\(\int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [290]

Optimal result

Integrand size = 34, antiderivative size = 79 \[ \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=-\frac {a B x}{b^2}+\frac {2 a^2 B \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b} d}+\frac {B \sin (c+d x)}{b d} \]

[Out]

-a*B*x/b^2+B*sin(d*x+c)/b/d+2*a^2*B*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 = 0.15 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {21, 2825, 12, 2814, 2738, 211} \[ \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {2 a^2 B \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d \sqrt {a-b} \sqrt {a+b}}-\frac {a B x}{b^2}+\frac {B \sin (c+d x)}{b d} \]

[In]

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

[Out]

-((a*B*x)/b^2) + (2*a^2*B*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^2*Sqrt[a + b]*d)
+ (B*Sin[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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + 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}& = B \int \frac {\cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx \\ & = \frac {B \sin (c+d x)}{b d}-\frac {B \int \frac {a \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{b} \\ & = \frac {B \sin (c+d x)}{b d}-\frac {(a B) \int \frac {\cos (c+d x)}{a+b \cos (c+d x)} \, dx}{b} \\ & = -\frac {a B x}{b^2}+\frac {B \sin (c+d x)}{b d}+\frac {\left (a^2 B\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{b^2} \\ & = -\frac {a B x}{b^2}+\frac {B \sin (c+d x)}{b d}+\frac {\left (2 a^2 B\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 B x}{b^2}+\frac {2 a^2 B \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 {B \sin (c+d x)}{b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {B \left (-a (c+d x)-\frac {2 a^2 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+b \sin (c+d x)\right )}{b^2 d} \]

[In]

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

[Out]

(B*(-(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 = 98, normalized size of antiderivative = 1.24

method	result	size
derivativedivides	\(\frac {2 B \left (\frac {a^{2} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}\right )}{d}\)	\(98\)
default	\(\frac {2 B \left (\frac {a^{2} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}\right )}{d}\)	\(98\)
risch	\(-\frac {a B x}{b^{2}}-\frac {i B \,{\mathrm e}^{i \left (d x +c \right )}}{2 d b}+\frac {i B \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d b}-\frac {a^{2} B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {a^{2} B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}\)	\(197\)

[In]

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

[Out]

2/d*B*(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))-1/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 = 0.30 (sec) , antiderivative size = 281, normalized size of antiderivative = 3.56 \[ \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} B a^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 2 \, {\left (B a^{3} - B a b^{2}\right )} d x - 2 \, {\left (B a^{2} b - B b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d}, \frac {\sqrt {a^{2} - b^{2}} B a^{2} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (B a^{3} - B a b^{2}\right )} d x + {\left (B a^{2} b - B b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{2} - b^{4}\right )} d}\right ] \]

[In]

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

[Out]

[-1/2*(sqrt(-a^2 + b^2)*B*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*c
os(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*(B*a^3 - B*a
*b^2)*d*x - 2*(B*a^2*b - B*b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d), (sqrt(a^2 - b^2)*B*a^2*arctan(-(a*cos(d*x +
c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (B*a^3 - B*a*b^2)*d*x + (B*a^2*b - B*b^3)*sin(d*x + c))/((a^2*b^2 -
b^4)*d)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

integrate(cos(d*x+c)^2*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^2,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 = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.62 \[ \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} B a^{2}}{\sqrt {a^{2} - b^{2}} b^{2}} - \frac {{\left (d x + c\right )} B a}{b^{2}} + \frac {2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b}}{d} \]

[In]

integrate(cos(d*x+c)^2*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^2,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)))*B*a^2/(sqrt(a^2 - b^2)*b^2) - (d*x + c)*B*a/b^2 + 2*B*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 = 0.90 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.44 \[ \int \frac {\cos ^2(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {B\,\sin \left (c+d\,x\right )}{b\,d}-\frac {2\,B\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d}-\frac {B\,a^2\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b-2{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2+1{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (b^2-a^2\right )}^{3/2}+a^2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )\,2{}\mathrm {i}}{b^2\,d\,\sqrt {b^2-a^2}} \]

[In]

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

[Out]

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