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[c + d*x])^2)

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Rubi [A]  time = 0.0312509, 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 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])

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]

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*}

Mathematica [B]  time = 0.118288, 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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Maple [A]  time = 0.168, size = 21, normalized size = 1. \begin{align*} -{\frac{1}{2\,db \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

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 = 1.1391, size = 231, normalized size = 10.5 \begin{align*} \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} \end{align*}

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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Fricas [B]  time = 0.482848, size = 313, normalized size = 14.23 \begin{align*} -\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 )}} \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.23095, size = 27, normalized size = 1.23 \begin{align*} -\frac{1}{2 \,{\left (b \tan \left (d x + c\right ) + a\right )}^{2} b d} \end{align*}

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)