3.6 \(\int \frac {A+B \tan (x)}{a+b \cos (x)} \, dx\)

Optimal. Leaf size=65 \[ \frac {2 A \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+\frac {B \log (a+b \cos (x))}{a}-\frac {B \log (\cos (x))}{a} \]

[Out]

-B*ln(cos(x))/a+B*ln(a+b*cos(x))/a+2*A*arctan((a-b)^(1/2)*tan(1/2*x)/(a+b)^(1/2))/(a-b)^(1/2)/(a+b)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {4401, 2659, 205, 2721, 36, 29, 31} \[ \frac {2 A \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+\frac {B \log (a+b \cos (x))}{a}-\frac {B \log (\cos (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[x])/(a + b*Cos[x]),x]

[Out]

(2*A*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]) - (B*Log[Cos[x]])/a + (B*Log[a + b*
Cos[x]])/a

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2659

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 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.15, size = 61, normalized size = 0.94 \[ \frac {B (\log (a+b \cos (x))-\log (\cos (x)))}{a}-\frac {2 A \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[x])/(a + b*Cos[x]),x]

[Out]

(-2*A*ArcTanh[((a - b)*Tan[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + (B*(-Log[Cos[x]] + Log[a + b*Cos[x]]))/
a

________________________________________________________________________________________

fricas [A]  time = 0.79, size = 263, normalized size = 4.05 \[ \left [-\frac {\sqrt {-a^{2} + b^{2}} A a \log \left (\frac {2 \, a b \cos \relax (x) + {\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \relax (x) + b\right )} \sin \relax (x) - a^{2} + 2 \, b^{2}}{b^{2} \cos \relax (x)^{2} + 2 \, a b \cos \relax (x) + a^{2}}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (b^{2} \cos \relax (x)^{2} + 2 \, a b \cos \relax (x) + a^{2}\right ) + 2 \, {\left (B a^{2} - B b^{2}\right )} \log \left (-\cos \relax (x)\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}, \frac {2 \, \sqrt {a^{2} - b^{2}} A a \arctan \left (-\frac {a \cos \relax (x) + b}{\sqrt {a^{2} - b^{2}} \sin \relax (x)}\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (b^{2} \cos \relax (x)^{2} + 2 \, a b \cos \relax (x) + a^{2}\right ) - 2 \, {\left (B a^{2} - B b^{2}\right )} \log \left (-\cos \relax (x)\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(-a^2 + b^2)*A*a*log((2*a*b*cos(x) + (2*a^2 - b^2)*cos(x)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(x) + b)*sin
(x) - a^2 + 2*b^2)/(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2)) - (B*a^2 - B*b^2)*log(b^2*cos(x)^2 + 2*a*b*cos(x) + a^
2) + 2*(B*a^2 - B*b^2)*log(-cos(x)))/(a^3 - a*b^2), 1/2*(2*sqrt(a^2 - b^2)*A*a*arctan(-(a*cos(x) + b)/(sqrt(a^
2 - b^2)*sin(x))) + (B*a^2 - B*b^2)*log(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2) - 2*(B*a^2 - B*b^2)*log(-cos(x)))/(
a^3 - a*b^2)]

________________________________________________________________________________________

giac [B]  time = 3.96, size = 121, normalized size = 1.86 \[ -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} A}{\sqrt {a^{2} - b^{2}}} + \frac {B \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + b \tan \left (\frac {1}{2} \, x\right )^{2} - a - b\right )}{a} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{a} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*tan(1/2*x))/sqrt(a^2 - b^2)))*A/sqrt(
a^2 - b^2) + B*log(-a*tan(1/2*x)^2 + b*tan(1/2*x)^2 - a - b)/a - B*log(abs(tan(1/2*x) + 1))/a - B*log(abs(tan(
1/2*x) - 1))/a

________________________________________________________________________________________

maple [B]  time = 0.10, size = 129, normalized size = 1.98 \[ \frac {\ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) b +a +b \right ) B}{a -b}-\frac {\ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) b +a +b \right ) B b}{a \left (a -b \right )}+\frac {2 A \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{a}-\frac {B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(x))/(a+b*cos(x)),x)

[Out]

1/(a-b)*ln(a*tan(1/2*x)^2-tan(1/2*x)^2*b+a+b)*B-1/a/(a-b)*ln(a*tan(1/2*x)^2-tan(1/2*x)^2*b+a+b)*B*b+2*A/((a-b)
*(a+b))^(1/2)*arctan(tan(1/2*x)*(a-b)/((a-b)*(a+b))^(1/2))-B/a*ln(tan(1/2*x)-1)-B/a*ln(tan(1/2*x)+1)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(x))/(a+b*cos(x)),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 details)Is 4*b^2-4*a^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 5.66, size = 1540, normalized size = 23.69 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(x))/(a + b*cos(x)),x)

[Out]

