\(\int \frac {x}{a+b \sin ^2(x)} \, dx\) [1]

Optimal result

Integrand size = 12, antiderivative size = 203 \[ \int \frac {x}{a+b \sin ^2(x)} \, dx=-\frac {i x \log \left (1-\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{2 \sqrt {a} \sqrt {a+b}}+\frac {i x \log \left (1-\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{2 \sqrt {a} \sqrt {a+b}}-\frac {\operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{4 \sqrt {a} \sqrt {a+b}}+\frac {\operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{4 \sqrt {a} \sqrt {a+b}} \] Output:

-1/2*I*x*ln(1-b*exp(2*I*x)/(2*a+b-2*a^(1/2)*(a+b)^(1/2)))/a^(1/2)/(a+b)^(1 
/2)+1/2*I*x*ln(1-b*exp(2*I*x)/(2*a+b+2*a^(1/2)*(a+b)^(1/2)))/a^(1/2)/(a+b) 
^(1/2)-1/4*polylog(2,b*exp(2*I*x)/(2*a+b-2*a^(1/2)*(a+b)^(1/2)))/a^(1/2)/( 
a+b)^(1/2)+1/4*polylog(2,b*exp(2*I*x)/(2*a+b+2*a^(1/2)*(a+b)^(1/2)))/a^(1/ 
2)/(a+b)^(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. \(545\) vs. \(2(203)=406\).

Time = 0.41 (sec) , antiderivative size = 545, normalized size of antiderivative = 2.68 \[ \int \frac {x}{a+b \sin ^2(x)} \, dx=\frac {4 x \text {arctanh}\left (\frac {a \cot (x)}{\sqrt {-a (a+b)}}\right )-2 \arccos \left (1+\frac {2 a}{b}\right ) \text {arctanh}\left (\frac {\sqrt {-a (a+b)} \tan (x)}{a}\right )+\left (\arccos \left (1+\frac {2 a}{b}\right )-2 i \text {arctanh}\left (\frac {a \cot (x)}{\sqrt {-a (a+b)}}\right )+2 i \text {arctanh}\left (\frac {\sqrt {-a (a+b)} \tan (x)}{a}\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-a (a+b)} e^{-i x}}{\sqrt {-b} \sqrt {2 a+b-b \cos (2 x)}}\right )+\left (\arccos \left (1+\frac {2 a}{b}\right )+2 i \left (\text {arctanh}\left (\frac {a \cot (x)}{\sqrt {-a (a+b)}}\right )-\text {arctanh}\left (\frac {\sqrt {-a (a+b)} \tan (x)}{a}\right )\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-a (a+b)} e^{i x}}{\sqrt {-b} \sqrt {2 a+b-b \cos (2 x)}}\right )-\left (\arccos \left (1+\frac {2 a}{b}\right )+2 i \text {arctanh}\left (\frac {\sqrt {-a (a+b)} \tan (x)}{a}\right )\right ) \log \left (\frac {2 a \left (a+b-i \sqrt {-a (a+b)}\right ) (1-i \tan (x))}{b \left (a+\sqrt {-a (a+b)} \tan (x)\right )}\right )-\left (\arccos \left (1+\frac {2 a}{b}\right )-2 i \text {arctanh}\left (\frac {\sqrt {-a (a+b)} \tan (x)}{a}\right )\right ) \log \left (\frac {2 a \left (a+b+i \sqrt {-a (a+b)}\right ) (1+i \tan (x))}{b \left (a+\sqrt {-a (a+b)} \tan (x)\right )}\right )+i \left (\operatorname {PolyLog}\left (2,\frac {\left (2 a+b-2 i \sqrt {-a (a+b)}\right ) \left (-a+\sqrt {-a (a+b)} \tan (x)\right )}{b \left (a+\sqrt {-a (a+b)} \tan (x)\right )}\right )-\operatorname {PolyLog}\left (2,\frac {\left (2 a+b+2 i \sqrt {-a (a+b)}\right ) \left (-a+\sqrt {-a (a+b)} \tan (x)\right )}{b \left (a+\sqrt {-a (a+b)} \tan (x)\right )}\right )\right )}{4 \sqrt {-a (a+b)}} \] Input:

Integrate[x/(a + b*Sin[x]^2),x]
 

Output:

(4*x*ArcTanh[(a*Cot[x])/Sqrt[-(a*(a + b))]] - 2*ArcCos[1 + (2*a)/b]*ArcTan 
h[(Sqrt[-(a*(a + b))]*Tan[x])/a] + (ArcCos[1 + (2*a)/b] - (2*I)*ArcTanh[(a 
*Cot[x])/Sqrt[-(a*(a + b))]] + (2*I)*ArcTanh[(Sqrt[-(a*(a + b))]*Tan[x])/a 
])*Log[(Sqrt[2]*Sqrt[-(a*(a + b))])/(Sqrt[-b]*E^(I*x)*Sqrt[2*a + b - b*Cos 
[2*x]])] + (ArcCos[1 + (2*a)/b] + (2*I)*(ArcTanh[(a*Cot[x])/Sqrt[-(a*(a + 
b))]] - ArcTanh[(Sqrt[-(a*(a + b))]*Tan[x])/a]))*Log[(Sqrt[2]*Sqrt[-(a*(a 
+ b))]*E^(I*x))/(Sqrt[-b]*Sqrt[2*a + b - b*Cos[2*x]])] - (ArcCos[1 + (2*a) 
/b] + (2*I)*ArcTanh[(Sqrt[-(a*(a + b))]*Tan[x])/a])*Log[(2*a*(a + b - I*Sq 
rt[-(a*(a + b))])*(1 - I*Tan[x]))/(b*(a + Sqrt[-(a*(a + b))]*Tan[x]))] - ( 
ArcCos[1 + (2*a)/b] - (2*I)*ArcTanh[(Sqrt[-(a*(a + b))]*Tan[x])/a])*Log[(2 
*a*(a + b + I*Sqrt[-(a*(a + b))])*(1 + I*Tan[x]))/(b*(a + Sqrt[-(a*(a + b) 
)]*Tan[x]))] + I*(PolyLog[2, ((2*a + b - (2*I)*Sqrt[-(a*(a + b))])*(-a + S 
qrt[-(a*(a + b))]*Tan[x]))/(b*(a + Sqrt[-(a*(a + b))]*Tan[x]))] - PolyLog[ 
2, ((2*a + b + (2*I)*Sqrt[-(a*(a + b))])*(-a + Sqrt[-(a*(a + b))]*Tan[x])) 
/(b*(a + Sqrt[-(a*(a + b))]*Tan[x]))]))/(4*Sqrt[-(a*(a + b))])
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5096, 3042, 3802, 25, 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.

