\(\int (a+b \sec ^2(e+f x))^{3/2} \tan ^5(e+f x) \, dx\) [389]

Optimal result

Integrand size = 25, antiderivative size = 135 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {a \sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f}-\frac {(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{7/2}}{7 b^2 f} \] Output:

-a^(3/2)*arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/f+a*(a+b*sec(f*x+e)^2)^ 
(1/2)/f+1/3*(a+b*sec(f*x+e)^2)^(3/2)/f-1/5*(a+2*b)*(a+b*sec(f*x+e)^2)^(5/2 
)/b^2/f+1/7*(a+b*sec(f*x+e)^2)^(7/2)/b^2/f
 

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=\frac {\sqrt {a+b \sec ^2(e+f x)} \left (35 b^2 \left (a+b \sec ^2(e+f x)\right )-42 b \left (a+b \sec ^2(e+f x)\right )^2+15 \left (a+b \sec ^2(e+f x)\right )^3-\frac {105 a b^2 \text {arctanh}\left (\sqrt {\frac {a+b \sec ^2(e+f x)}{a}}\right )}{\sqrt {\frac {a+b \sec ^2(e+f x)}{a}}}+21 a \left (5 b^2-\left (a+b \sec ^2(e+f x)\right )^2\right )\right )}{105 b^2 f} \] Input:

Integrate[(a + b*Sec[e + f*x]^2)^(3/2)*Tan[e + f*x]^5,x]
 

Output:

(Sqrt[a + b*Sec[e + f*x]^2]*(35*b^2*(a + b*Sec[e + f*x]^2) - 42*b*(a + b*S 
ec[e + f*x]^2)^2 + 15*(a + b*Sec[e + f*x]^2)^3 - (105*a*b^2*ArcTanh[Sqrt[( 
a + b*Sec[e + f*x]^2)/a]])/Sqrt[(a + b*Sec[e + f*x]^2)/a] + 21*a*(5*b^2 - 
(a + b*Sec[e + f*x]^2)^2)))/(105*b^2*f)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4627, 354, 99, 2009}

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.

\(\displaystyle \int \tan ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^5 \left (a+b \sec (e+f x)^2\right )^{3/2}dx\)

\(\Big \downarrow \) 4627

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 99

\(\displaystyle \frac {\int \left (\frac {\left (b \sec ^2(e+f x)+a\right )^{5/2}}{b}+\cos (e+f x) \left (b \sec ^2(e+f x)+a\right )^{3/2}+\frac {(-a-2 b) \left (b \sec ^2(e+f x)+a\right )^{3/2}}{b}\right )d\sec ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )+\frac {2 \left (a+b \sec ^2(e+f x)\right )^{7/2}}{7 b^2}-\frac {2 (a+2 b) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b^2}+\frac {2}{3} \left (a+b \sec ^2(e+f x)\right )^{3/2}+2 a \sqrt {a+b \sec ^2(e+f x)}}{2 f}\)

Input:

Int[(a + b*Sec[e + f*x]^2)^(3/2)*Tan[e + f*x]^5,x]
 

Output:

(-2*a^(3/2)*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]] + 2*a*Sqrt[a + b*S 
ec[e + f*x]^2] + (2*(a + b*Sec[e + f*x]^2)^(3/2))/3 - (2*(a + 2*b)*(a + b* 
Sec[e + f*x]^2)^(5/2))/(5*b^2) + (2*(a + b*Sec[e + f*x]^2)^(7/2))/(7*b^2)) 
/(2*f)
 

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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. \(437\) vs. \(2(115)=230\).

