\(\int \frac {\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx\) [452]
Optimal result
Integrand size = 21, antiderivative size = 76 \[
\int \frac {\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {a x}{b^2}+\frac {2 a^2 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b} d}+\frac {\sin (c+d x)}{b d}
\]
[Out]
-a*x/b^2+sin(d*x+c)/b/d+2*a^2*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/b^2/d/(a-b)^(1/2)/(a+b)^(1/2)
Rubi [A] (verified)
Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of
steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2825, 12, 2814, 2738, 211}
\[
\int \frac {\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 a^2 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d \sqrt {a-b} \sqrt {a+b}}-\frac {a x}{b^2}+\frac {\sin (c+d x)}{b d}
\]
[In]
Int[Cos[c + d*x]^2/(a + b*Cos[c + d*x]),x]
[Out]
-((a*x)/b^2) + (2*a^2*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^2*Sqrt[a + b]*d) + Si
n[c + d*x]/(b*d)
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 2814
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]
Rule 2825
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b^2
)*(Cos[e + f*x]/(d*f)), x] + Dist[1/d, Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(c + d*Sin[e + f*x]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Rubi steps \begin{align*}
\text {integral}& = \frac {\sin (c+d x)}{b d}-\frac {\int \frac {a \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{b} \\ & = \frac {\sin (c+d x)}{b d}-\frac {a \int \frac {\cos (c+d x)}{a+b \cos (c+d x)} \, dx}{b} \\ & = -\frac {a x}{b^2}+\frac {\sin (c+d x)}{b d}+\frac {a^2 \int \frac {1}{a+b \cos (c+d x)} \, dx}{b^2} \\ & = -\frac {a x}{b^2}+\frac {\sin (c+d x)}{b d}+\frac {\left (2 a^2\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 )}{b^2 d} \\ & = -\frac {a x}{b^2}+\frac {2 a^2 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b} d}+\frac {\sin (c+d x)}{b d} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.95
\[
\int \frac {\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-a (c+d x)-\frac {2 a^2 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+b \sin (c+d x)}{b^2 d}
\]
[In]
Integrate[Cos[c + d*x]^2/(a + b*Cos[c + d*x]),x]
[Out]
(-(a*(c + d*x)) - (2*a^2*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + b*Sin[c + d*
x])/(b^2*d)
Maple [A] (verified)
Time = 0.98 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.28
| | |
| method | result | size |
| | |
| derivativedivides |
\(\frac {-\frac {2 \left (-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{2}}+\frac {2 a^{2} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) |
\(97\) |
| default |
\(\frac {-\frac {2 \left (-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{2}}+\frac {2 a^{2} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) |
\(97\) |
| risch |
\(-\frac {a x}{b^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}\) |
\(194\) |
| | |
|
|
|
|
[In]
int(cos(d*x+c)^2/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)
[Out]
1/d*(-2/b^2*(-b*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+a*arctan(tan(1/2*d*x+1/2*c)))+2*a^2/b^2/((a-b)*(a+
b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))
Fricas [A] (verification not implemented)
none
Time = 0.28 (sec) , antiderivative size = 269, normalized size of antiderivative = 3.54
\[
\int \frac {\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} a^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} d x - 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d}, \frac {\sqrt {a^{2} - b^{2}} a^{2} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (a^{3} - a b^{2}\right )} d x + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{2} - b^{4}\right )} d}\right ]
\]
[In]
integrate(cos(d*x+c)^2/(a+b*cos(d*x+c)),x, algorithm="fricas")
[Out]
[-1/2*(sqrt(-a^2 + b^2)*a^2*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-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*(a^3 - a*b^2)*
d*x - 2*(a^2*b - b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d), (sqrt(a^2 - b^2)*a^2*arctan(-(a*cos(d*x + c) + b)/(sq
rt(a^2 - b^2)*sin(d*x + c))) - (a^3 - a*b^2)*d*x + (a^2*b - b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d)]
Sympy [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 1744 vs. \(2 (65) = 130\).
