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

Optimal. Leaf size=225 \[ -\frac{\text{PolyLog}\left (2,\frac{i b e^{2 i x}}{2 a-\sqrt{4 a^2-b^2}}\right )}{2 \sqrt{4 a^2-b^2}}+\frac{\text{PolyLog}\left (2,\frac{i b e^{2 i x}}{\sqrt{4 a^2-b^2}+2 a}\right )}{2 \sqrt{4 a^2-b^2}}-\frac{i x \log \left (1-\frac{i b e^{2 i x}}{2 a-\sqrt{4 a^2-b^2}}\right )}{\sqrt{4 a^2-b^2}}+\frac{i x \log \left (1-\frac{i b e^{2 i x}}{\sqrt{4 a^2-b^2}+2 a}\right )}{\sqrt{4 a^2-b^2}} \]

[Out]

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

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Rubi [A]  time = 0.318976, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4584, 3323, 2264, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,\frac{i b e^{2 i x}}{2 a-\sqrt{4 a^2-b^2}}\right )}{2 \sqrt{4 a^2-b^2}}+\frac{\text{PolyLog}\left (2,\frac{i b e^{2 i x}}{\sqrt{4 a^2-b^2}+2 a}\right )}{2 \sqrt{4 a^2-b^2}}-\frac{i x \log \left (1-\frac{i b e^{2 i x}}{2 a-\sqrt{4 a^2-b^2}}\right )}{\sqrt{4 a^2-b^2}}+\frac{i x \log \left (1-\frac{i b e^{2 i x}}{\sqrt{4 a^2-b^2}+2 a}\right )}{\sqrt{4 a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Cos[x]*Sin[x]),x]

[Out]

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

Rule 4584

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]

Rule 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[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 2264

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]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2391

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]

Rubi steps

\begin{align*} \int \frac{x}{a+b \cos (x) \sin (x)} \, dx &=\int \frac{x}{a+\frac{1}{2} b \sin (2 x)} \, dx\\ &=2 \int \frac{e^{2 i x} x}{\frac{i b}{2}+2 a e^{2 i x}-\frac{1}{2} i b e^{4 i x}} \, dx\\ &=-\frac{(2 i b) \int \frac{e^{2 i x} x}{2 a-\sqrt{4 a^2-b^2}-i b e^{2 i x}} \, dx}{\sqrt{4 a^2-b^2}}+\frac{(2 i b) \int \frac{e^{2 i x} x}{2 a+\sqrt{4 a^2-b^2}-i b e^{2 i x}} \, dx}{\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{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 \int \log \left (1-\frac{i b e^{2 i x}}{2 a-\sqrt{4 a^2-b^2}}\right ) \, dx}{\sqrt{4 a^2-b^2}}-\frac{i \int \log \left (1-\frac{i b e^{2 i x}}{2 a+\sqrt{4 a^2-b^2}}\right ) \, dx}{\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{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{Subst}\left (\int \frac{\log \left (1-\frac{i b x}{2 a-\sqrt{4 a^2-b^2}}\right )}{x} \, dx,x,e^{2 i x}\right )}{2 \sqrt{4 a^2-b^2}}-\frac{\operatorname{Subst}\left (\int \frac{\log \left (1-\frac{i b x}{2 a+\sqrt{4 a^2-b^2}}\right )}{x} \, dx,x,e^{2 i x}\right )}{2 \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{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{\text{Li}_2\left (\frac{i b e^{2 i x}}{2 a-\sqrt{4 a^2-b^2}}\right )}{2 \sqrt{4 a^2-b^2}}+\frac{\text{Li}_2\left (\frac{i b e^{2 i x}}{2 a+\sqrt{4 a^2-b^2}}\right )}{2 \sqrt{4 a^2-b^2}}\\ \end{align*}

