\(\int \tan ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) [300]

Optimal result

Integrand size = 25, antiderivative size = 169 \[ \int \tan ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\sqrt {a-b} \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (a^2+4 a b-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 b^{3/2} f}+\frac {(a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 b f}+\frac {\tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 6.16 (sec) , antiderivative size = 767, normalized size of antiderivative = 4.54 \[ \int \tan ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {-\frac {b \left (a^2-4 b^2\right ) \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{a (a+b+(a-b) \cos (2 (e+f x)))}-\frac {4 b \left (-4 a b+4 b^2\right ) \sqrt {1+\cos (2 (e+f x))} \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{4 a \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}-\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{2 (a-b) \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}\right )}{\sqrt {a+b+(a-b) \cos (2 (e+f x))}}}{4 b f}+\frac {\sqrt {\frac {a+b+a \cos (2 (e+f x))-b \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (\frac {\sec (e+f x) (a \sin (e+f x)-6 b \sin (e+f x))}{8 b}+\frac {1}{4} \sec ^2(e+f x) \tan (e+f x)\right )}{f} \] Input:

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

Output:

-1/4*(-((b*(a^2 - 4*b^2)*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[ 
2*(e + f*x)])]*Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x 
)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f 
*x]^2)/b]*Csc[2*(e + f*x)]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*( 
e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(a*(a + b + (a 
- b)*Cos[2*(e + f*x)]))) - (4*b*(-4*a*b + 4*b^2)*Sqrt[1 + Cos[2*(e + f*x)] 
]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*((Sqrt[- 
((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b 
)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(e + 
f*x)]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f* 
x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(4*a*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt 
[a + b + (a - b)*Cos[2*(e + f*x)]]) - (Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[ 
-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos 
[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticPi[-(b/(a - b)) 
, ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[ 
2]], 1]*Sin[e + f*x]^4)/(2*(a - b)*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + 
 (a - b)*Cos[2*(e + f*x)]])))/Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])/(b*f 
) + (Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e 
+ f*x)])]*((Sec[e + f*x]*(a*Sin[e + f*x] - 6*b*Sin[e + f*x]))/(8*b) + (Sec 
[e + f*x]^2*Tan[e + f*x])/4))/f
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4153, 380, 444, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 380

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

\(\Big \downarrow \) 444

\(\displaystyle \frac {\frac {1}{4} \left (\frac {\int -\frac {\left (a^2+4 b a-8 b^2\right ) \tan ^2(e+f x)+a (a-4 b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{2 b}+\frac {(a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}\right )+\frac {1}{4} \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{4} \left (\frac {(a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {\int \frac {\left (a^2+4 b a-8 b^2\right ) \tan ^2(e+f x)+a (a-4 b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{2 b}\right )+\frac {1}{4} \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 398

\(\displaystyle \frac {\frac {1}{4} \left (\frac {(a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {\left (a^2+4 a b-8 b^2\right ) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-8 b (a-b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{2 b}\right )+\frac {1}{4} \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{4} \left (\frac {(a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {\left (a^2+4 a b-8 b^2\right ) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}-8 b (a-b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{2 b}\right )+\frac {1}{4} \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{4} \left (\frac {(a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {\frac {\left (a^2+4 a b-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b}}-8 b (a-b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{2 b}\right )+\frac {1}{4} \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {1}{4} \left (\frac {(a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {\frac {\left (a^2+4 a b-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b}}-8 b (a-b) \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}}{2 b}\right )+\frac {1}{4} \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{4} \left (\frac {(a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {\frac {\left (a^2+4 a b-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b}}-8 b \sqrt {a-b} \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 b}\right )+\frac {1}{4} \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

Input:

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

Output:

((Tan[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/4 + (-1/2*(-8*Sqrt[a - b]*b*A 
rcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]] + ((a^2 + 4*a 
*b - 8*b^2)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/Sq 
rt[b])/b + ((a - 4*b)*Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(2*b))/4)/f
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 444

