3.143 \(\int \frac{\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=30 \[ -\frac{\cot ^3(c+d x)}{3 b d (a \cot (c+d x)+b)^3} \]

[Out]

-Cot[c + d*x]^3/(3*b*d*(b + a*Cot[c + d*x])^3)

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Rubi [A]  time = 0.0528595, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {3088, 37} \[ -\frac{\cot ^3(c+d x)}{3 b d (a \cot (c+d x)+b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

-Cot[c + d*x]^3/(3*b*d*(b + a*Cot[c + d*x])^3)

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 ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^2}{(b+a x)^4} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac{\cot ^3(c+d x)}{3 b d (b+a \cot (c+d x))^3}\\ \end{align*}

Mathematica [B]  time = 0.662484, size = 124, normalized size = 4.13 \[ \frac{-6 a b \left (a^2+b^2\right ) \cos (c+d x)+\left (2 a b^3-6 a^3 b\right ) \cos (3 (c+d x))+2 \left (a^2-b^2\right ) \sin (c+d x) \left (\left (3 a^2-b^2\right ) \cos (2 (c+d x))+3 a^2+b^2\right )}{12 a d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(-6*a*b*(a^2 + b^2)*Cos[c + d*x] + (-6*a^3*b + 2*a*b^3)*Cos[3*(c + d*x)] + 2*(a^2 - b^2)*(3*a^2 + b^2 + (3*a^2
- b^2)*Cos[2*(c + d*x)])*Sin[c + d*x])/(12*a*(a^2 + b^2)^2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)

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Maple [A]  time = 0.2, size = 21, normalized size = 0.7 \begin{align*} -{\frac{1}{3\,db \left ( a+b\tan \left ( dx+c \right ) \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

-1/3/d/b/(a+b*tan(d*x+c))^3

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Maxima [A]  time = 1.17786, size = 72, normalized size = 2.4 \begin{align*} -\frac{1}{3 \,{\left (b^{4} \tan \left (d x + c\right )^{3} + 3 \, a b^{3} \tan \left (d x + c\right )^{2} + 3 \, a^{2} b^{2} \tan \left (d x + c\right ) + a^{3} b\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/3/((b^4*tan(d*x + c)^3 + 3*a*b^3*tan(d*x + c)^2 + 3*a^2*b^2*tan(d*x + c) + a^3*b)*d)

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Fricas [B]  time = 0.526365, size = 552, normalized size = 18.4 \begin{align*} -\frac{{\left (9 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (a^{4} b - 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right ) -{\left (a^{3} b^{2} - 3 \, a b^{4} +{\left (3 \, a^{5} - 10 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \,{\left ({\left (a^{9} - 6 \, a^{5} b^{4} - 8 \, a^{3} b^{6} - 3 \, a b^{8}\right )} d \cos \left (d x + c\right )^{3} + 3 \,{\left (a^{7} b^{2} + 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right ) +{\left ({\left (3 \, a^{8} b + 8 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - b^{9}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9}\right )} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/3*((9*a^4*b - 6*a^2*b^3 + b^5)*cos(d*x + c)^3 - 3*(a^4*b - 3*a^2*b^3)*cos(d*x + c) - (a^3*b^2 - 3*a*b^4 + (
3*a^5 - 10*a^3*b^2 + 3*a*b^4)*cos(d*x + c)^2)*sin(d*x + c))/((a^9 - 6*a^5*b^4 - 8*a^3*b^6 - 3*a*b^8)*d*cos(d*x
+ c)^3 + 3*(a^7*b^2 + 3*a^5*b^4 + 3*a^3*b^6 + a*b^8)*d*cos(d*x + c) + ((3*a^8*b + 8*a^6*b^3 + 6*a^4*b^5 - b^9
)*d*cos(d*x + c)^2 + (a^6*b^3 + 3*a^4*b^5 + 3*a^2*b^7 + b^9)*d)*sin(d*x + c))

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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)**2/(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.11551, size = 27, normalized size = 0.9 \begin{align*} -\frac{1}{3 \,{\left (b \tan \left (d x + c\right ) + a\right )}^{3} b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

-1/3/((b*tan(d*x + c) + a)^3*b*d)