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

3.1.6.1 Optimal result

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

output

-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)
 

3.1.6.2 Mathematica [A] (verified)

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

input

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

output

(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]))
 

3.1.6.3 Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4901, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4901

\(\displaystyle \int \left (\frac {A}{a+b \sin (x)}+\frac {B \tan (x)}{a+b \sin (x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 A \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {b B \text {arctanh}(\sin (x))}{a^2-b^2}+\frac {a B \log (a+b \sin (x))}{a^2-b^2}-\frac {a B \log (\cos (x))}{a^2-b^2}\)

input

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

output

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

3.1.6.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4901

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

3.1.6.4 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	\(-\frac {2 i x B \,a^{3}}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {2 i x B a \,b^{2}}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {i x B}{a -b}+\frac {i x B}{a +b}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A +\sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) B a}{\left (a +b \right ) \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A +\sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) \sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{\left (a +b \right ) \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A -\sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) B a}{\left (a +b \right ) \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A -\sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) \sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{\left (a +b \right ) \left (a -b \right )}-\frac {B \ln \left ({\mathrm e}^{i x}+i\right )}{a -b}-\frac {B \ln \left ({\mathrm e}^{i x}-i\right )}{a +b}\)	\(360\)

input

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

output

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)
 

3.1.6.5 Fricas [A] (verification not implemented)

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

input

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

output

[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)*si 
n(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)]
 

3.1.6.6 Sympy [F]

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

input

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

output

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

3.1.6.7 Maxima [F(-2)]

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

input

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

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

3.1.6.8 Giac [A] (verification not implemented)

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

input

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

output

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)
 

3.1.6.9 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} \]

input

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

output

(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 + 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))*(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(tan(x/2) - 1))/(a + b) - (B*log(tan(x/2) + 1))/(a 
 - b)