\(\int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx\) [81]

Optimal result

Integrand size = 24, antiderivative size = 211 \[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}-\frac {\operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{4 \sqrt {a} \sqrt {b} d^2}+\frac {\operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{4 \sqrt {a} \sqrt {b} d^2} \] Output:

-1/2*I*x*ln(1+(a-b)*exp(2*I*(d*x+c))/(a^(1/2)-b^(1/2))^2)/a^(1/2)/b^(1/2)/ 
d+1/2*I*x*ln(1+(a-b)*exp(2*I*(d*x+c))/(a^(1/2)+b^(1/2))^2)/a^(1/2)/b^(1/2) 
/d-1/4*polylog(2,-(a-b)*exp(2*I*(d*x+c))/(a^(1/2)-b^(1/2))^2)/a^(1/2)/b^(1 
/2)/d^2+1/4*polylog(2,-(a-b)*exp(2*I*(d*x+c))/(a^(1/2)+b^(1/2))^2)/a^(1/2) 
/b^(1/2)/d^2
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(512\) vs. \(2(211)=422\).

Time = 1.89 (sec) , antiderivative size = 512, normalized size of antiderivative = 2.43 \[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {x \left (4 i \sqrt {-a} c \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )-\sqrt {a} \log (1+i \tan (c+d x)) \log \left (\frac {\sqrt {-a}-\sqrt {b} \tan (c+d x)}{\sqrt {-a}-i \sqrt {b}}\right )+\sqrt {a} \log (1-i \tan (c+d x)) \log \left (\frac {\sqrt {-a}-\sqrt {b} \tan (c+d x)}{\sqrt {-a}+i \sqrt {b}}\right )-\sqrt {a} \log (1-i \tan (c+d x)) \log \left (\frac {\sqrt {-a}+\sqrt {b} \tan (c+d x)}{\sqrt {-a}-i \sqrt {b}}\right )+\sqrt {a} \log (1+i \tan (c+d x)) \log \left (\frac {\sqrt {-a}+\sqrt {b} \tan (c+d x)}{\sqrt {-a}+i \sqrt {b}}\right )-\sqrt {a} \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (1-i \tan (c+d x))}{i \sqrt {-a}+\sqrt {b}}\right )-\sqrt {a} \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (1+i \tan (c+d x))}{i \sqrt {-a}+\sqrt {b}}\right )+\sqrt {a} \operatorname {PolyLog}\left (2,-\frac {\sqrt {b} (-i+\tan (c+d x))}{\sqrt {-a}+i \sqrt {b}}\right )+\sqrt {a} \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (i+\tan (c+d x))}{\sqrt {-a}+i \sqrt {b}}\right )\right )}{2 \sqrt {-a^2} \sqrt {b} d (2 i c+\log (1-i \tan (c+d x))-\log (1+i \tan (c+d x)))} \] Input:

Integrate[(x*Sec[c + d*x]^2)/(a + b*Tan[c + d*x]^2),x]
 

Output:

(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]]))
 

Rubi [A] (verified)

Time = 0.92 (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

\(\displaystyle 4 \left (\frac {(a-b) \left (-\frac {\operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{4 d^2 (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 d (a-b)}\right )}{4 \sqrt {a} \sqrt {b}}-\frac {(a-b) \left (-\frac {\operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{4 d^2 (a-b)}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 d (a-b)}\right )}{4 \sqrt {a} \sqrt {b}}\right )\)

Input:

Int[(x*Sec[c + d*x]^2)/(a + b*Tan[c + d*x]^2),x]
 

Output:

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]))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2620

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]
 

rule 2694

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]
 

rule 2715

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]
 

rule 2838

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3802

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]
 

rule 5099

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]
 

Maple [B] (verified)

