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]

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

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Rubi [A]  time = 0.11, antiderivative size = 141, normalized size of antiderivative = 1.00, 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 12

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

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])

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 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 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
]

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*}

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Mathematica [A]  time = 0.73, size = 128, normalized size = 0.91 \[ \frac {\frac {6 a \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac {3 \left (a^3-a b^2\right ) \sin (2 (c+d x))-4 b \left (a^2+b^2\right )-6 a^2 b \cos (2 (c+d x))}{2 \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3}}{6 d} \]

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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fricas [B]  time = 0.74, size = 420, normalized size = 2.98 \[ \frac {2 \, a^{4} b - 2 \, a^{2} b^{3} - 4 \, b^{5} - 12 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (a^{5} - a b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, {\left (3 \, a^{2} b^{2} \cos \left (d x + c\right ) + {\left (a^{4} - 3 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (a b^{3} + {\left (3 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right )}{12 \, {\left ({\left (a^{9} - 6 \, a^{5} b^{4} - 8 \, a^{3} b^{6} - 3 \, a b^{8}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{7} b^{2} + 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right ) + {\left ({\left (3 \, a^{8} b + 8 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - b^{9}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9}\right )} d\right )} \sin \left (d x + c\right )\right )}} \]

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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giac [B]  time = 4.08, size = 426, normalized size = 3.02 \[ -\frac {\frac {3 \, a \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (3 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 30 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a^{5} b + 2 \, a^{3} b^{3}\right )}}{{\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}^{3}}}{6 \, d} \]

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

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maple [B]  time = 0.30, size = 383, normalized size = 2.72 \[ \frac {-\frac {2 \left (-\frac {\left (a^{4}+4 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (a^{4}-8 a^{2} b^{2}-4 b^{4}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2} \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {b^{2} \left (15 a^{4}-4 a^{2} b^{2}-4 b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a^{3} \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {b \left (2 a^{4}-5 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (a^{4}-6 a^{2} b^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (5 a^{2}+2 b^{2}\right )}{6 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}\right )}{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{3}}+\frac {a \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}}}{d} \]

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*b*t
an(1/2*d*x+1/2*c)-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 [B]  time = 0.45, size = 606, normalized size = 4.30 \[ -\frac {\frac {3 \, a \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (5 \, a^{5} b + 2 \, a^{3} b^{3} - \frac {3 \, {\left (a^{6} - 6 \, a^{4} b^{2} - 2 \, a^{2} b^{4}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {6 \, {\left (2 \, a^{5} b - 5 \, a^{3} b^{3} - 2 \, a b^{5}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, {\left (15 \, a^{4} b^{2} - 4 \, a^{2} b^{4} - 4 \, b^{6}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, {\left (a^{5} b - 8 \, a^{3} b^{3} - 4 \, a b^{5}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {3 \, {\left (a^{6} + 4 \, a^{4} b^{2} + 2 \, a^{2} b^{4}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{10} + 2 \, a^{8} b^{2} + a^{6} b^{4} + \frac {6 \, {\left (a^{9} b + 2 \, a^{7} b^{3} + a^{5} b^{5}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, {\left (a^{10} - 2 \, a^{8} b^{2} - 7 \, a^{6} b^{4} - 4 \, a^{4} b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, {\left (3 \, a^{9} b + 4 \, a^{7} b^{3} - a^{5} b^{5} - 2 \, a^{3} b^{7}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, {\left (a^{10} - 2 \, a^{8} b^{2} - 7 \, a^{6} b^{4} - 4 \, a^{4} b^{6}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {6 \, {\left (a^{9} b + 2 \, a^{7} b^{3} + a^{5} b^{5}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {{\left (a^{10} + 2 \, a^{8} b^{2} + a^{6} b^{4}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}}}{6 \, 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))^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.82, size = 505, normalized size = 3.58 \[ -\frac {\frac {\frac {5\,a^2\,b}{3}+\frac {2\,b^3}{3}}{a^4+2\,a^2\,b^2+b^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-a^4\,b+8\,a^2\,b^3+4\,b^5\right )}{a^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-4\,a^4\,b+10\,a^2\,b^3+4\,b^5\right )}{a^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-a^4+6\,a^2\,b^2+2\,b^4\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (a^4+4\,a^2\,b^2+2\,b^4\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (5\,a^2\,b+2\,b^3\right )\,\left (3\,a^2-2\,b^2\right )}{3\,a^3\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (12\,a\,b^2-3\,a^3\right )-a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (12\,a\,b^2-3\,a^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (12\,a^2\,b-8\,b^3\right )+a^3+6\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}-\frac {a\,\mathrm {atanh}\left (\frac {a^4\,b+2\,a^2\,b^3+b^5}{{\left (a^2+b^2\right )}^{5/2}}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{{\left (a^2+b^2\right )}^{5/2}}\right )}{d\,{\left (a^2+b^2\right )}^{5/2}} \]

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))^4,x)

[Out]

- (((5*a^2*b)/3 + (2*b^3)/3)/(a^4 + b^4 + 2*a^2*b^2) - (tan(c/2 + (d*x)/2)^4*(4*b^5 - a^4*b + 8*a^2*b^3))/(a^2
*(a^4 + b^4 + 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^2*(4*b^5 - 4*a^4*b + 10*a^2*b^3))/(a^2*(a^4 + b^4 + 2*a^2*b^2)
) + (tan(c/2 + (d*x)/2)*(2*b^4 - a^4 + 6*a^2*b^2))/(a*(a^4 + b^4 + 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^5*(a^4 +
2*b^4 + 4*a^2*b^2))/(a*(a^4 + b^4 + 2*a^2*b^2)) - (2*b*tan(c/2 + (d*x)/2)^3*(5*a^2*b + 2*b^3)*(3*a^2 - 2*b^2))
/(3*a^3*(a^4 + b^4 + 2*a^2*b^2)))/(d*(tan(c/2 + (d*x)/2)^2*(12*a*b^2 - 3*a^3) - a^3*tan(c/2 + (d*x)/2)^6 - tan
(c/2 + (d*x)/2)^4*(12*a*b^2 - 3*a^3) - tan(c/2 + (d*x)/2)^3*(12*a^2*b - 8*b^3) + a^3 + 6*a^2*b*tan(c/2 + (d*x)
/2) + 6*a^2*b*tan(c/2 + (d*x)/2)^5)) - (a*atanh((a^4*b + b^5 + 2*a^2*b^3)/(a^2 + b^2)^(5/2) - (a*tan(c/2 + (d*
x)/2)*(a^4 + b^4 + 2*a^2*b^2))/(a^2 + b^2)^(5/2)))/(d*(a^2 + b^2)^(5/2))

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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))**4,x)

[Out]

Timed out

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