3.373 \(\int \frac{1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^4} \, dx\)

Optimal. Leaf size=168 \[ -\frac{\log \left (\tan \left (\frac{1}{2} (d+e x)\right )+1\right )}{4 a^4 e}-\frac{19 (a \cos (d+e x)-a \sin (d+e x))}{96 e \left (a^5 \sin (d+e x)+a^5 \cos (d+e x)+a^5\right )}+\frac{5 (\cos (d+e x)-\sin (d+e x))}{96 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}-\frac{\cos (d+e x)-\sin (d+e x)}{48 a e (a \sin (d+e x)+a \cos (d+e x)+a)^3} \]

[Out]

-Log[1 + Tan[(d + e*x)/2]]/(4*a^4*e) - (Cos[d + e*x] - Sin[d + e*x])/(48*a*e*(a + a*Cos[d + e*x] + a*Sin[d + e
*x])^3) + (5*(Cos[d + e*x] - Sin[d + e*x]))/(96*e*(a^2 + a^2*Cos[d + e*x] + a^2*Sin[d + e*x])^2) - (19*(a*Cos[
d + e*x] - a*Sin[d + e*x]))/(96*e*(a^5 + a^5*Cos[d + e*x] + a^5*Sin[d + e*x]))

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Rubi [A]  time = 0.186493, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3129, 3156, 3153, 3124, 31} \[ -\frac{\log \left (\tan \left (\frac{1}{2} (d+e x)\right )+1\right )}{4 a^4 e}-\frac{19 (a \cos (d+e x)-a \sin (d+e x))}{96 e \left (a^5 \sin (d+e x)+a^5 \cos (d+e x)+a^5\right )}+\frac{5 (\cos (d+e x)-\sin (d+e x))}{96 e \left (a^2 \sin (d+e x)+a^2 \cos (d+e x)+a^2\right )^2}-\frac{\cos (d+e x)-\sin (d+e x)}{48 a e (a \sin (d+e x)+a \cos (d+e x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(2*a + 2*a*Cos[d + e*x] + 2*a*Sin[d + e*x])^(-4),x]

[Out]

-Log[1 + Tan[(d + e*x)/2]]/(4*a^4*e) - (Cos[d + e*x] - Sin[d + e*x])/(48*a*e*(a + a*Cos[d + e*x] + a*Sin[d + e
*x])^3) + (5*(Cos[d + e*x] - Sin[d + e*x]))/(96*e*(a^2 + a^2*Cos[d + e*x] + a^2*Sin[d + e*x])^2) - (19*(a*Cos[
d + e*x] - a*Sin[d + e*x]))/(96*e*(a^5 + a^5*Cos[d + e*x] + a^5*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 3156

