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

Optimal. Leaf size=141 \[ -\frac{b}{3 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^3}-\frac{a (b \cos (c+d x)-a \sin (c+d x))}{2 d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{a \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{2 d \left (a^2+b^2\right )^{5/2}} \]

[Out]

-(a*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(2*(a^2 + b^2)^(5/2)*d) - b/(3*(a^2 + b^2)*d*(
a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (a*(b*Cos[c + d*x] - a*Sin[c + d*x]))/(2*(a^2 + b^2)^2*d*(a*Cos[c + d*x]
+ b*Sin[c + d*x])^2)

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Rubi [A]  time = 0.110976, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3158, 12, 3076, 3074, 206} \[ -\frac{b}{3 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^3}-\frac{a (b \cos (c+d x)-a \sin (c+d x))}{2 d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{a \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{2 d \left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(2*(a^2 + b^2)^(5/2)*d) - b/(3*(a^2 + b^2)*d*(
a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (a*(b*Cos[c + d*x] - a*Sin[c + d*x]))/(2*(a^2 + b^2)^2*d*(a*Cos[c + d*x]
+ b*Sin[c + d*x])^2)

Rule 3158

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.))*((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(
x_)])^(n_), x_Symbol] :> -Simp[((c*B + c*A*Cos[d + e*x] + (a*B - b*A)*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Si
n[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 + b*Cos[d
+ e*x] + c*Sin[d + e*x])^(n + 1)*Simp[(n + 1)*(a*A - b*B) + (n + 2)*(a*B - b*A)*Cos[d + e*x] - (n + 2)*c*A*Sin
[d + e*x], x], x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && LtQ[n, -1] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[n, -2
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3076

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

Rule 3074

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.699489, size = 128, normalized size = 0.91 \[ \frac{\frac{3 \left (a^3-a b^2\right ) \sin (2 (c+d x))-4 b \left (a^2+b^2\right )-6 a^2 b \cos (2 (c+d x))}{2 \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{6 a \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*a*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) + (-4*b*(a^2 + b^2) - 6*a^2*b*Cos[
2*(c + d*x)] + 3*(a^3 - a*b^2)*Sin[2*(c + d*x)])/(2*(a^2 + b^2)^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3))/(6*d)

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Maple [B]  time = 0.233, size = 383, normalized size = 2.7 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) ^{3}} \left ( -1/2\,{\frac{ \left ({a}^{4}+4\,{a}^{2}{b}^{2}+2\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{a \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}-1/2\,{\frac{b \left ({a}^{4}-8\,{a}^{2}{b}^{2}-4\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{{a}^{2} \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}+1/3\,{\frac{{b}^{2} \left ( 15\,{a}^{4}-4\,{a}^{2}{b}^{2}-4\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{{a}^{3} \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}+{\frac{b \left ( 2\,{a}^{4}-5\,{a}^{2}{b}^{2}-2\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{{a}^{2} \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}+1/2\,{\frac{ \left ({a}^{4}-6\,{a}^{2}{b}^{2}-2\,{b}^{4} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{a \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}-1/6\,{\frac{b \left ( 5\,{a}^{2}+2\,{b}^{2} \right ) }{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}} \right ) }+{\frac{a}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}{\it Artanh} \left ({\frac{1}{2} \left ( 2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ) } \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))^4,x)

[Out]

1/d*(-2*(-1/2*(a^4+4*a^2*b^2+2*b^4)/a/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)^5-1/2*b*(a^4-8*a^2*b^2-4*b^4)/a^2
/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)^4+1/3/a^3*b^2*(15*a^4-4*a^2*b^2-4*b^4)/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x
+1/2*c)^3+1/a^2*b*(2*a^4-5*a^2*b^2-2*b^4)/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)^2+1/2/a*(a^4-6*a^2*b^2-2*b^4)
/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)-1/6*b*(5*a^2+2*b^2)/(a^4+2*a^2*b^2+b^4))/(tan(1/2*d*x+1/2*c)^2*a-2*tan
(1/2*d*x+1/2*c)*b-a)^3+a/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2
)^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.566791, size = 945, normalized size = 6.7 \begin{align*} \frac{2 \, a^{4} b - 2 \, a^{2} b^{3} - 4 \, b^{5} - 12 \,{\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} + 6 \,{\left (a^{5} - a b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \,{\left (3 \, a^{2} b^{2} \cos \left (d x + c\right ) +{\left (a^{4} - 3 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (a b^{3} +{\left (3 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (-\frac{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} - 2 \, a^{2} - b^{2} + 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{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 )}{12 \,{\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)/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/12*(2*a^4*b - 2*a^2*b^3 - 4*b^5 - 12*(a^4*b + a^2*b^3)*cos(d*x + c)^2 + 6*(a^5 - a*b^4)*cos(d*x + c)*sin(d*x
+ c) + 3*(3*a^2*b^2*cos(d*x + c) + (a^4 - 3*a^2*b^2)*cos(d*x + c)^3 + (a*b^3 + (3*a^3*b - a*b^3)*cos(d*x + c)
^2)*sin(d*x + c))*sqrt(a^2 + b^2)*log(-(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 -
b^2 + 2*sqrt(a^2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos
(d*x + c)^2 + b^2)))/((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)/(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Timed out

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Giac [B]  time = 1.30217, size = 575, normalized size = 4.08 \begin{align*} -\frac{\frac{3 \, a \log \left (\frac{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} - \frac{2 \,{\left (3 \, a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, a^{4} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, a^{2} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a^{5} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 24 \, a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 12 \, a b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 30 \, a^{4} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, a^{2} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, a^{5} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 30 \, a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, a b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, a^{4} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a^{2} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, a^{5} b + 2 \, a^{3} b^{3}\right )}}{{\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}^{3}}}{6 \, 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))^4,x, algorithm="giac")

[Out]

-1/6*(3*a*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b + 2*s
qrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2*(3*a^6*tan(1/2*d*x + 1/2*c)^5 + 12*a^4*b^2*tan(
1/2*d*x + 1/2*c)^5 + 6*a^2*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*a^5*b*tan(1/2*d*x + 1/2*c)^4 - 24*a^3*b^3*tan(1/2*d*
x + 1/2*c)^4 - 12*a*b^5*tan(1/2*d*x + 1/2*c)^4 - 30*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 + 8*a^2*b^4*tan(1/2*d*x + 1
/2*c)^3 + 8*b^6*tan(1/2*d*x + 1/2*c)^3 - 12*a^5*b*tan(1/2*d*x + 1/2*c)^2 + 30*a^3*b^3*tan(1/2*d*x + 1/2*c)^2 +
12*a*b^5*tan(1/2*d*x + 1/2*c)^2 - 3*a^6*tan(1/2*d*x + 1/2*c) + 18*a^4*b^2*tan(1/2*d*x + 1/2*c) + 6*a^2*b^4*ta
n(1/2*d*x + 1/2*c) + 5*a^5*b + 2*a^3*b^3)/((a^7 + 2*a^5*b^2 + a^3*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2
*d*x + 1/2*c) - a)^3))/d