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

Optimal. Leaf size=123 \[ \frac {\log \left (\tan \left (\frac {1}{2} (d+e x)\right )+1\right )}{4 a^3 e}+\frac {3 (\cos (d+e x)-\sin (d+e x))}{16 e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}-\frac {a \cos (d+e x)-a \sin (d+e x)}{16 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2} \]

[Out]

1/4*ln(1+tan(1/2*e*x+1/2*d))/a^3/e+1/16*(-a*cos(e*x+d)+a*sin(e*x+d))/e/(a^2+a^2*cos(e*x+d)+a^2*sin(e*x+d))^2+3
/16*(cos(e*x+d)-sin(e*x+d))/e/(a^3+a^3*cos(e*x+d)+a^3*sin(e*x+d))

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Rubi [A]  time = 0.11, antiderivative size = 123, normalized size of antiderivative = 1.00, 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, 3124, 31} \[ \frac {\log \left (\tan \left (\frac {1}{2} (d+e x)\right )+1\right )}{4 a^3 e}+\frac {3 (\cos (d+e x)-\sin (d+e x))}{16 e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )}-\frac {a \cos (d+e x)-a \sin (d+e x)}{16 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 + Tan[(d + e*x)/2]]/(4*a^3*e) - (a*Cos[d + e*x] - a*Sin[d + e*x])/(16*e*(a^2 + a^2*Cos[d + e*x] + a^2*Si
n[d + e*x])^2) + (3*(Cos[d + e*x] - Sin[d + e*x]))/(16*e*(a^3 + a^3*Cos[d + e*x] + a^3*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 3124

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.57, size = 135, normalized size = 1.10 \[ \frac {\sec ^2\left (\frac {1}{2} (d+e x)\right )+2 \left (-3 \tan \left (\frac {1}{2} (d+e x)\right )-8 \log \left (\cos \left (\frac {1}{2} (d+e x)\right )\right )-\frac {6 \sin \left (\frac {1}{2} (d+e x)\right )}{\sin \left (\frac {1}{2} (d+e x)\right )+\cos \left (\frac {1}{2} (d+e x)\right )}-\frac {1}{\left (\sin \left (\frac {1}{2} (d+e x)\right )+\cos \left (\frac {1}{2} (d+e x)\right )\right )^2}+8 \log \left (\sin \left (\frac {1}{2} (d+e x)\right )+\cos \left (\frac {1}{2} (d+e x)\right )\right )\right )}{64 a^3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[(d + e*x)/2]^2 + 2*(-8*Log[Cos[(d + e*x)/2]] + 8*Log[Cos[(d + e*x)/2] + Sin[(d + e*x)/2]] - (Cos[(d + e*x
)/2] + Sin[(d + e*x)/2])^(-2) - (6*Sin[(d + e*x)/2])/(Cos[(d + e*x)/2] + Sin[(d + e*x)/2]) - 3*Tan[(d + e*x)/2
]))/(64*a^3*e)

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fricas [A]  time = 1.76, size = 143, normalized size = 1.16 \[ \frac {6 \, \cos \left (e x + d\right )^{2} - 4 \, {\left ({\left (\cos \left (e x + d\right ) + 1\right )} \sin \left (e x + d\right ) + \cos \left (e x + d\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (e x + d\right ) + \frac {1}{2}\right ) + 4 \, {\left ({\left (\cos \left (e x + d\right ) + 1\right )} \sin \left (e x + d\right ) + \cos \left (e x + d\right ) + 1\right )} \log \left (\sin \left (e x + d\right ) + 1\right ) + 2 \, \cos \left (e x + d\right ) - 2 \, \sin \left (e x + d\right ) - 3}{32 \, {\left (a^{3} e \cos \left (e x + d\right ) + a^{3} e + {\left (a^{3} e \cos \left (e x + d\right ) + a^{3} e\right )} \sin \left (e x + d\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 107, normalized size = 0.87 \[ \frac {1}{64} \, {\left (\frac {16 \, \log \left ({\left | \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 1 \right |}\right )}{a^{3}} - \frac {4 \, {\left (6 \, \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 8 \, \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 3\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 1\right )}^{2}} + \frac {a^{3} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} - 6 \, a^{3} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )}{a^{6}}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(16*log(abs(tan(1/2*x*e + 1/2*d) + 1))/a^3 - 4*(6*tan(1/2*x*e + 1/2*d)^2 + 8*tan(1/2*x*e + 1/2*d) + 3)/(a
^3*(tan(1/2*x*e + 1/2*d) + 1)^2) + (a^3*tan(1/2*x*e + 1/2*d)^2 - 6*a^3*tan(1/2*x*e + 1/2*d))/a^6)*e^(-1)