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

\(\Big \downarrow \) 5096

\(\displaystyle 2 \int \frac {x}{2 a+b-b \cos (2 x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int \frac {x}{2 a+b-b \sin \left (2 x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 3802

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

Int[x/(a + b*Sin[x]^2),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3802

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[2   Int[(c + d*x)^m*E^(I*Pi*(k - 1/2))*(E^(I*(e + 
f*x))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(2*I*( 
e + f*x)))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && NeQ 
[a^2 - b^2, 0] && IGtQ[m, 0]
 

rule 5096

Int[(x_)^(m_.)*((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]^2)^(n_), x_Symbol] :> 
Simp[1/2^n   Int[x^m*(2*a + b - b*Cos[2*c + 2*d*x])^n, x], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a + b, 0] && IGtQ[m, 0] && ILtQ[n, 0] && (EqQ[n, -1] || 
 (EqQ[m, 1] && EqQ[n, -2]))
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 464 vs. \(2 (153 ) = 306\).

Time = 0.33 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.29

method	result	size
risch	\(\frac {i \ln \left (1-\frac {b \,{\mathrm e}^{2 i x}}{2 \sqrt {a \left (a +b \right )}+2 a +b}\right ) x}{2 \sqrt {a \left (a +b \right )}+2 a +b}+\frac {i \ln \left (1-\frac {b \,{\mathrm e}^{2 i x}}{2 \sqrt {a \left (a +b \right )}+2 a +b}\right ) a x}{\sqrt {a \left (a +b \right )}\, \left (2 \sqrt {a \left (a +b \right )}+2 a +b \right )}+\frac {i \ln \left (1-\frac {b \,{\mathrm e}^{2 i x}}{2 \sqrt {a \left (a +b \right )}+2 a +b}\right ) b x}{2 \sqrt {a \left (a +b \right )}\, \left (2 \sqrt {a \left (a +b \right )}+2 a +b \right )}+\frac {x^{2}}{2 \sqrt {a \left (a +b \right )}+2 a +b}+\frac {a \,x^{2}}{\sqrt {a \left (a +b \right )}\, \left (2 \sqrt {a \left (a +b \right )}+2 a +b \right )}+\frac {b \,x^{2}}{2 \sqrt {a \left (a +b \right )}\, \left (2 \sqrt {a \left (a +b \right )}+2 a +b \right )}+\frac {\operatorname {polylog}\left (2, \frac {b \,{\mathrm e}^{2 i x}}{2 \sqrt {a \left (a +b \right )}+2 a +b}\right )}{4 \sqrt {a \left (a +b \right )}+4 a +2 b}+\frac {\operatorname {polylog}\left (2, \frac {b \,{\mathrm e}^{2 i x}}{2 \sqrt {a \left (a +b \right )}+2 a +b}\right ) a}{2 \sqrt {a \left (a +b \right )}\, \left (2 \sqrt {a \left (a +b \right )}+2 a +b \right )}+\frac {\operatorname {polylog}\left (2, \frac {b \,{\mathrm e}^{2 i x}}{2 \sqrt {a \left (a +b \right )}+2 a +b}\right ) b}{4 \sqrt {a \left (a +b \right )}\, \left (2 \sqrt {a \left (a +b \right )}+2 a +b \right )}-\frac {i x \ln \left (1-\frac {b \,{\mathrm e}^{2 i x}}{-2 \sqrt {a \left (a +b \right )}+2 a +b}\right )}{2 \sqrt {a \left (a +b \right )}}-\frac {x^{2}}{2 \sqrt {a \left (a +b \right )}}-\frac {\operatorname {polylog}\left (2, \frac {b \,{\mathrm e}^{2 i x}}{-2 \sqrt {a \left (a +b \right )}+2 a +b}\right )}{4 \sqrt {a \left (a +b \right )}}\)	\(465\)

Input:

int(x/(a+sin(x)^2*b),x,method=_RETURNVERBOSE)
 

Output:

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

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. 1608 vs. \(2 (149) = 298\).

Time = 1.11 (sec) , antiderivative size = 1608, normalized size of antiderivative = 7.92 \[ \int \frac {x}{a+b \sin ^2(x)} \, dx=\text {Too large to display} \] Input:

integrate(x/(a+b*sin(x)^2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(x/(a+b*sin(x)**2),x)
 

Output:

Integral(x/(a + b*sin(x)**2), x)
 

Maxima [F]

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

integrate(x/(a+b*sin(x)^2),x, algorithm="maxima")
 

Output:

integrate(x/(b*sin(x)^2 + a), x)
 

Giac [F]

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

integrate(x/(a+b*sin(x)^2),x, algorithm="giac")
 

Output:

integrate(x/(b*sin(x)^2 + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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