Integrand size = 14, antiderivative size = 411 \[ \int \frac {x^3}{a+b \sin ^2(x)} \, dx=-\frac {i x^3 \log \left (1-\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{2 \sqrt {a} \sqrt {a+b}}+\frac {i x^3 \log \left (1-\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{2 \sqrt {a} \sqrt {a+b}}-\frac {3 x^2 \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{4 \sqrt {a} \sqrt {a+b}}+\frac {3 x^2 \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{4 \sqrt {a} \sqrt {a+b}}-\frac {3 i x \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{4 \sqrt {a} \sqrt {a+b}}+\frac {3 i x \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{4 \sqrt {a} \sqrt {a+b}}+\frac {3 \operatorname {PolyLog}\left (4,\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{8 \sqrt {a} \sqrt {a+b}}-\frac {3 \operatorname {PolyLog}\left (4,\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{8 \sqrt {a} \sqrt {a+b}} \]
-1/2*I*x^3*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^3*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)-3/4*x^2*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)+3/4*x^2*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)-3/4*I*x*polylog(3,b*exp(2*I*x)/(2*a+b-2*a^(1/2) *(a+b)^(1/2)))/a^(1/2)/(a+b)^(1/2)+3/4*I*x*polylog(3,b*exp(2*I*x)/(2*a+b+2 *a^(1/2)*(a+b)^(1/2)))/a^(1/2)/(a+b)^(1/2)+3/8*polylog(4,b*exp(2*I*x)/(2*a +b-2*a^(1/2)*(a+b)^(1/2)))/a^(1/2)/(a+b)^(1/2)-3/8*polylog(4,b*exp(2*I*x)/ (2*a+b+2*a^(1/2)*(a+b)^(1/2)))/a^(1/2)/(a+b)^(1/2)
Time = 3.77 (sec) , antiderivative size = 288, normalized size of antiderivative = 0.70 \[ \int \frac {x^3}{a+b \sin ^2(x)} \, dx=\frac {-4 i x^3 \log \left (1-\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a (a+b)}}\right )+4 i x^3 \log \left (1-\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a (a+b)}}\right )-6 x^2 \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a (a+b)}}\right )+6 x^2 \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a (a+b)}}\right )-6 i x \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a (a+b)}}\right )+6 i x \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a (a+b)}}\right )+3 \operatorname {PolyLog}\left (4,\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a (a+b)}}\right )-3 \operatorname {PolyLog}\left (4,\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a (a+b)}}\right )}{8 \sqrt {a (a+b)}} \]
((-4*I)*x^3*Log[1 - (b*E^((2*I)*x))/(2*a + b - 2*Sqrt[a*(a + b)])] + (4*I) *x^3*Log[1 - (b*E^((2*I)*x))/(2*a + b + 2*Sqrt[a*(a + b)])] - 6*x^2*PolyLo g[2, (b*E^((2*I)*x))/(2*a + b - 2*Sqrt[a*(a + b)])] + 6*x^2*PolyLog[2, (b* E^((2*I)*x))/(2*a + b + 2*Sqrt[a*(a + b)])] - (6*I)*x*PolyLog[3, (b*E^((2* I)*x))/(2*a + b - 2*Sqrt[a*(a + b)])] + (6*I)*x*PolyLog[3, (b*E^((2*I)*x)) /(2*a + b + 2*Sqrt[a*(a + b)])] + 3*PolyLog[4, (b*E^((2*I)*x))/(2*a + b - 2*Sqrt[a*(a + b)])] - 3*PolyLog[4, (b*E^((2*I)*x))/(2*a + b + 2*Sqrt[a*(a + b)])])/(8*Sqrt[a*(a + b)])
Time = 1.33 (sec) , antiderivative size = 393, normalized size of antiderivative = 0.96, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {5096, 3042, 3802, 25, 2694, 27, 2620, 3011, 7163, 2720, 7143}
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.
