Integrand size = 16, antiderivative size = 327 \[ \int \frac {x}{\left (a+b \cos ^2(c+d x)\right )^2} \, dx=-\frac {i (2 a+b) x \log \left (1+\frac {b e^{2 i (c+d x)}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{4 a^{3/2} (a+b)^{3/2} d}+\frac {i (2 a+b) x \log \left (1+\frac {b e^{2 i (c+d x)}}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{4 a^{3/2} (a+b)^{3/2} d}-\frac {\log (2 a+b+b \cos (2 c+2 d x))}{4 a (a+b) d^2}-\frac {(2 a+b) \operatorname {PolyLog}\left (2,-\frac {b e^{2 i (c+d x)}}{2 a+b-2 \sqrt {a} \sqrt {a+b}}\right )}{8 a^{3/2} (a+b)^{3/2} d^2}+\frac {(2 a+b) \operatorname {PolyLog}\left (2,-\frac {b e^{2 i (c+d x)}}{2 a+b+2 \sqrt {a} \sqrt {a+b}}\right )}{8 a^{3/2} (a+b)^{3/2} d^2}-\frac {b x \sin (2 c+2 d x)}{2 a (a+b) d (2 a+b+b \cos (2 c+2 d x))} \] Output:
-1/4*I*(2*a+b)*x*ln(1+b*exp(2*I*(d*x+c))/(2*a+b-2*a^(1/2)*(a+b)^(1/2)))/a^ (3/2)/(a+b)^(3/2)/d+1/4*I*(2*a+b)*x*ln(1+b*exp(2*I*(d*x+c))/(2*a+b+2*a^(1/ 2)*(a+b)^(1/2)))/a^(3/2)/(a+b)^(3/2)/d-1/4*ln(2*a+b+b*cos(2*d*x+2*c))/a/(a +b)/d^2-1/8*(2*a+b)*polylog(2,-b*exp(2*I*(d*x+c))/(2*a+b-2*a^(1/2)*(a+b)^( 1/2)))/a^(3/2)/(a+b)^(3/2)/d^2+1/8*(2*a+b)*polylog(2,-b*exp(2*I*(d*x+c))/( 2*a+b+2*a^(1/2)*(a+b)^(1/2)))/a^(3/2)/(a+b)^(3/2)/d^2-1/2*b*x*sin(2*d*x+2* c)/a/(a+b)/d/(2*a+b+b*cos(2*d*x+2*c))
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(825\) vs. \(2(327)=654\).
Time = 14.52 (sec) , antiderivative size = 825, normalized size of antiderivative = 2.52 \[ \int \frac {x}{\left (a+b \cos ^2(c+d x)\right )^2} \, dx=\frac {b c \sin (2 (c+d x))-b (c+d x) \sin (2 (c+d x))}{2 a (a+b) d^2 (2 a+b+b \cos (2 (c+d x)))}+\frac {\cos ^2(c+d x) \left (-\frac {4 (2 a+b) c \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+b}}\right )}{\sqrt {a} \sqrt {a+b}}+2 \log \left (\sec ^2(c+d x)\right )-2 \log \left (a+b+a \tan ^2(c+d x)\right )-\frac {i (2 a+b) \left (\log (1-i \tan (c+d x)) \log \left (\frac {\sqrt {-a-b}+\sqrt {a} \tan (c+d x)}{-i \sqrt {a}+\sqrt {-a-b}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {a} (1-i \tan (c+d x))}{\sqrt {a}+i \sqrt {-a-b}}\right )\right )}{\sqrt {a} \sqrt {-a-b}}+\frac {i (2 a+b) \left (\log (1+i \tan (c+d x)) \log \left (\frac {\sqrt {-a-b}+\sqrt {a} \tan (c+d x)}{i \sqrt {a}+\sqrt {-a-b}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {a} (1+i \tan (c+d x))}{\sqrt {a}-i \sqrt {-a-b}}\right )\right )}{\sqrt {a} \sqrt {-a-b}}-\frac {i (2 a+b) \left (\log (1+i \tan (c+d