3.7 \(\int \frac {x^2}{a+b \cos ^2(x)} \, dx\)
Optimal. Leaf size=311 \[ -\frac {x \text {Li}_2\left (-\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )}{2 \sqrt {a} \sqrt {a+b}}+\frac {x \text {Li}_2\left (-\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )}{2 \sqrt {a} \sqrt {a+b}}-\frac {i \text {Li}_3\left (-\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )}{4 \sqrt {a} \sqrt {a+b}}+\frac {i \text {Li}_3\left (-\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )}{4 \sqrt {a} \sqrt {a+b}}-\frac {i x^2 \log \left (1+\frac {b e^{2 i x}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 \sqrt {a} \sqrt {a+b}}+\frac {i x^2 \log \left (1+\frac {b e^{2 i x}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 \sqrt {a} \sqrt {a+b}} \]
[Out]
-1/2*I*x^2*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^2*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*x*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)+1/2*x*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*I*pol
ylog(3,-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*I*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)
________________________________________________________________________________________
Rubi [A] time = 0.56, antiderivative size = 311, normalized size of antiderivative = 1.00,
number of steps used = 11, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used
= {4586, 3321, 2264, 2190, 2531, 2282, 6589} \[ -\frac {x \text {PolyLog}\left (2,-\frac {b e^{2 i x}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 \sqrt {a} \sqrt {a+b}}+\frac {x \text {PolyLog}\left (2,-\frac {b e^{2 i x}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 \sqrt {a} \sqrt {a+b}}-\frac {i \text {PolyLog}\left (3,-\frac {b e^{2 i x}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{4 \sqrt {a} \sqrt {a+b}}+\frac {i \text {PolyLog}\left (3,-\frac {b e^{2 i x}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{4 \sqrt {a} \sqrt {a+b}}-\frac {i x^2 \log \left (1+\frac {b e^{2 i x}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 \sqrt {a} \sqrt {a+b}}+\frac {i x^2 \log \left (1+\frac {b e^{2 i x}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 \sqrt {a} \sqrt {a+b}} \]
Antiderivative was successfully verified.
[In]
Int[x^2/(a + b*Cos[x]^2),x]
[Out]
((-I/2)*x^2*Log[1 + (b*E^((2*I)*x))/(2*a + b - 2*Sqrt[a]*Sqrt[a + b])])/(Sqrt[a]*Sqrt[a + b]) + ((I/2)*x^2*Log
[1 + (b*E^((2*I)*x))/(2*a + b + 2*Sqrt[a]*Sqrt[a + b])])/(Sqrt[a]*Sqrt[a + b]) - (x*PolyLog[2, -((b*E^((2*I)*x
))/(2*a + b - 2*Sqrt[a]*Sqrt[a + b]))])/(2*Sqrt[a]*Sqrt[a + b]) + (x*PolyLog[2, -((b*E^((2*I)*x))/(2*a + b + 2
*Sqrt[a]*Sqrt[a + b]))])/(2*Sqrt[a]*Sqrt[a + b]) - ((I/4)*PolyLog[3, -((b*E^((2*I)*x))/(2*a + b - 2*Sqrt[a]*Sq
rt[a + b]))])/(Sqrt[a]*Sqrt[a + b]) + ((I/4)*PolyLog[3, -((b*E^((2*I)*x))/(2*a + b + 2*Sqrt[a]*Sqrt[a + b]))])
/(Sqrt[a]*Sqrt[a + b])
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 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 2282
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]
Rule 2531
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] + Dist[(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]
Rule 3321
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 4586
Int[(Cos[(c_.) + (d_.)*(x_)]^2*(b_.) + (a_))^(n_)*(x_)^(m_.), 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]))
Rule 6589
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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]
Rubi steps
