\(\int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx\) [481]

Optimal result

Integrand size = 21, antiderivative size = 206 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {a \left (a^2+4 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2869, 2833, 12, 2738, 211} \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {a \left (a^2+4 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac {a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{6 b d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))} \]

[In]

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

[Out]

(a*(a^2 + 4*b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a + b)^(7/2)*d) - (a^2*Si
n[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) + (a*(a^2 - 6*b^2)*Sin[c + d*x])/(6*b*(a^2 - b^2)^2*d*(
a + b*Cos[c + d*x])^2) + ((a^4 - 10*a^2*b^2 - 6*b^4)*Sin[c + d*x])/(6*b*(a^2 - b^2)^3*d*(a + b*Cos[c + d*x]))

Rule 12

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2738

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

Rule 2833

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

Rule 2869

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

Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\int \frac {3 a b+\left (a^2-3 b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx}{3 b \left (a^2-b^2\right )} \\ & = -\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac {\int \frac {-2 b \left (2 a^2+3 b^2\right )-a \left (a^2-6 b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{6 b \left (a^2-b^2\right )^2} \\ & = -\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac {\int \frac {3 a b \left (a^2+4 b^2\right )}{a+b \cos (c+d x)} \, dx}{6 b \left (a^2-b^2\right )^3} \\ & = -\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac {\left (a \left (a^2+4 b^2\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3} \\ & = -\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac {\left (a \left (a^2+4 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d} \\ & = \frac {a \left (a^2+4 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.79 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {\frac {6 a \left (a^2+4 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{7/2}}+\frac {\left (-13 a^4 b-2 a^2 b^3+3 a \left (a^4-9 a^2 b^2-2 b^4\right ) \cos (c+d x)+b \left (a^4-10 a^2 b^2-6 b^4\right ) \cos ^2(c+d x)\right ) \sin (c+d x)}{(a-b)^3 (a+b)^3 (a+b \cos (c+d x))^3}}{6 d} \]

[In]

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

[Out]

((6*a*(a^2 + 4*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(7/2) + ((-13*a^4*b - 2
*a^2*b^3 + 3*a*(a^4 - 9*a^2*b^2 - 2*b^4)*Cos[c + d*x] + b*(a^4 - 10*a^2*b^2 - 6*b^4)*Cos[c + d*x]^2)*Sin[c + d
*x])/((a - b)^3*(a + b)^3*(a + b*Cos[c + d*x])^3))/(6*d)

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.41

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (a^{3}+6 a^{2} b +2 a \,b^{2}+2 b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {4 \left (7 a^{2}+3 b^{2}\right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (a^{3}-6 a^{2} b +2 a \,b^{2}-2 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{3}}+\frac {a \left (a^{2}+4 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(290\)
default	\(\frac {\frac {-\frac {\left (a^{3}+6 a^{2} b +2 a \,b^{2}+2 b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {4 \left (7 a^{2}+3 b^{2}\right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (a^{3}-6 a^{2} b +2 a \,b^{2}-2 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{3}}+\frac {a \left (a^{2}+4 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(290\)
risch	\(\frac {i \left (-3 a^{3} b^{4} {\mathrm e}^{5 i \left (d x +c \right )}-12 a \,b^{6} {\mathrm e}^{5 i \left (d x +c \right )}+6 a^{6} b \,{\mathrm e}^{4 i \left (d x +c \right )}-33 a^{4} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-42 a^{2} b^{5} {\mathrm e}^{4 i \left (d x +c \right )}-6 b^{7} {\mathrm e}^{4 i \left (d x +c \right )}+4 a^{7} {\mathrm e}^{3 i \left (d x +c \right )}-34 a^{5} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-84 a^{3} b^{4} {\mathrm e}^{3 i \left (d x +c \right )}-36 a \,b^{6} {\mathrm e}^{3 i \left (d x +c \right )}+6 a^{6} b \,{\mathrm e}^{2 i \left (d x +c \right )}-84 a^{4} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-60 a^{2} b^{5} {\mathrm e}^{2 i \left (d x +c \right )}-12 b^{7} {\mathrm e}^{2 i \left (d x +c \right )}+6 a^{5} b^{2} {\mathrm e}^{i \left (d x +c \right )}-57 a^{3} b^{4} {\mathrm e}^{i \left (d x +c \right )}-24 a \,b^{6} {\mathrm e}^{i \left (d x +c \right )}+a^{4} b^{3}-10 a^{2} b^{5}-6 b^{7}\right )}{3 b^{2} \left (a^{2}-b^{2}\right )^{3} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {2 a \,b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {2 a \,b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\)	\(678\)

[In]

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

[Out]

1/d*(2*(-1/2*(a^3+6*a^2*b+2*a*b^2+2*b^3)/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5-2/3*(7*a^2+3*b^2
)*b/(a^2-2*a*b+b^2)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+1/2*(a^3-6*a^2*b+2*a*b^2-2*b^3)/(a+b)/(a^3-3*a^2*b+3*
a*b^2-b^3)*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-b*tan(1/2*d*x+1/2*c)^2+a+b)^3+a*(a^2+4*b^2)/(a^6-3*a^4*
b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (191) = 382\).