x)) \log \left (\frac {\sqrt {-a-b}-\sqrt {a} \tan (c+d x)}{-i \sqrt {a}+\sqrt {-a-b}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {a} (1+i \tan (c+d x))}{\sqrt {a}+i \sqrt {-a-b}}\right )\right )}{\sqrt {a} \sqrt {-a-b}}+\frac {i (2 a+b) \left (\log (1-i \tan (c+d x)) \log \left (\frac {\sqrt {-a-b}-\sqrt {a} \tan (c+d x)}{i \sqrt {a}+\sqrt {-a-b}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {a} (i+\tan (c+d x))}{i \sqrt {a}+\sqrt {-a-b}}\right )\right )}{\sqrt {a} \sqrt {-a-b}}\right ) (2 (2 a+b) d x+b \sin (2 (c+d x))) \left (-\sqrt {-a-b}+\sqrt {a} \tan (c+d x)\right ) \left (\sqrt {-a-b}+\sqrt {a} \tan (c+d x)\right )}{4 a (a+b) d^2 (2 a+b+b \cos (2 (c+d x))) (-((2 a+b) (2 c-i \log (1-i \tan (c+d x))+i \log (1+i \tan (c+d x))))+b \sin (2 (c+d x)))} \] Input:
Integrate[x/(a + b*Cos[c + d*x]^2)^2,x]
Output:
(b*c*Sin[2*(c + d*x)] - b*(c + d*x)*Sin[2*(c + d*x)])/(2*a*(a + b)*d^2*(2* a + b + b*Cos[2*(c + d*x)])) + (Cos[c + d*x]^2*((-4*(2*a + b)*c*ArcTan[(Sq rt[a]*Tan[c + d*x])/Sqrt[a + b]])/(Sqrt[a]*Sqrt[a + b]) + 2*Log[Sec[c + d* x]^2] - 2*Log[a + b + a*Tan[c + d*x]^2] - (I*(2*a + b)*(Log[1 - I*Tan[c + d*x]]*Log[(Sqrt[-a - b] + Sqrt[a]*Tan[c + d*x])/((-I)*Sqrt[a] + Sqrt[-a - b])] + PolyLog[2, (Sqrt[a]*(1 - I*Tan[c + d*x]))/(Sqrt[a] + I*Sqrt[-a - b] )]))/(Sqrt[a]*Sqrt[-a - b]) + (I*(2*a + b)*(Log[1 + I*Tan[c + d*x]]*Log[(S qrt[-a - b] + Sqrt[a]*Tan[c + d*x])/(I*Sqrt[a] + Sqrt[-a - b])] + PolyLog[ 2, (Sqrt[a]*(1 + I*Tan[c + d*x]))/(Sqrt[a] - I*Sqrt[-a - b])]))/(Sqrt[a]*S qrt[-a - b]) - (I*(2*a + b)*(Log[1 + I*Tan[c + d*x]]*Log[(Sqrt[-a - b] - S qrt[a]*Tan[c + d*x])/((-I)*Sqrt[a] + Sqrt[-a - b])] + PolyLog[2, (Sqrt[a]* (1 + I*Tan[c + d*x]))/(Sqrt[a] + I*Sqrt[-a - b])]))/(Sqrt[a]*Sqrt[-a - b]) + (I*(2*a + b)*(Log[1 - I*Tan[c + d*x]]*Log[(Sqrt[-a - b] - Sqrt[a]*Tan[c + d*x])/(I*Sqrt[a] + Sqrt[-a - b])] + PolyLog[2, (Sqrt[a]*(I + Tan[c + d* x]))/(I*Sqrt[a] + Sqrt[-a - b])]))/(Sqrt[a]*Sqrt[-a - b]))*(2*(2*a + b)*d* x + b*Sin[2*(c + d*x)])*(-Sqrt[-a - b] + Sqrt[a]*Tan[c + d*x])*(Sqrt[-a - b] + Sqrt[a]*Tan[c + d*x]))/(4*a*(a + b)*d^2*(2*a + b + b*Cos[2*(c + d*x)] )*(-((2*a + b)*(2*c - I*Log[1 - I*Tan[c + d*x]] + I*Log[1 + I*Tan[c + d*x] ])) + b*Sin[2*(c + d*x)]))
Time = 1.19 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.812, Rules used = {5097, 3042, 3805, 25, 3042, 3147, 16, 3802, 2694, 27, 2620, 2715, 2838}
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.
