\(\int \frac {\tan ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [409]

Optimal result

Integrand size = 25, antiderivative size = 120 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{\sqrt {a} f}-\frac {(a+3 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 b^{3/2} f}+\frac {\tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 b f} \] Output:

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

Mathematica [A] (verified)

Time = 1.56 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.63 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\left (\frac {2 b \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {a}}-\frac {(a+3 b) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {b}}\right ) \sqrt {a+2 b+a \cos (2 e+2 f x)} \sec (e+f x)}{2 \sqrt {2} b f \sqrt {a+b \sec ^2(e+f x)}}+\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \tan (e+f x)}{4 b f \sqrt {a+b \sec ^2(e+f x)}} \] Input:

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

Output:

(((2*b*ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]])/Sqrt 
[a] - ((a + 3*b)*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x 
]^2]])/Sqrt[b])*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]*Sec[e + f*x])/(2*Sqrt[2 
]*b*f*Sqrt[a + b*Sec[e + f*x]^2]) + ((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e 
+ f*x]^2*Tan[e + f*x])/(4*b*f*Sqrt[a + b*Sec[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.02, 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, 381, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 381

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

(-1/2*((-2*b*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]] 
)/Sqrt[a] + ((a + 3*b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e 
 + f*x]^2]])/Sqrt[b])/b + (Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(2 
*b))/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 381

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q 
+ 1)/(b*d*(m + 2*(p + q) + 1))), x] - Simp[e^4/(b*d*(m + 2*(p + q) + 1)) 
Int[(e*x)^(m - 4)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 
2*q - 1) + b*c*(m + 2*p - 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q 
}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 3] && 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] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(731\) vs. \(2(102)=204\).

Time = 4.36 (sec) , antiderivative size = 732, normalized size of antiderivative = 6.10

method	result	size
default	\(\frac {\frac {\sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b \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 ) \left (-1-\sec \left (f x +e \right )\right )}{4}+\frac {\sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \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 ) \left (-3-3 \sec \left (f x +e \right )\right )}{4}+\frac {\sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b \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 ) \left (-1-\sec \left (f x +e \right )\right )}{4}+\frac {\sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \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 ) \left (-3-3 \sec \left (f x +e \right )\right )}{4}+\frac {b^{\frac {5}{2}} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \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 ) \left (4+4 \sec \left (f x +e \right )\right )}{4}+\frac {b^{\frac {3}{2}} \sqrt {-a}\, a \tan \left (f x +e \right )}{2}+\frac {\sqrt {-a}\, b^{\frac {5}{2}} \tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}}{2}}{f \sqrt {-a}\, b^{\frac {5}{2}} \sqrt {a +b \sec \left (f x +e \right )^{2}}}\)	\(732\)

Input:

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

Output:

1/f*(1/4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b*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)) 
*(-1-sec(f*x+e))+1/4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 
)*b^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) 
+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)/(si 
n(f*x+e)+1))*(-3-3*sec(f*x+e))+1/4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f 
*x+e))^2)^(1/2)*a*b*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))*(-1-sec(f*x+e))+1/4*(-a)^(1/2)*((b+a*cos(f*x+e 
)^2)/(1+cos(f*x+e))^2)^(1/2)*b^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)+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))*(-3-3*sec(f*x+e))+1/4*b^(5/2)*((b 
+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*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)*(4+4*sec(f*x+e))+1/2*b^(3/2)*(-a)^(1/ 
2)*a*tan(f*x+e)+1/2*(-a)^(1/2)*b^(5/2)*tan(f*x+e)*sec(f*x+e)^2)/(-a)^(1/2) 
/b^(5/2)/(a+b*sec(f*x+e)^2)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (102) = 204\).

Time = 0.49 (sec) , antiderivative size = 1507, normalized size of antiderivative = 12.56 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/8*(sqrt(-a)*b^2*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)*c 
os(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)/c 
os(f*x + e)^2)*sin(f*x + e)) - (a^2 + 3*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)/c 
os(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) - 4*a*b*sqrt((a*cos(f 
*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a*b^2*f*cos(f*x + e)), -1/8* 
(sqrt(-a)*b^2*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)) + 2*(a^2 + 3*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) - 4*...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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