\(\int \frac {1}{(a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \, dx\) [1254]

Optimal result

Integrand size = 27, antiderivative size = 244 \[ \int \frac {1}{(a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(a-i b)^2 \sqrt {c-i d} f}+\frac {i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(a+i b)^2 \sqrt {c+i d} f}-\frac {b^{3/2} \left (4 a b c-5 a^2 d-b^2 d\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\left (a^2+b^2\right )^2 (b c-a d)^{3/2} f}-\frac {b^2 \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))} \] Output:

-I*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(a-I*b)^2/(c-I*d)^(1/2)/f 
+I*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(a+I*b)^2/(c+I*d)^(1/2)/f 
-b^(3/2)*(-5*a^2*d+4*a*b*c-b^2*d)*arctanh(b^(1/2)*(c+d*tan(f*x+e))^(1/2)/( 
-a*d+b*c)^(1/2))/(a^2+b^2)^2/(-a*d+b*c)^(3/2)/f-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))
 

Mathematica [A] (verified)

Time = 1.69 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \, dx=\frac {-\frac {i \left (\frac {(a+i b)^2 (b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {(a-i b)^2 (-b c+a d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}\right )}{a^2+b^2}+\frac {b^{3/2} \left (-4 a b c+5 a^2 d+b^2 d\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\left (a^2+b^2\right ) \sqrt {b c-a d}}-\frac {b^2 \sqrt {c+d \tan (e+f x)}}{a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d) f} \] Input:

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

Output:

(((-I)*(((a + I*b)^2*(b*c - a*d)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - 
 I*d]])/Sqrt[c - I*d] + ((a - I*b)^2*(-(b*c) + a*d)*ArcTanh[Sqrt[c + d*Tan 
[e + f*x]]/Sqrt[c + I*d]])/Sqrt[c + I*d]))/(a^2 + b^2) + (b^(3/2)*(-4*a*b* 
c + 5*a^2*d + b^2*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - 
 a*d]])/((a^2 + b^2)*Sqrt[b*c - a*d]) - (b^2*Sqrt[c + d*Tan[e + f*x]])/(a 
+ b*Tan[e + f*x]))/((a^2 + b^2)*(b*c - a*d)*f)
 

Rubi [A] (warning: unable to verify)

Time = 1.80 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.12, 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, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

\(\displaystyle \frac {\frac {b^2 \left (-5 a^2 d+4 a b c-b^2 d\right ) \int \frac {\tan ^2(e+f x)+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\int \frac {2 \left (\left (a^2-b^2\right ) (b c-a d)-2 a b (b c-a d) \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}}{2 \left (a^2+b^2\right ) (b c-a d)}-\frac {b^2 \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {b^2 \left (-5 a^2 d+4 a b c-b^2 d\right ) \int \frac {\tan ^2(e+f x)+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {2 \int \frac {\left (a^2-b^2\right ) (b c-a d)-2 a b (b c-a d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}}{2 \left (a^2+b^2\right ) (b c-a d)}-\frac {b^2 \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {b^2 \left (-5 a^2 d+4 a b c-b^2 d\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {2 \int \frac {\left (a^2-b^2\right ) (b c-a d)-2 a b (b c-a d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}}{2 \left (a^2+b^2\right ) (b c-a d)}-\frac {b^2 \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}\)

\(\Big \downarrow \) 4022

\(\displaystyle -\frac {b^2 \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}+\frac {\frac {b^2 \left (-5 a^2 d+4 a b c-b^2 d\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {2 \left (\frac {1}{2} (a-i b)^2 (b c-a d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b)^2 (b c-a d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx\right )}{a^2+b^2}}{2 \left (a^2+b^2\right ) (b c-a d)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {b^2 \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}+\frac {\frac {b^2 \left (-5 a^2 d+4 a b c-b^2 d\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {2 \left (\frac {1}{2} (a-i b)^2 (b c-a d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b)^2 (b c-a d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx\right )}{a^2+b^2}}{2 \left (a^2+b^2\right ) (b c-a d)}\)

\(\Big \downarrow \) 4020

\(\displaystyle -\frac {b^2 \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}+\frac {\frac {b^2 \left (-5 a^2 d+4 a b c-b^2 d\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {2 \left (\frac {i (a+i b)^2 (b c-a d) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (a-i b)^2 (b c-a d) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}\right )}{a^2+b^2}}{2 \left (a^2+b^2\right ) (b c-a d)}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {b^2 \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}+\frac {\frac {b^2 \left (-5 a^2 d+4 a b c-b^2 d\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {2 \left (\frac {i (a-i b)^2 (b c-a d) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {i (a+i b)^2 (b c-a d) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}\right )}{a^2+b^2}}{2 \left (a^2+b^2\right ) (b c-a d)}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {b^2 \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}+\frac {\frac {b^2 \left (-5 a^2 d+4 a b c-b^2 d\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {2 \left (\frac {(a-i b)^2 (b c-a d) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(a+i b)^2 (b c-a d) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}\right )}{a^2+b^2}}{2 \left (a^2+b^2\right ) (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {b^2 \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}+\frac {\frac {b^2 \left (-5 a^2 d+4 a b c-b^2 d\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {2 \left (\frac {(a-i b)^2 (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}+\frac {(a+i b)^2 (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}\right )}{a^2+b^2}}{2 \left (a^2+b^2\right ) (b c-a d)}\)