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 = 0.83 (sec) , antiderivative size = 1003, normalized size of antiderivative = 4.75

method	result	size
risch	\(-\frac {\operatorname {polylog}\left (2, \frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {b a}-a -b}\right ) b}{4 d^{2} \sqrt {b a}\, \left (-2 \sqrt {b a}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {b a}-a -b}\right ) c}{d^{2} \left (-2 \sqrt {b a}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{2 \sqrt {b a}-a -b}\right ) x}{2 d \sqrt {b a}}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {b a}-a -b}\right ) x}{d \left (-2 \sqrt {b a}-a -b \right )}-\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 {b a}}\right )}{d^{2} \sqrt {b a}}-\frac {c x}{d \sqrt {b a}}-\frac {2 c x}{d \left (-2 \sqrt {b a}-a -b \right )}-\frac {a \,x^{2}}{2 \sqrt {b a}\, \left (-2 \sqrt {b a}-a -b \right )}-\frac {b \,x^{2}}{2 \sqrt {b a}\, \left (-2 \sqrt {b a}-a -b \right )}-\frac {a \,c^{2}}{2 d^{2} \sqrt {b a}\, \left (-2 \sqrt {b a}-a -b \right )}-\frac {b \,c^{2}}{2 d^{2} \sqrt {b a}\, \left (-2 \sqrt {b a}-a -b \right )}-\frac {\operatorname {polylog}\left (2, \frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {b a}-a -b}\right ) a}{4 d^{2} \sqrt {b a}\, \left (-2 \sqrt {b a}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {b a}-a -b}\right ) a x}{2 d \sqrt {b a}\, \left (-2 \sqrt {b a}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {b a}-a -b}\right ) b x}{2 d \sqrt {b a}\, \left (-2 \sqrt {b a}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {b a}-a -b}\right ) a c}{2 d^{2} \sqrt {b a}\, \left (-2 \sqrt {b a}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {b a}-a -b}\right ) b c}{2 d^{2} \sqrt {b a}\, \left (-2 \sqrt {b a}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{2 \sqrt {b a}-a -b}\right ) c}{2 d^{2} \sqrt {b a}}-\frac {a c x}{d \sqrt {b a}\, \left (-2 \sqrt {b a}-a -b \right )}-\frac {b c x}{d \sqrt {b a}\, \left (-2 \sqrt {b a}-a -b \right )}-\frac {x^{2}}{-2 \sqrt {b a}-a -b}-\frac {x^{2}}{2 \sqrt {b a}}-\frac {\operatorname {polylog}\left (2, \frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{2 \sqrt {b a}-a -b}\right )}{4 d^{2} \sqrt {b a}}-\frac {\operatorname {polylog}\left (2, \frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {b a}-a -b}\right )}{2 d^{2} \left (-2 \sqrt {b a}-a -b \right )}-\frac {c^{2}}{2 d^{2} \sqrt {b a}}-\frac {c^{2}}{d^{2} \left (-2 \sqrt {b a}-a -b \right )}\)	\(1003\)

Input:

int(x*sec(d*x+c)^2/(a+tan(d*x+c)^2*b),x,method=_RETURNVERBOSE)
 

Output:

-1/d/(b*a)^(1/2)*c*x-2/d/(-2*(b*a)^(1/2)-a-b)*c*x-1/2/(b*a)^(1/2)/(-2*(b*a 
)^(1/2)-a-b)*a*x^2-1/2/(b*a)^(1/2)/(-2*(b*a)^(1/2)-a-b)*b*x^2-1/2/d^2/(b*a 
)^(1/2)/(-2*(b*a)^(1/2)-a-b)*a*c^2-1/2/d^2/(b*a)^(1/2)/(-2*(b*a)^(1/2)-a-b 
)*b*c^2-1/4/d^2/(b*a)^(1/2)/(-2*(b*a)^(1/2)-a-b)*polylog(2,(a-b)*exp(2*I*( 
d*x+c))/(-2*(b*a)^(1/2)-a-b))*a-1/4/d^2/(b*a)^(1/2)/(-2*(b*a)^(1/2)-a-b)*p 
olylog(2,(a-b)*exp(2*I*(d*x+c))/(-2*(b*a)^(1/2)-a-b))*b-I/d^2/(-2*(b*a)^(1 
/2)-a-b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(b*a)^(1/2)-a-b))*c-1/2*I/d/(b*a) 
^(1/2)*ln(1-(a-b)*exp(2*I*(d*x+c))/(2*(b*a)^(1/2)-a-b))*x-I/d/(-2*(b*a)^(1 
/2)-a-b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(b*a)^(1/2)-a-b))*x-I/d^2*c/(b*a) 
^(1/2)*arctanh(1/4*(2*(a-b)*exp(2*I*(d*x+c))+2*a+2*b)/(b*a)^(1/2))-1/2*I/d 
^2/(b*a)^(1/2)*ln(1-(a-b)*exp(2*I*(d*x+c))/(2*(b*a)^(1/2)-a-b))*c-1/2*I/d/ 
(b*a)^(1/2)/(-2*(b*a)^(1/2)-a-b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(b*a)^(1/ 
2)-a-b))*a*x-1/2*I/d/(b*a)^(1/2)/(-2*(b*a)^(1/2)-a-b)*ln(1-(a-b)*exp(2*I*( 
d*x+c))/(-2*(b*a)^(1/2)-a-b))*b*x-1/2*I/d^2/(b*a)^(1/2)/(-2*(b*a)^(1/2)-a- 
b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(b*a)^(1/2)-a-b))*a*c-1/2*I/d^2/(b*a)^( 
1/2)/(-2*(b*a)^(1/2)-a-b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(b*a)^(1/2)-a-b) 
)*b*c-1/d/(b*a)^(1/2)/(-2*(b*a)^(1/2)-a-b)*a*c*x-1/d/(b*a)^(1/2)/(-2*(b*a) 
^(1/2)-a-b)*b*c*x-1/(-2*(b*a)^(1/2)-a-b)*x^2-1/2/(b*a)^(1/2)*x^2-1/2/d^2/( 
b*a)^(1/2)*c^2-1/4/d^2/(b*a)^(1/2)*polylog(2,(a-b)*exp(2*I*(d*x+c))/(2*(b* 
a)^(1/2)-a-b))-1/d^2/(-2*(b*a)^(1/2)-a-b)*c^2-1/2/d^2/(-2*(b*a)^(1/2)-a...
 