| \(\displaystyle \int \frac {x}{\left (a+b \cos ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 5097 |
| \(\displaystyle 4 \int \frac {x}{(2 a+b+b \cos (2 c+2 d x))^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 4 \int \frac {x}{\left (2 a+b+b \sin \left (2 c+2 d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3805 |
| \(\displaystyle 4 \left (\frac {(2 a+b) \int \frac {x}{2 a+b+b \cos (2 c+2 d x)}dx}{4 a (a+b)}-\frac {b \int -\frac {\sin (2 c+2 d x)}{2 a+b+b \cos (2 c+2 d x)}dx}{8 a d (a+b)}-\frac {b x \sin (2 c+2 d x)}{8 a d (a+b) (2 a+b \cos (2 c+2 d x)+b)}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 4 \left (\frac {(2 a+b) \int \frac {x}{2 a+b+b \cos (2 c+2 d x)}dx}{4 a (a+b)}+\frac {b \int \frac {\sin (2 c+2 d x)}{2 a+b+b \cos (2 c+2 d x)}dx}{8 a d (a+b)}-\frac {b x \sin (2 c+2 d x)}{8 a d (a+b) (2 a+b \cos (2 c+2 d x)+b)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 4 \left (\frac {(2 a+b) \int \frac {x}{2 a+b+b \sin \left (2 c+2 d x+\frac {\pi }{2}\right )}dx}{4 a (a+b)}+\frac {b \int \frac {\cos \left (2 c+2 d x-\frac {\pi }{2}\right )}{2 a+b-b \sin \left (2 c+2 d x-\frac {\pi }{2}\right )}dx}{8 a d (a+b)}-\frac {b x \sin (2 c+2 d x)}{8 a d (a+b) (2 a+b \cos (2 c+2 d x)+b)}\right )\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle 4 \left (-\frac {\int \frac {1}{2 a+b+b \cos (2 c+2 d x)}d(b \cos (2 c+2 d x))}{16 a d^2 (a+b)}+\frac {(2 a+b) \int \frac {x}{2 a+b+b \sin \left (2 c+2 d x+\frac {\pi }{2}\right )}dx}{4 a (a+b)}-\frac {b x \sin (2 c+2 d x)}{8 a d (a+b) (2 a+b \cos (2 c+2 d x)+b)}\right )\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle 4 \left (\frac {(2 a+b) \int \frac {x}{2 a+b+b \sin \left (2 c+2 d x+\frac {\pi }{2}\right )}dx}{4 a (a+b)}-\frac {\log (2 a+b \cos (2 c+2 d x)+b)}{16 a d^2 (a+b)}-\frac {b x \sin (2 c+2 d x)}{8 a d (a+b) (2 a+b \cos (2 c+2 d x)+b)}\right )\) |
| \(\Big \downarrow \) 3802 |
| \(\displaystyle 4 \left (\frac {(2 a+b) \int \frac {e^{2 i (c+d x)} x}{e^{4 i (c+d x)} b+b+2 (2 a+b) e^{2 i (c+d x)}}dx}{2 a (a+b)}-\frac {\log (2 a+b \cos (2 c+2 d x)+b)}{16 a d^2 (a+b)}-\frac {b x \sin (2 c+2 d x)}{8 a d (a+b) (2 a+b \cos (2 c+2 d x)+b)}\right )\) |
| \(\Big \downarrow \) 2694 |
