Integrand size = 16, antiderivative size = 329 \[ \int \frac {x^2}{a+b \cos (c+d x)} \, dx=-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}+\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}-\frac {2 x \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}+\frac {2 x \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}-\frac {2 i \operatorname {PolyLog}\left (3,-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^3}+\frac {2 i \operatorname {PolyLog}\left (3,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^3} \]
-I*x^2*ln(1+b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/d/(a^2-b^2)^(1/2)+I*x^2* ln(1+b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/d/(a^2-b^2)^(1/2)-2*x*polylog(2 ,-b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/d^2/(a^2-b^2)^(1/2)+2*x*polylog(2, -b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/d^2/(a^2-b^2)^(1/2)-2*I*polylog(3,- b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/d^3/(a^2-b^2)^(1/2)+2*I*polylog(3,-b *exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/d^3/(a^2-b^2)^(1/2)
Time = 0.61 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.78 \[ \int \frac {x^2}{a+b \cos (c+d x)} \, dx=\frac {-2 d x \operatorname {PolyLog}\left (2,\frac {b e^{i (c+d x)}}{-a+\sqrt {a^2-b^2}}\right )-i \left (d^2 x^2 \log \left (1-\frac {b e^{i (c+d x)}}{-a+\sqrt {a^2-b^2}}\right )-d^2 x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )+2 i d x \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )+2 \operatorname {PolyLog}\left (3,\frac {b e^{i (c+d x)}}{-a+\sqrt {a^2-b^2}}\right )-2 \operatorname {PolyLog}\left (3,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )\right )}{\sqrt {a^2-b^2} d^3} \]
(-2*d*x*PolyLog[2, (b*E^(I*(c + d*x)))/(-a + Sqrt[a^2 - b^2])] - I*(d^2*x^ 2*Log[1 - (b*E^(I*(c + d*x)))/(-a + Sqrt[a^2 - b^2])] - d^2*x^2*Log[1 + (b *E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])] + (2*I)*d*x*PolyLog[2, -((b*E^(I* (c + d*x)))/(a + Sqrt[a^2 - b^2]))] + 2*PolyLog[3, (b*E^(I*(c + d*x)))/(-a + Sqrt[a^2 - b^2])] - 2*PolyLog[3, -((b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2]))]))/(Sqrt[a^2 - b^2]*d^3)
Time = 1.24 (sec) , antiderivative size = 319, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3802, 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.
| \(\displaystyle \int \frac {x^2}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {x^2}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3802 |
| \(\displaystyle 2 \int \frac {e^{i (c+d x)} x^2}{2 e^{i (c+d x)} a+b e^{2 i (c+d x)}+b}dx\) |
| \(\Big \downarrow \) 2694 |
| \(\displaystyle 2 \left (\frac {b \int \frac {e^{i (c+d x)} x^2}{2 \left (a+b e^{i (c+d x)}-\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}-\frac {b \int \frac {e^{i (c+d x)} x^2}{2 \left (a+b e^{i (c+d x)}+\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \left (\frac {b \int \frac {e^{i (c+d x)} x^2}{a+b e^{i (c+d x)}-\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}-\frac {b \int \frac {e^{i (c+d x)} x^2}{a+b e^{i (c+d x)}+\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}\right )\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle 2 \left (\frac {b \left (\frac {2 i \int x \log \left (\frac {e^{i (c+d x)} b}{a-\sqrt {a^2-b^2}}+1\right )dx}{b d}-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {b \left (\frac {2 i \int x \log \left (\frac {e^{i (c+d x)} b}{a+\sqrt {a^2-b^2}}+1\right )dx}{b d}-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle 2 \left (\frac {b \left (\frac {2 i \left (\frac {i x \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {i \int \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {b \left (\frac {2 i \left (\frac {i x \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {i \int \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle 2 \left (\frac {b \left (\frac {2 i \left (\frac {i x \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {\int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}\right )}{b d}-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {b \left (\frac {2 i \left (\frac {i x \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {\int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}\right )}{b d}-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle 2 \left (\frac {b \left (\frac {2 i \left (\frac {i x \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {\operatorname {PolyLog}\left (3,-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {b \left (\frac {2 i \left (\frac {i x \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {\operatorname {PolyLog}\left (3,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )\) |
2*((b*(((-I)*x^2*Log[1 + (b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d) + ((2*I)*((I*x*PolyLog[2, -((b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2]))])/ d - PolyLog[3, -((b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2]))]/d^2))/(b*d))) /(2*Sqrt[a^2 - b^2]) - (b*(((-I)*x^2*Log[1 + (b*E^(I*(c + d*x)))/(a + Sqrt [a^2 - b^2])])/(b*d) + ((2*I)*((I*x*PolyLog[2, -((b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2]))])/d - PolyLog[3, -((b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b ^2]))]/d^2))/(b*d)))/(2*Sqrt[a^2 - b^2]))
3.2.86.3.1 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[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]
\[\int \frac {x^{2}}{a +\cos \left (d x +c \right ) b}d x\]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1263 vs. \(2 (283) = 566\).
Time = 0.43 (sec) , antiderivative size = 1263, normalized size of antiderivative = 3.84 \[ \int \frac {x^2}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]
-1/2*(2*b*d*x*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(d*x + c) + I*a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1 ) - 2*b*d*x*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(d*x + c) + I*a*sin(d*x + c ) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) + 2*b*d*x*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(d*x + c) - I*a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) - 2*b*d*x*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(d*x + c) - I*a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) - I* b*c^2*sqrt((a^2 - b^2)/b^2)*log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2* b*sqrt((a^2 - b^2)/b^2) + 2*a) + I*b*c^2*sqrt((a^2 - b^2)/b^2)*log(2*b*cos (d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) - I*b*c^ 2*sqrt((a^2 - b^2)/b^2)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*s qrt((a^2 - b^2)/b^2) - 2*a) + I*b*c^2*sqrt((a^2 - b^2)/b^2)*log(-2*b*cos(d *x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) - 2*a) + (I*b*d^2 *x^2 - I*b*c^2)*sqrt((a^2 - b^2)/b^2)*log((a*cos(d*x + c) + I*a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b) + ( -I*b*d^2*x^2 + I*b*c^2)*sqrt((a^2 - b^2)/b^2)*log((a*cos(d*x + c) + I*a*si n(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b )/b) + (-I*b*d^2*x^2 + I*b*c^2)*sqrt((a^2 - b^2)/b^2)*log((a*cos(d*x + c) - I*a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^...
\[ \int \frac {x^2}{a+b \cos (c+d x)} \, dx=\int \frac {x^{2}}{a + b \cos {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {x^2}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
\[ \int \frac {x^2}{a+b \cos (c+d x)} \, dx=\int { \frac {x^{2}}{b \cos \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {x^2}{a+b \cos (c+d x)} \, dx=\int \frac {x^2}{a+b\,\cos \left (c+d\,x\right )} \,d x \]