Int[((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_)*((A_.) + cos[(d_.) + (e_.)*(x
_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> -Simp[((c*B - b*C - (a*C - c*A)*Cos[d + e*x] + (a*B -
b*A)*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 + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1)*Simp[(n + 1)*(a*A - b*B - c*C)
+ (n + 2)*(a*B - b*A)*Cos[d + e*x] + (n + 2)*(a*C - c*A)*Sin[d + e*x], x], x], x] /; FreeQ[{a, b, c, d, e, A,
B, C}, x] && LtQ[n, -1] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[n, -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 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 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 a \sin (d+e x))^4} \, dx &=-\frac{\cos (d+e x)-\sin (d+e x)}{48 a e (a+a \cos (d+e x)+a \sin (d+e x))^3}+\frac{\int \frac{-6 a+4 a \cos (d+e x)+4 a \sin (d+e x)}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^3} \, dx}{12 a^2}\\ &=-\frac{\cos (d+e x)-\sin (d+e x)}{48 a e (a+a \cos (d+e x)+a \sin (d+e x))^3}+\frac{5 (\cos (d+e x)-\sin (d+e x))}{96 e \left (a^2+a^2 \cos (d+e x)+a^2 \sin (d+e x)\right )^2}+\frac{\int \frac{56 a^2-20 a^2 \cos (d+e x)-20 a^2 \sin (d+e x)}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx}{96 a^4}\\ &=-\frac{\cos (d+e x)-\sin (d+e x)}{48 a e (a+a \cos (d+e x)+a \sin (d+e x))^3}+\frac{5 (\cos (d+e x)-\sin (d+e x))}{96 e \left (a^2+a^2 \cos (d+e x)+a^2 \sin (d+e x)\right )^2}-\frac{19 \left (a^3 \cos (d+e x)-a^3 \sin (d+e x)\right )}{96 e \left (a^7+a^7 \cos (d+e x)+a^7 \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^3}\\ &=-\frac{\cos (d+e x)-\sin (d+e x)}{48 a e (a+a \cos (d+e x)+a \sin (d+e x))^3}+\frac{5 (\cos (d+e x)-\sin (d+e x))}{96 e \left (a^2+a^2 \cos (d+e x)+a^2 \sin (d+e x)\right )^2}-\frac{19 \left (a^3 \cos (d+e x)-a^3 \sin (d+e x)\right )}{96 e \left (a^7+a^7 \cos (d+e x)+a^7 \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^3 e}\\ &=-\frac{\log \left (1+\tan \left (\frac{1}{2} (d+e x)\right )\right )}{4 a^4 e}-\frac{\cos (d+e x)-\sin (d+e x)}{48 a e (a+a \cos (d+e x)+a \sin (d+e x))^3}+\frac{5 (\cos (d+e x)-\sin (d+e x))}{96 e \left (a^2+a^2 \cos (d+e x)+a^2 \sin (d+e x)\right )^2}-\frac{19 \left (a^3 \cos (d+e x)-a^3 \sin (d+e x)\right )}{96 e \left (a^7+a^7 \cos (d+e x)+a^7 \sin (d+e x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.973845, size = 247, normalized size = 1.47 \[ \frac{19 \tan \left (\frac{1}{2} (d+e x)\right )}{192 a^4 e}-\frac{\sec ^2\left (\frac{1}{2} (d+e x)\right )}{64 a^4 e}+\frac{\log \left (\cos \left (\frac{1}{2} (d+e x)\right )\right )}{4 a^4 e}+\frac{19 \sin \left (\frac{1}{2} (d+e x)\right )}{96 a^4 e \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right )}+\frac{5}{192 a^4 e \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right )^2}+\frac{\sin \left (\frac{1}{2} (d+e x)\right )}{96 a^4 e \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right )^3}+\frac{\tan \left (\frac{1}{2} (d+e x)\right ) \sec ^2\left (\frac{1}{2} (d+e x)\right )}{384 a^4 e}-\frac{\log \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right )}{4 a^4 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(2*a + 2*a*Cos[d + e*x] + 2*a*Sin[d + e*x])^(-4),x]

[Out]

Log[Cos[(d + e*x)/2]]/(4*a^4*e) - Log[Cos[(d + e*x)/2] + Sin[(d + e*x)/2]]/(4*a^4*e) - Sec[(d + e*x)/2]^2/(64*
a^4*e) + Sin[(d + e*x)/2]/(96*a^4*e*(Cos[(d + e*x)/2] + Sin[(d + e*x)/2])^3) + 5/(192*a^4*e*(Cos[(d + e*x)/2]
+ Sin[(d + e*x)/2])^2) + (19*Sin[(d + e*x)/2])/(96*a^4*e*(Cos[(d + e*x)/2] + Sin[(d + e*x)/2])) + (19*Tan[(d +
e*x)/2])/(192*a^4*e) + (Sec[(d + e*x)/2]^2*Tan[(d + e*x)/2])/(384*a^4*e)

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Maple [A]  time = 0.112, size = 140, normalized size = 0.8 \begin{align*}{\frac{1}{384\,e{a}^{4}} \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{3}}-{\frac{1}{64\,e{a}^{4}} \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{2}}+{\frac{13}{128\,e{a}^{4}}\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) }-{\frac{9}{32\,e{a}^{4}} \left ( 1+\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-1}}+{\frac{3}{32\,e{a}^{4}} \left ( 1+\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-2}}-{\frac{1}{4\,e{a}^{4}}\ln \left ( 1+\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) }-{\frac{1}{48\,e{a}^{4}} \left ( 1+\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-3}} \end{align*}

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))^4,x)

[Out]

1/384/e/a^4*tan(1/2*d+1/2*e*x)^3-1/64/e/a^4*tan(1/2*d+1/2*e*x)^2+13/128/e/a^4*tan(1/2*d+1/2*e*x)-9/32/e/a^4/(1
+tan(1/2*d+1/2*e*x))+3/32/e/a^4/(1+tan(1/2*d+1/2*e*x))^2-1/4*ln(1+tan(1/2*d+1/2*e*x))/a^4/e-1/48/e/a^4/(1+tan(
1/2*d+1/2*e*x))^3

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Maxima [A]  time = 1.12096, size = 281, normalized size = 1.67 \begin{align*} -\frac{\frac{4 \,{\left (\frac{45 \, \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac{27 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + 20\right )}}{a^{4} + \frac{3 \, a^{4} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac{3 \, a^{4} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac{a^{4} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}} - \frac{\frac{39 \, \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - \frac{6 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac{\sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}}{a^{4}} + \frac{96 \, \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right )}{a^{4}}}{384 \, 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*a*sin(e*x+d))^4,x, algorithm="maxima")

