3.110 \(\int \frac{\cos ^5(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=227 \[ \frac{b^3 \cos ^2(c+d x)}{2 d \left (a^2+b^2\right )^2}+\frac{b \cos ^4(c+d x)}{4 d \left (a^2+b^2\right )}+\frac{a \sin (c+d x) \cos ^3(c+d x)}{4 d \left (a^2+b^2\right )}+\frac{a b^2 \sin (c+d x) \cos (c+d x)}{2 d \left (a^2+b^2\right )^2}+\frac{3 a \sin (c+d x) \cos (c+d x)}{8 d \left (a^2+b^2\right )}+\frac{b^5 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{a b^4 x}{\left (a^2+b^2\right )^3}+\frac{a b^2 x}{2 \left (a^2+b^2\right )^2}+\frac{3 a x}{8 \left (a^2+b^2\right )} \]

[Out]

(a*b^4*x)/(a^2 + b^2)^3 + (a*b^2*x)/(2*(a^2 + b^2)^2) + (3*a*x)/(8*(a^2 + b^2)) + (b^3*Cos[c + d*x]^2)/(2*(a^2
+ b^2)^2*d) + (b*Cos[c + d*x]^4)/(4*(a^2 + b^2)*d) + (b^5*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)^
3*d) + (a*b^2*Cos[c + d*x]*Sin[c + d*x])/(2*(a^2 + b^2)^2*d) + (3*a*Cos[c + d*x]*Sin[c + d*x])/(8*(a^2 + b^2)*
d) + (a*Cos[c + d*x]^3*Sin[c + d*x])/(4*(a^2 + b^2)*d)

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Rubi [A]  time = 0.214251, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3100, 2635, 8, 3098, 3133} \[ \frac{b^3 \cos ^2(c+d x)}{2 d \left (a^2+b^2\right )^2}+\frac{b \cos ^4(c+d x)}{4 d \left (a^2+b^2\right )}+\frac{a \sin (c+d x) \cos ^3(c+d x)}{4 d \left (a^2+b^2\right )}+\frac{a b^2 \sin (c+d x) \cos (c+d x)}{2 d \left (a^2+b^2\right )^2}+\frac{3 a \sin (c+d x) \cos (c+d x)}{8 d \left (a^2+b^2\right )}+\frac{b^5 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{a b^4 x}{\left (a^2+b^2\right )^3}+\frac{a b^2 x}{2 \left (a^2+b^2\right )^2}+\frac{3 a x}{8 \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

(a*b^4*x)/(a^2 + b^2)^3 + (a*b^2*x)/(2*(a^2 + b^2)^2) + (3*a*x)/(8*(a^2 + b^2)) + (b^3*Cos[c + d*x]^2)/(2*(a^2
+ b^2)^2*d) + (b*Cos[c + d*x]^4)/(4*(a^2 + b^2)*d) + (b^5*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)^
3*d) + (a*b^2*Cos[c + d*x]*Sin[c + d*x])/(2*(a^2 + b^2)^2*d) + (3*a*Cos[c + d*x]*Sin[c + d*x])/(8*(a^2 + b^2)*
d) + (a*Cos[c + d*x]^3*Sin[c + d*x])/(4*(a^2 + b^2)*d)

Rule 3100

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3098

Int[cos[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp
[(a*x)/(a^2 + b^2), x] + Dist[b/(a^2 + b^2), Int[(b*Cos[c + d*x] - a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c +
d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3133

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[((b*B + c*C)*x)/(b^2 + c^2), x] + Simp[((c*B - b*C)*L
og[a + b*Cos[d + e*x] + c*Sin[d + e*x]])/(e*(b^2 + c^2)), x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2
+ c^2, 0] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]