\(\Big \downarrow \) 4117

\(\displaystyle -\frac {b^2 \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}+\frac {\frac {b^2 \left (-5 a^2 d+4 a b c-b^2 d\right ) \int \frac {1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{f \left (a^2+b^2\right )}+\frac {2 \left (\frac {(a-i b)^2 (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}+\frac {(a+i b)^2 (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}\right )}{a^2+b^2}}{2 \left (a^2+b^2\right ) (b c-a d)}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {b^2 \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}+\frac {\frac {2 b^2 \left (-5 a^2 d+4 a b c-b^2 d\right ) \int \frac {1}{a+\frac {b (c+d \tan (e+f x))}{d}-\frac {b c}{d}}d\sqrt {c+d \tan (e+f x)}}{d f \left (a^2+b^2\right )}+\frac {2 \left (\frac {(a-i b)^2 (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}+\frac {(a+i b)^2 (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}\right )}{a^2+b^2}}{2 \left (a^2+b^2\right ) (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {b^2 \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}+\frac {-\frac {2 b^{3/2} \left (-5 a^2 d+4 a b c-b^2 d\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{f \left (a^2+b^2\right ) \sqrt {b c-a d}}+\frac {2 \left (\frac {(a-i b)^2 (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}+\frac {(a+i b)^2 (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}\right )}{a^2+b^2}}{2 \left (a^2+b^2\right ) (b c-a d)}\)

Input:

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

Output:

((2*(((a + I*b)^2*(b*c - a*d)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c 
- I*d]*f) + ((a - I*b)^2*(b*c - a*d)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/( 
Sqrt[c + I*d]*f)))/(a^2 + b^2) - (2*b^(3/2)*(4*a*b*c - 5*a^2*d - b^2*d)*Ar 
cTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/((a^2 + b^2)*Sq 
rt[b*c - a*d]*f))/(2*(a^2 + b^2)*(b*c - a*d)) - (b^2*Sqrt[c + d*Tan[e + f* 
x]])/((a^2 + b^2)*(b*c - a*d)*f*(a + b*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. \(6547\) vs. \(2(212)=424\).

Time = 0.40 (sec) , antiderivative size = 6548, normalized size of antiderivative = 26.84

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(6548\)
default	\(\text {Expression too large to display}\)	\(6548\)

Input:

int(1/(a+b*tan(f*x+e))^2/(c+d*tan(f*x+e))^(1/2),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. 7092 vs. \(2 (204) = 408\).

Time = 34.17 (sec) , antiderivative size = 14222, normalized size of antiderivative = 58.29 \[ \int \frac {1}{(a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[0,14,6]%%%}+%%%{6,[0,12,6]%%%}+%%%{15,[0,10,6]%%%}+ 
%%%{20,[0
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 12.55 (sec) , antiderivative size = 51069, normalized size of antiderivative = 209.30 \[ \int \frac {1}{(a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan(((((((16*(2*b^13*d^11*f^2 - 24*a^2*b^11*d^11*f^2 - 196*a^4*b^9*d^11* 
f^2 - 120*a^6*b^7*d^11*f^2 + 50*a^8*b^5*d^11*f^2 + 8*b^13*c^2*d^9*f^2 - 8* 
a^2*b^11*c^2*d^9*f^2 + 64*a^3*b^10*c^3*d^8*f^2 - 232*a^4*b^9*c^2*d^9*f^2 + 
 96*a^5*b^8*c^3*d^8*f^2 - 216*a^6*b^7*c^2*d^9*f^2 - 32*a*b^12*c^3*d^8*f^2 
+ 208*a^3*b^10*c*d^10*f^2 + 288*a^5*b^8*c*d^10*f^2 + 80*a^7*b^6*c*d^10*f^2 
))/(a^10*d^2*f^5 + b^10*c^2*f^5 + 4*a^2*b^8*c^2*f^5 + 6*a^4*b^6*c^2*f^5 + 
4*a^6*b^4*c^2*f^5 + a^8*b^2*c^2*f^5 + a^2*b^8*d^2*f^5 + 4*a^4*b^6*d^2*f^5 
+ 6*a^6*b^4*d^2*f^5 + 4*a^8*b^2*d^2*f^5 - 2*a*b^9*c*d*f^5 - 2*a^9*b*c*d*f^ 
5 - 8*a^3*b^7*c*d*f^5 - 12*a^5*b^5*c*d*f^5 - 8*a^7*b^3*c*d*f^5) + (((16*(c 
 + d*tan(e + f*x))^(1/2)*(8*a*b^14*d^11*f^2 + 4*b^15*c*d^10*f^2 + 36*a^3*b 
^12*d^11*f^2 + 316*a^5*b^10*d^11*f^2 + 552*a^7*b^8*d^11*f^2 + 256*a^9*b^6* 
d^11*f^2 - 12*a^11*b^4*d^11*f^2 - 4*a^13*b^2*d^11*f^2 - 20*b^15*c^3*d^8*f^ 
2 + 116*a^2*b^13*c^3*d^8*f^2 - 220*a^3*b^12*c^2*d^9*f^2 + 216*a^4*b^11*c^3 
*d^8*f^2 - 104*a^5*b^10*c^2*d^9*f^2 + 8*a^6*b^9*c^3*d^8*f^2 + 232*a^7*b^8* 
c^2*d^9*f^2 - 68*a^8*b^7*c^3*d^8*f^2 + 156*a^9*b^6*c^2*d^9*f^2 + 4*a^10*b^ 
5*c^3*d^8*f^2 - 12*a^11*b^4*c^2*d^9*f^2 - 52*a*b^14*c^2*d^9*f^2 + 80*a^2*b 
^13*c*d^10*f^2 - 156*a^4*b^11*c*d^10*f^2 - 640*a^6*b^9*c*d^10*f^2 - 500*a^ 
8*b^7*c*d^10*f^2 - 80*a^10*b^5*c*d^10*f^2 + 12*a^12*b^3*c*d^10*f^2))/(a^10 
*d^2*f^4 + b^10*c^2*f^4 + 4*a^2*b^8*c^2*f^4 + 6*a^4*b^6*c^2*f^4 + 4*a^6*b^ 
4*c^2*f^4 + a^8*b^2*c^2*f^4 + a^2*b^8*d^2*f^4 + 4*a^4*b^6*d^2*f^4 + 6*a...
 