\begin {align*} \int \frac {x^2}{a+b \cos ^2(x)} \, dx &=2 \int \frac {x^2}{2 a+b+b \cos (2 x)} \, dx\\ &=4 \int \frac {e^{2 i x} x^2}{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^2}{-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^2}{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^2 \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^2 \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 x \log \left (1+\frac {2 b e^{2 i x}}{-4 \sqrt {a} \sqrt {a+b}+2 (2 a+b)}\right ) \, dx}{\sqrt {a} \sqrt {a+b}}-\frac {i \int x \log \left (1+\frac {2 b e^{2 i x}}{4 \sqrt {a} \sqrt {a+b}+2 (2 a+b)}\right ) \, dx}{\sqrt {a} \sqrt {a+b}}\\ &=-\frac {i x^2 \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^2 \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 {x \text {Li}_2\left (-\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{2 \sqrt {a} \sqrt {a+b}}+\frac {x \text {Li}_2\left (-\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{2 \sqrt {a} \sqrt {a+b}}+\frac {\int \text {Li}_2\left (-\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 {\int \text {Li}_2\left (-\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^2 \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^2 \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 {x \text {Li}_2\left (-\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{2 \sqrt {a} \sqrt {a+b}}+\frac {x \text {Li}_2\left (-\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{2 \sqrt {a} \sqrt {a+b}}-\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {b x}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{x} \, dx,x,e^{2 i x}\right )}{4 \sqrt {a} \sqrt {a+b}}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {b x}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{x} \, dx,x,e^{2 i x}\right )}{4 \sqrt {a} \sqrt {a+b}}\\ &=-\frac {i x^2 \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^2 \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 {x \text {Li}_2\left (-\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{2 \sqrt {a} \sqrt {a+b}}+\frac {x \text {Li}_2\left (-\frac {b e^{2 i x}}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{2 \sqrt {a} \sqrt {a+b}}-\frac {i \text {Li}_3\left (-\frac {b e^{2 i x}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{4 \sqrt {a} \sqrt {a+b}}+\frac {i \text {Li}_3\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 [A] time = 1.90, size = 239, normalized size = 0.77 \[ -\frac {i \left (-2 i x \text {Li}_2\left (-\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )+2 i x \text {Li}_2\left (-\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )+\text {Li}_3\left (-\frac {b e^{2 i x}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )-\text {Li}_3\left (-\frac {b e^{2 i x}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )+2 x^2 \log \left (1+\frac {b e^{2 i x}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )-2 x^2 \log \left (1+\frac {b e^{2 i x}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )\right )}{4 \sqrt {a} \sqrt {a+b}} \]
Antiderivative was successfully verified.
[In]
Integrate[x^2/(a + b*Cos[x]^2),x]
[Out]
((-1/4*I)*(2*x^2*Log[1 + (b*E^((2*I)*x))/(2*a + b - 2*Sqrt[a]*Sqrt[a + b])] - 2*x^2*Log[1 + (b*E^((2*I)*x))/(2
*a + b + 2*Sqrt[a]*Sqrt[a + b])] - (2*I)*x*PolyLog[2, -((b*E^((2*I)*x))/(2*a + b - 2*Sqrt[a]*Sqrt[a + b]))] +
(2*I)*x*PolyLog[2, -((b*E^((2*I)*x))/(2*a + b + 2*Sqrt[a]*Sqrt[a + b]))] + PolyLog[3, -((b*E^((2*I)*x))/(2*a +
b - 2*Sqrt[a]*Sqrt[a + b]))] - PolyLog[3, -((b*E^((2*I)*x))/(2*a + b + 2*Sqrt[a]*Sqrt[a + b]))]))/(Sqrt[a]*Sq
rt[a + b])
________________________________________________________________________________________
fricas [C] time = 2.12, size = 2452, normalized size = 7.88 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/(a+b*cos(x)^2),x, algorithm="fricas")
[Out]
1/16*(4*I*b*x^2*sqrt((a^2 + a*b)/b^2)*log(1/2*((2*(2*a + b)*cos(x) + (4*I*a + 2*I*b)*sin(x) - 4*(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) + 2*b)/b) - 4*I*b*x^2*sqrt((a^
2 + a*b)/b^2)*log(-1/2*((2*(2*a + b)*cos(x) - (4*I*a + 2*I*b)*sin(x) - 4*(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) - 2*b)/b) - 4*I*b*x^2*sqrt((a^2 + a*b)/b^2)*log(1/2*(