Time = 0.31 (sec) , antiderivative size = 893, normalized size of antiderivative = 4.33 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\left [\frac {3 \, {\left (a^{6} + 4 \, a^{4} b^{2} + {\left (a^{3} b^{3} + 4 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{5} b + 4 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (13 \, a^{6} b - 11 \, a^{4} b^{3} - 2 \, a^{2} b^{5} - {\left (a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + 6 \, b^{7}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{7} - 10 \, a^{5} b^{2} + 7 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left ({\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d\right )}}, \frac {3 \, {\left (a^{6} + 4 \, a^{4} b^{2} + {\left (a^{3} b^{3} + 4 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{5} b + 4 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (13 \, a^{6} b - 11 \, a^{4} b^{3} - 2 \, a^{2} b^{5} - {\left (a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + 6 \, b^{7}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{7} - 10 \, a^{5} b^{2} + 7 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d\right )}}\right ] \]

[In]

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

[Out]

[1/12*(3*(a^6 + 4*a^4*b^2 + (a^3*b^3 + 4*a*b^5)*cos(d*x + c)^3 + 3*(a^4*b^2 + 4*a^2*b^4)*cos(d*x + c)^2 + 3*(a
^5*b + 4*a^3*b^3)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*sq
rt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2
)) - 2*(13*a^6*b - 11*a^4*b^3 - 2*a^2*b^5 - (a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + 6*b^7)*cos(d*x + c)^2 - 3*(a^7 -
10*a^5*b^2 + 7*a^3*b^4 + 2*a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 +
b^11)*d*cos(d*x + c)^3 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c)^2 + 3*(a^10*
b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c) + (a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 +
a^3*b^8)*d), 1/6*(3*(a^6 + 4*a^4*b^2 + (a^3*b^3 + 4*a*b^5)*cos(d*x + c)^3 + 3*(a^4*b^2 + 4*a^2*b^4)*cos(d*x +
c)^2 + 3*(a^5*b + 4*a^3*b^3)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d
*x + c))) - (13*a^6*b - 11*a^4*b^3 - 2*a^2*b^5 - (a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + 6*b^7)*cos(d*x + c)^2 - 3*(
a^7 - 10*a^5*b^2 + 7*a^3*b^4 + 2*a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*
b^9 + b^11)*d*cos(d*x + c)^3 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c)^2 + 3*(
a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c) + (a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b
^6 + a^3*b^8)*d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (191) = 382\).

Time = 0.33 (sec) , antiderivative size = 427, normalized size of antiderivative = 2.07 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=-\frac {\frac {3 \, {\left (a^{3} + 4 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 28 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{3}}}{3 \, d} \]

[In]

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

[Out]

-1/3*(3*(a^3 + 4*a*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) -
b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + (3*a^5*tan(1
/2*d*x + 1/2*c)^5 + 12*a^4*b*tan(1/2*d*x + 1/2*c)^5 - 27*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 12*a^2*b^3*tan(1/2*d
*x + 1/2*c)^5 - 6*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*b^5*tan(1/2*d*x + 1/2*c)^5 + 28*a^4*b*tan(1/2*d*x + 1/2*c)^
3 - 16*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 12*b^5*tan(1/2*d*x + 1/2*c)^3 - 3*a^5*tan(1/2*d*x + 1/2*c) + 12*a^4*b*
tan(1/2*d*x + 1/2*c) + 27*a^3*b^2*tan(1/2*d*x + 1/2*c) + 12*a^2*b^3*tan(1/2*d*x + 1/2*c) + 6*a*b^4*tan(1/2*d*x
+ 1/2*c) + 6*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan
(1/2*d*x + 1/2*c)^2 + a + b)^3))/d

Mupad [B] (verification not implemented)

Time = 18.37 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.85 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+4\,b^2\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}\,\left (a^3+4\,a\,b^2\right )}\right )\,\left (a^2+4\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (a^3+6\,a^2\,b+2\,a\,b^2+2\,b^3\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,a^2\,b+3\,b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^3-6\,a^2\,b+2\,a\,b^2-2\,b^3\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )} \]

[In]

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

[Out]

(a*atan((a*tan(c/2 + (d*x)/2)*(a^2 + 4*b^2)*(2*a - 2*b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3))/(2*(a + b)^(1/2)*(a -
b)^(7/2)*(4*a*b^2 + a^3)))*(a^2 + 4*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2)) - ((tan(c/2 + (d*x)/2)^5*(2*a*b^2 +
6*a^2*b + a^3 + 2*b^3))/((a + b)^3*(a - b)) + (4*tan(c/2 + (d*x)/2)^3*(7*a^2*b + 3*b^3))/(3*(a + b)^2*(a^2 -
2*a*b + b^2)) - (tan(c/2 + (d*x)/2)*(2*a*b^2 - 6*a^2*b + a^3 - 2*b^3))/((a + b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3
)))/(d*(3*a*b^2 - tan(c/2 + (d*x)/2)^4*(3*a*b^2 + 3*a^2*b - 3*a^3 - 3*b^3) - tan(c/2 + (d*x)/2)^2*(3*a*b^2 - 3
*a^2*b - 3*a^3 + 3*b^3) + 3*a^2*b + a^3 + b^3 + tan(c/2 + (d*x)/2)^6*(3*a*b^2 - 3*a^2*b + a^3 - b^3)))