Reduce [F]

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

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

Output:

( - 2*sqrt(tan(e + f*x)*d + c) + int(sqrt(tan(e + f*x)*d + c)/(tan(e + f*x 
)**3*a**2*b**2*d**3 - 4*tan(e + f*x)**3*a*b**3*c*d**2 + 4*tan(e + f*x)**3* 
b**4*c**2*d + 2*tan(e + f*x)**2*a**3*b*d**3 - 7*tan(e + f*x)**2*a**2*b**2* 
c*d**2 + 4*tan(e + f*x)**2*a*b**3*c**2*d + 4*tan(e + f*x)**2*b**4*c**3 + t 
an(e + f*x)*a**4*d**3 - 2*tan(e + f*x)*a**3*b*c*d**2 - 4*tan(e + f*x)*a**2 
*b**2*c**2*d + 8*tan(e + f*x)*a*b**3*c**3 + a**4*c*d**2 - 4*a**3*b*c**2*d 
+ 4*a**2*b**2*c**3),x)*tan(e + f*x)*a**3*b*d**3*f - 6*int(sqrt(tan(e + f*x 
)*d + c)/(tan(e + f*x)**3*a**2*b**2*d**3 - 4*tan(e + f*x)**3*a*b**3*c*d**2 
 + 4*tan(e + f*x)**3*b**4*c**2*d + 2*tan(e + f*x)**2*a**3*b*d**3 - 7*tan(e 
 + f*x)**2*a**2*b**2*c*d**2 + 4*tan(e + f*x)**2*a*b**3*c**2*d + 4*tan(e + 
f*x)**2*b**4*c**3 + tan(e + f*x)*a**4*d**3 - 2*tan(e + f*x)*a**3*b*c*d**2 
- 4*tan(e + f*x)*a**2*b**2*c**2*d + 8*tan(e + f*x)*a*b**3*c**3 + a**4*c*d* 
*2 - 4*a**3*b*c**2*d + 4*a**2*b**2*c**3),x)*tan(e + f*x)*a**2*b**2*c*d**2* 
f + 12*int(sqrt(tan(e + f*x)*d + c)/(tan(e + f*x)**3*a**2*b**2*d**3 - 4*ta 
n(e + f*x)**3*a*b**3*c*d**2 + 4*tan(e + f*x)**3*b**4*c**2*d + 2*tan(e + f* 
x)**2*a**3*b*d**3 - 7*tan(e + f*x)**2*a**2*b**2*c*d**2 + 4*tan(e + f*x)**2 
*a*b**3*c**2*d + 4*tan(e + f*x)**2*b**4*c**3 + tan(e + f*x)*a**4*d**3 - 2* 
tan(e + f*x)*a**3*b*c*d**2 - 4*tan(e + f*x)*a**2*b**2*c**2*d + 8*tan(e + f 
*x)*a*b**3*c**3 + a**4*c*d**2 - 4*a**3*b*c**2*d + 4*a**2*b**2*c**3),x)*tan 
(e + f*x)*a*b**3*c**2*d*f - 8*int(sqrt(tan(e + f*x)*d + c)/(tan(e + f*x...