Integrand size = 34, antiderivative size = 267 \[ \int \frac {x \sec ^2(c+d x)}{a+c \sec ^2(c+d x)+b \tan ^2(c+d x)} \, dx=-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 c-2 \sqrt {a+c} \sqrt {b+c}}\right )}{2 \sqrt {a+c} \sqrt {b+c} d}+\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 \left (c+\sqrt {a+c} \sqrt {b+c}\right )}\right )}{2 \sqrt {a+c} \sqrt {b+c} d}-\frac {\operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 c-2 \sqrt {a+c} \sqrt {b+c}}\right )}{4 \sqrt {a+c} \sqrt {b+c} d^2}+\frac {\operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 \left (c+\sqrt {a+c} \sqrt {b+c}\right )}\right )}{4 \sqrt {a+c} \sqrt {b+c} d^2} \]
-1/2*I*x*ln(1+(a-b)*exp(2*I*(d*x+c))/(a+b+2*c-2*(a+c)^(1/2)*(b+c)^(1/2)))/ d/(a+c)^(1/2)/(b+c)^(1/2)+1/2*I*x*ln(1+(a-b)*exp(2*I*(d*x+c))/(a+b+2*c+2*( a+c)^(1/2)*(b+c)^(1/2)))/d/(a+c)^(1/2)/(b+c)^(1/2)-1/4*polylog(2,-(a-b)*ex p(2*I*(d*x+c))/(a+b+2*c-2*(a+c)^(1/2)*(b+c)^(1/2)))/d^2/(a+c)^(1/2)/(b+c)^ (1/2)+1/4*polylog(2,-(a-b)*exp(2*I*(d*x+c))/(a+b+2*c+2*(a+c)^(1/2)*(b+c)^( 1/2)))/d^2/(a+c)^(1/2)/(b+c)^(1/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(751\) vs. \(2(267)=534\).
Time = 4.10 (sec) , antiderivative size = 751, normalized size of antiderivative = 2.81 \[ \int \frac {x \sec ^2(c+d x)}{a+c \sec ^2(c+d x)+b \tan ^2(c+d x)} \, dx=\frac {x \left (4 \sqrt {-b-c} c \arctan \left (\frac {\sqrt {b+c} \tan (c+d x)}{\sqrt {a+c}}\right )-i \sqrt {b+c} \log (1+i \tan (c+d x)) \log \left (\frac {i \left (\sqrt {a+c}-\sqrt {-b-c} \tan (c+d x)\right )}{\sqrt {-b-c}+i \sqrt {a+c}}\right )+i \sqrt {b+c} \log (1-i \tan (c+d x)) \log \left (\frac {i \left (-\sqrt {a+c}+\sqrt {-b-c} \tan (c+d x)\right )}{\sqrt {-b-c}-i \sqrt {a+c}}\right )+i \sqrt {b+c} \log (1+i \tan (c+d x)) \log \left (-\frac {i \left (\sqrt {a+c}+\sqrt {-b-c} \tan (c+d x)\right )}{\sqrt {-b-c}-i \sqrt {a+c}}\right )-i \sqrt {b+c} \log (1-i \tan (c+d x)) \log \left (\frac {i \left (\sqrt {a+c}+\sqrt {-b-c} \tan (c+d x)\right )}{\sqrt {-b-c}+i \sqrt {a+c}}\right )+i \sqrt {b+c} \operatorname {PolyLog}\left (2,\frac {\sqrt {-b-c} (1-i \tan (c+d x))}{\sqrt {-b-c}-i \sqrt {a+c}}\right )-i \sqrt {b+c} \operatorname {PolyLog}\left (2,\frac {\sqrt {-b-c} (1-i \tan (c+d x))}{\sqrt {-b-c}+i \sqrt {a+c}}\right )+i \sqrt {b+c} \operatorname {PolyLog}\left (2,\frac {\sqrt {-b-c} (1+i \tan (c+d x))}{\sqrt {-b-c}-i \sqrt {a+c}}\right )-i \sqrt {b+c} \operatorname {PolyLog}\left (2,\frac {\sqrt {-b-c} (1+i \tan (c+d x))}{\sqrt {-b-c}+i \sqrt {a+c}}\right )\right ) \left (\sqrt {a+c}-\sqrt {-b-c} \tan (c+d x)\right ) \left (\sqrt {a+c}+\sqrt {-b-c} \tan (c+d x)\right )}{2 \sqrt {a+c} \sqrt {-(b+c)^2} d (2 c-i \log (1-i \tan (c+d x))+i \log (1+i \tan (c+d x))) \left (a+c \sec ^2(c+d x)+b \tan ^2(c+d x)\right )} \]
(x*(4*Sqrt[-b - c]*c*ArcTan[(Sqrt[b + c]*Tan[c + d*x])/Sqrt[a + c]] - I*Sq rt[b + c]*Log[1 + I*Tan[c + d*x]]*Log[(I*(Sqrt[a + c] - Sqrt[-b - c]*Tan[c + d*x]))/(Sqrt[-b - c] + I*Sqrt[a + c])] + I*Sqrt[b + c]*Log[1 - I*Tan[c + d*x]]*Log[(I*(-Sqrt[a + c] + Sqrt[-b - c]*Tan[c + d*x]))/(Sqrt[-b - c] - I*Sqrt[a + c])] + I*Sqrt[b + c]*Log[1 + I*Tan[c + d*x]]*Log[((-I)*(Sqrt[a + c] + Sqrt[-b - c]*Tan[c + d*x]))/(Sqrt[-b - c] - I*Sqrt[a + c])] - I*Sq rt[b + c]*Log[1 - I*Tan[c + d*x]]*Log[(I*(Sqrt[a + c] + Sqrt[-b - c]*Tan[c + d*x]))/(Sqrt[-b - c] + I*Sqrt[a + c])] + I*Sqrt[b + c]*PolyLog[2, (Sqrt [-b - c]*(1 - I*Tan[c + d*x]))/(Sqrt[-b - c] - I*Sqrt[a + c])] - I*Sqrt[b + c]*PolyLog[2, (Sqrt[-b - c]*(1 - I*Tan[c + d*x]))/(Sqrt[-b - c] + I*Sqrt [a + c])] + I*Sqrt[b + c]*PolyLog[2, (Sqrt[-b - c]*(1 + I*Tan[c + d*x]))/( Sqrt[-b - c] - I*Sqrt[a + c])] - I*Sqrt[b + c]*PolyLog[2, (Sqrt[-b - c]*(1 + I*Tan[c + d*x]))/(Sqrt[-b - c] + I*Sqrt[a + c])])*(Sqrt[a + c] - Sqrt[- b - c]*Tan[c + d*x])*(Sqrt[a + c] + Sqrt[-b - c]*Tan[c + d*x]))/(2*Sqrt[a + c]*Sqrt[-(b + c)^2]*d*(2*c - I*Log[1 - I*Tan[c + d*x]] + I*Log[1 + I*Tan [c + d*x]])*(a + c*Sec[c + d*x]^2 + b*Tan[c + d*x]^2))
Time = 1.10 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {5100, 3042, 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.
