\(\int \frac {\tan ^2(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [459]

Optimal result

Integrand size = 25, antiderivative size = 224 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {2 b \cos (e+f x) \sin (e+f x)}{(a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b)^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

-2*b*cos(f*x+e)*sin(f*x+e)/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(1/2)-2*(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)+(cos(f*x+e)^2)^(1/2)*Elliptic 
F(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)+tan(f*x+e)/(a+b)/f/(a+b*sin(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.33 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.65 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {-2 \sqrt {2} a \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+\sqrt {2} (a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )+2 (a-b \cos (2 (e+f x))) \tan (e+f x)}{\sqrt {2} (a+b)^2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \] Input:

Integrate[Tan[e + f*x]^2/(a + b*Sin[e + f*x]^2)^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3675, 373, 402, 25, 27, 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.

\(\displaystyle \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (e+f x)^2}{\left (a+b \sin (e+f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 3675

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

\(\Big \downarrow \) 373

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 399

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

\(\Big \downarrow \) 323

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

\(\Big \downarrow \) 321

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

\(\Big \downarrow \) 330

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

\(\Big \downarrow \) 327

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

Input:

Int[Tan[e + f*x]^2/(a + b*Sin[e + f*x]^2)^(3/2),x]
 

Output:

(Sqrt[Cos[e + f*x]^2]*Sec[e + f*x]*(Sin[e + f*x]/((a + b)*Sqrt[1 - Sin[e + 
 f*x]^2]*Sqrt[a + b*Sin[e + f*x]^2]) - ((2*b*Sin[e + f*x]*Sqrt[1 - Sin[e + 
 f*x]^2])/((a + b)*Sqrt[a + b*Sin[e + f*x]^2]) + ((2*EllipticE[ArcSin[Sin[ 
e + f*x]], -(b/a)]*Sqrt[a + b*Sin[e + f*x]^2])/Sqrt[1 + (b*Sin[e + f*x]^2) 
/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))/(a + b)))/f
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 373

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 + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[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 = 3.93 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.24

method	result	size
default	\(\frac {\sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, \left (a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )+b \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-2 a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-2 \cos \left (f x +e \right )^{2} b \sin \left (f x +e \right )+a \sin \left (f x +e \right )+b \sin \left (f x +e \right )\right )}{\left (a +b \right )^{2} \sqrt {-\left (a +b \sin \left (f x +e \right )^{2}\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\)	\(278\)

Input:

int(tan(f*x+e)^2/(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(a*(cos(f*x+e)^2)^(1/2)*(-b/a*c 
os(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-b/a)^(1/2))+b*(cos(f*x+e 
)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-b/a)^( 
1/2))-2*a*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE 
(sin(f*x+e),(-b/a)^(1/2))-2*cos(f*x+e)^2*b*sin(f*x+e)+a*sin(f*x+e)+b*sin(f 
*x+e))/(a+b)^2/(-(a+b*sin(f*x+e)^2)*(sin(f*x+e)-1)*(1+sin(f*x+e)))^(1/2)/c 
os(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 = 0.14 (sec) , antiderivative size = 1048, normalized size of antiderivative = 4.68 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(tan(f*x+e)^2/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

((2*(-I*b^3*cos(f*x + e)^3 + (I*a*b^2 + I*b^3)*cos(f*x + e))*sqrt(-b)*sqrt 
((a^2 + a*b)/b^2) - ((2*I*a*b^2 + I*b^3)*cos(f*x + e)^3 + (-2*I*a^2*b - 3* 
I*a*b^2 - I*b^3)*cos(f*x + e))*sqrt(-b))*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*(I*b^3*cos(f*x + e)^3 + (-I*a*b^2 - I*b 
^3)*cos(f*x + e))*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - ((-2*I*a*b^2 - I*b^3)*c 
os(f*x + e)^3 + (2*I*a^2*b + 3*I*a*b^2 + I*b^3)*cos(f*x + e))*sqrt(-b))*sq 
rt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sq 
rt((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*((I*a*b^ 
2 + I*b^3)*cos(f*x + e)^3 + (-I*a^2*b - 2*I*a*b^2 - I*b^3)*cos(f*x + e))*s 
qrt(-b)*sqrt((a^2 + a*b)/b^2) - ((2*I*a^2*b - I*a*b^2 - I*b^3)*cos(f*x + e 
)^3 + (-2*I*a^3 - I*a^2*b + 2*I*a*b^2 + I*b^3)*cos(f*x + e))*sqrt(-b))*sqr 
t((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqrt((2*b*sqr 
t((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*((-I*a*b^ 
2 - I*b^3)*cos(f*x + e)^3 + (I*a^2*b + 2*I*a*b^2 + I*b^3)*cos(f*x + e))*sq 
rt(-b)*sqrt((a^2 + a*b)/b^2) - ((-2*I*a^2*b + I*a*b^2 + I*b^3)*cos(f*x + e 
)^3 + (2*I*a^3 + I*a^2*b - 2*I*a*b^2 - I*b^3)*cos(f*x + e))*sqrt(-b))*s...
 

Sympy [F]

\[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(tan(f*x+e)**2/(a+b*sin(f*x+e)**2)**(3/2),x)
                                                                                    
                                                                                    
 

Output:

Integral(tan(e + f*x)**2/(a + b*sin(e + f*x)**2)**(3/2), x)
 

Maxima [F]

\[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(tan(f*x+e)^2/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate(tan(f*x + e)^2/(b*sin(f*x + e)^2 + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(tan(f*x+e)^2/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

integrate(tan(f*x + e)^2/(b*sin(f*x + e)^2 + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \] Input:

int(tan(e + f*x)^2/(a + b*sin(e + f*x)^2)^(3/2),x)
 

Output:

int(tan(e + f*x)^2/(a + b*sin(e + f*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{4} b^{2}+2 \sin \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:

int(tan(f*x+e)^2/(a+b*sin(f*x+e)^2)^(3/2),x)
 

Output:

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