\(\int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx\) [1259]

Optimal result

Integrand size = 27, antiderivative size = 211 \[ \int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(i a+b) (c-i d)^{3/2} f}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(i a-b) (c+i d)^{3/2} f}-\frac {2 b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\left (a^2+b^2\right ) (b c-a d)^{3/2} f}+\frac {2 d^2}{(b c-a d) \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.17 \[ \int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\frac {-\frac {i \left (\frac {(a+i b) (c+i d) (-b c+a d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {(a-i b) (c-i d) (b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}\right )}{a^2+b^2}+\frac {2 b^{5/2} \left (c^2+d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\left (a^2+b^2\right ) \sqrt {b c-a d}}-\frac {2 d^2}{\sqrt {c+d \tan (e+f x)}}}{(-b c+a d) \left (c^2+d^2\right ) f} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 1.75 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.23, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.593, Rules used = {3042, 4052, 27, 3042, 4136, 25, 3042, 4022, 3042, 4020, 25, 73, 221, 4117, 73, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-((-((((a + I*b)*(c + I*d)*(b*c - a*d)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]]) 
/(Sqrt[c - I*d]*f) + ((a - I*b)*(c - I*d)*(b*c - a*d)*ArcTan[Tan[e + f*x]/ 
Sqrt[c + I*d]])/(Sqrt[c + I*d]*f))/(a^2 + b^2)) + (2*b^(5/2)*(c^2 + d^2)*A 
rcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/((a^2 + b^2)*S 
qrt[b*c - a*d]*f))/((b*c - a*d)*(c^2 + d^2))) + (2*d^2)/((b*c - a*d)*(c^2 
+ d^2)*f*Sqrt[c + d*Tan[e + f*x]])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, 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 4020

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

rule 4022

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

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 4117

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

rule 4136

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^ 
n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ 
(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ 
e + f*x]^2)/(a + b*Tan[e + f*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] & 
&  !GtQ[n, 0] &&  !LeQ[n, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(8701\) vs. \(2(181)=362\).

Time = 0.40 (sec) , antiderivative size = 8702, normalized size of antiderivative = 41.24

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(8702\)
default	\(\text {Expression too large to display}\)	\(8702\)

Input:

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

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8967 vs. \(2 (175) = 350\).

Time = 32.69 (sec) , antiderivative size = 17973, normalized size of antiderivative = 85.18 \[ \int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 17.77 (sec) , antiderivative size = 115510, normalized size of antiderivative = 547.44 \[ \int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

