Integrand size = 25, antiderivative size = 118 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^2(e+f x) \, dx=-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}+\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 \sqrt {b} f}+\frac {\tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 f} \]
-arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))*a^(1/2)/f+1/2*(a-b) *arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f/b^(1/2)+1/2*(a+b +b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/f
Result contains complex when optimal does not.
Time = 4.61 (sec) , antiderivative size = 526, normalized size of antiderivative = 4.46 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^2(e+f x) \, dx=\frac {e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \cos (e+f x) \left (-\frac {i \left (-1+e^{2 i (e+f x)}\right )}{\left (1+e^{2 i (e+f x)}\right )^2}+\frac {-2 \sqrt {a} \sqrt {b} f x+i \sqrt {a} \sqrt {b} \log \left (a+2 b+a e^{2 i (e+f x)}+\sqrt {a} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )-i \sqrt {a} \sqrt {b} \log \left (a+a e^{2 i (e+f x)}+2 b e^{2 i (e+f x)}+\sqrt {a} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )-a \log \left (\frac {2 \left (\sqrt {b} \left (-1+e^{2 i (e+f x)}\right )-i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right ) f}{(a-b) \left (1+e^{2 i (e+f x)}\right )}\right )+b \log \left (\frac {2 \left (\sqrt {b} \left (-1+e^{2 i (e+f x)}\right )-i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right ) f}{(a-b) \left (1+e^{2 i (e+f x)}\right )}\right )}{\sqrt {b} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \sqrt {a+b \sec ^2(e+f x)}}{\sqrt {2} f \sqrt {a+2 b+a \cos (2 e+2 f x)}} \]
(E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + f*x))]*Cos[e + f*x]*(((-I)*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))^2 + (-2*Sqrt[a]*Sqrt[b]*f*x + I*Sqrt[a]*Sqrt[b]*Log[a + 2*b + a*E^( (2*I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)* (e + f*x)))^2]] - I*Sqrt[a]*Sqrt[b]*Log[a + a*E^((2*I)*(e + f*x)) + 2*b*E^ ((2*I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I) *(e + f*x)))^2]] - a*Log[(2*(Sqrt[b]*(-1 + E^((2*I)*(e + f*x))) - I*Sqrt[4 *b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2])*f)/((a - b)*(1 + E^((2*I)*(e + f*x))))] + b*Log[(2*(Sqrt[b]*(-1 + E^((2*I)*(e + f*x))) - I* Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2])*f)/((a - b) *(1 + E^((2*I)*(e + f*x))))])/(Sqrt[b]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]))*Sqrt[a + b*Sec[e + f*x]^2])/(Sqrt[2]*f*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]])
Time = 0.36 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4629, 2075, 380, 398, 224, 219, 291, 216}
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.
\(\displaystyle \int \tan ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^2 \sqrt {a+b \sec (e+f x)^2}dx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \frac {\int \frac {\tan ^2(e+f x) \sqrt {a+b \left (\tan ^2(e+f x)+1\right )}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {\tan ^2(e+f x) \sqrt {b \tan ^2(e+f x)+a+b}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 380 |
\(\displaystyle \frac {\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {1}{2} \int \frac {-\left ((a-b) \tan ^2(e+f x)\right )+a+b}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 398 |
\(\displaystyle \frac {\frac {1}{2} \left ((a-b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-2 a \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)\right )+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {1}{2} \left ((a-b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-2 a \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)\right )+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-2 a \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)\right )+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-2 a \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}\right )+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )\right )+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
((-2*Sqrt[a]*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]] + ((a - b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]] )/Sqrt[b])/2 + (Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/2)/f
3.4.84.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b* (m + 2*(p + q) + 1))), x] - Simp[e^2/(b*(m + 2*(p + q) + 1)) Int[(e*x)^(m - 2)*(a + b*x^2)^p*(c + d*x^2)^(q - 1)*Simp[a*c*(m - 1) + (a*d*(m - 1) - 2 *q*(b*c - a*d))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 0] && GtQ[m, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f _.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/f Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 )), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Leaf count of result is larger than twice the leaf count of optimal. \(660\) vs. \(2(100)=200\).
