\(\int \frac {1}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx\) [1304]

Optimal result

Integrand size = 29, antiderivative size = 295 \[ \int \frac {1}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{5/2} \sqrt {c-i d} f}+\frac {i \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2} \sqrt {c+i d} f}-\frac {2 b^2 \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac {4 b^2 \left (3 a b c-4 a^2 d-b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)}} \] Output:

-I*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)/(c-I*d)^(1/2)/f+I*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)/(c+I*d) 
^(1/2)/f-2/3*b^2*(c+d*tan(f*x+e))^(1/2)/(a^2+b^2)/(-a*d+b*c)/f/(a+b*tan(f* 
x+e))^(3/2)-4/3*b^2*(-4*a^2*d+3*a*b*c-b^2*d)*(c+d*tan(f*x+e))^(1/2)/(a^2+b 
^2)^2/(-a*d+b*c)^2/f/(a+b*tan(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 1.50 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=\frac {3 i \left (\frac {(a+i b)^2 \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+i b} \sqrt {-c+i d}}+\frac {(a-i b)^2 \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} \sqrt {c+i d}}\right )+\frac {2 b^2 \left (a^2+b^2\right ) \sqrt {c+d \tan (e+f x)}}{(-b c+a d) (a+b \tan (e+f x))^{3/2}}+\frac {4 b^2 \left (-3 a b c+4 a^2 d+b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{(b c-a d)^2 \sqrt {a+b \tan (e+f x)}}}{3 \left (a^2+b^2\right )^2 f} \] Input:

Integrate[1/((a + b*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]]),x]
 

Output:

((3*I)*(((a + I*b)^2*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sq 
rt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[-a + I*b]*Sqrt[-c + I*d]) + 
 ((a - I*b)^2*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]*Sqrt[c + I*d])) + (2*b^2*(a 
^2 + b^2)*Sqrt[c + d*Tan[e + f*x]])/((-(b*c) + a*d)*(a + b*Tan[e + f*x])^( 
3/2)) + (4*b^2*(-3*a*b*c + 4*a^2*d + b^2*d)*Sqrt[c + d*Tan[e + f*x]])/((b* 
c - a*d)^2*Sqrt[a + b*Tan[e + f*x]]))/(3*(a^2 + b^2)^2*f)
 

Rubi [A] (verified)

Time = 1.82 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.28, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 4052, 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.

\(\displaystyle \int \frac {1}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4099

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4098

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 b^2 \sqrt {c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))^{3/2}}+\frac {-\frac {4 b^2 \left (-4 a^2 d+3 a b c-b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)}}+\frac {3 \left (\frac {i (a-i b)^2 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a+i b} \sqrt {c+i d}}-\frac {i (a+i b)^2 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a-i b} \sqrt {c-i d}}\right )}{\left (a^2+b^2\right ) (b c-a d)}}{3 \left (a^2+b^2\right ) (b c-a d)}\)

Input:

Int[1/((a + b*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]]),x]
 

Output:

(-2*b^2*Sqrt[c + d*Tan[e + f*x]])/(3*(a^2 + b^2)*(b*c - a*d)*f*(a + b*Tan[ 
e + f*x])^(3/2)) + ((3*(((-I)*(a + I*b)^2*(b*c - a*d)^2*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]*Sqrt[c - I*d]*f) + (I*(a - I*b)^2*(b*c - a*d)^2*ArcTanh[(Sq 
rt[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]*Sqrt[c + I*d]*f)))/((a^2 + b^2)*(b*c - a*d)) - (4*b^ 
2*(3*a*b*c - 4*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)*(b*c - a*d))
 

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 4052

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

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.

\[\int \frac {1}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {5}{2}} \sqrt {c +d \tan \left (f x +e \right )}}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 25872 vs. \(2 (239) = 478\).

Time = 23.70 (sec) , antiderivative size = 25872, normalized size of antiderivative = 87.70 \[ \int \frac {1}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {1}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=\int { \frac {1}{{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {d \tan \left (f x + e\right ) + c}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {1}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=\int { \frac {1}{{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {d \tan \left (f x + e\right ) + c}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \sqrt {\tan \left (f x +e \right ) b +a}}{\tan \left (f x +e \right )^{4} b^{3} d +3 \tan \left (f x +e \right )^{3} a \,b^{2} d +\tan \left (f x +e \right )^{3} b^{3} c +3 \tan \left (f x +e \right )^{2} a^{2} b d +3 \tan \left (f x +e \right )^{2} a \,b^{2} c +\tan \left (f x +e \right ) a^{3} d +3 \tan \left (f x +e \right ) a^{2} b c +a^{3} c}d x \] Input:

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

Output:

int((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a))/(tan(e + f*x)**4*b 
**3*d + 3*tan(e + f*x)**3*a*b**2*d + tan(e + f*x)**3*b**3*c + 3*tan(e + f* 
x)**2*a**2*b*d + 3*tan(e + f*x)**2*a*b**2*c + tan(e + f*x)*a**3*d + 3*tan( 
e + f*x)*a**2*b*c + a**3*c),x)