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q 
_.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[f*g*(g*x)^(m - 1)*(a + b*x^2)^ 
(p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q + 1) + 1))), x] - Simp[g^2/ 
(b*d*(m + 2*(p + q + 1) + 1))   Int[(g*x)^(m - 2)*(a + b*x^2)^p*(c + d*x^2) 
^q*Simp[a*f*c*(m - 1) + (a*f*d*(m + 2*q + 1) + b*(f*c*(m + 2*p + 1) - e*d*( 
m + 2*(p + q + 1) + 1)))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, 
q}, x] && GtQ[m, 1]
 

rule 3042

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

rule 4153

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], 
 x]}, Simp[c*(ff/f)   Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, 
n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio 
nalQ[n]))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(306\) vs. \(2(147)=294\).

Time = 0.90 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.82

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (f x +e \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {\tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2}+\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 \sqrt {b}}\right )}{4 b}+b \left (\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{\sqrt {b}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{b^{2} \left (a -b \right )}\right )+\frac {a \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{b^{2} \left (a -b \right )}-\frac {\tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2}-\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 \sqrt {b}}}{f}\)	\(307\)
default	\(\frac {\frac {\tan \left (f x +e \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {\tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2}+\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 \sqrt {b}}\right )}{4 b}+b \left (\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{\sqrt {b}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{b^{2} \left (a -b \right )}\right )+\frac {a \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{b^{2} \left (a -b \right )}-\frac {\tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2}-\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 \sqrt {b}}}{f}\)	\(307\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 655, normalized size of antiderivative = 3.88 \[ \int \tan ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/16*(8*sqrt(-a + b)*b^2*log(-((a - 2*b)*tan(f*x + e)^2 + 2*sqrt(b*tan(f* 
x + e)^2 + a)*sqrt(-a + b)*tan(f*x + e) - a)/(tan(f*x + e)^2 + 1)) - (a^2 
+ 4*a*b - 8*b^2)*sqrt(b)*log(2*b*tan(f*x + e)^2 + 2*sqrt(b*tan(f*x + e)^2 
+ a)*sqrt(b)*tan(f*x + e) + a) + 2*(2*b^2*tan(f*x + e)^3 + (a*b - 4*b^2)*t 
an(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/(b^2*f), 1/16*(16*sqrt(a - b)*b^2 
*arctan(sqrt(a - b)*tan(f*x + e)/sqrt(b*tan(f*x + e)^2 + a)) - (a^2 + 4*a* 
b - 8*b^2)*sqrt(b)*log(2*b*tan(f*x + e)^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*s 
qrt(b)*tan(f*x + e) + a) + 2*(2*b^2*tan(f*x + e)^3 + (a*b - 4*b^2)*tan(f*x 
 + e))*sqrt(b*tan(f*x + e)^2 + a))/(b^2*f), 1/8*(4*sqrt(-a + b)*b^2*log(-( 
(a - 2*b)*tan(f*x + e)^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)*tan(f 
*x + e) - a)/(tan(f*x + e)^2 + 1)) + (a^2 + 4*a*b - 8*b^2)*sqrt(-b)*arctan 
(sqrt(-b)*tan(f*x + e)/sqrt(b*tan(f*x + e)^2 + a)) + (2*b^2*tan(f*x + e)^3 
 + (a*b - 4*b^2)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/(b^2*f), 1/8*(8 
*sqrt(a - b)*b^2*arctan(sqrt(a - b)*tan(f*x + e)/sqrt(b*tan(f*x + e)^2 + a 
)) + (a^2 + 4*a*b - 8*b^2)*sqrt(-b)*arctan(sqrt(-b)*tan(f*x + e)/sqrt(b*ta 
n(f*x + e)^2 + a)) + (2*b^2*tan(f*x + e)^3 + (a*b - 4*b^2)*tan(f*x + e))*s 
qrt(b*tan(f*x + e)^2 + a))/(b^2*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \tan ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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