| \(\displaystyle 4 \left (\frac {(2 a+b) \left (\frac {b \int \frac {e^{2 i (c+d x)} x}{2 \left (2 a-2 \sqrt {a+b} \sqrt {a}+b e^{2 i (c+d x)}+b\right )}dx}{2 \sqrt {a} \sqrt {a+b}}-\frac {b \int \frac {e^{2 i (c+d x)} x}{2 \left (2 a+2 \sqrt {a+b} \sqrt {a}+b e^{2 i (c+d x)}+b\right )}dx}{2 \sqrt {a} \sqrt {a+b}}\right )}{2 a (a+b)}-\frac {\log (2 a+b \cos (2 c+2 d x)+b)}{16 a d^2 (a+b)}-\frac {b x \sin (2 c+2 d x)}{8 a d (a+b) (2 a+b \cos (2 c+2 d x)+b)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \left (\frac {(2 a+b) \left (\frac {b \int \frac {e^{2 i (c+d x)} x}{2 a-2 \sqrt {a+b} \sqrt {a}+b e^{2 i (c+d x)}+b}dx}{4 \sqrt {a} \sqrt {a+b}}-\frac {b \int \frac {e^{2 i (c+d x)} x}{2 a+2 \sqrt {a+b} \sqrt {a}+b e^{2 i (c+d x)}+b}dx}{4 \sqrt {a} \sqrt {a+b}}\right )}{2 a (a+b)}-\frac {\log (2 a+b \cos (2 c+2 d x)+b)}{16 a d^2 (a+b)}-\frac {b x \sin (2 c+2 d x)}{8 a d (a+b) (2 a+b \cos (2 c+2 d x)+b)}\right )\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle 4 \left (\frac {(2 a+b) \left (\frac {b \left (\frac {i \int \log \left (\frac {e^{2 i (c+d x)} b}{2 a-2 \sqrt {a+b} \sqrt {a}+b}+1\right )dx}{2 b d}-\frac {i x \log \left (1+\frac {b e^{2 i (c+d x)}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b d}\right )}{4 \sqrt {a} \sqrt {a+b}}-\frac {b \left (\frac {i \int \log \left (\frac {e^{2 i (c+d x)} b}{2 a+2 \sqrt {a+b} \sqrt {a}+b}+1\right )dx}{2 b d}-\frac {i x \log \left (1+\frac {b e^{2 i (c+d x)}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b d}\right )}{4 \sqrt {a} \sqrt {a+b}}\right )}{2 a (a+b)}-\frac {\log (2 a+b \cos (2 c+2 d x)+b)}{16 a d^2 (a+b)}-\frac {b x \sin (2 c+2 d x)}{8 a d (a+b) (2 a+b \cos (2 c+2 d x)+b)}\right )\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle 4 \left (\frac {(2 a+b) \left (\frac {b \left (\frac {\int e^{-2 i (c+d x)} \log \left (\frac {e^{2 i (c+d x)} b}{2 a-2 \sqrt {a+b} \sqrt {a}+b}+1\right )de^{2 i (c+d x)}}{4 b d^2}-\frac {i x \log \left (1+\frac {b e^{2 i (c+d x)}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b d}\right )}{4 \sqrt {a} \sqrt {a+b}}-\frac {b \left (\frac {\int e^{-2 i (c+d x)} \log \left (\frac {e^{2 i (c+d x)} b}{2 a+2 \sqrt {a+b} \sqrt {a}+b}+1\right )de^{2 i (c+d x)}}{4 b d^2}-\frac {i x \log \left (1+\frac {b e^{2 i (c+d x)}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b d}\right )}{4 \sqrt {a} \sqrt {a+b}}\right )}{2 a (a+b)}-\frac {\log (2 a+b \cos (2 c+2 d x)+b)}{16 a d^2 (a+b)}-\frac {b x \sin (2 c+2 d x)}{8 a d (a+b) (2 a+b \cos (2 c+2 d x)+b)}\right )\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle 4 \left (\frac {(2 a+b) \left (\frac {b \left (-\frac {\operatorname {PolyLog}\left (2,-\frac {b e^{2 i (c+d x)}}{2 a-2 \sqrt {a+b} \sqrt {a}+b}\right )}{4 b d^2}-\frac {i x \log \left (1+\frac {b e^{2 i (c+d