\(\int \frac {\cos (c+d x) (1-\cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\) [595]

Optimal result

Integrand size = 31, antiderivative size = 109 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=-\frac {\left (2 a^2-b^2\right ) x}{2 b^3}+\frac {2 a \sqrt {a-b} \sqrt {a+b} \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d}+\frac {a \sin (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3129, 3102, 2814, 2738, 211} \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=-\frac {x \left (2 a^2-b^2\right )}{2 b^3}+\frac {2 a \sqrt {a-b} \sqrt {a+b} \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d}+\frac {a \sin (c+d x)}{b^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d} \]

[In]

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

[Out]

-1/2*((2*a^2 - b^2)*x)/b^3 + (2*a*Sqrt[a - b]*Sqrt[a + b]*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/
(b^3*d) + (a*Sin[c + d*x])/(b^2*d) - (Cos[c + d*x]*Sin[c + d*x])/(2*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 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (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) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3129

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)
*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^
(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e +
f*x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a,
0] && NeQ[c, 0])))

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (c+d x) \sin (c+d x)}{2 b d}+\frac {\int \frac {-a+b \cos (c+d x)+2 a \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 b} \\ & = \frac {a \sin (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d}+\frac {\int \frac {-a b-\left (2 a^2-b^2\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^2} \\ & = -\frac {\left (2 a^2-b^2\right ) x}{2 b^3}+\frac {a \sin (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d}+\frac {\left (a \left (a^2-b^2\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{b^3} \\ & = -\frac {\left (2 a^2-b^2\right ) x}{2 b^3}+\frac {a \sin (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d}+\frac {\left (2 a \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^3 d} \\ & = -\frac {\left (2 a^2-b^2\right ) x}{2 b^3}+\frac {2 a \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^3 d}+\frac {a \sin (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {-2 \left (2 a^2-b^2\right ) (c+d x)+8 a \sqrt {-a^2+b^2} \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )+4 a b \sin (c+d x)-b^2 \sin (2 (c+d x))}{4 b^3 d} \]

[In]

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

[Out]

(-2*(2*a^2 - b^2)*(c + d*x) + 8*a*Sqrt[-a^2 + b^2]*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]] + 4*a*
b*Sin[c + d*x] - b^2*Sin[2*(c + d*x)])/(4*b^3*d)

Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.34

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (\frac {\left (-a b -\frac {1}{2} b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a b +\frac {1}{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{3}}+\frac {2 \left (a -b \right ) a \left (a +b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(146\)
default	\(\frac {-\frac {2 \left (\frac {\left (-a b -\frac {1}{2} b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a b +\frac {1}{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{3}}+\frac {2 \left (a -b \right ) a \left (a +b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(146\)
risch	\(-\frac {x \,a^{2}}{b^{3}}+\frac {x}{2 b}-\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}+\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}+\frac {\sqrt {-a^{2}+b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right )}{d \,b^{3}}-\frac {\sqrt {-a^{2}+b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right )}{d \,b^{3}}-\frac {\sin \left (2 d x +2 c \right )}{4 d b}\)	\(176\)

[In]

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

[Out]

1/d*(-2/b^3*(((-a*b-1/2*b^2)*tan(1/2*d*x+1/2*c)^3+(-a*b+1/2*b^2)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^
2+1/2*(2*a^2-b^2)*arctan(tan(1/2*d*x+1/2*c)))+2*(a-b)*a*(a+b)/b^3/((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 = 0.29 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.30 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\left [-\frac {{\left (2 \, a^{2} - b^{2}\right )} d x - \sqrt {-a^{2} + b^{2}} a \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + {\left (b^{2} \cos \left (d x + c\right ) - 2 \, a b\right )} \sin \left (d x + c\right )}{2 \, b^{3} d}, -\frac {{\left (2 \, a^{2} - b^{2}\right )} d x - 2 \, \sqrt {a^{2} - b^{2}} a \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) + {\left (b^{2} \cos \left (d x + c\right ) - 2 \, a b\right )} \sin \left (d x + c\right )}{2 \, b^{3} d}\right ] \]

[In]

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

[Out]

[-1/2*((2*a^2 - b^2)*d*x - sqrt(-a^2 + b^2)*a*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*cos(d*x + c) - 2*a*b)*sin(d*x + c))/(b^3*d), -1/2*((2*a^2 - b^2)*d*x - 2*sqrt(a^2 - b^2)*a*arctan(-(a*c
os(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) + (b^2*cos(d*x + c) - 2*a*b)*sin(d*x + c))/(b^3*d)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2958 vs. \(2 (92) = 184\).

