a^3\,b^2-6\,a^2\,b^3+a\,b^4\right )}{2\,b^4\,{\left (a-b\right )}^2}}{e\,\left (8\,a\,b+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (24\,a^2-8\,b^2\right )-{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (16\,a\,b-16\,a^2\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4\,\left (4\,a^2-8\,a\,b+4\,b^2\right )+4\,a^2+4\,b^2+\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (16\,a^2+16\,b\,a\right )\right )}-\frac {\mathrm {atanh}\left (\frac {2\,a+\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,a-2\,b\right )}{2\,b}\right )\,\left (3\,a^2+b^2\right )}{8\,b^5\,e} \]

[In]

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

[Out]

- ((a*b^4 - 3*a^5 + 4*a^3*b^2)/(2*b^4*(a - b)^2) + (tan(d/2 + (e*x)/2)*(5*a*b^4 + 9*a^4*b - 9*a^5 - b^5 - 2*a^
2*b^3 + 2*a^3*b^2))/(2*b^4*(a - b)^2) + (tan(d/2 + (e*x)/2)^3*(2*a*b^3 + 6*a^3*b - 3*a^4 + b^4 - 4*a^2*b^2))/(
2*b^4*(a - b)) - (tan(d/2 + (e*x)/2)^2*(a*b^4 - 18*a^4*b + 9*a^5 - 6*a^2*b^3 + 12*a^3*b^2))/(2*b^4*(a - b)^2))
/(e*(8*a*b + tan(d/2 + (e*x)/2)^2*(24*a^2 - 8*b^2) - tan(d/2 + (e*x)/2)^3*(16*a*b - 16*a^2) + tan(d/2 + (e*x)/
2)^4*(4*a^2 - 8*a*b + 4*b^2) + 4*a^2 + 4*b^2 + tan(d/2 + (e*x)/2)*(16*a*b + 16*a^2))) - (atanh((2*a + tan(d/2
+ (e*x)/2)*(2*a - 2*b))/(2*b))*(3*a^2 + b^2))/(8*b^5*e)