Rubi steps

\begin{align*} \int \frac{\cos ^5(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx &=\frac{b \cos ^4(c+d x)}{4 \left (a^2+b^2\right ) d}+\frac{a \int \cos ^4(c+d x) \, dx}{a^2+b^2}+\frac{b^2 \int \frac{\cos ^3(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{a^2+b^2}\\ &=\frac{b^3 \cos ^2(c+d x)}{2 \left (a^2+b^2\right )^2 d}+\frac{b \cos ^4(c+d x)}{4 \left (a^2+b^2\right ) d}+\frac{a \cos ^3(c+d x) \sin (c+d x)}{4 \left (a^2+b^2\right ) d}+\frac{\left (a b^2\right ) \int \cos ^2(c+d x) \, dx}{\left (a^2+b^2\right )^2}+\frac{b^4 \int \frac{\cos (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{\left (a^2+b^2\right )^2}+\frac{(3 a) \int \cos ^2(c+d x) \, dx}{4 \left (a^2+b^2\right )}\\ &=\frac{a b^4 x}{\left (a^2+b^2\right )^3}+\frac{b^3 \cos ^2(c+d x)}{2 \left (a^2+b^2\right )^2 d}+\frac{b \cos ^4(c+d x)}{4 \left (a^2+b^2\right ) d}+\frac{a b^2 \cos (c+d x) \sin (c+d x)}{2 \left (a^2+b^2\right )^2 d}+\frac{3 a \cos (c+d x) \sin (c+d x)}{8 \left (a^2+b^2\right ) d}+\frac{a \cos ^3(c+d x) \sin (c+d x)}{4 \left (a^2+b^2\right ) d}+\frac{b^5 \int \frac{b \cos (c+d x)-a \sin (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{\left (a^2+b^2\right )^3}+\frac{\left (a b^2\right ) \int 1 \, dx}{2 \left (a^2+b^2\right )^2}+\frac{(3 a) \int 1 \, dx}{8 \left (a^2+b^2\right )}\\ &=\frac{a b^4 x}{\left (a^2+b^2\right )^3}+\frac{a b^2 x}{2 \left (a^2+b^2\right )^2}+\frac{3 a x}{8 \left (a^2+b^2\right )}+\frac{b^3 \cos ^2(c+d x)}{2 \left (a^2+b^2\right )^2 d}+\frac{b \cos ^4(c+d x)}{4 \left (a^2+b^2\right ) d}+\frac{b^5 \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac{a b^2 \cos (c+d x) \sin (c+d x)}{2 \left (a^2+b^2\right )^2 d}+\frac{3 a \cos (c+d x) \sin (c+d x)}{8 \left (a^2+b^2\right ) d}+\frac{a \cos ^3(c+d x) \sin (c+d x)}{4 \left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 0.409526, size = 218, normalized size = 0.96 \[ \frac{24 a^3 b^2 \sin (2 (c+d x))+2 a^3 b^2 \sin (4 (c+d x))+4 b \left (4 a^2 b^2+a^4+3 b^4\right ) \cos (2 (c+d x))+b \left (a^2+b^2\right )^2 \cos (4 (c+d x))+40 a^3 b^2 c+40 a^3 b^2 d x+8 a^5 \sin (2 (c+d x))+a^5 \sin (4 (c+d x))+12 a^5 c+12 a^5 d x+16 a b^4 \sin (2 (c+d x))+a b^4 \sin (4 (c+d x))+32 b^5 \log (a \cos (c+d x)+b \sin (c+d x))+60 a b^4 c+60 a b^4 d x}{32 d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

(12*a^5*c + 40*a^3*b^2*c + 60*a*b^4*c + 12*a^5*d*x + 40*a^3*b^2*d*x + 60*a*b^4*d*x + 4*b*(a^4 + 4*a^2*b^2 + 3*
b^4)*Cos[2*(c + d*x)] + b*(a^2 + b^2)^2*Cos[4*(c + d*x)] + 32*b^5*Log[a*Cos[c + d*x] + b*Sin[c + d*x]] + 8*a^5
*Sin[2*(c + d*x)] + 24*a^3*b^2*Sin[2*(c + d*x)] + 16*a*b^4*Sin[2*(c + d*x)] + a^5*Sin[4*(c + d*x)] + 2*a^3*b^2
*Sin[4*(c + d*x)] + a*b^4*Sin[4*(c + d*x)])/(32*(a^2 + b^2)^3*d)

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Maple [B]  time = 0.123, size = 524, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5/(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

