\(\int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx\) [307]

Optimal result

Integrand size = 25, antiderivative size = 208 \[ \int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx=\frac {3 b^{5/2} d^3 \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{32 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}-\frac {3 b^{5/2} d^3 \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{32 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}-\frac {3 b d^2 \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{16 f}+\frac {b (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}{4 f} \] Output:

3/32*b^(5/2)*d^3*arctan((b*sin(f*x+e))^(1/2)/b^(1/2))*(b*tan(f*x+e))^(1/2) 
/f/(d*sec(f*x+e))^(1/2)/(b*sin(f*x+e))^(1/2)-3/32*b^(5/2)*d^3*arctanh((b*s 
in(f*x+e))^(1/2)/b^(1/2))*(b*tan(f*x+e))^(1/2)/f/(d*sec(f*x+e))^(1/2)/(b*s 
in(f*x+e))^(1/2)-3/16*b*d^2*(d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(3/2)/f+1/ 
4*b*(d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(3/2)/f
 

Mathematica [A] (verified)

Time = 2.13 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.74 \[ \int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx=\frac {d^4 (b \tan (e+f x))^{5/2} \left (3 \arctan \left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right ) \sec ^2(e+f x)^{3/4}-3 \text {arctanh}\left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right ) \sec ^2(e+f x)^{3/4}+2 \sec ^2(e+f x) \left (-3+4 \sec ^2(e+f x)\right ) \tan ^{\frac {3}{2}}(e+f x)\right )}{32 f (d \sec (e+f x))^{3/2} \tan ^{\frac {5}{2}}(e+f x)} \] Input:

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

Output:

(d^4*(b*Tan[e + f*x])^(5/2)*(3*ArcTan[Sqrt[Tan[e + f*x]]/(Sec[e + f*x]^2)^ 
(1/4)]*(Sec[e + f*x]^2)^(3/4) - 3*ArcTanh[Sqrt[Tan[e + f*x]]/(Sec[e + f*x] 
^2)^(1/4)]*(Sec[e + f*x]^2)^(3/4) + 2*Sec[e + f*x]^2*(-3 + 4*Sec[e + f*x]^ 
2)*Tan[e + f*x]^(3/2)))/(32*f*(d*Sec[e + f*x])^(3/2)*Tan[e + f*x]^(5/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.66 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.82, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 3091, 3042, 3093, 3042, 3096, 3042, 3044, 27, 266, 827, 216, 219}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int (b \tan (e+f x))^{5/2} (d \sec (e+f x))^{5/2}dx\)

\(\Big \downarrow \) 3091

\(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \int (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \int (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}dx\)

\(\Big \downarrow \) 3093

\(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {1}{4} d^2 \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}dx+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {1}{4} d^2 \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}dx+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\)

\(\Big \downarrow \) 3096

\(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {d^3 \sqrt {b \tan (e+f x)} \int \sec (e+f x) \sqrt {b \sin (e+f x)}dx}{4 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {d^3 \sqrt {b \tan (e+f x)} \int \frac {\sqrt {b \sin (e+f x)}}{\cos (e+f x)}dx}{4 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\)

\(\Big \downarrow \) 3044

\(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {d^3 \sqrt {b \tan (e+f x)} \int \frac {b^2 \sqrt {b \sin (e+f x)}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{4 b f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {b d^3 \sqrt {b \tan (e+f x)} \int \frac {\sqrt {b \sin (e+f x)}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{4 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {b d^3 \sqrt {b \tan (e+f x)} \int \frac {b^2 \sin ^2(e+f x)}{b^2-b^4 \sin ^4(e+f x)}d\sqrt {b \sin (e+f x)}}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {b d^3 \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}-\frac {1}{2} \int \frac {1}{b^2 \sin ^2(e+f x)+b}d\sqrt {b \sin (e+f x)}\right )}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {b d^3 \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}-\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}\right )}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f}-\frac {3}{8} b^2 \left (\frac {b d^3 \sqrt {b \tan (e+f x)} \left (\frac {\text {arctanh}\left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}\right )}{2 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^2 (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 b f}\right )\)

Input:

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

Output:

(b*(d*Sec[e + f*x])^(5/2)*(b*Tan[e + f*x])^(3/2))/(4*f) - (3*b^2*((b*d^3*( 
-1/2*ArcTan[Sqrt[b]*Sin[e + f*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Sin[e + f*x]]/ 
(2*Sqrt[b]))*Sqrt[b*Tan[e + f*x]])/(2*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e 
+ f*x]]) + (d^2*Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(3/2))/(2*b*f)))/8
 

Defintions of rubi rules used

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 3042

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

rule 3044

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ 
Symbol] :> Simp[1/(a*f)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a 
*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&  !(I 
ntegerQ[(m - 1)/2] && LtQ[0, m, n])
 

rule 3091

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m 
 + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1))   Int[(a*Sec[e + f*x])^m*( 
b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & 
& NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
 

rule 3093

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[a^2*(a*Sec[e + f*x])^(m - 2)*((b*Tan[e + f*x])^(n + 
1)/(b*f*(m + n - 1))), x] + Simp[a^2*((m - 2)/(m + n - 1))   Int[(a*Sec[e + 
 f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && ( 
GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[ 
2*m, 2*n]
 

rule 3096

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[a^(m + n)*((b*Tan[e + f*x])^n/((a*Sec[e + f*x])^n*(b* 
Sin[e + f*x])^n))   Int[(b*Sin[e + f*x])^n/Cos[e + f*x]^(m + n), x], x] /; 
FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[n + 1/2] && IntegerQ[m + 1/2]
 

