\(\int \frac {\tan (e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [430]

Optimal result

Integrand size = 23, antiderivative size = 83 \[ \int \frac {\tan (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{5/2} f}+\frac {1}{3 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {1}{a^2 f \sqrt {a+b \sec ^2(e+f x)}} \] Output:

-arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/a^(5/2)/f+1/3/a/f/(a+b*sec(f*x+ 
e)^2)^(3/2)+1/a^2/f/(a+b*sec(f*x+e)^2)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 5.54 (sec) , antiderivative size = 521, normalized size of antiderivative = 6.28 \[ \int \frac {\tan (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^{5/2} \sec ^4(e+f x) \left (-\frac {4 (a+3 b+a \cos (2 (e+f x)))}{b^2 (a+2 b+a \cos (2 (e+f x)))^{3/2}}+\frac {6 (a+b+(a-2 b) \cos (2 (e+f x)))}{b^2 (a+2 b+a \cos (2 (e+f x)))^{3/2}}+\frac {\sqrt {2} e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \left (-\frac {\sqrt {a} \left (1+e^{2 i (e+f x)}\right ) \left (-96 b^3 e^{2 i (e+f x)}+a^3 \left (1+e^{2 i (e+f x)}\right )^2-32 a b^2 \left (1+e^{2 i (e+f x)}\right )^2-6 a^2 b \left (1+e^{2 i (e+f x)}+e^{4 i (e+f x)}\right )\right )}{b^2 \left (4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2\right )^2}+\frac {24 i f x-12 \log \left (a+2 b+a e^{2 i (e+f x)}+\sqrt {a} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )-12 \log \left (a+a e^{2 i (e+f x)}+2 b e^{2 i (e+f x)}+\sqrt {a} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )}{\sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \sec (e+f x)}{a^{5/2}}\right )}{192 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \] Input:

Integrate[Tan[e + f*x]/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

((a + 2*b + a*Cos[2*(e + f*x)])^(5/2)*Sec[e + f*x]^4*((-4*(a + 3*b + a*Cos 
[2*(e + f*x)]))/(b^2*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2)) + (6*(a + b + ( 
a - 2*b)*Cos[2*(e + f*x)]))/(b^2*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2)) + ( 
Sqrt[2]*E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I 
)*(e + f*x))]*(-((Sqrt[a]*(1 + E^((2*I)*(e + f*x)))*(-96*b^3*E^((2*I)*(e + 
 f*x)) + a^3*(1 + E^((2*I)*(e + f*x)))^2 - 32*a*b^2*(1 + E^((2*I)*(e + f*x 
)))^2 - 6*a^2*b*(1 + E^((2*I)*(e + f*x)) + E^((4*I)*(e + f*x)))))/(b^2*(4* 
b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2)^2)) + ((24*I)*f*x - 
 12*Log[a + 2*b + a*E^((2*I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e + f 
*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]] - 12*Log[a + a*E^((2*I)*(e + f*x)) 
+ 2*b*E^((2*I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + 
E^((2*I)*(e + f*x)))^2]])/Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*( 
e + f*x)))^2])*Sec[e + f*x])/a^(5/2)))/(192*f*(a + b*Sec[e + f*x]^2)^(5/2) 
)
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4627, 243, 61, 61, 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.

\(\displaystyle \int \frac {\tan (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (e+f x)}{\left (a+b \sec (e+f x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 4627

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[Tan[e + f*x]/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

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

Defintions of rubi rules used

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, 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 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 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(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 [A] (verified)

Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {1}{3 a f \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {1}{a^{2} f \sqrt {a +b \sec \left (f x +e \right )^{2}}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}{\sec \left (f x +e \right )}\right )}{f \,a^{\frac {5}{2}}}\)	\(86\)
default	\(\frac {1}{3 a f \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {1}{a^{2} f \sqrt {a +b \sec \left (f x +e \right )^{2}}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}{\sec \left (f x +e \right )}\right )}{f \,a^{\frac {5}{2}}}\)	\(86\)

Input:

int(tan(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/3/a/f/(a+b*sec(f*x+e)^2)^(3/2)+1/a^2/f/(a+b*sec(f*x+e)^2)^(1/2)-1/f/a^(5 
/2)*ln((2*a+2*a^(1/2)*(a+b*sec(f*x+e)^2)^(1/2))/sec(f*x+e))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (71) = 142\).