[Out]

-1/384*(4*(45*sin(e*x + d)/(cos(e*x + d) + 1) + 27*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 + 20)/(a^4 + 3*a^4*sin(
e*x + d)/(cos(e*x + d) + 1) + 3*a^4*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 + a^4*sin(e*x + d)^3/(cos(e*x + d) + 1
)^3) - (39*sin(e*x + d)/(cos(e*x + d) + 1) - 6*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 + sin(e*x + d)^3/(cos(e*x +
d) + 1)^3)/a^4 + 96*log(sin(e*x + d)/(cos(e*x + d) + 1) + 1)/a^4)/e

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Fricas [A]  time = 2.14268, size = 645, normalized size = 3.84 \begin{align*} \frac{38 \, \cos \left (e x + d\right )^{3} + 66 \, \cos \left (e x + d\right )^{2} + 24 \,{\left (\cos \left (e x + d\right )^{3} -{\left (\cos \left (e x + d\right )^{2} + 3 \, \cos \left (e x + d\right ) + 2\right )} \sin \left (e x + d\right ) - 3 \, \cos \left (e x + d\right ) - 2\right )} \log \left (\frac{1}{2} \, \cos \left (e x + d\right ) + \frac{1}{2}\right ) - 24 \,{\left (\cos \left (e x + d\right )^{3} -{\left (\cos \left (e x + d\right )^{2} + 3 \, \cos \left (e x + d\right ) + 2\right )} \sin \left (e x + d\right ) - 3 \, \cos \left (e x + d\right ) - 2\right )} \log \left (\sin \left (e x + d\right ) + 1\right ) +{\left (38 \, \cos \left (e x + d\right )^{2} - 35\right )} \sin \left (e x + d\right ) - 3 \, \cos \left (e x + d\right ) - 33}{192 \,{\left (a^{4} e \cos \left (e x + d\right )^{3} - 3 \, a^{4} e \cos \left (e x + d\right ) - 2 \, a^{4} e -{\left (a^{4} e \cos \left (e x + d\right )^{2} + 3 \, a^{4} e \cos \left (e x + d\right ) + 2 \, a^{4} 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*a*sin(e*x+d))^4,x, algorithm="fricas")

[Out]

1/192*(38*cos(e*x + d)^3 + 66*cos(e*x + d)^2 + 24*(cos(e*x + d)^3 - (cos(e*x + d)^2 + 3*cos(e*x + d) + 2)*sin(
e*x + d) - 3*cos(e*x + d) - 2)*log(1/2*cos(e*x + d) + 1/2) - 24*(cos(e*x + d)^3 - (cos(e*x + d)^2 + 3*cos(e*x
+ d) + 2)*sin(e*x + d) - 3*cos(e*x + d) - 2)*log(sin(e*x + d) + 1) + (38*cos(e*x + d)^2 - 35)*sin(e*x + d) - 3
*cos(e*x + d) - 33)/(a^4*e*cos(e*x + d)^3 - 3*a^4*e*cos(e*x + d) - 2*a^4*e - (a^4*e*cos(e*x + d)^2 + 3*a^4*e*c
os(e*x + d) + 2*a^4*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*a*sin(e*x+d))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.1494, size = 188, normalized size = 1.12 \begin{align*} -\frac{1}{384} \,{\left (\frac{96 \, \log \left ({\left | \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 1 \right |}\right )}{a^{4}} - \frac{4 \,{\left (44 \, \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} + 105 \, \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 87 \, \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 24\right )}}{a^{4}{\left (\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 1\right )}^{3}} - \frac{a^{8} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} - 6 \, a^{8} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 39 \, a^{8} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )}{a^{12}}\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*a*sin(e*x+d))^4,x, algorithm="giac")

[Out]

-1/384*(96*log(abs(tan(1/2*x*e + 1/2*d) + 1))/a^4 - 4*(44*tan(1/2*x*e + 1/2*d)^3 + 105*tan(1/2*x*e + 1/2*d)^2
+ 87*tan(1/2*x*e + 1/2*d) + 24)/(a^4*(tan(1/2*x*e + 1/2*d) + 1)^3) - (a^8*tan(1/2*x*e + 1/2*d)^3 - 6*a^8*tan(1
/2*x*e + 1/2*d)^2 + 39*a^8*tan(1/2*x*e + 1/2*d))/a^12)*e^(-1)