| \(\displaystyle \int \frac {x^3}{a+b \sin ^2(x)} \, dx\) |
| \(\Big \downarrow \) 5096 |
| \(\displaystyle 2 \int \frac {x^3}{2 a+b-b \cos (2 x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \int \frac {x^3}{2 a+b-b \sin \left (2 x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3802 |
| \(\displaystyle 4 \int -\frac {e^{2 i x} x^3}{e^{4 i x} b+b-2 (2 a+b) e^{2 i x}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -4 \int \frac {e^{2 i x} x^3}{e^{4 i x} b+b-2 (2 a+b) e^{2 i x}}dx\) |
| \(\Big \downarrow \) 2694 |
| \(\displaystyle -4 \left (\frac {b \int -\frac {e^{2 i x} x^3}{2 \left (2 a+2 \sqrt {a+b} \sqrt {a}-b e^{2 i x}+b\right )}dx}{2 \sqrt {a} \sqrt {a+b}}-\frac {b \int -\frac {e^{2 i x} x^3}{2 \left (2 a-2 \sqrt {a+b} \sqrt {a}-b e^{2 i x}+b\right )}dx}{2 \sqrt {a} \sqrt {a+b}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -4 \left (\frac {b \int \frac {e^{2 i x} x^3}{2 a-2 \sqrt {a+b} \sqrt {a}-b e^{2 i x}+b}dx}{4 \sqrt {a} \sqrt {a+b}}-\frac {b \int \frac {e^{2 i x} x^3}{2 a+2 \sqrt {a+b} \sqrt {a}-b e^{2 i x}+b}dx}{4 \sqrt {a} \sqrt {a+b}}\right )\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle -4 \left (\frac {b \left (\frac {i x^3 \log \left (1-\frac {b e^{2 i x}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b}-\frac {3 i \int x^2 \log \left (1-\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )dx}{2 b}\right )}{4 \sqrt {a} \sqrt {a+b}}-\frac {b \left (\frac {i x^3 \log \left (1-\frac {b e^{2 i x}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b}-\frac {3 i \int x^2 \log \left (1-\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )dx}{2 b}\right )}{4 \sqrt {a} \sqrt {a+b}}\right )\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle -4 \left (\frac {b \left (\frac {i x^3 \log \left (1-\frac {b e^{2 i x}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b}-\frac {3 i \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )-i \int x \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )dx\right )}{2 b}\right )}{4 \sqrt {a} \sqrt {a+b}}-\frac {b \left (\frac {i x^3 \log \left (1-\frac {b e^{2 i x}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b}-\frac {3 i \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )-i \int x \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )dx\right )}{2 b}\right )}{4 \sqrt {a} \sqrt {a+b}}\right )\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle -4 \left (\frac {b \left (\frac {i x^3 \log \left (1-\frac {b e^{2 i x}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b}-\frac {3 i \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )-i \left (\frac {1}{2} i \int \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )dx-\frac {1}{2} i x \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )\right )\right )}{2 b}\right )}{4 \sqrt {a} \sqrt {a+b}}-\frac {b \left (\frac {i x^3 \log \left (1-\frac {b e^{2 i x}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b}-\frac {3 i \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )-i \left (\frac {1}{2} i \int \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )dx-\frac {1}{2} i x \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )\right )\right )}{2 b}\right )}{4 \sqrt {a} \sqrt {a+b}}\right )\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle -4 \left (\frac {b \left (\frac {i x^3 \log \left (1-\frac {b e^{2 i x}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b}-\frac {3 i \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )-i \left (\frac {1}{4} \int e^{-2 i x} \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )de^{2 i x}-\frac {1}{2} i x \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )\right )\right )}{2 b}\right )}{4 \sqrt {a} \sqrt {a+b}}-\frac {b \left (\frac {i x^3 \log \left (1-\frac {b e^{2 i x}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b}-\frac {3 i \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )-i \left (\frac {1}{4} \int e^{-2 i x} \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )de^{2 i x}-\frac {1}{2} i x \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )\right )\right )}{2 b}\right )}{4 \sqrt {a} \sqrt {a+b}}\right )\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle -4 \left (\frac {b \left (\frac {i x^3 \log \left (1-\frac {b e^{2 i x}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b}-\frac {3 i \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )-i \left (\frac {1}{4} \operatorname {PolyLog}\left (4,\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )-\frac {1}{2} i x \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )\right )\right )}{2 b}\right )}{4 \sqrt {a} \sqrt {a+b}}-\frac {b \left (\frac {i x^3 \log \left (1-\frac {b e^{2 i x}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b}-\frac {3 i \left (\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )-i \left (\frac {1}{4} \operatorname {PolyLog}\left (4,\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )-\frac {1}{2} i x \operatorname {PolyLog}\left (3,\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )\right )\right )}{2 b}\right )}{4 \sqrt {a} \sqrt {a+b}}\right )\) |
-4*((b*(((I/2)*x^3*Log[1 - (b*E^((2*I)*x))/(2*a + b - 2*Sqrt[a]*Sqrt[a + b ])])/b - (((3*I)/2)*((I/2)*x^2*PolyLog[2, (b*E^((2*I)*x))/(2*a + b - 2*Sqr t[a]*Sqrt[a + b])] - I*((-1/2*I)*x*PolyLog[3, (b*E^((2*I)*x))/(2*a + b - 2 *Sqrt[a]*Sqrt[a + b])] + PolyLog[4, (b*E^((2*I)*x))/(2*a + b - 2*Sqrt[a]*S qrt[a + b])]/4)))/b))/(4*Sqrt[a]*Sqrt[a + b]) - (b*(((I/2)*x^3*Log[1 - (b* E^((2*I)*x))/(2*a + b + 2*Sqrt[a]*Sqrt[a + b])])/b - (((3*I)/2)*((I/2)*x^2 *PolyLog[2, (b*E^((2*I)*x))/(2*a + b + 2*Sqrt[a]*Sqrt[a + b])] - I*((-1/2* I)*x*PolyLog[3, (b*E^((2*I)*x))/(2*a + b + 2*Sqrt[a]*Sqrt[a + b])] + PolyL og[4, (b*E^((2*I)*x))/(2*a + b + 2*Sqrt[a]*Sqrt[a + b])]/4)))/b))/(4*Sqrt[ a]*Sqrt[a + b]))
3.1.3.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
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]
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]))
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. )*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F])) Int[(e + f*x) ^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c , d, e, f, n, p}, x] && GtQ[m, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 852 vs. \(2 (311 ) = 622\).
