\(\int \frac {\tan ^3(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [416]

Optimal result

Integrand size = 25, antiderivative size = 63 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2} f}-\frac {a+b}{a b f \sqrt {a+b \sec ^2(e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {a+b}{a \sqrt {a+b \sec ^2(e+f x)}}}{b f} \] Input:

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

Output:

((b*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/a^(3/2) - (a + b)/(a*Sqrt 
[a + b*Sec[e + f*x]^2]))/(b*f)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4627, 25, 354, 87, 73, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4627

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 3042

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

rule 4627

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si 
mp[1/f   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] 
, x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( 
m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers 
Q[2*n, p])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(873\) vs. \(2(55)=110\).

Time = 0.44 (sec) , antiderivative size = 874, normalized size of antiderivative = 13.87

method	result	size
default	\(\text {Expression too large to display}\)	\(874\)

Input:

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

Output:

-1/f/a^(3/2)/((-a*b)^(1/2)-a)/((-a*b)^(1/2)+a)/b/(cos(f*x+e)^2*a^2+(cos(f* 
x+e)^2+1)*a*b+b^2)/(a+b*sec(f*x+e)^2)^(3/2)*(ln(4*((b+a*cos(f*x+e)^2)/(1+c 
os(f*x+e))^2)^(1/2)*a^(1/2)*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+co 
s(f*x+e))^2)^(1/2)+4*cos(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^( 
1/2)*a^5*b*(cos(f*x+e)^2+cos(f*x+e))+ln(4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e 
))^2)^(1/2)*a^(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*cos(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^4 
*b^2*(2*cos(f*x+e)^2+2*cos(f*x+e)+2+2*sec(f*x+e))+ln(4*((b+a*cos(f*x+e)^2) 
/(1+cos(f*x+e))^2)^(1/2)*a^(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*cos(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)) 
^2)^(1/2)*a^3*b^3*(cos(f*x+e)^2+cos(f*x+e)+4+4*sec(f*x+e)+sec(f*x+e)^2+sec 
(f*x+e)^3)+ln(4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(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*cos(f*x+e)*a) 
*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^4*(2+2*sec(f*x+e)+2*sec 
(f*x+e)^2+2*sec(f*x+e)^3)+ln(4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2) 
*a^(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*cos(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^5*(sec(f*x 
+e)^2+sec(f*x+e)^3)-a^(13/2)*cos(f*x+e)^2+(-3*cos(f*x+e)^2-2)*a^(11/2)*b+a 
^(9/2)*b^2*(-3*cos(f*x+e)^2-6-sec(f*x+e)^2)+a^(7/2)*b^3*(-cos(f*x+e)^2-6-3 
*sec(f*x+e)^2)+a^(5/2)*b^4*(-2-3*sec(f*x+e)^2)-a^(3/2)*b^5*sec(f*x+e)^2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (55) = 110\).

Time = 0.21 (sec) , antiderivative size = 417, normalized size of antiderivative = 6.62 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {8 \, {\left (a^{2} + a b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )^{2} - {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right )}{8 \, {\left (a^{3} b f \cos \left (f x + e\right )^{2} + a^{2} b^{2} f\right )}}, -\frac {4 \, {\left (a^{2} + a b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )^{2} + {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right )}{4 \, {\left (a^{3} b f \cos \left (f x + e\right )^{2} + a^{2} b^{2} f\right )}}\right ] \] Input:

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

Output:

[-1/8*(8*(a^2 + a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + 
 e)^2 - (a*b*cos(f*x + e)^2 + b^2)*sqrt(a)*log(128*a^4*cos(f*x + e)^8 + 25 
6*a^3*b*cos(f*x + e)^6 + 160*a^2*b^2*cos(f*x + e)^4 + 32*a*b^3*cos(f*x + e 
)^2 + b^4 + 8*(16*a^3*cos(f*x + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2* 
cos(f*x + e)^4 + b^3*cos(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/c 
os(f*x + e)^2)))/(a^3*b*f*cos(f*x + e)^2 + a^2*b^2*f), -1/4*(4*(a^2 + a*b) 
*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)^2 + (a*b*cos(f*x 
 + e)^2 + b^2)*sqrt(-a)*arctan(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + 
 e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(2*a^3*c 
os(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2)))/(a^3*b*f*cos(f*x + e)^2 
+ a^2*b^2*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2532 vs. \(2 (55) = 110\).