| \(\displaystyle \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)+c \sec ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 5100 |
| \(\displaystyle 2 \int \frac {x}{a+b+2 c+(a-b) \cos (2 c+2 d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \int \frac {x}{a+b+2 c+(a-b) \sin \left (2 c+2 d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3802 |
| \(\displaystyle 4 \int \frac {e^{2 i (c+d x)} x}{a+2 (a+b+2 c) e^{2 i (c+d x)}+(a-b) e^{4 i (c+d x)}-b}dx\) |
| \(\Big \downarrow \) 2694 |
| \(\displaystyle 4 \left (\frac {(a-b) \int \frac {e^{2 i (c+d x)} x}{2 \left (a+(a-b) e^{2 i (c+d x)}+b+2 c-2 \sqrt {a+c} \sqrt {b+c}\right )}dx}{2 \sqrt {a+c} \sqrt {b+c}}-\frac {(a-b) \int \frac {e^{2 i (c+d x)} x}{2 \left (a+(a-b) e^{2 i (c+d x)}+b+2 \left (c+\sqrt {a+c} \sqrt {b+c}\right )\right )}dx}{2 \sqrt {a+c} \sqrt {b+c}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \left (\frac {(a-b) \int \frac {e^{2 i (c+d x)} x}{a+(a-b) e^{2 i (c+d x)}+b+2 c-2 \sqrt {a+c} \sqrt {b+c}}dx}{4 \sqrt {a+c} \sqrt {b+c}}-\frac {(a-b) \int \frac {e^{2 i (c+d x)} x}{a+(a-b) e^{2 i (c+d x)}+b+2 \left (c+\sqrt {a+c} \sqrt {b+c}\right )}dx}{4 \sqrt {a+c} \sqrt {b+c}}\right )\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle 4 \left (\frac {(a-b) \left (\frac {i \int \log \left (\frac {e^{2 i (c+d x)} (a-b)}{a+b+2 c-2 \sqrt {a+c} \sqrt {b+c}}+1\right )dx}{2 d (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{-2 \sqrt {a+c} \sqrt {b+c}+a+b+2 c}\right )}{2 d (a-b)}\right )}{4 \sqrt {a+c} \sqrt {b+c}}-\frac {(a-b) \left (\frac {i \int \log \left (\frac {e^{2 i (c+d x)} (a-b)}{a+b+2 \left (c+\sqrt {a+c} \sqrt {b+c}\right )}+1\right )dx}{2 d (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{2 \left (\sqrt {a+c} \sqrt {b+c}+c\right )+a+b}\right )}{2 d (a-b)}\right )}{4 \sqrt {a+c} \sqrt {b+c}}\right )\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle 4 \left (\frac {(a-b) \left (\frac {\int e^{-2 i (c+d x)} \log \left (\frac {e^{2 i (c+d x)} (a-b)}{a+b+2 c-2 \sqrt {a+c} \sqrt {b+c}}+1\right )de^{2 i (c+d x)}}{4 d^2 (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{-2 \sqrt {a+c} \sqrt {b+c}+a+b+2 c}\right )}{2 d (a-b)}\right )}{4 \sqrt {a+c} \sqrt {b+c}}-\frac {(a-b) \left (\frac {\int e^{-2 i (c+d x)} \log \left (\frac {e^{2 i (c+d x)} (a-b)}{a+b+2 \left (c+\sqrt {a+c} \sqrt {b+c}\right )}+1\right )de^{2 i (c+d x)}}{4 d^2 (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{2 \left (\sqrt {a+c} \sqrt {b+c}+c\right )+a+b}\right )}{2 d (a-b)}\right )}{4 \sqrt {a+c} \sqrt {b+c}}\right )\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle 4 \left (\frac {(a-b) \left (-\frac {\operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 c-2 \sqrt {a+c} \sqrt {b+c}}\right )}{4 d^2 (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{-2 \sqrt {a+c} \sqrt {b+c}+a+b+2 c}\right )}{2 d (a-b)}\right )}{4 \sqrt {a+c} \sqrt {b+c}}-\frac {(a-b) \left (-\frac {\operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 \left (c+\sqrt {a+c} \sqrt {b+c}\right )}\right )}{4 d^2 (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{2 \left (\sqrt {a+c} \sqrt {b+c}+c\right )+a+b}\right )}{2 d (a-b)}\right )}{4 \sqrt {a+c} \sqrt {b+c}}\right )\) |
4*(((a - b)*(((-1/2*I)*x*Log[1 + ((a - b)*E^((2*I)*(c + d*x)))/(a + b + 2* c - 2*Sqrt[a + c]*Sqrt[b + c])])/((a - b)*d) - PolyLog[2, -(((a - b)*E^((2 *I)*(c + d*x)))/(a + b + 2*c - 2*Sqrt[a + c]*Sqrt[b + c]))]/(4*(a - b)*d^2 )))/(4*Sqrt[a + c]*Sqrt[b + c]) - ((a - b)*(((-1/2*I)*x*Log[1 + ((a - b)*E ^((2*I)*(c + d*x)))/(a + b + 2*(c + Sqrt[a + c]*Sqrt[b + c]))])/((a - b)*d ) - PolyLog[2, -(((a - b)*E^((2*I)*(c + d*x)))/(a + b + 2*(c + Sqrt[a + c] *Sqrt[b + c])))]/(4*(a - b)*d^2)))/(4*Sqrt[a + c]*Sqrt[b + c]))
3.2.63.3.1 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[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[((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[(((f_.) + (g_.)*(x_))^(m_.)*Sec[(d_.) + (e_.)*(x_)]^2)/((b_.) + (a_.)*S ec[(d_.) + (e_.)*(x_)]^2 + (c_.)*Tan[(d_.) + (e_.)*(x_)]^2), x_Symbol] :> S imp[2 Int[(f + g*x)^m/(2*a + b + c + (b - c)*Cos[2*d + 2*e*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[m, 0] && NeQ[a + b, 0] && NeQ[a + c, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1669 vs. \(2 (217 ) = 434\).
