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

Optimal. Leaf size=203 \[ -\frac {i x \log \left (1-\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{2 \sqrt {a} \sqrt {a+b}}+\frac {i x \log \left (1-\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{2 \sqrt {a} \sqrt {a+b}}-\frac {\text {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{4 \sqrt {a} \sqrt {a+b}}+\frac {\text {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{4 \sqrt {a} \sqrt {a+b}} \]

[Out]

-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)

________________________________________________________________________________________

Rubi [A]
time = 0.26, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4681, 3402, 2296, 2221, 2317, 2438} \begin {gather*} -\frac {\text {Li}_2\left (\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )}{4 \sqrt {a} \sqrt {a+b}}+\frac {\text {Li}_2\left (\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )}{4 \sqrt {a} \sqrt {a+b}}-\frac {i x \log \left (1-\frac {b e^{2 i x}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 \sqrt {a} \sqrt {a+b}}+\frac {i x \log \left (1-\frac {b e^{2 i x}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 \sqrt {a} \sqrt {a+b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Sin[x]^2),x]

[Out]

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

Rule 2221

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]
- 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 2296

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 2317

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 2438

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 3402

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[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 4681

Int[(x_)^(m_.)*((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]^2)^(n_), x_Symbol] :> Dist[1/2^n, Int[x^m*(2*a + b - b*Co
s[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]))

Rubi steps

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

________________________________________________________________________________________

Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(545\) vs. \(2(203)=406\).
time = 0.65, size = 545, normalized size = 2.68 \begin {gather*} \frac {4 x \tanh ^{-1}\left (\frac {a \cot (x)}{\sqrt {-a (a+b)}}\right )-2 \text {ArcCos}\left (1+\frac {2 a}{b}\right ) \tanh ^{-1}\left (\frac {\sqrt {-a (a+b)} \tan (x)}{a}\right )+\left (\text {ArcCos}\left (1+\frac {2 a}{b}\right )-2 i \tanh ^{-1}\left (\frac {a \cot (x)}{\sqrt {-a (a+b)}}\right )+2 i \tanh ^{-1}\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 (\text {ArcCos}\left (1+\frac {2 a}{b}\right )+2 i \left (\tanh ^{-1}\left (\frac {a \cot (x)}{\sqrt {-a (a+b)}}\right )-\tanh ^{-1}\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 (\text {ArcCos}\left (1+\frac {2 a}{b}\right )+2 i \tanh ^{-1}\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 (\text {ArcCos}\left (1+\frac {2 a}{b}\right )-2 i \tanh ^{-1}\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 (\text {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 )-\text {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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Sin[x]^2),x]

[Out]

(4*x*ArcTanh[(a*Cot[x])/Sqrt[-(a*(a + b))]] - 2*ArcCos[1 + (2*a)/b]*ArcTanh[(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*Sqrt[-(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 + Sqrt[-(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))])

________________________________________________________________________________________

Maple [B] 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 = 0.12, size = 465, normalized size = 2.29

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*sin(x)^2 + a), x)

________________________________________________________________________________________

Fricas [B] 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 = 3.87, size = 1608, normalized size = 7.92 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)*s
in(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) - b*sqrt((a^2 + a*b)/b^2)*dilog((((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 + 1) -
b*sqrt((a^2 + a*b)/b^2)*dilog(-(((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 + 1) - b*sqrt((a^2 + a*b)/b^2)*dilog((((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 + 1) - b*sqrt((a^2 + a*b)/b^2)*dilog(-(((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 + 1)
+ b*sqrt((a^2 + a*b)/b^2)*dilog((((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 + 1) + b*sqrt((a^2 + a*b)/b^2)*dilog(-((
(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 + 1) + b*sqrt((a^2 + a*b)/b^2)*dilog((((2*a + b)*cos(x) + (-2*I*a - I*b)*si
n(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
+ 1) + b*sqrt((a^2 + a*b)/b^2)*dilog(-(((2*a + b)*cos(x) - (-2*I*a - I*b)*sin(x) + 2*(b*cos(x) + I*b*sin(x))*s
qrt((a^2 + a*b)/b^2))*sqrt(-(2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b) + b)/b + 1))/(a^2 + a*b)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{a + b \sin ^{2}{\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(a + b*sin(x)**2), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*sin(x)^2 + a), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x}{b\,{\sin \left (x\right )}^2+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*sin(x)^2),x)

[Out]

int(x/(a + b*sin(x)^2), x)

________________________________________________________________________________________