\(\int \frac {\tan ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [447]

Optimal result

Integrand size = 25, antiderivative size = 246 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {2 (2 a+b) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 (a+b)^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {a \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 (a+b) f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (2 a+b) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{3 (a+b)^2 f}+\frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{3 (a+b) f} \] Output:

2/3*(2*a+b)*(cos(f*x+e)^2)^(1/2)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*sec(f* 
x+e)*(a+b*sin(f*x+e)^2)^(1/2)/(a+b)^2/f/(1+b*sin(f*x+e)^2/a)^(1/2)-1/3*a*( 
cos(f*x+e)^2)^(1/2)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(1+b*sin 
(f*x+e)^2/a)^(1/2)/(a+b)/f/(a+b*sin(f*x+e)^2)^(1/2)-2/3*(2*a+b)*(a+b*sin(f 
*x+e)^2)^(1/2)*tan(f*x+e)/(a+b)^2/f+1/3*sec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1 
/2)*tan(f*x+e)/(a+b)/f
 

Mathematica [A] (verified)

Time = 2.75 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.76 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {4 a (2 a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )-2 a (a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )-\frac {\left (2 \left (4 a^2+3 a b+b^2\right ) \cos (2 (e+f x))+(2 a+b) (2 a-b-b \cos (4 (e+f x)))\right ) \sec ^2(e+f x) \tan (e+f x)}{\sqrt {2}}}{6 (a+b)^2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \] Input:

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

Output:

(4*a*(2*a + b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, - 
(b/a)] - 2*a*(a + b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + 
f*x, -(b/a)] - ((2*(4*a^2 + 3*a*b + b^2)*Cos[2*(e + f*x)] + (2*a + b)*(2*a 
 - b - b*Cos[4*(e + f*x)]))*Sec[e + f*x]^2*Tan[e + f*x])/Sqrt[2])/(6*(a + 
b)^2*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3675, 372, 402, 25, 399, 323, 321, 330, 327}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3675

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \int \frac {\sin ^4(e+f x)}{\left (1-\sin ^2(e+f x)\right )^{5/2} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 399

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

\(\Big \downarrow \) 323

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

\(\Big \downarrow \) 321

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 (a+b) \left (1-\sin ^2(e+f x)\right )^{3/2}}-\frac {\frac {2 (2 a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \sqrt {1-\sin ^2(e+f x)}}-\frac {2 (2 a+b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)-\frac {a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{\sqrt {a+b \sin ^2(e+f x)}}}{a+b}}{3 (a+b)}\right )}{f}\)

\(\Big \downarrow \) 330

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 (a+b) \left (1-\sin ^2(e+f x)\right )^{3/2}}-\frac {\frac {2 (2 a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \sqrt {1-\sin ^2(e+f x)}}-\frac {\frac {2 (2 a+b) \sqrt {a+b \sin ^2(e+f x)} \int \frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{\sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{\sqrt {a+b \sin ^2(e+f x)}}}{a+b}}{3 (a+b)}\right )}{f}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 (a+b) \left (1-\sin ^2(e+f x)\right )^{3/2}}-\frac {\frac {2 (2 a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \sqrt {1-\sin ^2(e+f x)}}-\frac {\frac {2 (2 a+b) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{\sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{\sqrt {a+b \sin ^2(e+f x)}}}{a+b}}{3 (a+b)}\right )}{f}\)

Input:

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

Output:

(Sqrt[Cos[e + f*x]^2]*Sec[e + f*x]*((Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^ 
2])/(3*(a + b)*(1 - Sin[e + f*x]^2)^(3/2)) - ((2*(2*a + b)*Sin[e + f*x]*Sq 
rt[a + b*Sin[e + f*x]^2])/((a + b)*Sqrt[1 - Sin[e + f*x]^2]) - ((2*(2*a + 
b)*EllipticE[ArcSin[Sin[e + f*x]], -(b/a)]*Sqrt[a + b*Sin[e + f*x]^2])/Sqr 
t[1 + (b*Sin[e + f*x]^2)/a] - (a*(a + b)*EllipticF[ArcSin[Sin[e + f*x]], - 
(b/a)]*Sqrt[1 + (b*Sin[e + f*x]^2)/a])/Sqrt[a + b*Sin[e + f*x]^2])/(a + b) 
)/(3*(a + b))))/f
 

Defintions of rubi rules used

rule 25

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

rule 321

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c 
/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 
0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
 

rule 323

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*x^2]   Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + ( 
d/c)*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]
 

