\(\int \sqrt {a+b \sec ^2(e+f x)} \tan ^2(e+f x) \, dx\) [384]

Optimal result

Integrand size = 25, antiderivative size = 118 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^2(e+f x) \, dx=-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}+\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 \sqrt {b} f}+\frac {\tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 f} \] Output:

-a^(1/2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(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))/b^(1/2)/f+1/2*tan( 
f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/f
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.64 (sec) , antiderivative size = 526, normalized size of antiderivative = 4.46 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^2(e+f x) \, dx=\frac {e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \cos (e+f x) \left (-\frac {i \left (-1+e^{2 i (e+f x)}\right )}{\left (1+e^{2 i (e+f x)}\right )^2}+\frac {-2 \sqrt {a} \sqrt {b} f x+i \sqrt {a} \sqrt {b} \log \left (a+2 b+a e^{2 i (e+f x)}+\sqrt {a} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )-i \sqrt {a} \sqrt {b} \log \left (a+a e^{2 i (e+f x)}+2 b e^{2 i (e+f x)}+\sqrt {a} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )-a \log \left (\frac {2 \left (\sqrt {b} \left (-1+e^{2 i (e+f x)}\right )-i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right ) f}{(a-b) \left (1+e^{2 i (e+f x)}\right )}\right )+b \log \left (\frac {2 \left (\sqrt {b} \left (-1+e^{2 i (e+f x)}\right )-i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right ) f}{(a-b) \left (1+e^{2 i (e+f x)}\right )}\right )}{\sqrt {b} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \sqrt {a+b \sec ^2(e+f x)}}{\sqrt {2} f \sqrt {a+2 b+a \cos (2 e+2 f x)}} \] Input:

Integrate[Sqrt[a + b*Sec[e + f*x]^2]*Tan[e + f*x]^2,x]
 

Output:

(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]])
 

Rubi [A] (verified)

Time = 0.37 (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

\(\displaystyle \frac {\frac {1}{2} \left (\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )\right )+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

Input:

Int[Sqrt[a + b*Sec[e + f*x]^2]*Tan[e + f*x]^2,x]
 

Output:

((-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
 

Defintions of rubi rules used

rule 216

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])
 

rule 219

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])
 

rule 224

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]
 

rule 291

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]
 

rule 380

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]
 

rule 398

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]
 

rule 2075

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]
 

rule 3042

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

rule 4629

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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(635\) vs. \(2(100)=200\).

Time = 24.11 (sec) , antiderivative size = 636, normalized size of antiderivative = 5.39

method	result	size
default	\(\frac {\sqrt {a +b \sec \left (f x +e \right )^{2}}\, \left (-\cos \left (f x +e \right ) \sqrt {-a}\, \ln \left (\frac {-4 \sqrt {b}\, \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 \sin \left (f x +e \right ) a -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 a -4 b}{\sin \left (f x +e \right )-1}\right ) b^{\frac {3}{2}}+\cos \left (f x +e \right ) \sqrt {-a}\, \ln \left (\frac {-4 \sqrt {b}\, \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 \sin \left (f x +e \right ) a -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 a -4 b}{\sin \left (f x +e \right )-1}\right ) \sqrt {b}\, a -\cos \left (f x +e \right ) \sqrt {-a}\, \ln \left (-\frac {4 \left (-\sqrt {b}\, \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 )+\sin \left (f x +e \right ) a -\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+a +b \right )}{\sin \left (f x +e \right )+1}\right ) b^{\frac {3}{2}}+\cos \left (f x +e \right ) \sqrt {-a}\, \ln \left (-\frac {4 \left (-\sqrt {b}\, \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 )+\sin \left (f x +e \right ) a -\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+a +b \right )}{\sin \left (f x +e \right )+1}\right ) \sqrt {b}\, a -4 \cos \left (f x +e \right ) \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 ) a b +\sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b \left (2 \sin \left (f x +e \right )+2 \tan \left (f x +e \right )\right )\right )}{4 f \sqrt {-a}\, b \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\)	\(636\)

Input:

int((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x,method=_RETURNVERBOSE)
 

Output:

1/4/f/(-a)^(1/2)/b*(a+b*sec(f*x+e)^2)^(1/2)/(1+cos(f*x+e))/((b+a*cos(f*x+e 
)^2)/(1+cos(f*x+e))^2)^(1/2)*(-cos(f*x+e)*(-a)^(1/2)*ln(4*(-b^(1/2)*((b+a* 
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+sin(f*x+e)*a-b^(1/2)*((b+ 
a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-a-b)/(sin(f*x+e)-1))*b^(3/2)+cos(f 
*x+e)*(-a)^(1/2)*ln(4*(-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 
)*cos(f*x+e)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1 
/2)-a-b)/(sin(f*x+e)-1))*b^(1/2)*a-cos(f*x+e)*(-a)^(1/2)*ln(-4*(-b^(1/2)*( 
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+sin(f*x+e)*a-b^(1/2) 
*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+a+b)/(sin(f*x+e)+1))*b^(3/2)+ 
cos(f*x+e)*(-a)^(1/2)*ln(-4*(-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2 
)^(1/2)*cos(f*x+e)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)) 
^2)^(1/2)+a+b)/(sin(f*x+e)+1))*b^(1/2)*a-4*cos(f*x+e)*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)*a*b+(-a)^(1/2)*((b+a*cos 
(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b*(2*sin(f*x+e)+2*tan(f*x+e)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (100) = 200\).

Time = 0.43 (sec) , antiderivative size = 1471, normalized size of antiderivative = 12.47 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^2(e+f x) \, dx=\text {Too large to display} \] Input:

integrate((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x, algorithm="fricas")
 

Output:

[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...
 

Sympy [F]

\[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^2(e+f x) \, dx=\int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx \] Input:

integrate((a+b*sec(f*x+e)**2)**(1/2)*tan(f*x+e)**2,x)
 

Output:

Integral(sqrt(a + b*sec(e + f*x)**2)*tan(e + f*x)**2, x)
                                                                                    
                                                                                    
 

Maxima [F]

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

integrate((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x, algorithm="maxima")
 

Output:

integrate(sqrt(b*sec(f*x + e)^2 + a)*tan(f*x + e)^2, x)
 

Giac [F]

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

integrate((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x, algorithm="giac")
 

Output:

integrate(sqrt(b*sec(f*x + e)^2 + a)*tan(f*x + e)^2, x)
 

Mupad [F(-1)]

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

int(tan(e + f*x)^2*(a + b/cos(e + f*x)^2)^(1/2),x)
 

Output:

int(tan(e + f*x)^2*(a + b/cos(e + f*x)^2)^(1/2), x)
 

Reduce [F]

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

int((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x)
 

Output:

int(sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x)**2,x)