\(\int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) [235]

Optimal result

Integrand size = 25, antiderivative size = 122 \[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {(a-3 b) (a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 b^{3/2} f}-\frac {(a-3 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 b f}+\frac {\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 b f} \] Output:

-1/8*(a-3*b)*(a+b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/ 
b^(3/2)/f-1/8*(a-3*b)*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/b/f+1/4*tan(f* 
x+e)*(a+b+b*tan(f*x+e)^2)^(3/2)/b/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 9.29 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.86 \[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {i e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \left (-\sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} \left (a \left (1+e^{2 i (e+f x)}\right )^2+b \left (3+14 e^{2 i (e+f x)}+3 e^{4 i (e+f x)}\right )\right )+\left (a^2-2 a b-3 b^2\right ) \left (1+e^{2 i (e+f x)}\right )^4 \arctan \left (\frac {\sqrt {b} \left (-1+e^{2 i (e+f x)}\right )}{\sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right )\right ) \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 \sqrt {2} b^{3/2} \left (1+e^{2 i (e+f x)}\right )^4 \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} f \sqrt {a+2 b+a \cos (2 e+2 f x)}} \] Input:

Integrate[Sec[e + f*x]^4*Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

((I/4)*E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I) 
*(e + f*x))]*(-(Sqrt[b]*(-1 + E^((2*I)*(e + f*x)))*Sqrt[4*b*E^((2*I)*(e + 
f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]*(a*(1 + E^((2*I)*(e + f*x)))^2 + b* 
(3 + 14*E^((2*I)*(e + f*x)) + 3*E^((4*I)*(e + f*x))))) + (a^2 - 2*a*b - 3* 
b^2)*(1 + E^((2*I)*(e + f*x)))^4*ArcTan[(Sqrt[b]*(-1 + E^((2*I)*(e + f*x)) 
))/Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]])*Cos[e + 
 f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(Sqrt[2]*b^(3/2)*(1 + E^((2*I)*(e + f*x) 
))^4*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]*f*Sqrt[ 
a + 2*b + a*Cos[2*e + 2*f*x]])
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4634, 299, 211, 224, 219}

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.

\(\displaystyle \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (e+f x)^4 \sqrt {a+b \sec (e+f x)^2}dx\)

\(\Big \downarrow \) 4634

\(\displaystyle \frac {\int \left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {(a-3 b) \int \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{4 b}}{f}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {(a-3 b) \left (\frac {1}{2} (a+b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )}{4 b}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {(a-3 b) \left (\frac {1}{2} (a+b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )}{4 b}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {(a-3 b) \left (\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )}{4 b}}{f}\)

Input:

Int[Sec[e + f*x]^4*Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

((Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/(4*b) - ((a - 3*b)*(((a + 
 b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*Sqr 
t[b]) + (Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/2))/(4*b))/f
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 3042

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

rule 4634

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) 
)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f 
Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), 
x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ 
[m/2] && IntegerQ[n/2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(796\) vs. \(2(106)=212\).

Time = 31.99 (sec) , antiderivative size = 797, normalized size of antiderivative = 6.53

method	result	size
default	\(\frac {\sqrt {a +b \sec \left (f x +e \right )^{2}}\, \left (3 b^{\frac {5}{2}} \cos \left (f x +e \right ) \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 \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 )+2 b^{\frac {3}{2}} \cos \left (f x +e \right ) \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 \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 ) a -\sqrt {b}\, \cos \left (f x +e \right ) \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 \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 ) a^{2}+3 b^{\frac {5}{2}} \cos \left (f x +e \right ) \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 )+\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 )+2 b^{\frac {3}{2}} \cos \left (f x +e \right ) \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 )+\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 ) a -\sqrt {b}\, \cos \left (f x +e \right ) \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 )+\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 ) a^{2}+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b \left (2 \sin \left (f x +e \right )+2 \tan \left (f x +e \right )\right )+\left (6 \cos \left (f x +e \right )^{3}+6 \cos \left (f x +e \right )^{2}+4 \cos \left (f x +e \right )+4\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}\right )}{16 f \,b^{2} \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}}}}\)	\(797\)