Time = 159.03 (sec) , antiderivative size = 2958, normalized size of antiderivative = 27.14 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((zoo*x*(1 - cos(c)**2), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (-x*sin(c + d*x)**2/(2*b) - x*cos(c + d*x)*
*2/(2*b) - sin(c + d*x)*cos(c + d*x)/(2*b*d) + sin(c + d*x)/(b*d), Eq(a, b)), (-x*sin(c + d*x)**2/(2*b) - x*co
s(c + d*x)**2/(2*b) - sin(c + d*x)*cos(c + d*x)/(2*b*d) - sin(c + d*x)/(b*d), Eq(a, -b)), ((-2*sin(c + d*x)**3
/(3*d) - sin(c + d*x)*cos(c + d*x)**2/d + sin(c + d*x)/d)/a, Eq(b, 0)), (x*(1 - cos(c)**2)*cos(c)/(a + b*cos(c
)), Eq(d, 0)), (-2*a**2*d*x*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4/(2*b**3*d*sqrt(-a/(a - b) - b/(a
- b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a -
b) - b/(a - b))) - 4*a**2*d*x*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2/(2*b**3*d*sqrt(-a/(a - b) - b/(
a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a
- b) - b/(a - b))) - 2*a**2*d*x*sqrt(-a/(a - b) - b/(a - b))/(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 +
d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b)))
+ 2*a**2*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**4/(2*b**3*d*sqrt(-a/(a - b)
- b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-
a/(a - b) - b/(a - b))) + 4*a**2*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**2/(2*
b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/
2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) + 2*a**2*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/
(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d
*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) - 2*a**2*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2)
)*tan(c/2 + d*x/2)**4/(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) -
b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) - 4*a**2*log(sqrt(-a/(a - b) - b/(a -
b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**2/(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**
3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) - 2*a**2*log(sqr
t(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b
**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) + 4*a*b*sqrt(-
a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**3/(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3
*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) + 4*a*b*sqrt(-a/(
a - b) - b/(a - b))*tan(c/2 + d*x/2)/(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqr
t(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) + 2*a*b*log(-sqrt(-a/(a
- b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**4/(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*
x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) +
4*a*b*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**2/(2*b**3*d*sqrt(-a/(a - b) - b
/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(
a - b) - b/(a - b))) + 2*a*b*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(2*b**3*d*sqrt(-a/(a - b) -
b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a
/(a - b) - b/(a - b))) - 2*a*b*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**4/(2*b**
3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)*
*2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) - 4*a*b*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c
/2 + d*x/2)**2/(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a -
b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) - 2*a*b*log(sqrt(-a/(a - b) - b/(a - b)) + ta
n(c/2 + d*x/2))/(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a -
b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) + b**2*d*x*sqrt(-a/(a - b) - b/(a - b))*tan(
c/2 + d*x/2)**4/(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a -
b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) + 2*b**2*d*x*sqrt(-a/(a - b) - b/(a - b))*ta
n(c/2 + d*x/2)**2/(2*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a
- b))*tan(c/2 + d*x/2)**2 + 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))) + b**2*d*x*sqrt(-a/(a - b) - b/(a - b))/(2
*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x
/2)**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)**3/(2*b
**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2
)**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)/(2*b**3*d
*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**4 + 4*b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2
+ 2*b**3*d*sqrt(-a/(a - b) - b/(a - b))), True))

Maxima [F(-2)]

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

[In]

integrate(cos(d*x+c)*(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 = 186, normalized size of antiderivative = 1.71 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=-\frac {\frac {{\left (2 \, a^{2} - b^{2}\right )} {\left (d x + c\right )}}{b^{3}} + \frac {4 \, {\left (a^{3} - a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{3}} - \frac {2 \, {\left (2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \]

[In]

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

[Out]

-1/2*((2*a^2 - b^2)*(d*x + c)/b^3 + 4*(a^3 - 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)))/(sqrt(a^2 - b^2)*b^3) - 2*(2*a*tan(1/2*d
*x + 1/2*c)^3 + b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x +
1/2*c)^2 + 1)^2*b^2))/d

Mupad [B] (verification not implemented)

Time = 1.89 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.35 \[ \int \frac {\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {\sin \left (2\,c+2\,d\,x\right )}{4}}{b\,d}-\frac {2\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^3\,d}+\frac {a\,\sin \left (c+d\,x\right )}{b^2\,d}-\frac {2\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a+b\right )}\right )\,\sqrt {b^2-a^2}}{b^3\,d} \]

[In]

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

[Out]

(atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - sin(2*c + 2*d*x)/4)/(b*d) - (2*a^2*atan(sin(c/2 + (d*x)/2)/cos(
c/2 + (d*x)/2)))/(b^3*d) + (a*sin(c + d*x))/(b^2*d) - (2*a*atanh((sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(cos(c
/2 + (d*x)/2)*(a + b)))*(b^2 - a^2)^(1/2))/(b^3*d)