\(\int \frac {1-\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx\) [596]

Optimal result

Integrand size = 25, antiderivative size = 73 \[ \int \frac {1-\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {a x}{b^2}-\frac {2 \sqrt {a-b} \sqrt {a+b} \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d}-\frac {\sin (c+d x)}{b d} \]

[Out]

a*x/b^2-sin(d*x+c)/b/d-2*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))*(a-b)^(1/2)*(a+b)^(1/2)/b^2/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3103, 2814, 2738, 211} \[ \int \frac {1-\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 \sqrt {a-b} \sqrt {a+b} \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d}+\frac {a x}{b^2}-\frac {\sin (c+d x)}{b d} \]

[In]

Int[(1 - Cos[c + d*x]^2)/(a + b*Cos[c + d*x]),x]

[Out]

(a*x)/b^2 - (2*Sqrt[a - b]*Sqrt[a + b]*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(b^2*d) - Sin[c + d
*x]/(b*d)

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 3103

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[
(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*
x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !Lt
Q[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sin (c+d x)}{b d}+\frac {\int \frac {b+a \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 {\left (a^2-b^2\right ) \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 \left (a^2-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 )}{b^2 d} \\ & = \frac {a x}{b^2}-\frac {2 \sqrt {a-b} \sqrt {a+b} \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d}-\frac {\sin (c+d x)}{b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.95 \[ \int \frac {1-\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {a (c+d x)-2 \sqrt {-a^2+b^2} \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )-b \sin (c+d x)}{b^2 d} \]

[In]

Integrate[(1 - Cos[c + d*x]^2)/(a + b*Cos[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 1.69 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.40

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (a +b \right ) \left (a -b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {-\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}}{d}\)	\(102\)
default	\(\frac {-\frac {2 \left (a +b \right ) \left (a -b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {-\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}}{d}\)	\(102\)
risch	\(\frac {a x}{b^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right )}{d \,b^{2}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right )}{d \,b^{2}}\)	\(146\)

[In]

int((-cos(d*x+c)^2+1)/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)

[Out]

1/d*(-2*(a+b)*(a-b)/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))+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))))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.77 \[ \int \frac {1-\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\left [\frac {2 \, a d x - 2 \, b \sin \left (d x + c\right ) + \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right )}{2 \, b^{2} d}, \frac {a d x - b \sin \left (d x + c\right ) - \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right )}{b^{2} d}\right ] \]

[In]

integrate((1-cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

[1/2*(2*a*d*x - 2*b*sin(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
*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)))/(b^2*d), (a*d*x - b*sin(d*x + c) - sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*
x + c))))/(b^2*d)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1039 vs. \(2 (61) = 122\).

Time = 52.51 (sec) , antiderivative size = 1039, normalized size of antiderivative = 14.23 \[ \int \frac {1-\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((1-cos(d*x+c)**2)/(a+b*cos(d*x+c)),x)

[Out]

Piecewise((zoo*x*(1 - cos(c)**2)/cos(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (x/b - sin(c + d*x)/(b*d), Eq(a, b))
, (-x/b - sin(c + d*x)/(b*d), Eq(a, -b)), ((-x*sin(c + d*x)**2/2 - x*cos(c + d*x)**2/2 + x - sin(c + d*x)*cos(
c + d*x)/(2*d))/a, Eq(b, 0)), (x*(1 - cos(c)**2)/(a + b*cos(c)), Eq(d, 0)), (a*d*x*sqrt(-a/(a - b) - b/(a - b)
)*tan(c/2 + d*x/2)**2/(b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(-a/(a - b) - b/(a
- b))) + a*d*x*sqrt(-a/(a - b) - b/(a - b))/(b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + b**2*d
*sqrt(-a/(a - b) - b/(a - b))) - a*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**2/(
b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(-a/(a - b) - b/(a - b))) - a*log(-sqrt(-
a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + b**2*d*s
qrt(-a/(a - b) - b/(a - b))) + a*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**2/(b**
2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(-a/(a - b) - b/(a - b))) + a*log(sqrt(-a/(a
- b) - b/(a - b)) + tan(c/2 + d*x/2))/(b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(
-a/(a - b) - b/(a - b))) - 2*b*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)/(b**2*d*sqrt(-a/(a - b) - b/(a -
b))*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(-a/(a - b) - b/(a - b))) - b*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2
+ d*x/2))*tan(c/2 + d*x/2)**2/(b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(-a/(a -
b) - b/(a - b))) - b*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(b**2*d*sqrt(-a/(a - b) - b/(a - b)
)*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(-a/(a - b) - b/(a - b))) + b*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 +
d*x/2))*tan(c/2 + d*x/2)**2/(b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(-a/(a - b)
- b/(a - b))) + b*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(b**2*d*sqrt(-a/(a - b) - b/(a - b))*ta
n(c/2 + d*x/2)**2 + b**2*d*sqrt(-a/(a - b) - b/(a - b))), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {1-\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((1-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 = 0.31 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.67 \[ \int \frac {1-\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} a}{b^{2}} + \frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} \sqrt {a^{2} - b^{2}}}{b^{2}} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b}}{d} \]

[In]

integrate((1-cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

((d*x + c)*a/b^2 + 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*t
an(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))*sqrt(a^2 - b^2)/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 = 1.94 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.53 \[ \int \frac {1-\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}}{b^2\,d}-\frac {\sin \left (c+d\,x\right )}{b\,d}+\frac {2\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d} \]

[In]

int(-(cos(c + d*x)^2 - 1)/(a + b*cos(c + d*x)),x)

[Out]

(2*atanh((sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(a*cos(c/2 + (d*x)/2) + b*cos(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/
2))/(b^2*d) - sin(c + d*x)/(b*d) + (2*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^2*d)