Input:

int(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/16/f/b^2*(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)*(3*b^(5/2)*cos(f*x+e)*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^ 
2)/(1+cos(f*x+e))^2)^(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))+2*b^(3/2)*cos(f*x+e)*ln( 
4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(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-b^(1/2)*cos(f*x+e)*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 
2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-si 
n(f*x+e)*a-a-b)/(sin(f*x+e)+1))*a^2+3*b^(5/2)*cos(f*x+e)*ln(-4*(b^(1/2)*(( 
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(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))+2*b^(3/2) 
*cos(f*x+e)*ln(-4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(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-b^(1/2)*cos(f*x+e)*ln(-4*(b^(1/2)*((b+a*cos(f*x+e)^2 
)/(1+cos(f*x+e))^2)^(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^2+((b+a*cos(f*x+e)^2)/( 
1+cos(f*x+e))^2)^(1/2)*a*b*(2*sin(f*x+e)+2*tan(f*x+e))+(6*cos(f*x+e)^3+6*c 
os(f*x+e)^2+4*cos(f*x+e)+4)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^ 
2*tan(f*x+e)*sec(f*x+e)^2)
 

Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 390, normalized size of antiderivative = 3.20 \[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\left [-\frac {{\left (a^{2} - 2 \, a b - 3 \, b^{2}\right )} \sqrt {b} \cos \left (f x + e\right )^{3} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, {\left ({\left (a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{32 \, b^{2} f \cos \left (f x + e\right )^{3}}, -\frac {{\left (a^{2} - 2 \, a b - 3 \, b^{2}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{3} - 2 \, {\left ({\left (a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{16 \, b^{2} f \cos \left (f x + e\right )^{3}}\right ] \] Input:

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[-1/32*((a^2 - 2*a*b - 3*b^2)*sqrt(b)*cos(f*x + e)^3*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*((a*b + 3*b^2)*cos(f*x + e)^2 
+ 2*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(b^2*f* 
cos(f*x + e)^3), -1/16*((a^2 - 2*a*b - 3*b^2)*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)^3 
- 2*((a*b + 3*b^2)*cos(f*x + e)^2 + 2*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos 
(f*x + e)^2)*sin(f*x + e))/(b^2*f*cos(f*x + e)^3)]
 

Sympy [F]

\[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \sec ^{4}{\left (e + f x \right )}\, dx \] Input:

integrate(sec(f*x+e)**4*(a+b*sec(f*x+e)**2)**(1/2),x)
 

Output:

Integral(sqrt(a + b*sec(e + f*x)**2)*sec(e + f*x)**4, x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.42 \[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {{\left (a + b\right )} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} + \frac {{\left (a + b\right )} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - \frac {4 \, a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - 4 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) - 4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right ) - \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right )}{b} + \frac {\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )}{b}}{8 \, f} \] Input:

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

-1/8*((a + b)*a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(3/2) + (a + b)* 
arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) - 4*a*arcsinh(b*tan(f*x + 
e)/sqrt((a + b)*b))/sqrt(b) - 4*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b 
)*b)) - 4*sqrt(b*tan(f*x + e)^2 + a + b)*tan(f*x + e) - 2*(b*tan(f*x + e)^ 
2 + a + b)^(3/2)*tan(f*x + e)/b + sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*t 
an(f*x + e)/b)/f
 

Giac [F]

\[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{4} \,d x } \] Input:

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*sec(f*x + e)^2 + a)*sec(f*x + e)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{{\cos \left (e+f\,x\right )}^4} \,d x \] Input:

int((a + b/cos(e + f*x)^2)^(1/2)/cos(e + f*x)^4,x)
 

Output:

int((a + b/cos(e + f*x)^2)^(1/2)/cos(e + f*x)^4, x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{4}d x \] Input:

int(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^(1/2),x)
 

Output:

int(sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**4,x)