Time = 12.30 (sec) , antiderivative size = 661, normalized size of antiderivative = 5.60
method | result | size |
default | \(-\frac {\sqrt {a +b \sec \left (f x +e \right )^{2}}\, \left (\ln \left (\frac {4 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (f x +e \right )+4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a -4 a -4 b}{\sin \left (f x +e \right )+1}\right ) \sqrt {-a}\, \cos \left (f x +e \right ) b^{\frac {3}{2}}+b^{\frac {3}{2}} \ln \left (-\frac {4 \left (\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (f x +e \right )+\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-\sin \left (f x +e \right ) a +a +b \right )}{\sin \left (f x +e \right )-1}\right ) \sqrt {-a}\, \cos \left (f x +e \right )-\ln \left (\frac {4 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (f x +e \right )+4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a -4 a -4 b}{\sin \left (f x +e \right )+1}\right ) \sqrt {-a}\, \cos \left (f x +e \right ) \sqrt {b}\, a -\ln \left (-\frac {4 \left (\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (f x +e \right )+\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-\sin \left (f x +e \right ) a +a +b \right )}{\sin \left (f x +e \right )-1}\right ) \sqrt {-a}\, \cos \left (f x +e \right ) \sqrt {b}\, a -2 \sin \left (f x +e \right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {-a}\, b +4 \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \cos \left (f x +e \right ) a b -2 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b \tan \left (f x +e \right )\right )}{4 f \sqrt {-a}\, b \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (1+\cos \left (f x +e \right )\right )}\) | \(661\) |
-1/4/f/(-a)^(1/2)/b*(a+b*sec(f*x+e)^2)^(1/2)/((b+a*cos(f*x+e)^2)/(1+cos(f* x+e))^2)^(1/2)/(1+cos(f*x+e))*(ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) ^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1 /2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*(-a)^(1/2)*cos(f*x+e)*b^(3/2)+b^(3/2 )*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^ (1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f *x+e)-1))*(-a)^(1/2)*cos(f*x+e)-ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2 )^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^( 1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*(-a)^(1/2)*cos(f*x+e)*b^(1/2)*a-ln( -4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2) *((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e) -1))*(-a)^(1/2)*cos(f*x+e)*b^(1/2)*a-2*sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+c os(f*x+e))^2)^(1/2)*(-a)^(1/2)*b+4*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+ cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f* x+e))^2)^(1/2)-4*sin(f*x+e)*a)*cos(f*x+e)*a*b-2*(-a)^(1/2)*((b+a*cos(f*x+e )^2)/(1+cos(f*x+e))^2)^(1/2)*b*tan(f*x+e))
Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (100) = 200\).
Time = 0.62 (sec) , antiderivative size = 1471, normalized size of antiderivative = 12.47 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^2(e+f x) \, dx=\text {Too large to display} \]
[1/8*(sqrt(-a)*b*cos(f*x + e)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3* b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos( f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2* b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos( f*x + e)^2)*sin(f*x + e)) - (a - b)*sqrt(b)*cos(f*x + e)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 - 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) + 4*b*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(b*f*cos(f*x + e)), 1/8*(2*(a - b)*sqrt( -b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt( (a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e)))*cos(f*x + e) + sqrt(-a)*b*cos(f*x + e)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos (f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a ^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*( a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^ 3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 4*b*sqrt((a*cos(f*x + e)^2...
\[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^2(e+f x) \, dx=\int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx \]
\[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^2(e+f x) \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{2} \,d x } \]
\[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^2(e+f x) \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{2} \,d x } \]
Timed out. \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^2(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^2\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}} \,d x \]