Integrand size = 24, antiderivative size = 211 \[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}-\frac {\operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{4 \sqrt {a} \sqrt {b} d^2}+\frac {\operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{4 \sqrt {a} \sqrt {b} d^2} \]
-1/2*I*x*ln(1+(a-b)*exp(2*I*(d*x+c))/(a^(1/2)-b^(1/2))^2)/d/a^(1/2)/b^(1/2 )+1/2*I*x*ln(1+(a-b)*exp(2*I*(d*x+c))/(a^(1/2)+b^(1/2))^2)/d/a^(1/2)/b^(1/ 2)-1/4*polylog(2,-(a-b)*exp(2*I*(d*x+c))/(a^(1/2)-b^(1/2))^2)/d^2/a^(1/2)/ b^(1/2)+1/4*polylog(2,-(a-b)*exp(2*I*(d*x+c))/(a^(1/2)+b^(1/2))^2)/d^2/a^( 1/2)/b^(1/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(512\) vs. \(2(211)=422\).
Time = 2.20 (sec) , antiderivative size = 512, normalized size of antiderivative = 2.43 \[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {x \left (4 i \sqrt {-a} c \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )-\sqrt {a} \log (1+i \tan (c+d x)) \log \left (\frac {\sqrt {-a}-\sqrt {b} \tan (c+d x)}{\sqrt {-a}-i \sqrt {b}}\right )+\sqrt {a} \log (1-i \tan (c+d x)) \log \left (\frac {\sqrt {-a}-\sqrt {b} \tan (c+d x)}{\sqrt {-a}+i \sqrt {b}}\right )-\sqrt {a} \log (1-i \tan (c+d x)) \log \left (\frac {\sqrt {-a}+\sqrt {b} \tan (c+d x)}{\sqrt {-a}-i \sqrt {b}}\right )+\sqrt {a} \log (1+i \tan (c+d x)) \log \left (\frac {\sqrt {-a}+\sqrt {b} \tan (c+d x)}{\sqrt {-a}+i \sqrt {b}}\right )-\sqrt {a} \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (1-i \tan (c+d x))}{i \sqrt {-a}+\sqrt {b}}\right )-\sqrt {a} \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (1+i \tan (c+d x))}{i \sqrt {-a}+\sqrt {b}}\right )+\sqrt {a} \operatorname {PolyLog}\left (2,-\frac {\sqrt {b} (-i+\tan (c+d x))}{\sqrt {-a}+i \sqrt {b}}\right )+\sqrt {a} \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (i+\tan (c+d x))}{\sqrt {-a}+i \sqrt {b}}\right )\right )}{2 \sqrt {-a^2} \sqrt {b} d (2 i c+\log (1-i \tan (c+d x))-\log (1+i \tan (c+d x)))} \]
(x*((4*I)*Sqrt[-a]*c*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]] - Sqrt[a]*Log[ 1 + I*Tan[c + d*x]]*Log[(Sqrt[-a] - Sqrt[b]*Tan[c + d*x])/(Sqrt[-a] - I*Sq rt[b])] + Sqrt[a]*Log[1 - I*Tan[c + d*x]]*Log[(Sqrt[-a] - Sqrt[b]*Tan[c + d*x])/(Sqrt[-a] + I*Sqrt[b])] - Sqrt[a]*Log[1 - I*Tan[c + d*x]]*Log[(Sqrt[ -a] + Sqrt[b]*Tan[c + d*x])/(Sqrt[-a] - I*Sqrt[b])] + Sqrt[a]*Log[1 + I*Ta n[c + d*x]]*Log[(Sqrt[-a] + Sqrt[b]*Tan[c + d*x])/(Sqrt[-a] + I*Sqrt[b])] - Sqrt[a]*PolyLog[2, (Sqrt[b]*(1 - I*Tan[c + d*x]))/(I*Sqrt[-a] + Sqrt[b]) ] - Sqrt[a]*PolyLog[2, (Sqrt[b]*(1 + I*Tan[c + d*x]))/(I*Sqrt[-a] + Sqrt[b ])] + Sqrt[a]*PolyLog[2, -((Sqrt[b]*(-I + Tan[c + d*x]))/(Sqrt[-a] + I*Sqr t[b]))] + Sqrt[a]*PolyLog[2, (Sqrt[b]*(I + Tan[c + d*x]))/(Sqrt[-a] + I*Sq rt[b])]))/(2*Sqrt[-a^2]*Sqrt[b]*d*((2*I)*c + Log[1 - I*Tan[c + d*x]] - Log [1 + I*Tan[c + d*x]]))
Time = 0.91 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5099, 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.
