\(\int \frac {x}{a+b \cos (x) \sin (x)} \, dx\) [524]

Optimal result

Integrand size = 12, antiderivative size = 225 \[ \int \frac {x}{a+b \cos (x) \sin (x)} \, dx=-\frac {i x \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2}}+\frac {i x \log \left (1-\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2}}-\frac {\operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 \sqrt {4 a^2-b^2}}+\frac {\operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{2 \sqrt {4 a^2-b^2}} \] Output:

-I*x*ln(1-I*b*exp(2*I*x)/(2*a-(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)+I*x*ln 
(1-I*b*exp(2*I*x)/(2*a+(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)-1/2*polylog(2 
,I*b*exp(2*I*x)/(2*a-(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)+1/2*polylog(2,I 
*b*exp(2*I*x)/(2*a+(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(788\) vs. \(2(225)=450\).

Time = 1.16 (sec) , antiderivative size = 788, normalized size of antiderivative = 3.50 \[ \int \frac {x}{a+b \cos (x) \sin (x)} \, dx =\text {Too large to display} \] Input:

Integrate[x/(a + b*Cos[x]*Sin[x]),x]
 

Output:

((Pi*ArcTan[(b + 2*a*Tan[x])/Sqrt[4*a^2 - b^2]])/Sqrt[4*a^2 - b^2] + (2*Ar 
cCos[(-2*a)/b]*ArcTanh[((2*a - b)*Cot[Pi/4 + x])/Sqrt[-4*a^2 + b^2]] + (Pi 
 - 4*x)*ArcTanh[((2*a + b)*Tan[Pi/4 + x])/Sqrt[-4*a^2 + b^2]] - (ArcCos[(- 
2*a)/b] + (2*I)*ArcTanh[((2*a - b)*Cot[Pi/4 + x])/Sqrt[-4*a^2 + b^2]])*Log 
[((2*a + b)*(-2*a + b - I*Sqrt[-4*a^2 + b^2])*(1 + I*Cot[Pi/4 + x]))/(b*(2 
*a + b + Sqrt[-4*a^2 + b^2]*Cot[Pi/4 + x]))] - (ArcCos[(-2*a)/b] - (2*I)*A 
rcTanh[((2*a - b)*Cot[Pi/4 + x])/Sqrt[-4*a^2 + b^2]])*Log[((2*a + b)*((2*I 
)*a - I*b + Sqrt[-4*a^2 + b^2])*(I + Cot[Pi/4 + x]))/(b*(2*a + b + Sqrt[-4 
*a^2 + b^2]*Cot[Pi/4 + x]))] + (ArcCos[(-2*a)/b] + (2*I)*(ArcTanh[((2*a - 
b)*Cot[Pi/4 + x])/Sqrt[-4*a^2 + b^2]] + ArcTanh[((2*a + b)*Tan[Pi/4 + x])/ 
Sqrt[-4*a^2 + b^2]]))*Log[((-1)^(1/4)*Sqrt[-4*a^2 + b^2])/(2*Sqrt[b]*E^(I* 
x)*Sqrt[a + b*Cos[x]*Sin[x]])] + (ArcCos[(-2*a)/b] - (2*I)*ArcTanh[((2*a - 
 b)*Cot[Pi/4 + x])/Sqrt[-4*a^2 + b^2]] - (2*I)*ArcTanh[((2*a + b)*Tan[Pi/4 
 + x])/Sqrt[-4*a^2 + b^2]])*Log[-1/2*((-1)^(3/4)*Sqrt[-4*a^2 + b^2]*E^(I*x 
))/(Sqrt[b]*Sqrt[a + b*Cos[x]*Sin[x]])] + I*(PolyLog[2, ((2*a - I*Sqrt[-4* 
a^2 + b^2])*(2*a + b - Sqrt[-4*a^2 + b^2]*Cot[Pi/4 + x]))/(b*(2*a + b + Sq 
rt[-4*a^2 + b^2]*Cot[Pi/4 + x]))] - PolyLog[2, ((2*a + I*Sqrt[-4*a^2 + b^2 
])*(2*a + b - Sqrt[-4*a^2 + b^2]*Cot[Pi/4 + x]))/(b*(2*a + b + Sqrt[-4*a^2 
 + b^2]*Cot[Pi/4 + x]))]))/Sqrt[-4*a^2 + b^2])/2
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5095, 3042, 3804, 27, 2694, 27, 2620, 2715, 2838}

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.