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 330

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
Sqrt[a + b*x^2]/Sqrt[1 + (b/a)*x^2]   Int[Sqrt[1 + (b/a)*x^2]/Sqrt[c + d*x^ 
2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ[a, 
0]
 

rule 372

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

rule 399

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_) 
^2]), x_Symbol] :> Simp[f/b   Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] + 
 Simp[(b*e - a*f)/b   Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; Fr 
eeQ[{a, b, c, d, e, f}, x] &&  !((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && 
(PosQ[d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c])))))
 

rule 402

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

rule 3042

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

rule 3675

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ 
(m_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff^(m + 1 
)*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x]))   Subst[Int[x^m*((a + b*ff^2*x^2) 
^p/(1 - ff^2*x^2)^((m + 1)/2)), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b 
, e, f, p}, x] && IntegerQ[m/2] &&  !IntegerQ[p]
 

Maple [A] (verified)

Time = 4.24 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.53

method	result	size
default	\(-\frac {2 \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, b \left (2 a +b \right ) \cos \left (f x +e \right )^{4} \sin \left (f x +e \right )-\sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, \left (4 a^{2}+7 a b +3 b^{2}\right ) \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )-\sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, a \left (\operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a +\operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -4 \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a -2 \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b \right ) \cos \left (f x +e \right )^{2}+\sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, \left (a^{2}+2 a b +b^{2}\right ) \sin \left (f x +e \right )}{3 \left (1+\sin \left (f x +e \right )\right ) \sqrt {-\left (a +b \sin \left (f x +e \right )^{2}\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \left (\sin \left (f x +e \right )-1\right ) \left (a +b \right )^{2} \cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\)	\(377\)

Input:

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

Output:

-1/3*(2*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b*(2*a+b)*cos(f*x+e)^4* 
sin(f*x+e)-(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(4*a^2+7*a*b+3*b^2)* 
cos(f*x+e)^2*sin(f*x+e)-(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*(cos(f*x+e)^2)^( 
1/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*a*(EllipticF(sin(f*x+e),(- 
b/a)^(1/2))*a+EllipticF(sin(f*x+e),(-b/a)^(1/2))*b-4*EllipticE(sin(f*x+e), 
(-b/a)^(1/2))*a-2*EllipticE(sin(f*x+e),(-b/a)^(1/2))*b)*cos(f*x+e)^2+(-b*c 
os(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(a^2+2*a*b+b^2)*sin(f*x+e))/(1+sin(f 
*x+e))/(-(a+b*sin(f*x+e)^2)*(sin(f*x+e)-1)*(1+sin(f*x+e)))^(1/2)/(sin(f*x+ 
e)-1)/(a+b)^2/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 845, normalized size of antiderivative = 3.43 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/3*((2*(2*I*a*b^2 + I*b^3)*sqrt(-b)*sqrt((a^2 + a*b)/b^2)*cos(f*x + e)^3 
- (-4*I*a^2*b - 4*I*a*b^2 - I*b^3)*sqrt(-b)*cos(f*x + e)^3)*sqrt((2*b*sqrt 
((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 + a* 
b)/b^2) + 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e))), (8*a^2 + 8*a*b + b 
^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (2*(-2*I*a*b^2 - I*b^3) 
*sqrt(-b)*sqrt((a^2 + a*b)/b^2)*cos(f*x + e)^3 - (4*I*a^2*b + 4*I*a*b^2 + 
I*b^3)*sqrt(-b)*cos(f*x + e)^3)*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b) 
/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f 
*x + e) - I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a 
^2 + a*b)/b^2))/b^2) + (2*(-3*I*a^2*b - 5*I*a*b^2 - 2*I*b^3)*sqrt(-b)*sqrt 
((a^2 + a*b)/b^2)*cos(f*x + e)^3 - (-6*I*a^3 - 5*I*a^2*b - I*a*b^2)*sqrt(- 
b)*cos(f*x + e)^3)*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_ 
f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) + I*s 
in(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2 
))/b^2) + (2*(3*I*a^2*b + 5*I*a*b^2 + 2*I*b^3)*sqrt(-b)*sqrt((a^2 + a*b)/b 
^2)*cos(f*x + e)^3 - (6*I*a^3 + 5*I*a^2*b + I*a*b^2)*sqrt(-b)*cos(f*x + e) 
^3)*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqrt(( 
2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) - I*sin(f*x + e))), 
(8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (a*b^ 
2 + b^3 - 2*(2*a*b^2 + b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a ...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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