| \(\displaystyle \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 5099 |
| \(\displaystyle 2 \int \frac {x}{a+b+(a-b) \cos (2 c+2 d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \int \frac {x}{a+b+(a-b) \sin \left (2 c+2 d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3802 |
| \(\displaystyle 4 \int \frac {e^{2 i (c+d x)} x}{a+2 (a+b) e^{2 i (c+d x)}+(a-b) e^{4 i (c+d x)}-b}dx\) |
| \(\Big \downarrow \) 2694 |
| \(\displaystyle 4 \left (\frac {(a-b) \int \frac {e^{2 i (c+d x)} x}{2 \left (a-2 \sqrt {b} \sqrt {a}+(a-b) e^{2 i (c+d x)}+b\right )}dx}{2 \sqrt {a} \sqrt {b}}-\frac {(a-b) \int \frac {e^{2 i (c+d x)} x}{2 \left (\left (\sqrt {a}+\sqrt {b}\right )^2+(a-b) e^{2 i (c+d x)}\right )}dx}{2 \sqrt {a} \sqrt {b}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \left (\frac {(a-b) \int \frac {e^{2 i (c+d x)} x}{a-2 \sqrt {b} \sqrt {a}+(a-b) e^{2 i (c+d x)}+b}dx}{4 \sqrt {a} \sqrt {b}}-\frac {(a-b) \int \frac {e^{2 i (c+d x)} x}{\left (\sqrt {a}+\sqrt {b}\right )^2+(a-b) e^{2 i (c+d x)}}dx}{4 \sqrt {a} \sqrt {b}}\right )\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle 4 \left (\frac {(a-b) \left (\frac {i \int \log \left (\frac {e^{2 i (c+d x)} (a-b)}{\left (\sqrt {a}-\sqrt {b}\right )^2}+1\right )dx}{2 d (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 d (a-b)}\right )}{4 \sqrt {a} \sqrt {b}}-\frac {(a-b) \left (\frac {i \int \log \left (\frac {e^{2 i (c+d x)} (a-b)}{\left (\sqrt {a}+\sqrt {b}\right )^2}+1\right )dx}{2 d (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 d (a-b)}\right )}{4 \sqrt {a} \sqrt {b}}\right )\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle 4 \left (\frac {(a-b) \left (\frac {\int e^{-2 i (c+d x)} \log \left (\frac {e^{2 i (c+d x)} (a-b)}{\left (\sqrt {a}-\sqrt {b}\right )^2}+1\right )de^{2 i (c+d x)}}{4 d^2 (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 d (a-b)}\right )}{4 \sqrt {a} \sqrt {b}}-\frac {(a-b) \left (\frac {\int e^{-2 i (c+d x)} \log \left (\frac {e^{2 i (c+d x)} (a-b)}{\left (\sqrt {a}+\sqrt {b}\right )^2}+1\right )de^{2 i (c+d x)}}{4 d^2 (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 d (a-b)}\right )}{4 \sqrt {a} \sqrt {b}}\right )\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle 4 \left (\frac {(a-b) \left (-\frac {\operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{4 d^2 (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 d (a-b)}\right )}{4 \sqrt {a} \sqrt {b}}-\frac {(a-b) \left (-\frac {\operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{4 d^2 (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 d (a-b)}\right )}{4 \sqrt {a} \sqrt {b}}\right )\) |
4*(((a - b)*(((-1/2*I)*x*Log[1 + ((a - b)*E^((2*I)*(c + d*x)))/(Sqrt[a] - Sqrt[b])^2])/((a - b)*d) - PolyLog[2, -(((a - b)*E^((2*I)*(c + d*x)))/(Sqr t[a] - Sqrt[b])^2)]/(4*(a - b)*d^2)))/(4*Sqrt[a]*Sqrt[b]) - ((a - b)*(((-1 /2*I)*x*Log[1 + ((a - b)*E^((2*I)*(c + d*x)))/(Sqrt[a] + Sqrt[b])^2])/((a - b)*d) - PolyLog[2, -(((a - b)*E^((2*I)*(c + d*x)))/(Sqrt[a] + Sqrt[b])^2 )]/(4*(a - b)*d^2)))/(4*Sqrt[a]*Sqrt[b]))
3.2.60.3.1 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_) + (c_.)*Ta n[(d_.) + (e_.)*(x_)]^2), x_Symbol] :> Simp[2 Int[(f + g*x)^m/(b + c + (b - c)*Cos[2*d + 2*e*x]), x], x] /; FreeQ[{b, c, d, e, f, g}, x] && IGtQ[m, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1002 vs. \(2 (161 ) = 322\).