x)}}{-2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b d}\right )}{4 \sqrt {a} \sqrt {a+b}}-\frac {b \left (-\frac {\operatorname {PolyLog}\left (2,-\frac {b e^{2 i (c+d x)}}{2 a+2 \sqrt {a+b} \sqrt {a}+b}\right )}{4 b d^2}-\frac {i x \log \left (1+\frac {b e^{2 i (c+d x)}}{2 \sqrt {a} \sqrt {a+b}+2 a+b}\right )}{2 b d}\right )}{4 \sqrt {a} \sqrt {a+b}}\right )}{2 a (a+b)}-\frac {\log (2 a+b \cos (2 c+2 d x)+b)}{16 a d^2 (a+b)}-\frac {b x \sin (2 c+2 d x)}{8 a d (a+b) (2 a+b \cos (2 c+2 d x)+b)}\right )\) |
Input:
Int[x/(a + b*Cos[c + d*x]^2)^2,x]
Output:
4*(-1/16*Log[2*a + b + b*Cos[2*c + 2*d*x]]/(a*(a + b)*d^2) + ((2*a + b)*(( b*(((-1/2*I)*x*Log[1 + (b*E^((2*I)*(c + d*x)))/(2*a + b - 2*Sqrt[a]*Sqrt[a + b])])/(b*d) - PolyLog[2, -((b*E^((2*I)*(c + d*x)))/(2*a + b - 2*Sqrt[a] *Sqrt[a + b]))]/(4*b*d^2)))/(4*Sqrt[a]*Sqrt[a + b]) - (b*(((-1/2*I)*x*Log[ 1 + (b*E^((2*I)*(c + d*x)))/(2*a + b + 2*Sqrt[a]*Sqrt[a + b])])/(b*d) - Po lyLog[2, -((b*E^((2*I)*(c + d*x)))/(2*a + b + 2*Sqrt[a]*Sqrt[a + b]))]/(4* b*d^2)))/(4*Sqrt[a]*Sqrt[a + b])))/(2*a*(a + b)) - (b*x*Sin[2*c + 2*d*x])/ (8*a*(a + b)*d*(2*a + b + b*Cos[2*c + 2*d*x])))
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[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]
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]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 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[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_ Symbol] :> Simp[b*(c + d*x)^m*(Cos[e + f*x]/(f*(a^2 - b^2)*(a + b*Sin[e + f *x]))), x] + (Simp[a/(a^2 - b^2) Int[(c + d*x)^m/(a + b*Sin[e + f*x]), x] , x] - Simp[b*d*(m/(f*(a^2 - b^2))) Int[(c + d*x)^(m - 1)*(Cos[e + f*x]/( a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]
Int[(Cos[(c_.) + (d_.)*(x_)]^2*(b_.) + (a_))^(n_)*(x_)^(m_.), 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]))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2269 vs. \(2 (273 ) = 546\).
Time = 0.86 (sec) , antiderivative size = 2270, normalized size of antiderivative = 6.94
Input:
int(x/(a+cos(d*x+c)^2*b)^2,x,method=_RETURNVERBOSE)
Output:
-1/d^2/(a+b)/(-2*(a*(a+b))^(1/2)-2*a-b)*c^2-1/2/d^2/(a+b)/(-2*(a*(a+b))^(1 /2)-2*a-b)*polylog(2,b*exp(2*I*(d*x+c))/(-2*(a*(a+b))^(1/2)-2*a-b))-1/4/d^ 2/(a+b)/(a*(a+b))^(1/2)*polylog(2,b*exp(2*I*(d*x+c))/(2*(a*(a+b))^(1/2)-2* a-b))-1/2/d/(a+b)/a*b/(a*(a+b))^(1/2)*c*x-1/4/d^2/(a+b)/a*ln(b*exp(4*I*(d* x+c))+4*a*exp(2*I*(d*x+c))+2*b*exp(2*I*(d*x+c))+b)-1/2/d^2/(a+b)/(a*(a+b)) ^(1/2)*c^2-I*x*(2*a*exp(2*I*(d*x+c))+b*exp(2*I*(d*x+c))+b)/a/(a+b)/d/(b*ex