Time = 0.40 (sec) , antiderivative size = 2532, normalized size of antiderivative = 40.19 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(4*a*b*cos(1/2*arctan2(a*sin(4*f*x + 4*e) + 2*(a + 2*b)*sin(2*f*x + 2* 
e), a*cos(4*f*x + 4*e) + 2*(a + 2*b)*cos(2*f*x + 2*e) + a))^3 + 4*a*b*cos( 
1/2*arctan2(a*sin(4*f*x + 4*e) + 2*(a + 2*b)*sin(2*f*x + 2*e), a*cos(4*f*x 
 + 4*e) + 2*(a + 2*b)*cos(2*f*x + 2*e) + a))*sin(1/2*arctan2(a*sin(4*f*x + 
 4*e) + 2*(a + 2*b)*sin(2*f*x + 2*e), a*cos(4*f*x + 4*e) + 2*(a + 2*b)*cos 
(2*f*x + 2*e) + a))^2 - 4*(a^2 + a*b)*sin(2*f*x + 2*e)*sin(1/2*arctan2(a*s 
in(4*f*x + 4*e) + 2*(a + 2*b)*sin(2*f*x + 2*e), a*cos(4*f*x + 4*e) + 2*(a 
+ 2*b)*cos(2*f*x + 2*e) + a)) - 4*(a^2 + 2*a*b + (a^2 + a*b)*cos(2*f*x + 2 
*e))*cos(1/2*arctan2(a*sin(4*f*x + 4*e) + 2*(a + 2*b)*sin(2*f*x + 2*e), a* 
cos(4*f*x + 4*e) + 2*(a + 2*b)*cos(2*f*x + 2*e) + a)) + (a^2*cos(4*f*x + 4 
*e)^2 + a^2*sin(4*f*x + 4*e)^2 + 4*(a^2 + 4*a*b + 4*b^2)*cos(2*f*x + 2*e)^ 
2 + 4*(a^2 + 2*a*b)*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*(a^2 + 4*a*b + 4 
*b^2)*sin(2*f*x + 2*e)^2 + a^2 + 2*(a^2 + 2*(a^2 + 2*a*b)*cos(2*f*x + 2*e) 
)*cos(4*f*x + 4*e) + 4*(a^2 + 2*a*b)*cos(2*f*x + 2*e))^(1/4)*((b*cos(1/2*a 
rctan2(a*sin(4*f*x + 4*e) + 2*(a + 2*b)*sin(2*f*x + 2*e), a*cos(4*f*x + 4* 
e) + 2*(a + 2*b)*cos(2*f*x + 2*e) + a))^2 + b*sin(1/2*arctan2(a*sin(4*f*x 
+ 4*e) + 2*(a + 2*b)*sin(2*f*x + 2*e), a*cos(4*f*x + 4*e) + 2*(a + 2*b)*co 
s(2*f*x + 2*e) + a))^2)*log(4*a^2*cos(2*f*x + 2*e)^2 + 4*a^2*sin(2*f*x + 2 
*e)^2 + 4*a^2 + 16*a*b + 16*b^2 + 8*(a^2 + 2*a*b)*cos(2*f*x + 2*e) + 8*(a^ 
2*cos(4*f*x + 4*e)^2 + a^2*sin(4*f*x + 4*e)^2 + 4*(a^2 + 4*a*b + 4*b^2)...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (55) = 110\).

Time = 0.84 (sec) , antiderivative size = 251, normalized size of antiderivative = 3.98 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {\frac {{\left (a^{3} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + a^{2} b \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{3} b} - \frac {a^{3} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + a^{2} b \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{a^{3} b}}{\sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}} - \frac {2 \, \arctan \left (-\frac {\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b} + \sqrt {a + b}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}}{f} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x)**3)/(sec(e + f*x)**4*b**2 + 
2*sec(e + f*x)**2*a*b + a**2),x)