3.8.91 \(\int \frac {\frac {b B}{a}+B \cos (c+d x)}{a+b \cos (c+d x)} \, dx\) [791]

Optimal. Leaf size=63 \[ \frac {B x}{b}-\frac {2 \sqrt {a-b} \sqrt {a+b} B \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a b d} \]

[Out]

B*x/b-2*B*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)/a/b/d

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2814, 2738, 211} \begin {gather*} \frac {B x}{b}-\frac {2 B \sqrt {a-b} \sqrt {a+b} \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.12, size = 64, normalized size = 1.02 \begin {gather*} \frac {B \left (a (c+d x)+2 \sqrt {-a^2+b^2} \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )\right )}{a b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.17, size = 77, normalized size = 1.22

method	result	size
derivativedivides	\(\frac {2 B \left (-\frac {\left (a -b \right ) \left (a +b \right ) \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}\right )}{d a}\)	\(77\)
default	\(\frac {2 B \left (-\frac {\left (a -b \right ) \left (a +b \right ) \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}\right )}{d a}\)	\(77\)
risch	\(\frac {B x}{b}+\frac {\sqrt {-a^{2}+b^{2}}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right )}{d b a}-\frac {\sqrt {-a^{2}+b^{2}}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right )}{d b a}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Fricas [A]
time = 0.39, size = 194, normalized size = 3.08 \begin {gather*} \left [\frac {2 \, B a d x + \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 \, a b d}, \frac {B a d x - \sqrt {a^{2} - b^{2}} B \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right )}{a b d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (51) = 102\).
time = 15.99, size = 235, normalized size = 3.73 \begin {gather*} \begin {cases} \text {NaN} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {B x}{b} & \text {for}\: a = b \\\frac {x \left (B \cos {\left (c \right )} + \frac {B b}{a}\right )}{a + b \cos {\left (c \right )}} & \text {for}\: d = 0 \\\frac {B \sin {\left (c + d x \right )}}{a d} & \text {for}\: b = 0 \\\frac {B x}{b} & \text {for}\: a = - b \\\frac {B x}{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 )}}{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 )}}{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}}} + \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}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((nan, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (B*x/b, Eq(a, b)), (x*(B*cos(c) + B*b/a)/(a + b*cos(c)), Eq(d
, 0)), (B*sin(c + d*x)/(a*d), Eq(b, 0)), (B*x/b, Eq(a, -b)), (B*x/b - B*log(-sqrt(-a/(a - b) - b/(a - b)) + ta
n(c/2 + d*x/2))/(b*d*sqrt(-a/(a - b) - b/(a - b))) + B*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(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*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(a*d*sqrt(-a/(a - b) - b/(a - b))), T
rue))

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (54) = 108\).
time = 0.43, size = 281, normalized size = 4.46 \begin {gather*} -\frac {\frac {{\left (\sqrt {a^{2} - b^{2}} B {\left | a - b \right |} {\left | a \right |} {\left | b \right |} + {\left (2 \, a^{2} + a b\right )} \sqrt {a^{2} - b^{2}} B {\left | a - b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {a^{2} + \sqrt {a^{4} - {\left (a^{2} + a b\right )} {\left (a^{2} - a b\right )}}}{a^{2} - a b}}}\right )\right )}}{{\left (a - b\right )} a^{2} b^{2} + {\left (a^{3} - a^{2} b\right )} {\left | a \right |} {\left | b \right |}} + \frac {{\left (2 \, B a^{3} - B a^{2} b - B a b^{2} - B a {\left | a \right |} {\left | b \right |} + B b {\left | a \right |} {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {a^{2} - \sqrt {a^{4} - {\left (a^{2} + a b\right )} {\left (a^{2} - a b\right )}}}{a^{2} - a b}}}\right )\right )}}{a^{2} b^{2} - a^{2} {\left | a \right |} {\left | b \right |}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((sqrt(a^2 - b^2)*B*abs(a - b)*abs(a)*abs(b) + (2*a^2 + a*b)*sqrt(a^2 - b^2)*B*abs(a - b))*(pi*floor(1/2*(d*x
+ c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt((a^2 + sqrt(a^4 - (a^2 + a*b)*(a^2 - a*b)))/(a^2 - a*b))))/
((a - b)*a^2*b^2 + (a^3 - a^2*b)*abs(a)*abs(b)) + (2*B*a^3 - B*a^2*b - B*a*b^2 - B*a*abs(a)*abs(b) + B*b*abs(a
)*abs(b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt((a^2 - sqrt(a^4 - (a^2 + a*b)*(
a^2 - a*b)))/(a^2 - a*b))))/(a^2*b^2 - a^2*abs(a)*abs(b)))/d

________________________________________________________________________________________

Mupad [B]
time = 0.94, size = 93, normalized size = 1.48 \begin {gather*} \frac {2\,B\,\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\,d}+\frac {2\,B\,\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}}{a\,b\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________