(2*(2*a + b)*cos(x) + (-4*I*a - 2*I*b)*sin(x) - 4*(b*cos(x) - I*b*sin(x))*sqrt((a^2 + a*b)/b^2))*sqrt(-(2*b*sq
rt((a^2 + a*b)/b^2) + 2*a + b)/b) + 2*b)/b) + 4*I*b*x^2*sqrt((a^2 + a*b)/b^2)*log(-1/2*((2*(2*a + b)*cos(x) -
(-4*I*a - 2*I*b)*sin(x) - 4*(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) - 2*b)/b) - 4*I*b*x^2*sqrt((a^2 + a*b)/b^2)*log(1/2*((2*(2*a + b)*cos(x) + (4*I*a + 2*I*b)*sin(x)
+ 4*(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) + 2*b)/b) + 4
*I*b*x^2*sqrt((a^2 + a*b)/b^2)*log(-1/2*((2*(2*a + b)*cos(x) - (4*I*a + 2*I*b)*sin(x) + 4*(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) - 2*b)/b) + 4*I*b*x^2*sqrt((a^2 + a*b
)/b^2)*log(1/2*((2*(2*a + b)*cos(x) + (-4*I*a - 2*I*b)*sin(x) + 4*(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) + 2*b)/b) - 4*I*b*x^2*sqrt((a^2 + a*b)/b^2)*log(-1/2*((2*(2*a
+ b)*cos(x) - (-4*I*a - 2*I*b)*sin(x) + 4*(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) - 2*b)/b) + 8*b*x*sqrt((a^2 + a*b)/b^2)*dilog(-1/2*((2*(2*a + b)*cos(x) + (4*I*a + 2
*I*b)*sin(x) - 4*(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)
+ 2*b)/b + 1) + 8*b*x*sqrt((a^2 + a*b)/b^2)*dilog(1/2*((2*(2*a + b)*cos(x) - (4*I*a + 2*I*b)*sin(x) - 4*(b*co
s(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) - 2*b)/b + 1) + 8*b*x
*sqrt((a^2 + a*b)/b^2)*dilog(-1/2*((2*(2*a + b)*cos(x) + (-4*I*a - 2*I*b)*sin(x) - 4*(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) + 2*b)/b + 1) + 8*b*x*sqrt((a^2 + a*b)/b^
2)*dilog(1/2*((2*(2*a + b)*cos(x) - (-4*I*a - 2*I*b)*sin(x) - 4*(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) - 2*b)/b + 1) - 8*b*x*sqrt((a^2 + a*b)/b^2)*dilog(-1/2*((2*(2*
a + b)*cos(x) + (4*I*a + 2*I*b)*sin(x) + 4*(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) + 2*b)/b + 1) - 8*b*x*sqrt((a^2 + a*b)/b^2)*dilog(1/2*((2*(2*a + b)*cos(x) - (4*I*a
+ 2*I*b)*sin(x) + 4*(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) - 2*b)/b + 1) - 8*b*x*sqrt((a^2 + a*b)/b^2)*dilog(-1/2*((2*(2*a + b)*cos(x) + (-4*I*a - 2*I*b)*sin(x) + 4*(
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) + 2*b)/b + 1) - 8*
b*x*sqrt((a^2 + a*b)/b^2)*dilog(1/2*((2*(2*a + b)*cos(x) - (-4*I*a - 2*I*b)*sin(x) + 4*(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) - 2*b)/b + 1) + 8*I*b*sqrt((a^2 + a*b)/b
^2)*polylog(3, 1/2*(2*(2*a + b)*cos(x) + (4*I*a + 2*I*b)*sin(x) - 4*(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) - 8*I*b*sqrt((a^2 + a*b)/b^2)*polylog(3, -1/2*(2*(2*a +
b)*cos(x) - (4*I*a + 2*I*b)*sin(x) - 4*(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) - 8*I*b*sqrt((a^2 + a*b)/b^2)*polylog(3, 1/2*(2*(2*a + b)*cos(x) + (-4*I*a - 2*I*b)
*sin(x) - 4*(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) +
8*I*b*sqrt((a^2 + a*b)/b^2)*polylog(3, -1/2*(2*(2*a + b)*cos(x) - (-4*I*a - 2*I*b)*sin(x) - 4*(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) - 8*I*b*sqrt((a^2 + a*b)/b^2
)*polylog(3, 1/2*(2*(2*a + b)*cos(x) + (4*I*a + 2*I*b)*sin(x) + 4*(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) + 8*I*b*sqrt((a^2 + a*b)/b^2)*polylog(3, -1/2*(2*(2*a + b)
*cos(x) - (4*I*a + 2*I*b)*sin(x) + 4*(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) + 8*I*b*sqrt((a^2 + a*b)/b^2)*polylog(3, 1/2*(2*(2*a + b)*cos(x) + (-4*I*a - 2*I*b)*sin
(x) + 4*(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) - 8*I*
b*sqrt((a^2 + a*b)/b^2)*polylog(3, -1/2*(2*(2*a + b)*cos(x) - (-4*I*a - 2*I*b)*sin(x) + 4*(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))/(a^2 + a*b)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{b \cos \relax (x)^{2} + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/(a+b*cos(x)^2),x, algorithm="giac")
[Out]
integrate(x^2/(b*cos(x)^2 + a), x)