\(\displaystyle \int \frac {x}{a+b \sin (x) \cos (x)} \, dx\)

\(\Big \downarrow \) 5095

\(\displaystyle \int \frac {x}{a+\frac {1}{2} b \sin (2 x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {x}{a+\frac {1}{2} b \sin (2 x)}dx\)

\(\Big \downarrow \) 3804

\(\displaystyle 2 \int \frac {2 e^{2 i x} x}{4 e^{2 i x} a-i b e^{4 i x}+i b}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 4 \int \frac {e^{2 i x} x}{4 e^{2 i x} a-i b e^{4 i x}+i b}dx\)

\(\Big \downarrow \) 2694

\(\displaystyle 4 \left (\frac {i b \int \frac {e^{2 i x} x}{2 \left (2 a-i b e^{2 i x}+\sqrt {4 a^2-b^2}\right )}dx}{\sqrt {4 a^2-b^2}}-\frac {i b \int \frac {e^{2 i x} x}{2 \left (2 a-i b e^{2 i x}-\sqrt {4 a^2-b^2}\right )}dx}{\sqrt {4 a^2-b^2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 4 \left (\frac {i b \int \frac {e^{2 i x} x}{2 a-i b e^{2 i x}+\sqrt {4 a^2-b^2}}dx}{2 \sqrt {4 a^2-b^2}}-\frac {i b \int \frac {e^{2 i x} x}{2 a-i b e^{2 i x}-\sqrt {4 a^2-b^2}}dx}{2 \sqrt {4 a^2-b^2}}\right )\)

\(\Big \downarrow \) 2620

\(\displaystyle 4 \left (\frac {i b \left (\frac {x \log \left (1-\frac {i b e^{2 i x}}{\sqrt {4 a^2-b^2}+2 a}\right )}{2 b}-\frac {\int \log \left (1-\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )dx}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}-\frac {i b \left (\frac {x \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 b}-\frac {\int \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )dx}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}\right )\)

\(\Big \downarrow \) 2715

\(\displaystyle 4 \left (\frac {i b \left (\frac {i \int e^{-2 i x} \log \left (1-\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )de^{2 i x}}{4 b}+\frac {x \log \left (1-\frac {i b e^{2 i x}}{\sqrt {4 a^2-b^2}+2 a}\right )}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}-\frac {i b \left (\frac {i \int e^{-2 i x} \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )de^{2 i x}}{4 b}+\frac {x \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 b}\right )}{2 \sqrt {4 a^2-b^2}}\right )\)

\(\Big \downarrow \) 2838

\(\displaystyle 4 \left (\frac {i b \left (\frac {x \log \left (1-\frac {i b e^{2 i x}}{\sqrt {4 a^2-b^2}+2 a}\right )}{2 b}-\frac {i \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{4 b}\right )}{2 \sqrt {4 a^2-b^2}}-\frac {i b \left (\frac {x \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 b}-\frac {i \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{4 b}\right )}{2 \sqrt {4 a^2-b^2}}\right )\)

Input:

Int[x/(a + b*Cos[x]*Sin[x]),x]
 

Output:

4*(((-1/2*I)*b*((x*Log[1 - (I*b*E^((2*I)*x))/(2*a - Sqrt[4*a^2 - b^2])])/( 
2*b) - ((I/4)*PolyLog[2, (I*b*E^((2*I)*x))/(2*a - Sqrt[4*a^2 - b^2])])/b)) 
/Sqrt[4*a^2 - b^2] + ((I/2)*b*((x*Log[1 - (I*b*E^((2*I)*x))/(2*a + Sqrt[4* 
a^2 - b^2])])/(2*b) - ((I/4)*PolyLog[2, (I*b*E^((2*I)*x))/(2*a + Sqrt[4*a^ 
2 - b^2])])/b))/Sqrt[4*a^2 - b^2])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2694

