\(\int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{5/2}} \, dx\) [1286]

Optimal result

Integrand size = 29, antiderivative size = 280 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{5/2}} \, dx=-\frac {i \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{5/2} f}+\frac {i \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2} f}-\frac {2 b \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac {2 b \left (6 a b c-5 a^2 d+b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d) f \sqrt {a+b \tan (e+f x)}} \] Output:

-I*(c-I*d)^(1/2)*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2 
)/(c+d*tan(f*x+e))^(1/2))/(a-I*b)^(5/2)/f+I*(c+I*d)^(1/2)*arctanh((c+I*d)^ 
(1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))/(a+I*b) 
^(5/2)/f-2/3*b*(c+d*tan(f*x+e))^(1/2)/(a^2+b^2)/f/(a+b*tan(f*x+e))^(3/2)-2 
/3*b*(-5*a^2*d+6*a*b*c+b^2*d)*(c+d*tan(f*x+e))^(1/2)/(a^2+b^2)^2/(-a*d+b*c 
)/f/(a+b*tan(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 3.27 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{5/2}} \, dx=\frac {-\frac {3 i \sqrt {-c+i d} \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(-a+i b)^{5/2}}+\frac {3 i \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2}}+\frac {2 b \sqrt {c+d \tan (e+f x)} \left (7 a^2 b c+b^3 c-6 a^3 d+b \left (6 a b c-5 a^2 d+b^2 d\right ) \tan (e+f x)\right )}{\left (a^2+b^2\right )^2 (-b c+a d) (a+b \tan (e+f x))^{3/2}}}{3 f} \] Input:

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

Output:

(((-3*I)*Sqrt[-c + I*d]*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e + f*x]])/ 
(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(-a + I*b)^(5/2) + ((3*I)*Sqrt 
[c + I*d]*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]* 
Sqrt[c + d*Tan[e + f*x]])])/(a + I*b)^(5/2) + (2*b*Sqrt[c + d*Tan[e + f*x] 
]*(7*a^2*b*c + b^3*c - 6*a^3*d + b*(6*a*b*c - 5*a^2*d + b^2*d)*Tan[e + f*x 
]))/((a^2 + b^2)^2*(-(b*c) + a*d)*(a + b*Tan[e + f*x])^(3/2)))/(3*f)
 

Rubi [A] (verified)

Time = 1.76 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.24, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 4051, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 4051

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

\(\displaystyle \frac {-\frac {2 \int -\frac {3 \left ((b c-a d) \left (c a^2+2 b d a-b^2 c\right )-(b c-a d) \left (-d a^2+2 b c a+b^2 d\right ) \tan (e+f x)\right )}{2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (a^2+b^2\right ) (b c-a d)}-\frac {2 b \left (-5 a^2 d+6 a b c+b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)}}}{3 \left (a^2+b^2\right )}-\frac {2 b \sqrt {c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \frac {(b c-a d) \left (c a^2+2 b d a-b^2 c\right )-(b c-a d) \left (-d a^2+2 b c a+b^2 d\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (a^2+b^2\right ) (b c-a d)}-\frac {2 b \left (-5 a^2 d+6 a b c+b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)}}}{3 \left (a^2+b^2\right )}-\frac {2 b \sqrt {c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \int \frac {(b c-a d) \left (c a^2+2 b d a-b^2 c\right )-(b c-a d) \left (-d a^2+2 b c a+b^2 d\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (a^2+b^2\right ) (b c-a d)}-\frac {2 b \left (-5 a^2 d+6 a b c+b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)}}}{3 \left (a^2+b^2\right )}-\frac {2 b \sqrt {c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4099

\(\displaystyle -\frac {2 b \sqrt {c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}+\frac {-\frac {2 b \left (-5 a^2 d+6 a b c+b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)}}+\frac {3 \left (\frac {1}{2} (a-i b)^2 (c+i d) (b c-a d) \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b)^2 (c-i d) (b c-a d) \int \frac {i \tan (e+f x)+1}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx\right )}{\left (a^2+b^2\right ) (b c-a d)}}{3 \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 b \sqrt {c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}+\frac {-\frac {2 b \left (-5 a^2 d+6 a b c+b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)}}+\frac {3 \left (\frac {1}{2} (a-i b)^2 (c+i d) (b c-a d) \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b)^2 (c-i d) (b c-a d) \int \frac {i \tan (e+f x)+1}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx\right )}{\left (a^2+b^2\right ) (b c-a d)}}{3 \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 4098

