\(\int \frac {A+B \tan (x)}{a+b \sin (x)} \, dx\) [6]

Optimal result

Integrand size = 15, antiderivative size = 97 \[ \int \frac {A+B \tan (x)}{a+b \sin (x)} \, dx=\frac {2 A \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {B \log (1-\sin (x))}{2 (a+b)}-\frac {B \log (1+\sin (x))}{2 (a-b)}+\frac {a B \log (a+b \sin (x))}{a^2-b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4486, 2739, 632, 210, 2800, 815} \[ \int \frac {A+B \tan (x)}{a+b \sin (x)} \, dx=\frac {2 A \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {a B \log (a+b \sin (x))}{a^2-b^2}-\frac {B \log (1-\sin (x))}{2 (a+b)}-\frac {B \log (\sin (x)+1)}{2 (a-b)} \]

[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 210

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

Rule 632

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 815

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]

Rule 2739

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 2800

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 4486

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

Rubi steps \begin{align*} \text {integral}& = \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) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+B \text {Subst}\left (\int \frac {x}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (x)\right ) \\ & = -\left ((4 A) \text {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 \text {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 \arctan \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] (verified)

Time = 0.64 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.55 \[ \int \frac {A+B \tan (x)}{a+b \sin (x)} \, dx=\frac {\cos (x) \left (2 A \left (a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )-\sqrt {a^2-b^2} B \left ((a-b) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+(a+b) \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )-a \log (a+b \sin (x))\right )\right ) (A+B \tan (x))}{(a-b) (a+b) \sqrt {a^2-b^2} (A \cos (x)+B \sin (x))} \]

[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]))

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.30

method	result	size
default	\(\frac {B a \ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a \right )+\frac {2 \left (A \,a^{2}-A \,b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{\left (a -b \right ) \left (a +b \right )}-\frac {2 B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{2 a -2 b}-\frac {2 B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{2 a +2 b}\)	\(126\)
risch	\(-\frac {2 i x B \,a^{3}}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {2 i x B a \,b^{2}}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {i x B}{a -b}+\frac {i x B}{a +b}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A +\sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) B a}{\left (a +b \right ) \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A +\sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) \sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{\left (a +b \right ) \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A -\sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) B a}{\left (a +b \right ) \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A -\sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) \sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{\left (a +b \right ) \left (a -b \right )}-\frac {B \ln \left ({\mathrm e}^{i x}+i\right )}{a -b}-\frac {B \ln \left ({\mathrm e}^{i x}-i\right )}{a +b}\)	\(360\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 2.82 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.90 \[ \int \frac {A+B \tan (x)}{a+b \sin (x)} \, dx=\left [\frac {B a \log \left (-b^{2} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2}\right ) - \sqrt {-a^{2} + b^{2}} A \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - {\left (B a + B b\right )} \log \left (\sin \left (x\right ) + 1\right ) - {\left (B a - B b\right )} \log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )}}, \frac {B a \log \left (-b^{2} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2}\right ) - 2 \, \sqrt {a^{2} - b^{2}} A \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - {\left (B a + B b\right )} \log \left (\sin \left (x\right ) + 1\right ) - {\left (B a - B b\right )} \log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )}}\right ] \]

[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)]

Sympy [F]

\[ \int \frac {A+B \tan (x)}{a+b \sin (x)} \, dx=\int \frac {A + B \tan {\left (x \right )}}{a + b \sin {\left (x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (x)}{a+b \sin (x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((A+B*tan(x))/(a+b*sin(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 de

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.20 \[ \int \frac {A+B \tan (x)}{a+b \sin (x)} \, dx=\frac {B a \log \left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}{a^{2} - b^{2}} + \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} A}{\sqrt {a^{2} - b^{2}}} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{a - b} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{a + b} \]

[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)

Mupad [B] (verification not implemented)