Time = 0.57 (sec) , antiderivative size = 853, normalized size of antiderivative = 2.08
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^3-3/4*I/(a*(a+b))^(1/2)*x*polylog(3,b*exp(2*I*x)/(- 2*(a*(a+b))^(1/2)+2*a+b))+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^3+1/2/(2*(a*(a+b))^(1/2)+2*a+ b)*x^4+1/2/(a*(a+b))^(1/2)/(2*(a*(a+b))^(1/2)+2*a+b)*a*x^4+1/4/(a*(a+b))^( 1/2)/(2*(a*(a+b))^(1/2)+2*a+b)*b*x^4+3/2*I/(2*(a*(a+b))^(1/2)+2*a+b)*polyl og(3,b*exp(2*I*x)/(2*(a*(a+b))^(1/2)+2*a+b))*x+3/2/(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))*x^2-1/2*I/(a*(a+b))^( 1/2)*x^3*ln(1-b*exp(2*I*x)/(-2*(a*(a+b))^(1/2)+2*a+b))+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^3+3/2/(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*x^2+3/4/(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*x^2-3/4/(2*(a*(a+b))^(1/2)+2*a+b)*poly log(4,b*exp(2*I*x)/(2*(a*(a+b))^(1/2)+2*a+b))-3/4/(a*(a+b))^(1/2)/(2*(a*(a +b))^(1/2)+2*a+b)*polylog(4,b*exp(2*I*x)/(2*(a*(a+b))^(1/2)+2*a+b))*a-3/8/ (a*(a+b))^(1/2)/(2*(a*(a+b))^(1/2)+2*a+b)*polylog(4,b*exp(2*I*x)/(2*(a*(a+ b))^(1/2)+2*a+b))*b+3/2*I/(a*(a+b))^(1/2)/(2*(a*(a+b))^(1/2)+2*a+b)*polylo g(3,b*exp(2*I*x)/(2*(a*(a+b))^(1/2)+2*a+b))*a*x-1/4/(a*(a+b))^(1/2)*x^4-3/ 4/(a*(a+b))^(1/2)*x^2*polylog(2,b*exp(2*I*x)/(-2*(a*(a+b))^(1/2)+2*a+b))+3 /4*I/(a*(a+b))^(1/2)/(2*(a*(a+b))^(1/2)+2*a+b)*polylog(3,b*exp(2*I*x)/(...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3188 vs. \(2 (307) = 614\).
Time = 1.22 (sec) , antiderivative size = 3188, normalized size of antiderivative = 7.76 \[ \int \frac {x^3}{a+b \sin ^2(x)} \, dx=\text {Too large to display} \]
-1/4*(-I*b*x^3*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*sqr t((a^2 + a*b)/b^2) + 2*a + b)/b) - b)/b) + I*b*x^3*sqrt((a^2 + a*b)/b^2)*l og((((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 ) + I*b*x^3*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^3*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))*sq rt((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^3*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^3*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^3*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^3*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))*sqr t((a^2 + a*b)/b^2))*sqrt(-(2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b) + b)...
\[ \int \frac {x^3}{a+b \sin ^2(x)} \, dx=\int \frac {x^{3}}{a + b \sin ^{2}{\left (x \right )}}\, dx \]
\[ \int \frac {x^3}{a+b \sin ^2(x)} \, dx=\int { \frac {x^{3}}{b \sin \left (x\right )^{2} + a} \,d x } \]
\[ \int \frac {x^3}{a+b \sin ^2(x)} \, dx=\int { \frac {x^{3}}{b \sin \left (x\right )^{2} + a} \,d x } \]
Timed out. \[ \int \frac {x^3}{a+b \sin ^2(x)} \, dx=\int \frac {x^3}{b\,{\sin \left (x\right )}^2+a} \,d x \]