\(\int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx\) [1310]

Optimal result

Integrand size = 29, antiderivative size = 301 \[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{3/2} (c-i d)^{3/2} f}+\frac {i \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2} (c+i d)^{3/2} f}-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}} \] 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)^(3/2)/(c-I*d)^(3/2)/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)^(3/2)/(c+I*d) 
^(3/2)/f-2*b^2/(a^2+b^2)/(-a*d+b*c)/f/(a+b*tan(f*x+e))^(1/2)/(c+d*tan(f*x+ 
e))^(1/2)-2*d*(a^2*d^2+b^2*(c^2+2*d^2))*(a+b*tan(f*x+e))^(1/2)/(a^2+b^2)/( 
-a*d+b*c)^2/(c^2+d^2)/f/(c+d*tan(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 3.35 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {\frac {(b c-a d)^2 \left (\frac {i (a+i b) (c+i d) \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+i b} \sqrt {-c+i d}}+\frac {(i a+b) (c-i d) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} \sqrt {c+i d}}\right )}{(-b c+a d) \left (c^2+d^2\right )}+\frac {2 b^2}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{(b c-a d) \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{\left (a^2+b^2\right ) (b c-a d) f} \] Input:

Integrate[1/((a + b*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(3/2)),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.85 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.27, 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.

\(\displaystyle \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4099

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4098

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

Int[1/((a + b*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(3/2)),x]
 

Output:

(-2*b^2)/((a^2 + b^2)*(b*c - a*d)*f*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Ta 
n[e + f*x]]) + ((((-I)*(a + I*b)*(c + I*d)*(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)*(c - I*d)*(b*c - a*d)^2*Ar 
cTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*T 
an[e + f*x]])])/(Sqrt[a + I*b]*Sqrt[c + I*d]*f))/((b*c - a*d)*(c^2 + d^2)) 
 - (2*d*(a^2*d^2 + b^2*(c^2 + 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]]))/((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.

\[\int \frac {1}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {3}{2}} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32905 vs. \(2 (249) = 498\).

Time = 37.19 (sec) , antiderivative size = 32905, normalized size of antiderivative = 109.32 \[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="fr 
icas")
 

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="ma 
xima")
 

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]

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

integrate(1/(a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="gi 
ac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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