(log((a + b*cos(x))/(cos(x) + 1))*(2*B*a^3 - 2*B*a*b^2))/(2*(a^4 - a^2*b^2)) + (2*A*atan(((a^2 - b^2)*((A*(64*
A*B*a^3 + ((2*B*a^3 - 2*B*a*b^2)*(32*A*a^4 + 32*A*a^2*b^2 - 64*A*a^3*b))/(2*(a^4 - a^2*b^2)) + 64*A*B*a*b^2 -
128*A*B*a^2*b))/(a^2 - b^2)^(1/2) + (A*(2*B*a^3 - 2*B*a*b^2)*(32*A*a^4 + 32*A*a^2*b^2 - 64*A*a^3*b))/(2*(a^4 -
a^2*b^2)*(a^2 - b^2)^(1/2)))*(A^2*a^2 - 4*B^2*a^2 + 4*B^2*b^2))/((32*A*a - 32*A*b)*(a - b)*(A^2*a^2 + 4*B^2*a
^2 - 4*B^2*b^2)^2) - (tan(x/2)*(a^2 - b^2)^(3/2)*(((((2*B*a^3 - 2*B*a*b^2)*((A*(64*B*a^4 + 64*B*a^2*b^2 - 128*
B*a^3*b - ((2*B*a^3 - 2*B*a*b^2)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2))/(2*(a^4 - a^2*b^2))))/(a^2 - b^2)^(1/2
) - (A*(2*B*a^3 - 2*B*a*b^2)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2))/(2*(a^4 - a^2*b^2)*(a^2 - b^2)^(1/2))))/(2
*(a^4 - a^2*b^2)) + (A^3*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2))/(a^2 - b^2)^(3/2) + (A*(64*B^2*b^3 - 32*A^2*a^
3 + 32*A^2*a^2*b - 128*B^2*a*b^2 + 64*B^2*a^2*b + ((2*B*a^3 - 2*B*a*b^2)*(64*B*a^4 + 64*B*a^2*b^2 - 128*B*a^3*
b - ((2*B*a^3 - 2*B*a*b^2)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2))/(2*(a^4 - a^2*b^2))))/(2*(a^4 - a^2*b^2))))/
(a^2 - b^2)^(1/2))*(A^2*a^2 - 4*B^2*a^2 + 4*B^2*b^2))/((a^2 - b^2)^(1/2)*(a - b)*(A^2*a^2 + 4*B^2*a^2 - 4*B^2*
b^2)^2) - (4*A*B*a*(64*B^3*a^2 + 64*B^3*b^2 + 32*A^2*B*a^2 + (A*((A*(64*B*a^4 + 64*B*a^2*b^2 - 128*B*a^3*b - (
(2*B*a^3 - 2*B*a*b^2)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2))/(2*(a^4 - a^2*b^2))))/(a^2 - b^2)^(1/2) - (A*(2*B
*a^3 - 2*B*a*b^2)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2))/(2*(a^4 - a^2*b^2)*(a^2 - b^2)^(1/2))))/(a^2 - b^2)^(
1/2) - ((2*B*a^3 - 2*B*a*b^2)*(64*B^2*b^3 - 32*A^2*a^3 + 32*A^2*a^2*b - 128*B^2*a*b^2 + 64*B^2*a^2*b + ((2*B*a
^3 - 2*B*a*b^2)*(64*B*a^4 + 64*B*a^2*b^2 - 128*B*a^3*b - ((2*B*a^3 - 2*B*a*b^2)*(64*a^4*b + 64*a^2*b^3 - 128*a
^3*b^2))/(2*(a^4 - a^2*b^2))))/(2*(a^4 - a^2*b^2))))/(2*(a^4 - a^2*b^2)) - 128*B^3*a*b - 32*A^2*B*a*b - (A^2*(
2*B*a^3 - 2*B*a*b^2)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2))/(2*(a^4 - a^2*b^2)*(a^2 - b^2))))/((a - b)*(A^2*a^
2 + 4*B^2*a^2 - 4*B^2*b^2)^2)))/(32*A*a - 32*A*b) + (4*A*B*a*(a^2 - b^2)^(3/2)*(32*A*B^2*a^2 + 32*A*B^2*b^2 +
((2*B*a^3 - 2*B*a*b^2)*(64*A*B*a^3 + ((2*B*a^3 - 2*B*a*b^2)*(32*A*a^4 + 32*A*a^2*b^2 - 64*A*a^3*b))/(2*(a^4 -
a^2*b^2)) + 64*A*B*a*b^2 - 128*A*B*a^2*b))/(2*(a^4 - a^2*b^2)) - (A^2*(32*A*a^4 + 32*A*a^2*b^2 - 64*A*a^3*b))/
(a^2 - b^2) - 64*A*B^2*a*b))/((32*A*a - 32*A*b)*(a - b)*(A^2*a^2 + 4*B^2*a^2 - 4*B^2*b^2)^2)))/(a^2 - b^2)^(1/
2) - (B*log(cos(x)/(cos(x) + 1)))/a

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \tan {\relax (x )}}{a + b \cos {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(x))/(a+b*cos(x)),x)

[Out]

Integral((A + B*tan(x))/(a + b*cos(x)), x)

________________________________________________________________________________________