\(\int \frac {1}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx\) [1303]

Optimal result

Integrand size = 29, antiderivative size = 218 \[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{3/2} \sqrt {c-i d} f}+\frac {i \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2} \sqrt {c+i d} f}-\frac {2 b^2 \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)}} \] 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)^(3/2)/(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)^(3/2)/(c+I*d) 
^(1/2)/f-2*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))^(1/2)
 

Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=\frac {\frac {i (a+i b) \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+i b} \sqrt {-c+i d}}+\frac {(i a+b) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} \sqrt {c+i d}}+\frac {2 b^2 \sqrt {c+d \tan (e+f x)}}{(-b c+a d) \sqrt {a+b \tan (e+f x)}}}{\left (a^2+b^2\right ) f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.23, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 4052, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4099

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4098

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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]
 

Maple [F(-1)]

Timed out.

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

Input:

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

Output:

int(1/(a+b*tan(f*x+e))^(3/2)/(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. 17637 vs. \(2 (170) = 340\).

Time = 14.40 (sec) , antiderivative size = 17637, normalized size of antiderivative = 80.90 \[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

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

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