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maple [B] time = 0.09, size = 710, normalized size = 2.28 \[ -\frac {x \polylog \left (2, \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 )}-2 a -b}-\frac {i b \polylog \left (3, \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 )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {x \polylog \left (2, \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 {i x^{2} \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 )}-2 a -b}-\frac {a x \polylog \left (2, \frac {b \,{\mathrm e}^{2 i x}}{-2 \sqrt {a \left (a +b \right )}-2 a -b}\right )}{\sqrt {a \left (a +b \right )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {i a \,x^{2} \ln \left (1-\frac {b \,{\mathrm e}^{2 i x}}{-2 \sqrt {a \left (a +b \right )}-2 a -b}\right )}{\sqrt {a \left (a +b \right )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {2 x^{3}}{3 \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {i a \polylog \left (3, \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 )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {b \,x^{3}}{3 \sqrt {a \left (a +b \right )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {i \polylog \left (3, \frac {b \,{\mathrm e}^{2 i x}}{-2 \sqrt {a \left (a +b \right )}-2 a -b}\right )}{2 \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {b x \polylog \left (2, \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 )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {i x^{2} \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 {2 a \,x^{3}}{3 \sqrt {a \left (a +b \right )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {i b \,x^{2} \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 )}\, \left (-2 \sqrt {a \left (a +b \right )}-2 a -b \right )}-\frac {x^{3}}{3 \sqrt {a \left (a +b \right )}}-\frac {i \polylog \left (3, \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 )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2/(a+b*cos(x)^2),x)
[Out]
-1/(-2*(a*(a+b))^(1/2)-2*a-b)*x*polylog(2,b*exp(2*I*x)/(-2*(a*(a+b))^(1/2)-2*a-b))-1/2*I/(a*(a+b))^(1/2)/(-2*(
a*(a+b))^(1/2)-2*a-b)*a*polylog(3,b*exp(2*I*x)/(-2*(a*(a+b))^(1/2)-2*a-b))-1/2/(a*(a+b))^(1/2)*x*polylog(2,b*e
xp(2*I*x)/(2*(a*(a+b))^(1/2)-2*a-b))-1/4*I/(a*(a+b))^(1/2)/(-2*(a*(a+b))^(1/2)-2*a-b)*b*polylog(3,b*exp(2*I*x)
/(-2*(a*(a+b))^(1/2)-2*a-b))-1/(a*(a+b))^(1/2)/(-2*(a*(a+b))^(1/2)-2*a-b)*a*x*polylog(2,b*exp(2*I*x)/(-2*(a*(a
+b))^(1/2)-2*a-b))-1/2*I/(-2*(a*(a+b))^(1/2)-2*a-b)*polylog(3,b*exp(2*I*x)/(-2*(a*(a+b))^(1/2)-2*a-b))-2/3/(-2
*(a*(a+b))^(1/2)-2*a-b)*x^3-I/(-2*(a*(a+b))^(1/2)-2*a-b)*x^2*ln(1-b*exp(2*I*x)/(-2*(a*(a+b))^(1/2)-2*a-b))-1/3
/(a*(a+b))^(1/2)/(-2*(a*(a+b))^(1/2)-2*a-b)*b*x^3-I/(a*(a+b))^(1/2)/(-2*(a*(a+b))^(1/2)-2*a-b)*a*x^2*ln(1-b*ex
p(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)*b*x*polylog(2,b*exp(2*I*x)
/(-2*(a*(a+b))^(1/2)-2*a-b))-1/2*I/(a*(a+b))^(1/2)*x^2*ln(1-b*exp(2*I*x)/(2*(a*(a+b))^(1/2)-2*a-b))-2/3/(a*(a+
b))^(1/2)/(-2*(a*(a+b))^(1/2)-2*a-b)*a*x^3-1/2*I/(a*(a+b))^(1/2)/(-2*(a*(a+b))^(1/2)-2*a-b)*b*x^2*ln(1-b*exp(2
*I*x)/(-2*(a*(a+b))^(1/2)-2*a-b))-1/3/(a*(a+b))^(1/2)*x^3-1/4*I/(a*(a+b))^(1/2)*polylog(3,b*exp(2*I*x)/(2*(a*(
a+b))^(1/2)-2*a-b))
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{b \cos \relax (x)^{2} + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/(a+b*cos(x)^2),x, algorithm="maxima")
[Out]
integrate(x^2/(b*cos(x)^2 + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2}{b\,{\cos \relax (x)}^2+a} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2/(a + b*cos(x)^2),x)
[Out]
int(x^2/(a + b*cos(x)^2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{a + b \cos ^{2}{\relax (x )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2/(a+b*cos(x)**2),x)
[Out]
Integral(x**2/(a + b*cos(x)**2), x)
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