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.) 
*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[2*(c/q)   Int[(f + g*x) 
^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[ 
v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

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

rule 3804

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[2   Int[(c + d*x)^m*(E^(I*(e + f*x))/(I*b + 2*a*E^(I*(e + f*x 
)) - I*b*E^(2*I*(e + f*x)))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ 
[a^2 - b^2, 0] && IGtQ[m, 0]
 

rule 5095

Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + Cos[(c_.) + (d_.)*(x_)]*(b_.)*Sin[(c 
_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Int[(e + f*x)^m*(a + b*(Sin[2*c + 2* 
d*x]/2))^n, x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1283 vs. \(2 (191 ) = 382\).

Time = 0.15 (sec) , antiderivative size = 1284, normalized size of antiderivative = 5.71

method	result	size
risch	\(\text {Expression too large to display}\)	\(1284\)

Input:

int(x/(a+b*cos(x)*sin(x)),x,method=_RETURNVERBOSE)
 

Output:

2*I/(8*a^2-2*b^2)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*b^2*x^2-4/(8*a^2-2*b^2 
)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*(-(2*a-b)*(2*a+b))^(1/2)*a*x^2+8/(8*a^ 
2-2*b^2)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*ln(1-b*exp(2*I*x)/(-2*I*a-(-(2* 
a-b)*(2*a+b))^(1/2)))*a^2*x-2/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/ 
2))*ln(1-b*exp(2*I*x)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2)))*b^2*x+4/(8*a^2-2* 
b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*(-(2*a-b)*(2*a+b))^(1/2)*a*x^2+4*I/ 
(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*ln(1-b*exp(2*I*x)/(-2*I*a+ 
(-(2*a-b)*(2*a+b))^(1/2)))*(-(2*a-b)*(2*a+b))^(1/2)*a*x+I/(8*a^2-2*b^2)/(- 
2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*polylog(2,b*exp(2*I*x)/(-2*I*a-(-(2*a-b)*( 
2*a+b))^(1/2)))*b^2-4*I/(8*a^2-2*b^2)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*po 
lylog(2,b*exp(2*I*x)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2)))*a^2+2*I/(8*a^2-2*b 
^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*b^2*x^2-8*I/(8*a^2-2*b^2)/(-2*I*a-(- 
(2*a-b)*(2*a+b))^(1/2))*a^2*x^2+8/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b)) 
^(1/2))*ln(1-b*exp(2*I*x)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2)))*a^2*x-2/(8*a^ 
2-2*b^2)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*ln(1-b*exp(2*I*x)/(-2*I*a-(-(2* 
a-b)*(2*a+b))^(1/2)))*b^2*x-2/(8*a^2-2*b^2)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/ 
2))*(-(2*a-b)*(2*a+b))^(1/2)*polylog(2,b*exp(2*I*x)/(-2*I*a-(-(2*a-b)*(2*a 
+b))^(1/2)))*a+2/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*polylog(2 
,b*exp(2*I*x)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2)))*(-(2*a-b)*(2*a+b))^(1/2)* 
a+I/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*polylog(2,b*exp(2*I...
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1676 vs. \(2 (179) = 358\).

Time = 0.97 (sec) , antiderivative size = 1676, normalized size of antiderivative = 7.45 \[ \int \frac {x}{a+b \cos (x) \sin (x)} \, dx=\text {Too large to display} \] Input:

integrate(x/(a+b*cos(x)*sin(x)),x, algorithm="fricas")
 

Output:

-1/2*(b*x*sqrt(-(4*a^2 - b^2)/b^2)*log(-((2*I*a*cos(x) + 2*a*sin(x) - (b*c 
os(x) - I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt((b*sqrt(-(4*a^2 - b^2)/ 
b^2) + 2*I*a)/b) - b)/b) + b*x*sqrt(-(4*a^2 - b^2)/b^2)*log(-((-2*I*a*cos( 
x) - 2*a*sin(x) + (b*cos(x) - I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt(( 
b*sqrt(-(4*a^2 - b^2)/b^2) + 2*I*a)/b) - b)/b) - b*x*sqrt(-(4*a^2 - b^2)/b 
^2)*log(-((2*I*a*cos(x) - 2*a*sin(x) - (b*cos(x) + I*b*sin(x))*sqrt(-(4*a^ 
2 - b^2)/b^2))*sqrt(-(b*sqrt(-(4*a^2 - b^2)/b^2) + 2*I*a)/b) - b)/b) - b*x 
*sqrt(-(4*a^2 - b^2)/b^2)*log(-((-2*I*a*cos(x) + 2*a*sin(x) + (b*cos(x) + 
I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt(-(b*sqrt(-(4*a^2 - b^2)/b^2) + 
2*I*a)/b) - b)/b) + b*x*sqrt(-(4*a^2 - b^2)/b^2)*log(-((2*I*a*cos(x) - 2*a 
*sin(x) + (b*cos(x) + I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt((b*sqrt(- 
(4*a^2 - b^2)/b^2) - 2*I*a)/b) - b)/b) + b*x*sqrt(-(4*a^2 - b^2)/b^2)*log( 
-((-2*I*a*cos(x) + 2*a*sin(x) - (b*cos(x) + I*b*sin(x))*sqrt(-(4*a^2 - b^2 
)/b^2))*sqrt((b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b) - b)/b) - b*x*sqrt(-( 
4*a^2 - b^2)/b^2)*log(-((2*I*a*cos(x) + 2*a*sin(x) + (b*cos(x) - I*b*sin(x 
))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt(-(b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b) 
 - b)/b) - b*x*sqrt(-(4*a^2 - b^2)/b^2)*log(-((-2*I*a*cos(x) - 2*a*sin(x) 
- (b*cos(x) - I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt(-(b*sqrt(-(4*a^2 
- b^2)/b^2) - 2*I*a)/b) - b)/b) + I*b*sqrt(-(4*a^2 - b^2)/b^2)*dilog(((2*I 
*a*cos(x) + 2*a*sin(x) - (b*cos(x) - I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^...
 

Sympy [F]

\[ \int \frac {x}{a+b \cos (x) \sin (x)} \, dx=\int \frac {x}{a + b \sin {\left (x \right )} \cos {\left (x \right )}}\, dx \] Input:

integrate(x/(a+b*cos(x)*sin(x)),x)
 

Output:

Integral(x/(a + b*sin(x)*cos(x)), x)
 

Maxima [F]

\[ \int \frac {x}{a+b \cos (x) \sin (x)} \, dx=\int { \frac {x}{b \cos \left (x\right ) \sin \left (x\right ) + a} \,d x } \] Input:

integrate(x/(a+b*cos(x)*sin(x)),x, algorithm="maxima")
 

Output:

integrate(x/(b*cos(x)*sin(x) + a), x)
 

Giac [F]

\[ \int \frac {x}{a+b \cos (x) \sin (x)} \, dx=\int { \frac {x}{b \cos \left (x\right ) \sin \left (x\right ) + a} \,d x } \] Input:

integrate(x/(a+b*cos(x)*sin(x)),x, algorithm="giac")
 

Output:

integrate(x/(b*cos(x)*sin(x) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{a+b \cos (x) \sin (x)} \, dx=\int \frac {x}{a+b\,\cos \left (x\right )\,\sin \left (x\right )} \,d x \] Input:

int(x/(a + b*cos(x)*sin(x)),x)
 

Output:

int(x/(a + b*cos(x)*sin(x)), x)
 

Reduce [F]

\[ \int \frac {x}{a+b \cos (x) \sin (x)} \, dx=\frac {-2 \left (\int \frac {\cos \left (x \right ) \sin \left (x \right ) x}{\cos \left (x \right ) \sin \left (x \right ) b +a}d x \right ) b +x^{2}}{2 a} \] Input:

int(x/(a+b*cos(x)*sin(x)),x)
 

Output:

( - 2*int((cos(x)*sin(x)*x)/(cos(x)*sin(x)*b + a),x)*b + x**2)/(2*a)