Mathematica [B]  time = 1.43413, size = 788, normalized size = 3.5 \[ \frac{1}{2} \left (\frac{\pi \tan ^{-1}\left (\frac{2 a \tan (x)+b}{\sqrt{4 a^2-b^2}}\right )}{\sqrt{4 a^2-b^2}}+\frac{i \left (\text{PolyLog}\left (2,\frac{\left (2 a-i \sqrt{b^2-4 a^2}\right ) \left (-\sqrt{b^2-4 a^2} \cot \left (x+\frac{\pi }{4}\right )+2 a+b\right )}{b \left (\sqrt{b^2-4 a^2} \cot \left (x+\frac{\pi }{4}\right )+2 a+b\right )}\right )-\text{PolyLog}\left (2,\frac{\left (2 a+i \sqrt{b^2-4 a^2}\right ) \left (-\sqrt{b^2-4 a^2} \cot \left (x+\frac{\pi }{4}\right )+2 a+b\right )}{b \left (\sqrt{b^2-4 a^2} \cot \left (x+\frac{\pi }{4}\right )+2 a+b\right )}\right )\right )+(\pi -4 x) \tanh ^{-1}\left (\frac{(2 a+b) \tan \left (x+\frac{\pi }{4}\right )}{\sqrt{b^2-4 a^2}}\right )+2 \cos ^{-1}\left (-\frac{2 a}{b}\right ) \tanh ^{-1}\left (\frac{(2 a-b) \cot \left (x+\frac{\pi }{4}\right )}{\sqrt{b^2-4 a^2}}\right )-\log \left (\frac{(2 a+b) \left (-i \sqrt{b^2-4 a^2}-2 a+b\right ) \left (1+i \cot \left (x+\frac{\pi }{4}\right )\right )}{b \left (\sqrt{b^2-4 a^2} \cot \left (x+\frac{\pi }{4}\right )+2 a+b\right )}\right ) \left (\cos ^{-1}\left (-\frac{2 a}{b}\right )+2 i \tanh ^{-1}\left (\frac{(2 a-b) \cot \left (x+\frac{\pi }{4}\right )}{\sqrt{b^2-4 a^2}}\right )\right )-\log \left (\frac{(2 a+b) \left (\sqrt{b^2-4 a^2}+2 i a-i b\right ) \left (\cot \left (x+\frac{\pi }{4}\right )+i\right )}{b \left (\sqrt{b^2-4 a^2} \cot \left (x+\frac{\pi }{4}\right )+2 a+b\right )}\right ) \left (\cos ^{-1}\left (-\frac{2 a}{b}\right )-2 i \tanh ^{-1}\left (\frac{(2 a-b) \cot \left (x+\frac{\pi }{4}\right )}{\sqrt{b^2-4 a^2}}\right )\right )+\log \left (\frac{\sqrt [4]{-1} e^{-i x} \sqrt{b^2-4 a^2}}{2 \sqrt{b} \sqrt{a+b \sin (x) \cos (x)}}\right ) \left (\cos ^{-1}\left (-\frac{2 a}{b}\right )+2 i \left (\tanh ^{-1}\left (\frac{(2 a+b) \tan \left (x+\frac{\pi }{4}\right )}{\sqrt{b^2-4 a^2}}\right )+\tanh ^{-1}\left (\frac{(2 a-b) \cot \left (x+\frac{\pi }{4}\right )}{\sqrt{b^2-4 a^2}}\right )\right )\right )+\log \left (-\frac{(-1)^{3/4} e^{i x} \sqrt{b^2-4 a^2}}{2 \sqrt{b} \sqrt{a+b \sin (x) \cos (x)}}\right ) \left (-2 i \tanh ^{-1}\left (\frac{(2 a+b) \tan \left (x+\frac{\pi }{4}\right )}{\sqrt{b^2-4 a^2}}\right )-2 i \tanh ^{-1}\left (\frac{(2 a-b) \cot \left (x+\frac{\pi }{4}\right )}{\sqrt{b^2-4 a^2}}\right )+\cos ^{-1}\left (-\frac{2 a}{b}\right )\right )}{\sqrt{b^2-4 a^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Cos[x]*Sin[x]),x]

[Out]

