\(\int \frac {a B+b B \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [292]

Optimal result

Integrand size = 26, antiderivative size = 50 \[ \int \frac {a B+b B \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 B \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {21, 2738, 211} \[ \int \frac {a B+b B \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 B \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}} \]

[In]

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

[Out]

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

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]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98 \[ \int \frac {a B+b B \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 B \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2} d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {2 B \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \sqrt {\left (a -b \right ) \left (a +b \right )}}\)	\(45\)
default	\(\frac {2 B \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \sqrt {\left (a -b \right ) \left (a +b \right )}}\)	\(45\)
risch	\(-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d}+\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d}\)	\(141\)

[In]

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

[Out]

2/d*B/((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 = 0.30 (sec) , antiderivative size = 177, normalized size of antiderivative = 3.54 \[ \int \frac {a B+b B \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} B \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right )}{2 \, {\left (a^{2} - b^{2}\right )} d}, \frac {B \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right )}{\sqrt {a^{2} - b^{2}} d}\right ] \]

[In]

integrate((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*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))/((a^2 - b^2)*d), B*ar
ctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))/(sqrt(a^2 - b^2)*d)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (42) = 84\).

Time = 155.04 (sec) , antiderivative size = 190, normalized size of antiderivative = 3.80 \[ \int \frac {a B+b B \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\begin {cases} \frac {\tilde {\infty } B x}{\cos {\left (c \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b d} & \text {for}\: a = b \\\frac {B}{b d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}} & \text {for}\: a = - b \\\frac {x \left (B a + B b \cos {\left (c \right )}\right )}{\left (a + b \cos {\left (c \right )}\right )^{2}} & \text {for}\: d = 0 \\\frac {B \log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )}}{a d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - b d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {B \log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )}}{a d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - b d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo*B*x/cos(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (B*tan(c/2 + d*x/2)/(b*d), Eq(a, b)), (B/(b*d*tan(
c/2 + d*x/2)), Eq(a, -b)), (x*(B*a + B*b*cos(c))/(a + b*cos(c))**2, Eq(d, 0)), (B*log(-sqrt(-a/(a - b) - b/(a
- b)) + tan(c/2 + d*x/2))/(a*d*sqrt(-a/(a - b) - b/(a - b)) - b*d*sqrt(-a/(a - b) - b/(a - b))) - B*log(sqrt(-
a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(a*d*sqrt(-a/(a - b) - b/(a - b)) - b*d*sqrt(-a/(a - b) - b/(a - b)
)), True))

Maxima [F(-2)]

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

[In]

integrate((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 = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.56 \[ \int \frac {a B+b B \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} B}{\sqrt {a^{2} - b^{2}} d} \]

[In]

integrate((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/(sqrt(a^2 - b^2)*d)

Mupad [B] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.88 \[ \int \frac {a B+b B \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2\,B\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a-b\right )}{\sqrt {a^2-b^2}}\right )}{d\,\sqrt {a^2-b^2}} \]

[In]

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

[Out]

(2*B*atan((tan(c/2 + (d*x)/2)*(a - b))/(a^2 - b^2)^(1/2)))/(d*(a^2 - b^2)^(1/2))