Time = 36.38 (sec) , antiderivative size = 438, normalized size of antiderivative = 3.24

method	result	size
default	\(-\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (105 a^{\frac {3}{2}} \ln \left (4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \cos \left (f x +e \right ) a \right ) b^{2} \cos \left (f x +e \right )^{3}+6 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{3} \left (\cos \left (f x +e \right )^{3}+\cos \left (f x +e \right )^{2}\right )+3 \left (14 \cos \left (f x +e \right )^{3}+14 \cos \left (f x +e \right )^{2}-\cos \left (f x +e \right )-1\right ) a^{2} b \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 a \,b^{2} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (21-35 \cos \left (f x +e \right )^{3}-35 \cos \left (f x +e \right )^{2}+21 \cos \left (f x +e \right )-6 \sec \left (f x +e \right )-6 \sec \left (f x +e \right )^{2}\right )+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{3} \left (-35 \cos \left (f x +e \right )-35+42 \sec \left (f x +e \right )+42 \sec \left (f x +e \right )^{2}-15 \sec \left (f x +e \right )^{3}-15 \sec \left (f x +e \right )^{4}\right )\right )}{105 f \,b^{2} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (a \cos \left (f x +e \right )^{3}+a \cos \left (f x +e \right )^{2}+\cos \left (f x +e \right ) b +b \right )}\)	\(438\)

Input:

int((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x,method=_RETURNVERBOSE)
 

Output:

-1/105/f/b^2*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2 
)^(1/2)/(a*cos(f*x+e)^3+a*cos(f*x+e)^2+cos(f*x+e)*b+b)*(105*a^(3/2)*ln(4*a 
^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(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^2*cos(f*x+e)^3 
+6*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*(cos(f*x+e)^3+cos(f*x+e 
)^2)+3*(14*cos(f*x+e)^3+14*cos(f*x+e)^2-cos(f*x+e)-1)*a^2*b*((b+a*cos(f*x+ 
e)^2)/(1+cos(f*x+e))^2)^(1/2)+4*a*b^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2 
)^(1/2)*(21-35*cos(f*x+e)^3-35*cos(f*x+e)^2+21*cos(f*x+e)-6*sec(f*x+e)-6*s 
ec(f*x+e)^2)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^3*(-35*cos(f*x+ 
e)-35+42*sec(f*x+e)+42*sec(f*x+e)^2-15*sec(f*x+e)^3-15*sec(f*x+e)^4))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (115) = 230\).

Time = 6.88 (sec) , antiderivative size = 527, normalized size of antiderivative = 3.90 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=\left [\frac {105 \, a^{\frac {3}{2}} b^{2} \cos \left (f x + e\right )^{6} \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 (2 \, {\left (3 \, a^{3} + 21 \, a^{2} b - 70 \, a b^{2}\right )} \cos \left (f x + e\right )^{6} - {\left (3 \, a^{2} b - 84 \, a b^{2} + 35 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 15 \, b^{3} - 6 \, {\left (4 \, a b^{2} - 7 \, b^{3}\right )} \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}}}}{840 \, b^{2} f \cos \left (f x + e\right )^{6}}, \frac {105 \, \sqrt {-a} a b^{2} \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 ) \cos \left (f x + e\right )^{6} - 4 \, {\left (2 \, {\left (3 \, a^{3} + 21 \, a^{2} b - 70 \, a b^{2}\right )} \cos \left (f x + e\right )^{6} - {\left (3 \, a^{2} b - 84 \, a b^{2} + 35 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 15 \, b^{3} - 6 \, {\left (4 \, a b^{2} - 7 \, b^{3}\right )} \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}}}}{420 \, b^{2} f \cos \left (f x + e\right )^{6}}\right ] \] Input:

integrate((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x, algorithm="fricas")
 

Output:

[1/840*(105*a^(3/2)*b^2*cos(f*x + e)^6*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*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)/cos(f 
*x + e)^2)) - 8*(2*(3*a^3 + 21*a^2*b - 70*a*b^2)*cos(f*x + e)^6 - (3*a^2*b 
 - 84*a*b^2 + 35*b^3)*cos(f*x + e)^4 - 15*b^3 - 6*(4*a*b^2 - 7*b^3)*cos(f* 
x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(b^2*f*cos(f*x + e) 
^6), 1/420*(105*sqrt(-a)*a*b^2*arctan(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*co 
s(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))*cos(f*x + e)^6 - 4 
*(2*(3*a^3 + 21*a^2*b - 70*a*b^2)*cos(f*x + e)^6 - (3*a^2*b - 84*a*b^2 + 3 
5*b^3)*cos(f*x + e)^4 - 15*b^3 - 6*(4*a*b^2 - 7*b^3)*cos(f*x + e)^2)*sqrt( 
(a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(b^2*f*cos(f*x + e)^6)]
 