Maple [A] (verified)

Time = 267.10 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.04

method	result	size
default	\(-\frac {\sqrt {b \tan \left (f x +e \right )}\, b^{2} d^{2} \sqrt {d \sec \left (f x +e \right )}\, \left (-3 \cos \left (f x +e \right ) \operatorname {arctanh}\left (\frac {\sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sin \left (f x +e \right )}{\cos \left (f x +e \right )-1}\right )-3 \cos \left (f x +e \right ) \arctan \left (\frac {\sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sin \left (f x +e \right )}{\cos \left (f x +e \right )-1}\right )+\tan \left (f x +e \right ) \sec \left (f x +e \right )^{2} \left (6 \cos \left (f x +e \right )^{3}+6 \cos \left (f x +e \right )^{2}-8 \cos \left (f x +e \right )-8\right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\right )}{32 f \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\)	\(216\)

Input:

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

Output:

-1/32/f*(b*tan(f*x+e))^(1/2)*b^2*d^2*(d*sec(f*x+e))^(1/2)/(1+cos(f*x+e))/( 
sin(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*(-3*cos(f*x+e)*arctanh((sin(f*x+e)/(1+c 
os(f*x+e))^2)^(1/2)*sin(f*x+e)/(cos(f*x+e)-1))-3*cos(f*x+e)*arctan((sin(f* 
x+e)/(1+cos(f*x+e))^2)^(1/2)*sin(f*x+e)/(cos(f*x+e)-1))+tan(f*x+e)*sec(f*x 
+e)^2*(6*cos(f*x+e)^3+6*cos(f*x+e)^2-8*cos(f*x+e)-8)*(sin(f*x+e)/(1+cos(f* 
x+e))^2)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (168) = 336\).

Time = 0.29 (sec) , antiderivative size = 852, normalized size of antiderivative = 4.10 \[ \int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/256*(6*sqrt(-b*d)*b^2*d^2*arctan(1/4*(cos(f*x + e)^3 - 5*cos(f*x + e)^2 
 - (cos(f*x + e)^2 + 6*cos(f*x + e) + 4)*sin(f*x + e) - 2*cos(f*x + e) + 4 
)*sqrt(-b*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))/(b*d*c 
os(f*x + e)^2 - b*d - (b*d*cos(f*x + e) + b*d)*sin(f*x + e)))*cos(f*x + e) 
^3 + 3*sqrt(-b*d)*b^2*d^2*cos(f*x + e)^3*log((b*d*cos(f*x + e)^4 - 72*b*d* 
cos(f*x + e)^2 + 8*(7*cos(f*x + e)^3 - (cos(f*x + e)^3 - 8*cos(f*x + e))*s 
in(f*x + e) - 8*cos(f*x + e))*sqrt(-b*d)*sqrt(b*sin(f*x + e)/cos(f*x + e)) 
*sqrt(d/cos(f*x + e)) + 72*b*d + 28*(b*d*cos(f*x + e)^2 - 2*b*d)*sin(f*x + 
 e))/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - 2)*sin(f*x + 
 e) + 8)) - 16*(3*b^2*d^2*cos(f*x + e)^2 - 4*b^2*d^2)*sqrt(b*sin(f*x + e)/ 
cos(f*x + e))*sqrt(d/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^3), 1/256 
*(6*sqrt(b*d)*b^2*d^2*arctan(1/4*(cos(f*x + e)^3 - 5*cos(f*x + e)^2 + (cos 
(f*x + e)^2 + 6*cos(f*x + e) + 4)*sin(f*x + e) - 2*cos(f*x + e) + 4)*sqrt( 
b*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))/(b*d*cos(f*x + 
 e)^2 - b*d + (b*d*cos(f*x + e) + b*d)*sin(f*x + e)))*cos(f*x + e)^3 + 3*s 
qrt(b*d)*b^2*d^2*cos(f*x + e)^3*log((b*d*cos(f*x + e)^4 - 72*b*d*cos(f*x + 
 e)^2 + 8*(7*cos(f*x + e)^3 + (cos(f*x + e)^3 - 8*cos(f*x + e))*sin(f*x + 
e) - 8*cos(f*x + e))*sqrt(b*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/co 
s(f*x + e)) + 72*b*d - 28*(b*d*cos(f*x + e)^2 - 2*b*d)*sin(f*x + e))/(cos( 
f*x + e)^4 - 8*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8...
 

Sympy [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[4,14]%%%}+%%%{6,[4,12]%%%}+%%%{15,[4,10]%%%}+%%%{20 
,[4,8]%%%
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(d)*sqrt(b)*int(sqrt(tan(e + f*x))*sqrt(sec(e + f*x))*sec(e + f*x)**2* 
tan(e + f*x)**2,x)*b**2*d**2