\(\displaystyle -\frac {2 b \sqrt {c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}+\frac {-\frac {2 b \left (-5 a^2 d+6 a b c+b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)}}+\frac {3 \left (\frac {(a-i b)^2 (c+i d) (b c-a d) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{2 f}+\frac {(a+i b)^2 (c-i d) (b c-a d) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{2 f}\right )}{\left (a^2+b^2\right ) (b c-a d)}}{3 \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {2 b \sqrt {c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}+\frac {-\frac {2 b \left (-5 a^2 d+6 a b c+b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)}}+\frac {3 \left (\frac {(a-i b)^2 (c+i d) (b c-a d) \int \frac {1}{-i a+b+\frac {(i c-d) (a+b \tan (e+f x))}{c+d \tan (e+f x)}}d\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{f}+\frac {(a+i b)^2 (c-i d) (b c-a d) \int \frac {1}{i a+b-\frac {(i c+d) (a+b \tan (e+f x))}{c+d \tan (e+f x)}}d\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{f}\right )}{\left (a^2+b^2\right ) (b c-a d)}}{3 \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 b \sqrt {c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}+\frac {-\frac {2 b \left (-5 a^2 d+6 a b c+b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)}}+\frac {3 \left (\frac {i (a-i b)^2 \sqrt {c+i d} (b c-a d) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a+i b}}-\frac {i (a+i b)^2 \sqrt {c-i d} (b c-a d) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a-i b}}\right )}{\left (a^2+b^2\right ) (b c-a d)}}{3 \left (a^2+b^2\right )}\)

Input:

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

Output:

(-2*b*Sqrt[c + d*Tan[e + f*x]])/(3*(a^2 + b^2)*f*(a + b*Tan[e + f*x])^(3/2 
)) + ((3*(((-I)*(a + I*b)^2*Sqrt[c - I*d]*(b*c - a*d)*ArcTanh[(Sqrt[c - I* 
d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]])])/(S 
qrt[a - I*b]*f) + (I*(a - I*b)^2*Sqrt[c + I*d]*(b*c - a*d)*ArcTanh[(Sqrt[c 
 + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]]) 
])/(Sqrt[a + I*b]*f)))/((a^2 + b^2)*(b*c - a*d)) - (2*b*(6*a*b*c - 5*a^2*d 
 + b^2*d)*Sqrt[c + d*Tan[e + f*x]])/((a^2 + b^2)*(b*c - a*d)*f*Sqrt[a + b* 
Tan[e + f*x]]))/(3*(a^2 + b^2))
 

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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*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 3042

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

rule 4051

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

rule 4098

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[A^2/f   Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f* 
x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && 
 NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
 

rule 4099

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(A + I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T 
an[e + f*x]), x], x] + Simp[(A - I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d 
*Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A 
, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^2, 
0]
 

rule 4132

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ 
[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

Maple [F(-1)]

Timed out.

\[\int \frac {\sqrt {c +d \tan \left (f x +e \right )}}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {5}{2}}}d x\]

Input:

int((c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^(5/2),x)
 

Output:

int((c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^(5/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17607 vs. \(2 (224) = 448\).

Time = 14.24 (sec) , antiderivative size = 17607, normalized size of antiderivative = 62.88 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^(5/2),x, algorithm="fric 
as")
 

Output:

Too large to include
 

Sympy [F]

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

integrate((c+d*tan(f*x+e))**(1/2)/(a+b*tan(f*x+e))**(5/2),x)
 

Output:

Integral(sqrt(c + d*tan(e + f*x))/(a + b*tan(e + f*x))**(5/2), x)
 

Maxima [F(-2)]

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

integrate((c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^(5/2),x, algorithm="maxi 
ma")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(((2*b*d+2*a*c)^2>0)', see `assum 
e?` for mo
 

Giac [F]

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

integrate((c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^(5/2),x, algorithm="giac 
")
 

Output:

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

Mupad [F(-1)]

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

int((c + d*tan(e + f*x))^(1/2)/(a + b*tan(e + f*x))^(5/2),x)
 

Output:

\text{Hanged}
 

Reduce [F]

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

int((c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^(5/2),x)
 

Output:

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