Time = 65.92 (sec) , antiderivative size = 1744, normalized size of antiderivative = 22.95
\[
\int \frac {\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display}
\]
[In]
integrate(cos(d*x+c)**2/(a+b*cos(d*x+c)),x)
[Out]
Piecewise((zoo*x*cos(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (-d*x*tan(c/2 + d*x/2)**2/(b*d*tan(c/2 + d*x/2)**2 +
b*d) - d*x/(b*d*tan(c/2 + d*x/2)**2 + b*d) + tan(c/2 + d*x/2)**3/(b*d*tan(c/2 + d*x/2)**2 + b*d) + 3*tan(c/2
+ d*x/2)/(b*d*tan(c/2 + d*x/2)**2 + b*d), Eq(a, b)), (d*x*tan(c/2 + d*x/2)**3/(b*d*tan(c/2 + d*x/2)**3 + b*d*t
an(c/2 + d*x/2)) + d*x*tan(c/2 + d*x/2)/(b*d*tan(c/2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)) + 3*tan(c/2 + d*x/2)*
*2/(b*d*tan(c/2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)) + 1/(b*d*tan(c/2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)), Eq(a
, -b)), ((x*sin(c + d*x)**2/2 + x*cos(c + d*x)**2/2 + sin(c + d*x)*cos(c + d*x)/(2*d))/a, Eq(b, 0)), (x*cos(c)
**2/(a + b*cos(c)), Eq(d, 0)), (-a**2*d*x*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2/(a*b**2*d*sqrt(-a/(
a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(
a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) - a**2*d*x*sqrt(-a/(a - b) - b/(a - b))/(a*
b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(
-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) + a**2*log(-sqrt(-a/(a - b)
- b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**2/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)*
*2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*
sqrt(-a/(a - b) - b/(a - b))) + a**2*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(a*b**2*d*sqrt(-a/(
a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(
a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) - a**2*log(sqrt(-a/(a - b) - b/(a - b)) + t
an(c/2 + d*x/2))*tan(c/2 + d*x/2)**2/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqr
t(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) -
b/(a - b))) - a**2*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b)
)*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 +
d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) + a*b*d*x*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2/(a
*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt
(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) + a*b*d*x*sqrt(-a/(a - b)
- b/(a - b))/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)
) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) + 2*a*b*sqr
t(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2
*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a
- b) - b/(a - b))) - 2*b**2*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b
))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 +
d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))), True))
Maxima [F(-2)]
Exception generated. \[
\int \frac {\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError}
\]
[In]
integrate(cos(d*x+c)^2/(a+b*cos(d*x+c)),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 [A] (verification not implemented)
none
Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.66
\[
\int \frac {\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{2}}{\sqrt {a^{2} - b^{2}} b^{2}} + \frac {{\left (d x + c\right )} a}{b^{2}} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b}}{d}
\]
[In]
integrate(cos(d*x+c)^2/(a+b*cos(d*x+c)),x, algorithm="giac")
[Out]
-(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^2/(sqrt(a^2 - b^2)*b^2) + (d*x + c)*a/b^2 - 2*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2
*c)^2 + 1)*b))/d
Mupad [B] (verification not implemented)
Time = 14.41 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.50
\[
\int \frac {\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\sin \left (c+d\,x\right )}{b\,d}-\frac {2\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d}-\frac {a^2\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b-2{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2+1{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (b^2-a^2\right )}^{3/2}+a^2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )\,2{}\mathrm {i}}{b^2\,d\,\sqrt {b^2-a^2}}
\]
[In]
int(cos(c + d*x)^2/(a + b*cos(c + d*x)),x)
[Out]
sin(c + d*x)/(b*d) - (2*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^2*d) - (a^2*atan((b^3*sin(c/2 + (d*x
)/2)*1i - a*b^2*sin(c/2 + (d*x)/2)*2i + a^2*b*sin(c/2 + (d*x)/2)*1i)/(cos(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) + a
^2*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - a*b*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)))*2i)/(b^2*d*(b^2 - a^2)^(1
/2))