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maple [A]  time = 0.45, size = 100, normalized size = 0.81 \[ \frac {\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )}{64 a^{3} e}-\frac {3 \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{32 a^{3} e}-\frac {1}{16 a^{3} e \left (1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )\right )^{2}}+\frac {1}{4 a^{3} e \left (1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )\right )}+\frac {\ln \left (1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{4 a^{3} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 146, normalized size = 1.19 \[ \frac {\frac {4 \, {\left (\frac {4 \, \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 3\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {a^{3} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}} - \frac {\frac {6 \, \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - \frac {\sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}}{a^{3}} + \frac {16 \, \log \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right )}{a^{3}}}{64 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(4*(4*sin(e*x + d)/(cos(e*x + d) + 1) + 3)/(a^3 + 2*a^3*sin(e*x + d)/(cos(e*x + d) + 1) + a^3*sin(e*x + d
)^2/(cos(e*x + d) + 1)^2) - (6*sin(e*x + d)/(cos(e*x + d) + 1) - sin(e*x + d)^2/(cos(e*x + d) + 1)^2)/a^3 + 16
*log(sin(e*x + d)/(cos(e*x + d) + 1) + 1)/a^3)/e

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mupad [B]  time = 2.45, size = 90, normalized size = 0.73 \[ \frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2}{64\,a^3\,e}+\frac {\ln \left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1\right )}{4\,a^3\,e}-\frac {3\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{32\,a^3\,e}+\frac {\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{4}+\frac {3}{16}}{a^3\,e\,{\left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 6.49, size = 423, normalized size = 3.44 \[ \begin {cases} \frac {16 \log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )} \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )}}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} + \frac {32 \log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )} \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )}}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} + \frac {16 \log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )}}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} + \frac {\tan ^{4}{\left (\frac {d}{2} + \frac {e x}{2} \right )}}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} - \frac {4 \tan ^{3}{\left (\frac {d}{2} + \frac {e x}{2} \right )}}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} + \frac {32 \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )}}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} + \frac {23}{64 a^{3} e \tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 128 a^{3} e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} + 64 a^{3} e} & \text {for}\: e \neq 0 \\\frac {x}{\left (2 a \sin {\relax (d )} + 2 a \cos {\relax (d )} + 2 a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((16*log(tan(d/2 + e*x/2) + 1)*tan(d/2 + e*x/2)**2/(64*a**3*e*tan(d/2 + e*x/2)**2 + 128*a**3*e*tan(d/
2 + e*x/2) + 64*a**3*e) + 32*log(tan(d/2 + e*x/2) + 1)*tan(d/2 + e*x/2)/(64*a**3*e*tan(d/2 + e*x/2)**2 + 128*a
**3*e*tan(d/2 + e*x/2) + 64*a**3*e) + 16*log(tan(d/2 + e*x/2) + 1)/(64*a**3*e*tan(d/2 + e*x/2)**2 + 128*a**3*e
*tan(d/2 + e*x/2) + 64*a**3*e) + tan(d/2 + e*x/2)**4/(64*a**3*e*tan(d/2 + e*x/2)**2 + 128*a**3*e*tan(d/2 + e*x
/2) + 64*a**3*e) - 4*tan(d/2 + e*x/2)**3/(64*a**3*e*tan(d/2 + e*x/2)**2 + 128*a**3*e*tan(d/2 + e*x/2) + 64*a**
3*e) + 32*tan(d/2 + e*x/2)/(64*a**3*e*tan(d/2 + e*x/2)**2 + 128*a**3*e*tan(d/2 + e*x/2) + 64*a**3*e) + 23/(64*
a**3*e*tan(d/2 + e*x/2)**2 + 128*a**3*e*tan(d/2 + e*x/2) + 64*a**3*e), Ne(e, 0)), (x/(2*a*sin(d) + 2*a*cos(d)
+ 2*a)**3, True))

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