Integrand size = 14, antiderivative size = 340 \[ \int \frac {x^2}{a+b \cos (x) \sin (x)} \, dx=-\frac {i x^2 \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2}}+\frac {i x^2 \log \left (1-\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2}}-\frac {x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2}}+\frac {x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2}}-\frac {i \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 \sqrt {4 a^2-b^2}}+\frac {i \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{2 \sqrt {4 a^2-b^2}} \]
-I*x^2*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^ 2*ln(1-I*b*exp(2*I*x)/(2*a+(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)-x*polylog (2,I*b*exp(2*I*x)/(2*a-(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)+x*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*I*polylog(3,I *b*exp(2*I*x)/(2*a-(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)+1/2*I*polylog(3,I *b*exp(2*I*x)/(2*a+(4*a^2-b^2)^(1/2)))/(4*a^2-b^2)^(1/2)
Time = 0.90 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.75 \[ \int \frac {x^2}{a+b \cos (x) \sin (x)} \, dx=-\frac {i \left (2 x^2 \log \left (1+\frac {i b e^{2 i x}}{-2 a+\sqrt {4 a^2-b^2}}\right )-2 x^2 \log \left (1-\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )-2 i x \operatorname {PolyLog}\left (2,-\frac {i b e^{2 i x}}{-2 a+\sqrt {4 a^2-b^2}}\right )+2 i x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )+\operatorname {PolyLog}\left (3,-\frac {i b e^{2 i x}}{-2 a+\sqrt {4 a^2-b^2}}\right )-\operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )\right )}{2 \sqrt {4 a^2-b^2}} \]
((-1/2*I)*(2*x^2*Log[1 + (I*b*E^((2*I)*x))/(-2*a + Sqrt[4*a^2 - b^2])] - 2 *x^2*Log[1 - (I*b*E^((2*I)*x))/(2*a + Sqrt[4*a^2 - b^2])] - (2*I)*x*PolyLo g[2, ((-I)*b*E^((2*I)*x))/(-2*a + Sqrt[4*a^2 - b^2])] + (2*I)*x*PolyLog[2, (I*b*E^((2*I)*x))/(2*a + Sqrt[4*a^2 - b^2])] + PolyLog[3, ((-I)*b*E^((2*I )*x))/(-2*a + Sqrt[4*a^2 - b^2])] - PolyLog[3, (I*b*E^((2*I)*x))/(2*a + Sq rt[4*a^2 - b^2])]))/Sqrt[4*a^2 - b^2]
Time = 1.07 (sec) , antiderivative size = 321, normalized size of antiderivative = 0.94, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5095, 3042, 3804, 27, 2694, 27, 2620, 3011, 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.
\(\displaystyle \int \frac {x^2}{a+b \sin (x) \cos (x)} \, dx\) |
\(\Big \downarrow \) 5095 |
\(\displaystyle \int \frac {x^2}{a+\frac {1}{2} b \sin (2 x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {x^2}{a+\frac {1}{2} b \sin (2 x)}dx\) |
\(\Big \downarrow \) 3804 |
\(\displaystyle 2 \int \frac {2 e^{2 i x} x^2}{4 e^{2 i x} a-i b e^{4 i x}+i b}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int \frac {e^{2 i x} x^2}{4 e^{2 i x} a-i b e^{4 i x}+i b}dx\) |
\(\Big \downarrow \) 2694 |
\(\displaystyle 4 \left (\frac {i b \int \frac {e^{2 i x} x^2}{2 \left (2 a-i b e^{2 i x}+\sqrt {4 a^2-b^2}\right )}dx}{\sqrt {4 a^2-b^2}}-\frac {i b \int \frac {e^{2 i x} x^2}{2 \left (2 a-i b e^{2 i x}-\sqrt {4 a^2-b^2}\right )}dx}{\sqrt {4 a^2-b^2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \left (\frac {i b \int \frac {e^{2 i x} x^2}{2 a-i b e^{2 i x}+\sqrt {4 a^2-b^2}}dx}{2 \sqrt {4 a^2-b^2}}-\frac {i b \int \frac {e^{2 i x} x^2}{2 a-i b e^{2 i x}-\sqrt {4 a^2-b^2}}dx}{2 \sqrt {4 a^2-b^2}}\right )\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle 4 \left (\frac {i b \left (\frac {x^2 \log \left (1-\frac {i b e^{2 i x}}{\sqrt {4 a^2-b^2}+2 a}\right )}{2 b}-\frac {\int x \log \left (1-\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )dx}{b}\right )}{2 \sqrt {4 a^2-b^2}}-\frac {i b \left (\frac {x^2 \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 b}-\frac {\int x \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )dx}{b}\right )}{2 \sqrt {4 a^2-b^2}}\right )\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle 4 \left (\frac {i b \left (\frac {x^2 \log \left (1-\frac {i b e^{2 i x}}{\sqrt {4 a^2-b^2}+2 a}\right )}{2 b}-\frac {\frac {1}{2} i x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )-\frac {1}{2} i \int \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )dx}{b}\right )}{2 \sqrt {4 a^2-b^2}}-\frac {i b \left (\frac {x^2 \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 b}-\frac {\frac {1}{2} i x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )-\frac {1}{2} i \int \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )dx}{b}\right )}{2 \sqrt {4 a^2-b^2}}\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle 4 \left (\frac {i b \left (\frac {x^2 \log \left (1-\frac {i b e^{2 i x}}{\sqrt {4 a^2-b^2}+2 a}\right )}{2 b}-\frac {\frac {1}{2} i x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )-\frac {1}{4} \int e^{-2 i x} \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )de^{2 i x}}{b}\right )}{2 \sqrt {4 a^2-b^2}}-\frac {i b \left (\frac {x^2 \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 b}-\frac {\frac {1}{2} i x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )-\frac {1}{4} \int e^{-2 i x} \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )de^{2 i x}}{b}\right )}{2 \sqrt {4 a^2-b^2}}\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle 4 \left (\frac {i b \left (\frac {x^2 \log \left (1-\frac {i b e^{2 i x}}{\sqrt {4 a^2-b^2}+2 a}\right )}{2 b}-\frac {\frac {1}{2} i x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )-\frac {1}{4} \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{b}\right )}{2 \sqrt {4 a^2-b^2}}-\frac {i b \left (\frac {x^2 \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 b}-\frac {\frac {1}{2} i x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )-\frac {1}{4} \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{b}\right )}{2 \sqrt {4 a^2-b^2}}\right )\) |
4*(((-1/2*I)*b*((x^2*Log[1 - (I*b*E^((2*I)*x))/(2*a - Sqrt[4*a^2 - b^2])]) /(2*b) - ((I/2)*x*PolyLog[2, (I*b*E^((2*I)*x))/(2*a - Sqrt[4*a^2 - b^2])] - PolyLog[3, (I*b*E^((2*I)*x))/(2*a - Sqrt[4*a^2 - b^2])]/4)/b))/Sqrt[4*a^ 2 - b^2] + ((I/2)*b*((x^2*Log[1 - (I*b*E^((2*I)*x))/(2*a + Sqrt[4*a^2 - b^ 2])])/(2*b) - ((I/2)*x*PolyLog[2, (I*b*E^((2*I)*x))/(2*a + Sqrt[4*a^2 - b^ 2])] - PolyLog[3, (I*b*E^((2*I)*x))/(2*a + Sqrt[4*a^2 - b^2])]/4)/b))/Sqrt [4*a^2 - b^2])
3.6.80.3.1 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_.) + (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]
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]
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1781 vs. \(2 (290 ) = 580\).
Time = 0.96 (sec) , antiderivative size = 1782, normalized size of antiderivative = 5.24
-8/3/(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^3-16/3*I/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*a^2*x^3+4 /3*I/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*b^2*x^3-2*I/(8*a^2-2* b^2)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*polylog(3,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+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^2-2/(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^2-4/(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*x-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^2-8*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*x+2*I/( 8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*polylog(3,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+4/(8*a^2-2*b^2)/ (-2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*polylog(3,b*exp(2*I*x)/(-2*I*a-(-(2*a-b) *(2*a+b))^(1/2)))*a^2-1/(8*a^2-2*b^2)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*po lylog(3,b*exp(2*I*x)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2)))*b^2+8/3/(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^3+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)/(...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2492 vs. \(2 (272) = 544\).
Time = 0.89 (sec) , antiderivative size = 2492, normalized size of antiderivative = 7.33 \[ \int \frac {x^2}{a+b \cos (x) \sin (x)} \, dx=\text {Too large to display} \]
-1/2*(b*x^2*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^2*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))*sq rt((b*sqrt(-(4*a^2 - b^2)/b^2) + 2*I*a)/b) - b)/b) - b*x^2*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^2*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^2*sqrt(-(4*a^2 - b^2)/b^2)*log(-((2*I*a*co s(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^2*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^2*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^2*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) + 2*I*b*x*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))*sq...
\[ \int \frac {x^2}{a+b \cos (x) \sin (x)} \, dx=\int \frac {x^{2}}{a + b \sin {\left (x \right )} \cos {\left (x \right )}}\, dx \]
\[ \int \frac {x^2}{a+b \cos (x) \sin (x)} \, dx=\int { \frac {x^{2}}{b \cos \left (x\right ) \sin \left (x\right ) + a} \,d x } \]
\[ \int \frac {x^2}{a+b \cos (x) \sin (x)} \, dx=\int { \frac {x^{2}}{b \cos \left (x\right ) \sin \left (x\right ) + a} \,d x } \]
Timed out. \[ \int \frac {x^2}{a+b \cos (x) \sin (x)} \, dx=\int \frac {x^2}{a+b\,\cos \left (x\right )\,\sin \left (x\right )} \,d x \]