Time = 0.53 (sec) , antiderivative size = 494, normalized size of antiderivative = 5.95 \[ \int \frac {\tan (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} - 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) + 8 \, {\left (4 \, a^{2} \cos \left (f x + e\right )^{4} + 3 \, a b \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}, \frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) + 4 \, {\left (4 \, a^{2} \cos \left (f x + e\right )^{4} + 3 \, a b \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}\right ] \] Input:

integrate(tan(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

Output:

[1/24*(3*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(a)*log(128 
*a^4*cos(f*x + e)^8 + 256*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*co 
s(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)/cos(f*x + e)^2)) + 8*(4*a^2*cos(f*x + e)^4 + 3*a*b* 
cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a^5*f*cos(f* 
x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2*f), 1/12*(3*(a^2*cos(f*x + e 
)^4 + 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*cos(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2)) + 4*(4 
*a^2*cos(f*x + e)^4 + 3*a*b*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/co 
s(f*x + e)^2))/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2* 
f)]
 

Sympy [A] (verification not implemented)

Time = 10.06 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.20 \[ \int \frac {\tan (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b}{6 a f \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}} + \frac {b}{2 a^{2} f \sqrt {a + b \sec ^{2}{\left (e + f x \right )}}} + \frac {b \operatorname {atan}{\left (\frac {\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}{\sqrt {- a}} \right )}}{2 a^{2} f \sqrt {- a}}\right )}{b} & \text {for}\: b \neq 0 \\\frac {\log {\left (\sec ^{2}{\left (e + f x \right )} \right )}}{2 a^{\frac {5}{2}} f} & \text {otherwise} \end {cases} \] Input:

integrate(tan(f*x+e)/(a+b*sec(f*x+e)**2)**(5/2),x)
 

Output:

Piecewise((2*(b/(6*a*f*(a + b*sec(e + f*x)**2)**(3/2)) + b/(2*a**2*f*sqrt( 
a + b*sec(e + f*x)**2)) + b*atan(sqrt(a + b*sec(e + f*x)**2)/sqrt(-a))/(2* 
a**2*f*sqrt(-a)))/b, Ne(b, 0)), (log(sec(e + f*x)**2)/(2*a**(5/2)*f), True 
))
 

Maxima [F]

\[ \int \frac {\tan (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\tan \left (f x + e\right )}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(tan(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate(tan(f*x + e)/(b*sec(f*x + e)^2 + a)^(5/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (71) = 142\).

Time = 1.07 (sec) , antiderivative size = 370, normalized size of antiderivative = 4.46 \[ \int \frac {\tan (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {{\left ({\left (\frac {{\left (4 \, a^{9} b^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 3 \, a^{8} b^{3} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{10} b^{2}} - \frac {3 \, {\left (4 \, a^{9} b^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - a^{8} b^{3} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\right )}}{a^{10} b^{2}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \frac {3 \, {\left (4 \, a^{9} b^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - a^{8} b^{3} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\right )}}{a^{10} b^{2}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \frac {4 \, a^{9} b^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 3 \, a^{8} b^{3} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{a^{10} b^{2}}}{{\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b\right )}^{\frac {3}{2}}} - \frac {6 \, \arctan \left (-\frac {\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b} + \sqrt {a + b}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}}{3 \, f} \] Input:

integrate(tan(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
 

Output:

-1/3*(((((4*a^9*b^2*sgn(cos(f*x + e)) + 3*a^8*b^3*sgn(cos(f*x + e)))*tan(1 
/2*f*x + 1/2*e)^2/(a^10*b^2) - 3*(4*a^9*b^2*sgn(cos(f*x + e)) - a^8*b^3*sg 
n(cos(f*x + e)))/(a^10*b^2))*tan(1/2*f*x + 1/2*e)^2 + 3*(4*a^9*b^2*sgn(cos 
(f*x + e)) - a^8*b^3*sgn(cos(f*x + e)))/(a^10*b^2))*tan(1/2*f*x + 1/2*e)^2 
 - (4*a^9*b^2*sgn(cos(f*x + e)) + 3*a^8*b^3*sgn(cos(f*x + e)))/(a^10*b^2)) 
/(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)^(3/2) - 6*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^2*sgn(cos(f*x + e))))/f
 

Mupad [B] (verification not implemented)

Time = 19.21 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.82 \[ \int \frac {\tan (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}{a^2}+\frac {1}{3\,a}}{f\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{\sqrt {a}}\right )}{a^{5/2}\,f} \] Input:

int(tan(e + f*x)/(a + b/cos(e + f*x)^2)^(5/2),x)
 

Output:

((a + b/cos(e + f*x)^2)/a^2 + 1/(3*a))/(f*(a + b/cos(e + f*x)^2)^(3/2)) - 
atanh((a + b/cos(e + f*x)^2)^(1/2)/a^(1/2))/(a^(5/2)*f)
 

Reduce [F]

\[ \int \frac {\tan (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )}{\sec \left (f x +e \right )^{6} b^{3}+3 \sec \left (f x +e \right )^{4} a \,b^{2}+3 \sec \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

int(tan(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x)
 

Output:

int((sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x))/(sec(e + f*x)**6*b**3 + 3*s 
ec(e + f*x)**4*a*b**2 + 3*sec(e + f*x)**2*a**2*b + a**3),x)