\(\int \frac {\sqrt {a+b \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx\) [1316]

Optimal result

Integrand size = 29, antiderivative size = 283 \[ \int \frac {\sqrt {a+b \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {i \sqrt {a-i b} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c-i d)^{5/2} f}+\frac {i \sqrt {a+i b} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c+i d)^{5/2} f}-\frac {2 d \sqrt {a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 d \left (6 a c d-b \left (5 c^2-d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 3.33 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {a+b \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {-\frac {3 i \sqrt {-a+i b} \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(-c+i d)^{5/2}}+\frac {3 i \sqrt {a+i b} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c+i d)^{5/2}}+\frac {2 d \sqrt {a+b \tan (e+f x)} \left (-6 b c^3+a d \left (7 c^2+d^2\right )+d \left (6 a c d+b \left (-5 c^2+d^2\right )\right ) \tan (e+f x)\right )}{(b c-a d) \left (c^2+d^2\right )^2 (c+d \tan (e+f x))^{3/2}}}{3 f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.77 (sec) , antiderivative size = 351, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4051

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4099

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4098

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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

Input:

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

Output:

int((a+b*tan(f*x+e))^(1/2)/(c+d*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. 19053 vs. \(2 (227) = 454\).

Time = 40.33 (sec) , antiderivative size = 19053, normalized size of antiderivative = 67.33 \[ \int \frac {\sqrt {a+b \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((a+b*tan(f*x+e))^(1/2)/(c+d*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*d 
**3 + 3*tan(e + f*x)**2*c*d**2 + 3*tan(e + f*x)*c**2*d + c**3),x)