((Pi*ArcTan[(b + 2*a*Tan[x])/Sqrt[4*a^2 - b^2]])/Sqrt[4*a^2 - b^2] + (2*ArcCos[(-2*a)/b]*ArcTanh[((2*a - b)*Co
t[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)*ArcTanh[((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)^(3/4)*Sqrt[-4*a^2 + b^2]*E^(I*x))/(2*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 + Sqrt[-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 + S
qrt[-4*a^2 + b^2]*Cot[Pi/4 + x]))]))/Sqrt[-4*a^2 + b^2])/2

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Maple [B]  time = 0.111, size = 1284, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{b \cos \left (x\right ) \sin \left (x\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*cos(x)*sin(x) + a), x)

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Fricas [B]  time = 4.88477, size = 3969, normalized size = 17.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*b*x*sqrt(-(4*a^2 - b^2)/b^2)*log(1/2*((4*I*a*cos(x) + 4*a*sin(x) - 2*(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) + 2*b)/b) + 2*b*x*sqrt(-(4*a^2 - b^2)/b^2)*log(1
/2*((-4*I*a*cos(x) - 4*a*sin(x) + 2*(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) + 2*b)/b) - 2*b*x*sqrt(-(4*a^2 - b^2)/b^2)*log(1/2*((4*I*a*cos(x) - 4*a*sin(x) - 2*(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) + 2*b)/b) - 2*b*x*sq
rt(-(4*a^2 - b^2)/b^2)*log(1/2*((-4*I*a*cos(x) + 4*a*sin(x) + 2*(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) + 2*b)/b) + 2*b*x*sqrt(-(4*a^2 - b^2)/b^2)*log(1/2*((4*I*a*c
os(x) - 4*a*sin(x) + 2*(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) + 2*b)/b) + 2*b*x*sqrt(-(4*a^2 - b^2)/b^2)*log(1/2*((-4*I*a*cos(x) + 4*a*sin(x) - 2*(b*cos(x) + I*b*si
n(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt((b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b) + 2*b)/b) - 2*b*x*sqrt(-(4*a^2 -
b^2)/b^2)*log(1/2*((4*I*a*cos(x) + 4*a*sin(x) + 2*(b*cos(x) - I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt(-(b*s
qrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b) + 2*b)/b) - 2*b*x*sqrt(-(4*a^2 - b^2)/b^2)*log(1/2*((-4*I*a*cos(x) - 4*a*s
in(x) - 2*(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) + 2*
b)/b) + 2*I*b*sqrt(-(4*a^2 - b^2)/b^2)*dilog(-1/2*((4*I*a*cos(x) + 4*a*sin(x) - 2*(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) + 2*b)/b + 1) + 2*I*b*sqrt(-(4*a^2 - b^2)/b
^2)*dilog(-1/2*((-4*I*a*cos(x) - 4*a*sin(x) + 2*(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) + 2*b)/b + 1) + 2*I*b*sqrt(-(4*a^2 - b^2)/b^2)*dilog(-1/2*((4*I*a*cos(x) - 4*
a*sin(x) - 2*(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) +
2*b)/b + 1) + 2*I*b*sqrt(-(4*a^2 - b^2)/b^2)*dilog(-1/2*((-4*I*a*cos(x) + 4*a*sin(x) + 2*(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) + 2*b)/b + 1) - 2*I*b*sqrt(-(4*a^2
- b^2)/b^2)*dilog(-1/2*((4*I*a*cos(x) - 4*a*sin(x) + 2*(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) + 2*b)/b + 1) - 2*I*b*sqrt(-(4*a^2 - b^2)/b^2)*dilog(-1/2*((-4*I*a*co
s(x) + 4*a*sin(x) - 2*(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) + 2*b)/b + 1) - 2*I*b*sqrt(-(4*a^2 - b^2)/b^2)*dilog(-1/2*((4*I*a*cos(x) + 4*a*sin(x) + 2*(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) + 2*b)/b + 1) - 2*I*b*sqrt(
-(4*a^2 - b^2)/b^2)*dilog(-1/2*((-4*I*a*cos(x) - 4*a*sin(x) - 2*(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) + 2*b)/b + 1))/(4*a^2 - b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(a + b*sin(x)*cos(x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{b \cos \left (x\right ) \sin \left (x\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*cos(x)*sin(x) + a), x)