p(4*I*(d*x+c))+4*a*exp(2*I*(d*x+c))+2*b*exp(2*I*(d*x+c))+b)-1/4/d^2/(a+b)/ a*b^2/(a*(a+b))^(1/2)/(-2*(a*(a+b))^(1/2)-2*a-b)*c^2-2/d/(a+b)/(a*(a+b))^( 1/2)/(-2*(a*(a+b))^(1/2)-2*a-b)*c*x*b-1/8/d^2/(a+b)/a*b^2/(a*(a+b))^(1/2)/ (-2*(a*(a+b))^(1/2)-2*a-b)*polylog(2,b*exp(2*I*(d*x+c))/(-2*(a*(a+b))^(1/2 )-2*a-b))-2/d/(a+b)*a/(a*(a+b))^(1/2)/(-2*(a*(a+b))^(1/2)-2*a-b)*c*x-1/d/( a+b)/a*b/(-2*(a*(a+b))^(1/2)-2*a-b)*c*x-2/d/(a+b)/(-2*(a*(a+b))^(1/2)-2*a- b)*c*x-1/2/(a+b)/a*b/(-2*(a*(a+b))^(1/2)-2*a-b)*x^2-1/(a+b)*a/(a*(a+b))^(1 /2)/(-2*(a*(a+b))^(1/2)-2*a-b)*x^2-1/(a+b)/(a*(a+b))^(1/2)/(-2*(a*(a+b))^( 1/2)-2*a-b)*x^2*b+1/d^2/(a+b)/a*ln(exp(I*(d*x+c)))-1/2/d/(a+b)/a*b^2/(a*(a +b))^(1/2)/(-2*(a*(a+b))^(1/2)-2*a-b)*c*x-1/4*I/d/(a+b)/a*b/(a*(a+b))^(1/2 )*ln(1-b*exp(2*I*(d*x+c))/(2*(a*(a+b))^(1/2)-2*a-b))*x-1/2*I/d/(a+b)/a*b/( -2*(a*(a+b))^(1/2)-2*a-b)*ln(1-b*exp(2*I*(d*x+c))/(-2*(a*(a+b))^(1/2)-2*a- b))*x-I/d^2/(a+b)*a/(a*(a+b))^(1/2)/(-2*(a*(a+b))^(1/2)-2*a-b)*ln(1-b*exp( 2*I*(d*x+c))/(-2*(a*(a+b))^(1/2)-2*a-b))*c-I/d/(a+b)*a/(a*(a+b))^(1/2)/...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3759 vs. \(2 (277) = 554\).
Time = 1.69 (sec) , antiderivative size = 3759, normalized size of antiderivative = 11.50 \[ \int \frac {x}{\left (a+b \cos ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(x/(a+b*cos(d*x+c)^2)^2,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {x}{\left (a+b \cos ^2(c+d x)\right )^2} \, dx=\text {Timed out} \] Input:
integrate(x/(a+b*cos(d*x+c)**2)**2,x)
Output:
Timed out
Timed out. \[ \int \frac {x}{\left (a+b \cos ^2(c+d x)\right )^2} \, dx=\text {Timed out} \] Input:
integrate(x/(a+b*cos(d*x+c)^2)^2,x, algorithm="maxima")
Output:
Timed out
\[ \int \frac {x}{\left (a+b \cos ^2(c+d x)\right )^2} \, dx=\int { \frac {x}{{\left (b \cos \left (d x + c\right )^{2} + a\right )}^{2}} \,d x } \] Input:
integrate(x/(a+b*cos(d*x+c)^2)^2,x, algorithm="giac")
Output:
integrate(x/(b*cos(d*x + c)^2 + a)^2, x)
Timed out. \[ \int \frac {x}{\left (a+b \cos ^2(c+d x)\right )^2} \, dx=\int \frac {x}{{\left (b\,{\cos \left (c+d\,x\right )}^2+a\right )}^2} \,d x \] Input:
int(x/(a + b*cos(c + d*x)^2)^2,x)
Output:
int(x/(a + b*cos(c + d*x)^2)^2, x)
\[ \int \frac {x}{\left (a+b \cos ^2(c+d x)\right )^2} \, dx=\int \frac {x}{\cos \left (d x +c \right )^{4} b^{2}+2 \cos \left (d x +c \right )^{2} a b +a^{2}}d x \] Input:
int(x/(a+b*cos(d*x+c)^2)^2,x)
Output:
int(x/(cos(c + d*x)**4*b**2 + 2*cos(c + d*x)**2*a*b + a**2),x)