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

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

[Out]

((3*a^2 + c^2)*Log[a + c*Tan[(d + e*x)/2]])/(16*c^5*e) - (c*Cos[d + e*x] - a*Sin[d + e*x])/(16*c^2*e*(a + a*Co
s[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 = 0.111693, 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, 3124, 31} \[ \frac{\left (3 a^2+c^2\right ) \log \left (a+c \tan \left (\frac{1}{2} (d+e x)\right )\right )}{16 c^5 e}+\frac{3 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{16 c^4 e (a \cos (d+e x)+a+c \sin (d+e x))}-\frac{c \cos (d+e x)-a \sin (d+e x)}{16 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2} \]

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*Tan[(d + e*x)/2]])/(16*c^5*e) - (c*Cos[d + e*x] - a*Sin[d + e*x])/(16*c^2*e*(a + a*Co
s[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 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 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}{4 a+4 c x} \, dx,x,\tan \left (\frac{1}{2} (d+e x)\right )\right )}{4 c^4 e}\\ &=\frac{\left (3 a^2+c^2\right ) \log \left (a+c \tan \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 [A]  time = 3.0798, size = 186, normalized size = 1.39 \[ -\frac{4 \left (3 a^2+c^2\right ) \log \left (\cos \left (\frac{1}{2} (d+e x)\right )\right )+\frac{c^2 \left (a^2+c^2\right )}{\left (a \cos \left (\frac{1}{2} (d+e x)\right )+c \sin \left (\frac{1}{2} (d+e x)\right )\right )^2}+\frac{6 c \left (a^2+c^2\right ) \sin \left (\frac{1}{2} (d+e x)\right )}{a \cos \left (\frac{1}{2} (d+e x)\right )+c \sin \left (\frac{1}{2} (d+e x)\right )}-4 \left (3 a^2+c^2\right ) \log \left (a \cos \left (\frac{1}{2} (d+e x)\right )+c \sin \left (\frac{1}{2} (d+e x)\right )\right )+6 a c \tan \left (\frac{1}{2} (d+e x)\right )+c^2 \left (-\sec ^2\left (\frac{1}{2} (d+e x)\right )\right )}{64 c^5 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(4*(3*a^2 + c^2)*Log[Cos[(d + e*x)/2]] - 4*(3*a^2 + c^2)*Log[a*Cos[(d + e*x)/2] + c*Sin[(d + e*x)/2]] - c^2*S
ec[(d + e*x)/2]^2 + (c^2*(a^2 + c^2))/(a*Cos[(d + e*x)/2] + c*Sin[(d + e*x)/2])^2 + (6*c*(a^2 + c^2)*Sin[(d +
e*x)/2])/(a*Cos[(d + e*x)/2] + c*Sin[(d + e*x)/2]) + 6*a*c*Tan[(d + e*x)/2])/(64*c^5*e)

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Maple [A]  time = 0.189, size = 211, normalized size = 1.6 \begin{align*}{\frac{1}{64\,e{c}^{3}} \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{2}}-{\frac{3\,a}{32\,{c}^{4}e}\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) }-{\frac{{a}^{4}}{64\,e{c}^{5}} \left ( a+c\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-2}}-{\frac{{a}^{2}}{32\,e{c}^{3}} \left ( a+c\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-2}}-{\frac{1}{64\,ce} \left ( a+c\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-2}}+{\frac{3\,{a}^{2}}{16\,e{c}^{5}}\ln \left ( a+c\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) }+{\frac{1}{16\,e{c}^{3}}\ln \left ( a+c\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) }+{\frac{{a}^{3}}{8\,e{c}^{5}} \left ( a+c\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-1}}+{\frac{a}{8\,e{c}^{3}} \left ( a+c\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]

1/64/e/c^3*tan(1/2*d+1/2*e*x)^2-3/32/e/c^4*tan(1/2*d+1/2*e*x)*a-1/64/e/c^5/(a+c*tan(1/2*d+1/2*e*x))^2*a^4-1/32
/e/c^3/(a+c*tan(1/2*d+1/2*e*x))^2*a^2-1/64/e/c/(a+c*tan(1/2*d+1/2*e*x))^2+3/16/e/c^5*ln(a+c*tan(1/2*d+1/2*e*x)
)*a^2+1/16/e/c^3*ln(a+c*tan(1/2*d+1/2*e*x))+1/8/e*a^3/c^5/(a+c*tan(1/2*d+1/2*e*x))+1/8/e*a/c^3/(a+c*tan(1/2*d+
1/2*e*x))

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Maxima [A]  time = 1.06366, size = 257, normalized size = 1.92 \begin{align*} \frac{\frac{7 \, a^{4} + 6 \, a^{2} c^{2} - c^{4} + \frac{8 \,{\left (a^{3} c + a c^{3}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}}{a^{2} c^{5} + \frac{2 \, a c^{6} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac{c^{7} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}} - \frac{\frac{6 \, a \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - \frac{c \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}}{c^{4}} + \frac{4 \,{\left (3 \, a^{2} + c^{2}\right )} \log \left (a + \frac{c \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*((7*a^4 + 6*a^2*c^2 - c^4 + 8*(a^3*c + a*c^3)*sin(e*x + d)/(cos(e*x + d) + 1))/(a^2*c^5 + 2*a*c^6*sin(e*x
+ d)/(cos(e*x + d) + 1) + c^7*sin(e*x + d)^2/(cos(e*x + d) + 1)^2) - (6*a*sin(e*x + d)/(cos(e*x + d) + 1) - c
*sin(e*x + d)^2/(cos(e*x + d) + 1)^2)/c^4 + 4*(3*a^2 + c^2)*log(a + c*sin(e*x + d)/(cos(e*x + d) + 1))/c^5)/e

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Fricas [B]  time = 2.35285, size = 986, normalized size = 7.36 \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 = 1.1766, size = 231, normalized size = 1.72 \begin{align*} \frac{1}{64} \,{\left (\frac{4 \,{\left (3 \, a^{2} + c^{2}\right )} \log \left ({\left | c \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + a \right |}\right )}{c^{5}} + \frac{c^{3} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} - 6 \, a c^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )}{c^{6}} - \frac{18 \, a^{2} c^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 6 \, c^{4} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 28 \, a^{3} c \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 4 \, a c^{3} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 11 \, a^{4} + c^{4}}{{\left (c \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + a\right )}^{2} c^{5}}\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(c*tan(1/2*x*e + 1/2*d) + a))/c^5 + (c^3*tan(1/2*x*e + 1/2*d)^2 - 6*a*c^2*tan(1/2
*x*e + 1/2*d))/c^6 - (18*a^2*c^2*tan(1/2*x*e + 1/2*d)^2 + 6*c^4*tan(1/2*x*e + 1/2*d)^2 + 28*a^3*c*tan(1/2*x*e
+ 1/2*d) + 4*a*c^3*tan(1/2*x*e + 1/2*d) + 11*a^4 + c^4)/((c*tan(1/2*x*e + 1/2*d) + a)^2*c^5))*e^(-1)