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

[Out]

1/16*(3*a^2+c^2)*ln(a+c*tan(1/2*e*x+1/2*d))/c^5/e+1/16*(-c*cos(e*x+d)+a*sin(e*x+d))/c^2/e/(a+a*cos(e*x+d)+c*si
n(e*x+d))^2+3/16*(a*c*cos(e*x+d)-a^2*sin(e*x+d))/c^4/e/(a+a*cos(e*x+d)+c*sin(e*x+d))

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Rubi [A]  time = 0.11, antiderivative size = 134, 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} \[ \frac {\left (3 a^2+c^2\right ) \log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )}{16 c^5 e}+\frac {3 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{16 c^4 e (a \cos (d+e x)+a+c \sin (d+e x))}-\frac {c \cos (d+e x)-a \sin (d+e x)}{16 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2} \]

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 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 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*}

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Mathematica [A]  time = 3.00, 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]

-1/64*(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*Sec[(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])/(c^5*e)

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fricas [B]  time = 0.95, size = 433, normalized size = 3.23 \[ \frac {12 \, a^{2} c^{2} \cos \left (e x + d\right )^{2} - 6 \, a^{2} c^{2} + 2 \, {\left (3 \, a^{2} c^{2} - c^{4}\right )} \cos \left (e x + d\right ) + {\left (3 \, a^{4} + 4 \, a^{2} c^{2} + c^{4} + {\left (3 \, a^{4} - 2 \, a^{2} c^{2} - c^{4}\right )} \cos \left (e x + d\right )^{2} + 2 \, {\left (3 \, a^{4} + a^{2} c^{2}\right )} \cos \left (e x + d\right ) + 2 \, {\left (3 \, a^{3} c + a c^{3} + {\left (3 \, a^{3} c + a c^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )} \log \left (a c \sin \left (e x + d\right ) + \frac {1}{2} \, a^{2} + \frac {1}{2} \, c^{2} + \frac {1}{2} \, {\left (a^{2} - c^{2}\right )} \cos \left (e x + d\right )\right ) - {\left (3 \, a^{4} + 4 \, a^{2} c^{2} + c^{4} + {\left (3 \, a^{4} - 2 \, a^{2} c^{2} - c^{4}\right )} \cos \left (e x + d\right )^{2} + 2 \, {\left (3 \, a^{4} + a^{2} c^{2}\right )} \cos \left (e x + d\right ) + 2 \, {\left (3 \, a^{3} c + a c^{3} + {\left (3 \, a^{3} c + a c^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )} \log \left (\frac {1}{2} \, \cos \left (e x + d\right ) + \frac {1}{2}\right ) - 2 \, {\left (3 \, a^{3} c - a c^{3} + 3 \, {\left (a^{3} c - a c^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{32 \, {\left (2 \, a^{2} c^{5} e \cos \left (e x + d\right ) + {\left (a^{2} c^{5} - c^{7}\right )} e \cos \left (e x + d\right )^{2} + {\left (a^{2} c^{5} + c^{7}\right )} e + 2 \, {\left (a c^{6} e \cos \left (e x + d\right ) + a c^{6} e\right )} \sin \left (e x + d\right )\right )}} \]

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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giac [A]  time = 0.17, size = 171, normalized size = 1.28 \[ \frac {1}{64} \, {\left (\frac {4 \, {\left (3 \, a^{2} + c^{2}\right )} \log \left ({\left | c \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + a \right |}\right )}{c^{5}} + \frac {c^{3} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} - 6 \, a c^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )}{c^{6}} - \frac {18 \, a^{2} c^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 6 \, c^{4} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 28 \, a^{3} c \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 4 \, a c^{3} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 11 \, a^{4} + c^{4}}{{\left (c \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + a\right )}^{2} c^{5}}\right )} e^{\left (-1\right )} \]

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)

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maple [A]  time = 0.56, size = 211, normalized size = 1.57 \[ \frac {\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )}{64 e \,c^{3}}-\frac {3 \tan \left (\frac {d}{2}+\frac {e x}{2}\right ) a}{32 e \,c^{4}}-\frac {a^{4}}{64 e \,c^{5} \left (a +c \tan \left (\frac {d}{2}+\frac {e x}{2}\right )\right )^{2}}-\frac {a^{2}}{32 e \,c^{3} \left (a +c \tan \left (\frac {d}{2}+\frac {e x}{2}\right )\right )^{2}}-\frac {1}{64 e c \left (a +c \tan \left (\frac {d}{2}+\frac {e x}{2}\right )\right )^{2}}+\frac {a^{3}}{8 e \,c^{5} \left (a +c \tan \left (\frac {d}{2}+\frac {e x}{2}\right )\right )}+\frac {a}{8 e \,c^{3} \left (a +c \tan \left (\frac {d}{2}+\frac {e x}{2}\right )\right )}+\frac {3 \ln \left (a +c \tan \left (\frac {d}{2}+\frac {e x}{2}\right )\right ) a^{2}}{16 e \,c^{5}}+\frac {\ln \left (a +c \tan \left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{16 e \,c^{3}} \]

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+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))+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))

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maxima [A]  time = 0.34, size = 190, normalized size = 1.42 \[ \frac {\frac {7 \, a^{4} + 6 \, a^{2} c^{2} - c^{4} + \frac {8 \, {\left (a^{3} c + a c^{3}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}}{a^{2} c^{5} + \frac {2 \, a c^{6} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {c^{7} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}} - \frac {\frac {6 \, a \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - \frac {c \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}}{c^{4}} + \frac {4 \, {\left (3 \, a^{2} + c^{2}\right )} \log \left (a + \frac {c \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{c^{5}}}{64 \, e} \]

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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mupad [B]  time = 2.48, size = 162, normalized size = 1.21 \[ \frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2}{64\,c^3\,e}+\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (4\,a^3+4\,a\,c^2\right )+\frac {7\,a^4+6\,a^2\,c^2-c^4}{2\,c}}{e\,\left (32\,a^2\,c^4+64\,a\,c^5\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+32\,c^6\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\right )}-\frac {3\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{32\,c^4\,e}+\frac {\ln \left (a+c\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\right )\,\left (3\,a^2+c^2\right )}{16\,c^5\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(d/2 + (e*x)/2)^2/(64*c^3*e) + (tan(d/2 + (e*x)/2)*(4*a*c^2 + 4*a^3) + (7*a^4 - c^4 + 6*a^2*c^2)/(2*c))/(e*
(32*c^6*tan(d/2 + (e*x)/2)^2 + 32*a^2*c^4 + 64*a*c^5*tan(d/2 + (e*x)/2))) - (3*a*tan(d/2 + (e*x)/2))/(32*c^4*e
) + (log(a + c*tan(d/2 + (e*x)/2))*(3*a^2 + c^2))/(16*c^5*e)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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