(log(((((2*a^2*b^2*d^6*f^4 - b^4*d^6*f^4 - 4*a^2*b^2*c^6*f^4 - a^4*d^6*f^4 
 + 6*a^4*c^2*d^4*f^4 - 9*a^4*c^4*d^2*f^4 + 6*b^4*c^2*d^4*f^4 - 9*b^4*c^4*d 
^2*f^4 - 48*a^2*b^2*c^2*d^4*f^4 + 42*a^2*b^2*c^4*d^2*f^4 + 12*a*b^3*c*d^5* 
f^4 + 12*a*b^3*c^5*d*f^4 - 12*a^3*b*c*d^5*f^4 - 12*a^3*b*c^5*d*f^4 - 40*a* 
b^3*c^3*d^3*f^4 + 40*a^3*b*c^3*d^3*f^4)^(1/2) - a^2*c^3*f^2 + b^2*c^3*f^2 
+ 3*a^2*c*d^2*f^2 - 3*b^2*c*d^2*f^2 - 2*a*b*d^3*f^2 + 6*a*b*c^2*d*f^2)/(a^ 
4*c^6*f^4 + a^4*d^6*f^4 + b^4*c^6*f^4 + b^4*d^6*f^4 + 2*a^2*b^2*c^6*f^4 + 
2*a^2*b^2*d^6*f^4 + 3*a^4*c^2*d^4*f^4 + 3*a^4*c^4*d^2*f^4 + 3*b^4*c^2*d^4* 
f^4 + 3*b^4*c^4*d^2*f^4 + 6*a^2*b^2*c^2*d^4*f^4 + 6*a^2*b^2*c^4*d^2*f^4))^ 
(1/2)*(((((2*a^2*b^2*d^6*f^4 - b^4*d^6*f^4 - 4*a^2*b^2*c^6*f^4 - a^4*d^6*f 
^4 + 6*a^4*c^2*d^4*f^4 - 9*a^4*c^4*d^2*f^4 + 6*b^4*c^2*d^4*f^4 - 9*b^4*c^4 
*d^2*f^4 - 48*a^2*b^2*c^2*d^4*f^4 + 42*a^2*b^2*c^4*d^2*f^4 + 12*a*b^3*c*d^ 
5*f^4 + 12*a*b^3*c^5*d*f^4 - 12*a^3*b*c*d^5*f^4 - 12*a^3*b*c^5*d*f^4 - 40* 
a*b^3*c^3*d^3*f^4 + 40*a^3*b*c^3*d^3*f^4)^(1/2) - a^2*c^3*f^2 + b^2*c^3*f^ 
2 + 3*a^2*c*d^2*f^2 - 3*b^2*c*d^2*f^2 - 2*a*b*d^3*f^2 + 6*a*b*c^2*d*f^2)/( 
a^4*c^6*f^4 + a^4*d^6*f^4 + b^4*c^6*f^4 + b^4*d^6*f^4 + 2*a^2*b^2*c^6*f^4 
+ 2*a^2*b^2*d^6*f^4 + 3*a^4*c^2*d^4*f^4 + 3*a^4*c^4*d^2*f^4 + 3*b^4*c^2*d^ 
4*f^4 + 3*b^4*c^4*d^2*f^4 + 6*a^2*b^2*c^2*d^4*f^4 + 6*a^2*b^2*c^4*d^2*f^4) 
)^(1/2)*(((((2*a^2*b^2*d^6*f^4 - b^4*d^6*f^4 - 4*a^2*b^2*c^6*f^4 - a^4*d^6 
*f^4 + 6*a^4*c^2*d^4*f^4 - 9*a^4*c^4*d^2*f^4 + 6*b^4*c^2*d^4*f^4 - 9*b^...
 

Reduce [F]

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

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

Output:

(2*sqrt(tan(e + f*x)*d + c) + 2*int(sqrt(tan(e + f*x)*d + c)/(tan(e + f*x) 
**3*b*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 2*tan(e + 
f*x)*a*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*tan(e + f*x)*a*d**2*f + 2*in 
t(sqrt(tan(e + f*x)*d + c)/(tan(e + f*x)**3*b*d**2 + tan(e + f*x)**2*a*d** 
2 + 2*tan(e + f*x)**2*b*c*d + 2*tan(e + f*x)*a*c*d + tan(e + f*x)*b*c**2 + 
 a*c**2),x)*a*c*d*f + int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**3)/(tan( 
e + f*x)**3*b*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 2* 
tan(e + f*x)*a*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*tan(e + f*x)*b*d**2* 
f + int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**3)/(tan(e + f*x)**3*b*d**2 
 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 2*tan(e + f*x)*a*c*d 
 + tan(e + f*x)*b*c**2 + a*c**2),x)*b*c*d*f + int((sqrt(tan(e + f*x)*d + c 
)*tan(e + f*x)**2)/(tan(e + f*x)**3*b*d**2 + tan(e + f*x)**2*a*d**2 + 2*ta 
n(e + f*x)**2*b*c*d + 2*tan(e + f*x)*a*c*d + tan(e + f*x)*b*c**2 + a*c**2) 
,x)*tan(e + f*x)*a*d**2*f + int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2) 
/(tan(e + f*x)**3*b*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c* 
d + 2*tan(e + f*x)*a*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*a*c*d*f + int( 
(sqrt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)**3*b*d**2 + tan(e + 
f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 2*tan(e + f*x)*a*c*d + tan(e + 
f*x)*b*c**2 + a*c**2),x)*tan(e + f*x)*b*d**2*f + int((sqrt(tan(e + f*x)*d 
+ c)*tan(e + f*x))/(tan(e + f*x)**3*b*d**2 + tan(e + f*x)**2*a*d**2 + 2...