3.134 \(\int \frac {\cos (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=22 \[ -\frac {1}{2 b d (a+b \tan (c+d x))^2} \]

[Out]

-1/2/b/d/(a+b*tan(d*x+c))^2

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Rubi [A]  time = 0.03, antiderivative size = 30, normalized size of antiderivative = 1.36, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3088, 37} \[ -\frac {\cot ^2(c+d x)}{2 b d (a \cot (c+d x)+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]/(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

-Cot[c + d*x]^2/(2*b*d*(b + a*Cot[c + d*x])^2)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 3088

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Dist[d^(-1), Subst[Int[(x^m*(b + a*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Cot[c + d*x]], x] /; FreeQ[
{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rubi steps

\begin {align*} \int \frac {\cos (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x}{(b+a x)^3} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {\cot ^2(c+d x)}{2 b d (b+a \cot (c+d x))^2}\\ \end {align*}

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Mathematica [B]  time = 0.12, size = 57, normalized size = 2.59 \[ \frac {a \sin (2 (c+d x))-b \cos (2 (c+d x))}{2 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]/(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(-(b*Cos[2*(c + d*x)]) + a*Sin[2*(c + d*x)])/(2*(a^2 + b^2)*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)

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fricas [B]  time = 0.53, size = 142, normalized size = 6.45 \[ -\frac {4 \, a^{2} b \cos \left (d x + c\right )^{2} - a^{2} b + b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} + a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*(4*a^2*b*cos(d*x + c)^2 - a^2*b + b^3 - 2*(a^3 - a*b^2)*cos(d*x + c)*sin(d*x + c))/((a^6 + a^4*b^2 - a^2*
b^4 - b^6)*d*cos(d*x + c)^2 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*d*cos(d*x + c)*sin(d*x + c) + (a^4*b^2 + 2*a^2*b^4
+ b^6)*d)

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giac [A]  time = 0.44, size = 20, normalized size = 0.91 \[ -\frac {1}{2 \, {\left (b \tan \left (d x + c\right ) + a\right )}^{2} b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/((b*tan(d*x + c) + a)^2*b*d)

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maple [A]  time = 0.23, size = 21, normalized size = 0.95 \[ -\frac {1}{2 b d \left (a +b \tan \left (d x +c \right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)/(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

-1/2/b/d/(a+b*tan(d*x+c))^2

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maxima [B]  time = 0.36, size = 171, normalized size = 7.77 \[ \frac {2 \, {\left (\frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {b \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{{\left (a^{4} + \frac {4 \, a^{3} b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {4 \, a^{3} b \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

2*(a*sin(d*x + c)/(cos(d*x + c) + 1) + b*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - a*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3)/((a^4 + 4*a^3*b*sin(d*x + c)/(cos(d*x + c) + 1) - 4*a^3*b*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + a^4*si
n(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2*(a^4 - 2*a^2*b^2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*d)

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mupad [B]  time = 0.61, size = 85, normalized size = 3.86 \[ -\frac {b\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}-\frac {1}{2}\right )-a\,\sin \left (2\,c+2\,d\,x\right )}{a^2\,d\,\left (a^2+b^2+a^2\,\cos \left (2\,c+2\,d\,x\right )-b^2\,\cos \left (2\,c+2\,d\,x\right )+2\,a\,b\,\sin \left (2\,c+2\,d\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)/(a*cos(c + d*x) + b*sin(c + d*x))^3,x)

[Out]

-(b*(cos(2*c + 2*d*x)/2 - 1/2) - a*sin(2*c + 2*d*x))/(a^2*d*(a^2 + b^2 + a^2*cos(2*c + 2*d*x) - b^2*cos(2*c +
2*d*x) + 2*a*b*sin(2*c + 2*d*x)))

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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(cos(d*x+c)/(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Timed out

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