Time = 2.66 (sec) , antiderivative size = 1670, normalized size of antiderivative = 6.25
-1/2/((a+c)*(b+c))^(1/2)*x^2-1/(-2*((a+c)*(b+c))^(1/2)-a-b-2*c)*x^2-1/d^2/ (-2*((a+c)*(b+c))^(1/2)-a-b-2*c)/((a+c)*(b+c))^(1/2)*c^3-2/d/(-2*((a+c)*(b +c))^(1/2)-a-b-2*c)*c*x-1/d/((a+c)*(b+c))^(1/2)*c*x-1/2/(-2*((a+c)*(b+c))^ (1/2)-a-b-2*c)/((a+c)*(b+c))^(1/2)*a*x^2-1/2/(-2*((a+c)*(b+c))^(1/2)-a-b-2 *c)/((a+c)*(b+c))^(1/2)*x^2*b-1/(-2*((a+c)*(b+c))^(1/2)-a-b-2*c)/((a+c)*(b +c))^(1/2)*c*x^2-1/d/(-2*((a+c)*(b+c))^(1/2)-a-b-2*c)/((a+c)*(b+c))^(1/2)* a*c*x-1/d/(-2*((a+c)*(b+c))^(1/2)-a-b-2*c)/((a+c)*(b+c))^(1/2)*c*x*b-1/2*I /d/(-2*((a+c)*(b+c))^(1/2)-a-b-2*c)/((a+c)*(b+c))^(1/2)*ln(1-(a-b)*exp(2*I *(d*x+c))/(-2*((a+c)*(b+c))^(1/2)-a-b-2*c))*a*x-1/2*I/d/(-2*((a+c)*(b+c))^ (1/2)-a-b-2*c)/((a+c)*(b+c))^(1/2)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*((a+c)* (b+c))^(1/2)-a-b-2*c))*b*x-I/d/(-2*((a+c)*(b+c))^(1/2)-a-b-2*c)/((a+c)*(b+ c))^(1/2)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*((a+c)*(b+c))^(1/2)-a-b-2*c))*c* x-1/2*I/d^2/(-2*((a+c)*(b+c))^(1/2)-a-b-2*c)/((a+c)*(b+c))^(1/2)*ln(1-(a-b )*exp(2*I*(d*x+c))/(-2*((a+c)*(b+c))^(1/2)-a-b-2*c))*a*c-1/2*I/d^2/(-2*((a +c)*(b+c))^(1/2)-a-b-2*c)/((a+c)*(b+c))^(1/2)*ln(1-(a-b)*exp(2*I*(d*x+c))/ (-2*((a+c)*(b+c))^(1/2)-a-b-2*c))*b*c-1/2/d^2/((a+c)*(b+c))^(1/2)*c^2-1/4/ d^2/((a+c)*(b+c))^(1/2)*polylog(2,(a-b)*exp(2*I*(d*x+c))/(2*((a+c)*(b+c))^ (1/2)-a-b-2*c))-1/d^2/(-2*((a+c)*(b+c))^(1/2)-a-b-2*c)*c^2-1/2/d^2/(-2*((a +c)*(b+c))^(1/2)-a-b-2*c)*polylog(2,(a-b)*exp(2*I*(d*x+c))/(-2*((a+c)*(b+c ))^(1/2)-a-b-2*c))-2/d/(-2*((a+c)*(b+c))^(1/2)-a-b-2*c)/((a+c)*(b+c))^(...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 4100 vs. \(2 (213) = 426\).