Time = 2.15 (sec) , antiderivative size = 1003, normalized size of antiderivative = 4.75
| method | result | size |
| risch | \(-\frac {a \,c^{2}}{2 d^{2} \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {c^{2} b}{2 d^{2} \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {\operatorname {polylog}\left (2, \frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) a}{4 d^{2} \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {a \,x^{2}}{2 \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {x^{2} b}{2 \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {2 c x}{d \left (-2 \sqrt {a b}-a -b \right )}-\frac {c x}{d \sqrt {a b}}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) b c}{2 d^{2} \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {\operatorname {polylog}\left (2, \frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{2 \sqrt {a b}-a -b}\right )}{4 d^{2} \sqrt {a b}}-\frac {\operatorname {polylog}\left (2, \frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right )}{2 d^{2} \left (-2 \sqrt {a b}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) a x}{2 d \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) b x}{2 d \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) a c}{2 d^{2} \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {x^{2}}{-2 \sqrt {a b}-a -b}-\frac {x^{2}}{2 \sqrt {a b}}-\frac {a c x}{d \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {c x b}{d \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {c^{2}}{2 d^{2} \sqrt {a b}}-\frac {c^{2}}{d^{2} \left (-2 \sqrt {a b}-a -b \right )}-\frac {\operatorname {polylog}\left (2, \frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) b}{4 d^{2} \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) c}{d^{2} \left (-2 \sqrt {a b}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{2 \sqrt {a b}-a -b}\right ) x}{2 d \sqrt {a b}}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{2 \sqrt {a b}-a -b}\right ) c}{2 d^{2} \sqrt {a b}}-\frac {i c \,\operatorname {arctanh}\left (\frac {2 \left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}+2 a +2 b}{4 \sqrt {a b}}\right )}{d^{2} \sqrt {a b}}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) x}{d \left (-2 \sqrt {a b}-a -b \right )}\) | \(1003\) |
-1/2/d^2/(a*b)^(1/2)/(-2*(a*b)^(1/2)-a-b)*a*c^2-1/2/d^2/(a*b)^(1/2)/(-2*(a *b)^(1/2)-a-b)*c^2*b-1/4/d^2/(a*b)^(1/2)/(-2*(a*b)^(1/2)-a-b)*polylog(2,(a -b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b))*a-1/4/d^2/(a*b)^(1/2)/(-2*(a*b) ^(1/2)-a-b)*polylog(2,(a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b))*b-I/d^2 /(-2*(a*b)^(1/2)-a-b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b))*c- 1/2*I/d/(a*b)^(1/2)*ln(1-(a-b)*exp(2*I*(d*x+c))/(2*(a*b)^(1/2)-a-b))*x-1/2 *I/d^2/(a*b)^(1/2)*ln(1-(a-b)*exp(2*I*(d*x+c))/(2*(a*b)^(1/2)-a-b))*c-I/d^ 2*c/(a*b)^(1/2)*arctanh(1/4*(2*(a-b)*exp(2*I*(d*x+c))+2*a+2*b)/(a*b)^(1/2) )-I/d/(-2*(a*b)^(1/2)-a-b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b ))*x-1/2/(a*b)^(1/2)/(-2*(a*b)^(1/2)-a-b)*a*x^2-1/2/(a*b)^(1/2)/(-2*(a*b)^ (1/2)-a-b)*x^2*b-2/d/(-2*(a*b)^(1/2)-a-b)*c*x-1/d/(a*b)^(1/2)*c*x-1/2*I/d/ (a*b)^(1/2)/(-2*(a*b)^(1/2)-a-b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/ 2)-a-b))*a*x-1/2*I/d/(a*b)^(1/2)/(-2*(a*b)^(1/2)-a-b)*ln(1-(a-b)*exp(2*I*( d*x+c))/(-2*(a*b)^(1/2)-a-b))*b*x-1/2*I/d^2/(a*b)^(1/2)/(-2*(a*b)^(1/2)-a- b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b))*a*c-1/2*I/d^2/(a*b)^( 1/2)/(-2*(a*b)^(1/2)-a-b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b) )*b*c-1/(-2*(a*b)^(1/2)-a-b)*x^2-1/2/(a*b)^(1/2)*x^2-1/d/(a*b)^(1/2)/(-2*( a*b)^(1/2)-a-b)*a*c*x-1/d/(a*b)^(1/2)/(-2*(a*b)^(1/2)-a-b)*c*x*b-1/2/d^2/( a*b)^(1/2)*c^2-1/4/d^2/(a*b)^(1/2)*polylog(2,(a-b)*exp(2*I*(d*x+c))/(2*(a* b)^(1/2)-a-b))-1/d^2/(-2*(a*b)^(1/2)-a-b)*c^2-1/2/d^2/(-2*(a*b)^(1/2)-a...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3284 vs. \(2 (157) = 314\).
Time = 3.66 (sec) , antiderivative size = 3284, normalized size of antiderivative = 15.56 \[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\text {Too large to display} \]
-1/4*(I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c*log(2*sqrt(-(2*(a - b)*sqr t(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) + 2*cos(d*x + c) + 2*I*sin(d* x + c)) - I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c*log(2*sqrt(-(2*(a - b) *sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) + 2*cos(d*x + c) - 2*I*si n(d*x + c)) - I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c*log(2*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) - 2*cos(d*x + c) + 2* I*sin(d*x + c)) + I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c*log(2*sqrt(-(2 *(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) - 2*cos(d*x + c) - 2*I*sin(d*x + c)) - I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c*log(2*sqrt ((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) + 2*cos(d*x + c) + 2*I*sin(d*x + c)) + I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c*log(2*s qrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) + 2*cos(d*x + c) - 2*I*sin(d*x + c)) + I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c*log( 2*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) - 2*cos( d*x + c) + 2*I*sin(d*x + c)) - I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c*l og(2*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) - 2*c os(d*x + c) - 2*I*sin(d*x + c)) - (a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*di log(-(((a + b)*cos(d*x + c) + (I*a + I*b)*sin(d*x + c) - 2*((a - b)*cos(d* x + c) + (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2 *(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) + a - b)/(a - ...
\[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\int \frac {x \sec ^{2}{\left (c + d x \right )}}{a + b \tan ^{2}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\int { \frac {x \sec \left (d x + c\right )^{2}}{b \tan \left (d x + c\right )^{2} + a} \,d x } \]
\[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\int { \frac {x \sec \left (d x + c\right )^{2}}{b \tan \left (d x + c\right )^{2} + a} \,d x } \]
Timed out. \[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\int \frac {x}{{\cos \left (c+d\,x\right )}^2\,\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )} \,d x \]