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

Optimal. Leaf size=97 \[ \frac{2 A \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{a B \log (a+b \sin (x))}{a^2-b^2}-\frac{B \log (1-\sin (x))}{2 (a+b)}-\frac{B \log (\sin (x)+1)}{2 (a-b)} \]

[Out]

(2*A*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - (B*Log[1 - Sin[x]])/(2*(a + b)) - (B*Log[1 +
Sin[x]])/(2*(a - b)) + (a*B*Log[a + b*Sin[x]])/(a^2 - b^2)

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Rubi [A]  time = 0.168295, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4401, 2660, 618, 204, 2721, 801} \[ \frac{2 A \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{a B \log (a+b \sin (x))}{a^2-b^2}-\frac{B \log (1-\sin (x))}{2 (a+b)}-\frac{B \log (\sin (x)+1)}{2 (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[x])/(a + b*Sin[x]),x]

[Out]

(2*A*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - (B*Log[1 - Sin[x]])/(2*(a + b)) - (B*Log[1 +
Sin[x]])/(2*(a - b)) + (a*B*Log[a + b*Sin[x]])/(a^2 - b^2)

Rule 4401

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

\begin{align*} \int \frac{A+B \tan (x)}{a+b \sin (x)} \, dx &=\int \left (\frac{A}{a+b \sin (x)}+\frac{B \tan (x)}{a+b \sin (x)}\right ) \, dx\\ &=A \int \frac{1}{a+b \sin (x)} \, dx+B \int \frac{\tan (x)}{a+b \sin (x)} \, dx\\ &=(2 A) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )+B \operatorname{Subst}\left (\int \frac{x}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (x)\right )\\ &=-\left ((4 A) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{x}{2}\right )\right )\right )+B \operatorname{Subst}\left (\int \left (\frac{1}{2 (a+b) (b-x)}+\frac{a}{(a-b) (a+b) (a+x)}-\frac{1}{2 (a-b) (b+x)}\right ) \, dx,x,b \sin (x)\right )\\ &=\frac{2 A \tan ^{-1}\left (\frac{b+a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{B \log (1-\sin (x))}{2 (a+b)}-\frac{B \log (1+\sin (x))}{2 (a-b)}+\frac{a B \log (a+b \sin (x))}{a^2-b^2}\\ \end{align*}

Mathematica [A]  time = 0.36109, size = 150, normalized size = 1.55 \[ \frac{\cos (x) (A+B \tan (x)) \left (2 A \left (a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )-B \sqrt{a^2-b^2} \left (-a \log (a+b \sin (x))+(a-b) \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )+(a+b) \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )\right )}{(a-b) (a+b) \sqrt{a^2-b^2} (A \cos (x)+B \sin (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.099, size = 181, normalized size = 1.9 \begin{align*} -2\,{\frac{B\ln \left ( \tan \left ( x/2 \right ) -1 \right ) }{2\,a+2\,b}}+{\frac{Ba}{ \left ( a-b \right ) \left ( a+b \right ) }\ln \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) }+2\,{\frac{A{a}^{2}}{ \left ( a-b \right ) \left ( a+b \right ) \sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-2\,{\frac{A{b}^{2}}{ \left ( a-b \right ) \left ( a+b \right ) \sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-2\,{\frac{B\ln \left ( \tan \left ( x/2 \right ) +1 \right ) }{2\,a-2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 38.3097, size = 682, normalized size = 7.03 \begin{align*} \left [\frac{B a \log \left (-b^{2} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2}\right ) - \sqrt{-a^{2} + b^{2}} A \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) -{\left (B a + B b\right )} \log \left (\sin \left (x\right ) + 1\right ) -{\left (B a - B b\right )} \log \left (-\sin \left (x\right ) + 1\right )}{2 \,{\left (a^{2} - b^{2}\right )}}, \frac{B a \log \left (-b^{2} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2}\right ) - 2 \, \sqrt{a^{2} - b^{2}} A \arctan \left (-\frac{a \sin \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (x\right )}\right ) -{\left (B a + B b\right )} \log \left (\sin \left (x\right ) + 1\right ) -{\left (B a - B b\right )} \log \left (-\sin \left (x\right ) + 1\right )}{2 \,{\left (a^{2} - b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \tan{\left (x \right )}}{a + b \sin{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11793, size = 157, normalized size = 1.62 \begin{align*} \frac{B a \log \left (a \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\right )}{a^{2} - b^{2}} + \frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )} A}{\sqrt{a^{2} - b^{2}}} - \frac{B \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right )}{a - b} - \frac{B \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) - 1 \right |}\right )}{a + b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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