Time = 4.60 (sec) , antiderivative size = 4100, normalized size of antiderivative = 15.36 \[ \int \frac {x \sec ^2(c+d x)}{a+c \sec ^2(c+d x)+b \tan ^2(c+d x)} \, dx=\text {Too large to display} \]
-1/4*(I*(a - b)*c*sqrt((a*b + (a + b)*c + c^2)/(a^2 - 2*a*b + b^2))*log(2* sqrt(-(2*(a - b)*sqrt((a*b + (a + b)*c + c^2)/(a^2 - 2*a*b + b^2)) + a + b + 2*c)/(a - b)) + 2*cos(d*x + c) + 2*I*sin(d*x + c)) - I*(a - b)*c*sqrt(( a*b + (a + b)*c + c^2)/(a^2 - 2*a*b + b^2))*log(2*sqrt(-(2*(a - b)*sqrt((a *b + (a + b)*c + c^2)/(a^2 - 2*a*b + b^2)) + a + b + 2*c)/(a - b)) + 2*cos (d*x + c) - 2*I*sin(d*x + c)) - I*(a - b)*c*sqrt((a*b + (a + b)*c + c^2)/( a^2 - 2*a*b + b^2))*log(2*sqrt(-(2*(a - b)*sqrt((a*b + (a + b)*c + c^2)/(a ^2 - 2*a*b + b^2)) + a + b + 2*c)/(a - b)) - 2*cos(d*x + c) + 2*I*sin(d*x + c)) + I*(a - b)*c*sqrt((a*b + (a + b)*c + c^2)/(a^2 - 2*a*b + b^2))*log( 2*sqrt(-(2*(a - b)*sqrt((a*b + (a + b)*c + c^2)/(a^2 - 2*a*b + b^2)) + a + b + 2*c)/(a - b)) - 2*cos(d*x + c) - 2*I*sin(d*x + c)) - I*(a - b)*c*sqrt ((a*b + (a + b)*c + c^2)/(a^2 - 2*a*b + b^2))*log(2*sqrt((2*(a - b)*sqrt(( a*b + (a + b)*c + c^2)/(a^2 - 2*a*b + b^2)) - a - b - 2*c)/(a - b)) + 2*co s(d*x + c) + 2*I*sin(d*x + c)) + I*(a - b)*c*sqrt((a*b + (a + b)*c + c^2)/ (a^2 - 2*a*b + b^2))*log(2*sqrt((2*(a - b)*sqrt((a*b + (a + b)*c + c^2)/(a ^2 - 2*a*b + b^2)) - a - b - 2*c)/(a - b)) + 2*cos(d*x + c) - 2*I*sin(d*x + c)) + I*(a - b)*c*sqrt((a*b + (a + b)*c + c^2)/(a^2 - 2*a*b + b^2))*log( 2*sqrt((2*(a - b)*sqrt((a*b + (a + b)*c + c^2)/(a^2 - 2*a*b + b^2)) - a - b - 2*c)/(a - b)) - 2*cos(d*x + c) + 2*I*sin(d*x + c)) - I*(a - b)*c*sqrt( (a*b + (a + b)*c + c^2)/(a^2 - 2*a*b + b^2))*log(2*sqrt((2*(a - b)*sqrt...
\[ \int \frac {x \sec ^2(c+d x)}{a+c \sec ^2(c+d x)+b \tan ^2(c+d x)} \, dx=\int \frac {x \sec ^{2}{\left (c + d x \right )}}{a + b \tan ^{2}{\left (c + d x \right )} + c \sec ^{2}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {x \sec ^2(c+d x)}{a+c \sec ^2(c+d x)+b \tan ^2(c+d x)} \, dx=\int { \frac {x \sec \left (d x + c\right )^{2}}{c \sec \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a} \,d x } \]
\[ \int \frac {x \sec ^2(c+d x)}{a+c \sec ^2(c+d x)+b \tan ^2(c+d x)} \, dx=\int { \frac {x \sec \left (d x + c\right )^{2}}{c \sec \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a} \,d x } \]
Timed out. \[ \int \frac {x \sec ^2(c+d x)}{a+c \sec ^2(c+d x)+b \tan ^2(c+d x)} \, dx=\int \frac {x}{{\cos \left (c+d\,x\right )}^2\,\left (a+\frac {c}{{\cos \left (c+d\,x\right )}^2}+b\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )} \,d x \]