1/d*b^5/(a^2+b^2)^3*ln(a+b*tan(d*x+c))+3/8/d/(a^2+b^2)^3/(tan(d*x+c)^2+1)^2*tan(d*x+c)^3*a^5+5/4/d/(a^2+b^2)^3
/(tan(d*x+c)^2+1)^2*tan(d*x+c)^3*a^3*b^2+7/8/d/(a^2+b^2)^3/(tan(d*x+c)^2+1)^2*tan(d*x+c)^3*a*b^4+1/2/d/(a^2+b^
2)^3/(tan(d*x+c)^2+1)^2*tan(d*x+c)^2*a^2*b^3+1/2/d/(a^2+b^2)^3/(tan(d*x+c)^2+1)^2*tan(d*x+c)^2*b^5+7/4/d/(a^2+
b^2)^3/(tan(d*x+c)^2+1)^2*tan(d*x+c)*a^3*b^2+9/8/d/(a^2+b^2)^3/(tan(d*x+c)^2+1)^2*tan(d*x+c)*a*b^4+5/8/d/(a^2+
b^2)^3/(tan(d*x+c)^2+1)^2*tan(d*x+c)*a^5+1/4/d/(a^2+b^2)^3/(tan(d*x+c)^2+1)^2*a^4*b+1/d/(a^2+b^2)^3/(tan(d*x+c
)^2+1)^2*a^2*b^3+3/4/d/(a^2+b^2)^3/(tan(d*x+c)^2+1)^2*b^5-1/2/d/(a^2+b^2)^3*b^5*ln(tan(d*x+c)^2+1)+15/8/d/(a^2
+b^2)^3*arctan(tan(d*x+c))*a*b^4+3/8/d/(a^2+b^2)^3*arctan(tan(d*x+c))*a^5+5/4/d/(a^2+b^2)^3*arctan(tan(d*x+c))
*a^3*b^2

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Maxima [B]  time = 1.81493, size = 761, normalized size = 3.35 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/4*(4*b^5*log(-a - 2*b*sin(d*x + c)/(cos(d*x + c) + 1) + a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^6 + 3*a^4*
b^2 + 3*a^2*b^4 + b^6) - 4*b^5*log(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6
) + (3*a^5 + 10*a^3*b^2 + 15*a*b^4)*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6
) - (16*b^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - (5*a^3 + 9*a*b^2)*sin(d*x + c)/(cos(d*x + c) + 1) + 8*(a^2*b
+ 2*b^3)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + (3*a^3 - a*b^2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - (3*a^3 -
a*b^2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 8*(a^2*b + 2*b^3)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + (5*a^3 +
9*a*b^2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a^4 + 2*a^2*b^2 + b^4 + 4*(a^4 + 2*a^2*b^2 + b^4)*sin(d*x + c)
^2/(cos(d*x + c) + 1)^2 + 6*(a^4 + 2*a^2*b^2 + b^4)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*(a^4 + 2*a^2*b^2 +
b^4)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + (a^4 + 2*a^2*b^2 + b^4)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8))/d

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Fricas [A]  time = 0.562692, size = 471, normalized size = 2.07 \begin{align*} \frac{4 \, b^{5} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) + 2 \,{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} +{\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} d x + 4 \,{\left (a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} +{\left (2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{3} +{\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 7 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/8*(4*b^5*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) + 2*(a^4*b + 2*a^2*b^3 + b^
5)*cos(d*x + c)^4 + (3*a^5 + 10*a^3*b^2 + 15*a*b^4)*d*x + 4*(a^2*b^3 + b^5)*cos(d*x + c)^2 + (2*(a^5 + 2*a^3*b
^2 + a*b^4)*cos(d*x + c)^3 + (3*a^5 + 10*a^3*b^2 + 7*a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^6 + 3*a^4*b^2 + 3*
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)**5/(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.19883, size = 435, normalized size = 1.92 \begin{align*} \frac{\frac{8 \, b^{6} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac{4 \, b^{5} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{6 \, b^{5} \tan \left (d x + c\right )^{4} + 3 \, a^{5} \tan \left (d x + c\right )^{3} + 10 \, a^{3} b^{2} \tan \left (d x + c\right )^{3} + 7 \, a b^{4} \tan \left (d x + c\right )^{3} + 4 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} + 16 \, b^{5} \tan \left (d x + c\right )^{2} + 5 \, a^{5} \tan \left (d x + c\right ) + 14 \, a^{3} b^{2} \tan \left (d x + c\right ) + 9 \, a b^{4} \tan \left (d x + c\right ) + 2 \, a^{4} b + 8 \, a^{2} b^{3} + 12 \, b^{5}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/8*(8*b^6*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7) - 4*b^5*log(tan(d*x + c)^2 + 1)/
(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (3*a^5 + 10*a^3*b^2 + 15*a*b^4)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 +
b^6) + (6*b^5*tan(d*x + c)^4 + 3*a^5*tan(d*x + c)^3 + 10*a^3*b^2*tan(d*x + c)^3 + 7*a*b^4*tan(d*x + c)^3 + 4*
a^2*b^3*tan(d*x + c)^2 + 16*b^5*tan(d*x + c)^2 + 5*a^5*tan(d*x + c) + 14*a^3*b^2*tan(d*x + c) + 9*a*b^4*tan(d*
x + c) + 2*a^4*b + 8*a^2*b^3 + 12*b^5)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*(tan(d*x + c)^2 + 1)^2))/d