Time = 19.57 (sec) , antiderivative size = 752, normalized size of antiderivative = 7.75 \[ \int \frac {A+B \tan (x)}{a+b \sin (x)} \, dx=\frac {\ln \left (-32\,A\,B^2\,a^2-\frac {\left (B\,a^3+A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a\,b^2\right )\,\left (64\,A^2\,a^2\,b-32\,B^2\,a^2\,b-\frac {\left (B\,a^3+A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a\,b^2\right )\,\left (32\,A\,a^2\,b^2-32\,A\,a^4+32\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,B\,a^3-2\,A\,a^2\,b+B\,a\,b^2+2\,A\,b^3\right )+32\,B\,a\,b^3+64\,B\,a^3\,b-\frac {96\,a\,b\,\left (a+b\,\mathrm {tan}\left (\frac {x}{2}\right )\right )\,\left (B\,a^3+A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a\,b^2\right )}{a^2-b^2}\right )}{{\left (a^2-b^2\right )}^2}+32\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (A^2\,a^2+A^2\,b^2+4\,A\,B\,a\,b+2\,B^2\,b^2\right )+64\,A\,B\,a^3\right )}{{\left (a^2-b^2\right )}^2}-32\,A^2\,B\,a\,b-32\,B\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a\,A^2+2\,b\,A\,B+2\,a\,B^2\right )\right )\,\left (B\,a^3+A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {\ln \left (\frac {\left (A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^3+B\,a\,b^2\right )\,\left (64\,A^2\,a^2\,b-32\,B^2\,a^2\,b+\frac {\left (A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^3+B\,a\,b^2\right )\,\left (32\,A\,a^2\,b^2-32\,A\,a^4+32\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,B\,a^3-2\,A\,a^2\,b+B\,a\,b^2+2\,A\,b^3\right )+32\,B\,a\,b^3+64\,B\,a^3\,b+\frac {96\,a\,b\,\left (a+b\,\mathrm {tan}\left (\frac {x}{2}\right )\right )\,\left (A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^3+B\,a\,b^2\right )}{a^2-b^2}\right )}{{\left (a^2-b^2\right )}^2}+32\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (A^2\,a^2+A^2\,b^2+4\,A\,B\,a\,b+2\,B^2\,b^2\right )+64\,A\,B\,a^3\right )}{{\left (a^2-b^2\right )}^2}-32\,A\,B^2\,a^2-32\,A^2\,B\,a\,b-32\,B\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a\,A^2+2\,b\,A\,B+2\,a\,B^2\right )\right )\,\left (A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^3+B\,a\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {B\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )}{a+b}-\frac {B\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}{a-b} \]

[In]

int((A + B*tan(x))/(a + b*sin(x)),x)

[Out]

(log(- 32*A*B^2*a^2 - ((B*a^3 + A*(-(a + b)^3*(a - b)^3)^(1/2) - B*a*b^2)*(64*A^2*a^2*b - 32*B^2*a^2*b - ((B*a
^3 + A*(-(a + b)^3*(a - b)^3)^(1/2) - B*a*b^2)*(32*A*a^2*b^2 - 32*A*a^4 + 32*a*tan(x/2)*(2*A*b^3 + 2*B*a^3 - 2
*A*a^2*b + B*a*b^2) + 32*B*a*b^3 + 64*B*a^3*b - (96*a*b*(a + b*tan(x/2))*(B*a^3 + A*(-(a + b)^3*(a - b)^3)^(1/
2) - B*a*b^2))/(a^2 - b^2)))/(a^2 - b^2)^2 + 32*a*tan(x/2)*(A^2*a^2 + A^2*b^2 + 2*B^2*b^2 + 4*A*B*a*b) + 64*A*
B*a^3))/(a^2 - b^2)^2 - 32*A^2*B*a*b - 32*B*a*tan(x/2)*(A^2*a + 2*B^2*a + 2*A*B*b))*(B*a^3 + A*(-(a + b)^3*(a
- b)^3)^(1/2) - B*a*b^2))/(a^4 + b^4 - 2*a^2*b^2) - (log(((A*(-(a + b)^3*(a - b)^3)^(1/2) - B*a^3 + B*a*b^2)*(
64*A^2*a^2*b - 32*B^2*a^2*b + ((A*(-(a + b)^3*(a - b)^3)^(1/2) - B*a^3 + B*a*b^2)*(32*A*a^2*b^2 - 32*A*a^4 + 3
2*a*tan(x/2)*(2*A*b^3 + 2*B*a^3 - 2*A*a^2*b + B*a*b^2) + 32*B*a*b^3 + 64*B*a^3*b + (96*a*b*(a + b*tan(x/2))*(A
*(-(a + b)^3*(a - b)^3)^(1/2) - B*a^3 + B*a*b^2))/(a^2 - b^2)))/(a^2 - b^2)^2 + 32*a*tan(x/2)*(A^2*a^2 + A^2*b
^2 + 2*B^2*b^2 + 4*A*B*a*b) + 64*A*B*a^3))/(a^2 - b^2)^2 - 32*A*B^2*a^2 - 32*A^2*B*a*b - 32*B*a*tan(x/2)*(A^2*
a + 2*B^2*a + 2*A*B*b))*(A*(-(a + b)^3*(a - b)^3)^(1/2) - B*a^3 + B*a*b^2))/(a^4 + b^4 - 2*a^2*b^2) - (B*log(t
an(x/2) - 1))/(a + b) - (B*log(tan(x/2) + 1))/(a - b)