Sympy [F]

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

integrate((a+b*sec(f*x+e)**2)**(3/2)*tan(f*x+e)**5,x)
 

Output:

Integral((a + b*sec(e + f*x)**2)**(3/2)*tan(e + f*x)**5, x)
 

Maxima [F]

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

integrate((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*tan(f*x + e)^5, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2026 vs. \(2 (115) = 230\).

Time = 2.12 (sec) , antiderivative size = 2026, normalized size of antiderivative = 15.01 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=\text {Too large to display} \] Input:

integrate((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x, algorithm="giac")
 

Output:

2/105*(105*a^2*arctan(-1/2*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*ta 
n(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))*sgn(cos( 
f*x + e))/sqrt(-a) - 2*(105*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*t 
an(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))^13*a^2*sgn(cos(f*x + e)) - 1575 
*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*t 
an(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))^12*sqrt(a + b)*a^2*sgn(cos(f*x + e)) + 70*(129*a^3 + 21*a^ 
2*b - 96*a*b^2 - 32*b^3)*(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))^11*sgn(cos(f*x + e)) - 70*(387*a^3 
 - 417*a^2*b - 96*a*b^2 + 128*b^3)*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - s 
qrt(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))^10*sqrt(a + b)*sgn(cos(f 
*x + e)) + 7*(6525*a^4 - 11790*a^3*b - 2235*a^2*b^2 + 9088*a*b^3 - 1344*b^ 
4)*(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))^9*sgn(cos(f*x + e)) - 7*(5355*a^4 - 28290*a^3*b + 28995* 
a^2*b^2 - 3008*a*b^3 - 896*b^4)*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - s...
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^5\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \] Input:

int(tan(e + f*x)^5*(a + b/cos(e + f*x)^2)^(3/2),x)
 

Output:

int(tan(e + f*x)^5*(a + b/cos(e + f*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=\frac {60 \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{4} b -48 \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{2} b +32 \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2} b +105 \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{4} a -156 \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{2} a +344 \sqrt {\sec \left (f x +e \right )^{2} b +a}\, a -45 \left (\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{5}}{\sec \left (f x +e \right )^{2} b +a}d x \right ) a b f -108 \left (\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{3}}{\sec \left (f x +e \right )^{2} b +a}d x \right ) a^{2} f +312 \left (\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )}{\sec \left (f x +e \right )^{2} b +a}d x \right ) a^{2} f}{420 f} \] Input:

int((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x)
 

Output:

(60*sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**2*tan(e + f*x)**4*b - 48*sqr 
t(sec(e + f*x)**2*b + a)*sec(e + f*x)**2*tan(e + f*x)**2*b + 32*sqrt(sec(e 
 + f*x)**2*b + a)*sec(e + f*x)**2*b + 105*sqrt(sec(e + f*x)**2*b + a)*tan( 
e + f*x)**4*a - 156*sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x)**2*a + 344*sq 
rt(sec(e + f*x)**2*b + a)*a - 45*int((sqrt(sec(e + f*x)**2*b + a)*sec(e + 
f*x)**2*tan(e + f*x)**5)/(sec(e + f*x)**2*b + a),x)*a*b*f - 108*int((sqrt( 
sec(e + f*x)**2*b + a)*tan(e + f*x)**3)/(sec(e + f*x)**2*b + a),x)*a**2*f 
+ 312*int((sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x))/(sec(e + f*x)**2*b + 
a),x)*a**2*f)/(420*f)