Fricas [B] (verification not implemented)

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 = 3.28 (sec) , antiderivative size = 3284, normalized size of antiderivative = 15.56 \[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(x*sec(d*x+c)^2/(a+b*tan(d*x+c)^2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\int \frac {x \sec ^{2}{\left (c + d x \right )}}{a + b \tan ^{2}{\left (c + d x \right )}}\, dx \] Input:

integrate(x*sec(d*x+c)**2/(a+b*tan(d*x+c)**2),x)
 

Output:

Integral(x*sec(c + d*x)**2/(a + b*tan(c + d*x)**2), x)
 

Maxima [F]

\[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\int { \frac {x \sec \left (d x + c\right )^{2}}{b \tan \left (d x + c\right )^{2} + a} \,d x } \] Input:

integrate(x*sec(d*x+c)^2/(a+b*tan(d*x+c)^2),x, algorithm="maxima")
 

Output:

integrate(x*sec(d*x + c)^2/(b*tan(d*x + c)^2 + a), x)
 

Giac [F]

\[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\int { \frac {x \sec \left (d x + c\right )^{2}}{b \tan \left (d x + c\right )^{2} + a} \,d x } \] Input:

integrate(x*sec(d*x+c)^2/(a+b*tan(d*x+c)^2),x, algorithm="giac")
 

Output:

integrate(x*sec(d*x + c)^2/(b*tan(d*x + c)^2 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\int \frac {x}{{\cos \left (c+d\,x\right )}^2\,\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )} \,d x \] Input:

int(x/(cos(c + d*x)^2*(a + b*tan(c + d*x)^2)),x)
 

Output:

int(x/(cos(c + d*x)^2*(a + b*tan(c + d*x)^2)), x)
 

Reduce [F]

\[ \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {-2 \left (\int \frac {\sin \left (d x +c \right )^{2} x}{\sin \left (d x +c \right )^{2} a -\sin \left (d x +c \right )^{2} b -a}d x \right ) a +2 \left (\int \frac {\sin \left (d x +c \right )^{2} x}{\sin \left (d x +c \right )^{2} a -\sin \left (d x +c \right )^{2} b -a}d x \right ) b +x^{2}}{2 a} \] Input:

int(x*sec(d*x+c)^2/(a+b*tan(d*x+c)^2),x)
 

Output:

( - 2*int((sin(c + d*x)**2*x)/(sin(c + d*x)**2*a - sin(c + d*x)**2*b - a), 
x)*a + 2*int((sin(c + d*x)**2*x)/(sin(c + d*x)**2*a - sin(c + d*x)**2*b - 
a),x)*b + x**2)/(2*a)