\(\int \frac {(a+b \tan (e+f x))^2}{\sqrt {c+d \tan (e+f x)}} \, dx\) [1251]

Optimal result

Integrand size = 27, antiderivative size = 134 \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i (a-i b)^2 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d} f}+\frac {i (a+i b)^2 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d} f}+\frac {2 b^2 \sqrt {c+d \tan (e+f x)}}{d f} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {-\frac {i (a-i b)^2 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {i (a+i b)^2 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}+\frac {2 b^2 \sqrt {c+d \tan (e+f x)}}{d}}{f} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.63 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.84, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4026, 3042, 4022, 3042, 4020, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4026

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 4026

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3482\) vs. \(2(112)=224\).

Time = 0.46 (sec) , antiderivative size = 3483, normalized size of antiderivative = 25.99

method	result	size
parts	\(\text {Expression too large to display}\)	\(3483\)
derivativedivides	\(\text {Expression too large to display}\)	\(5680\)
default	\(\text {Expression too large to display}\)	\(5680\)

Input:

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

Output:

a^2*(1/4/f/d/(c^2+d^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^ 
2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2+1/4 
/f*d/(c^2+d^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2) 
+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-1/4/f/d/(c^2+d^ 
2)^(3/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^ 
(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3-1/4/f*d/(c^2+d^2) 
^(3/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1 
/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c-1/f/d/(c^2+d^2)^(1/2) 
/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^ 
2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2-1/f*d/(c^2+d^2)^(1 
/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2 
+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))+1/f/d/(c^2+d^2)^(3/ 
2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+ 
d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^4+3/f*d/(c^2+d^2)^ 
(3/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c 
^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2+2/f*d^3/(c^2+ 
d^2)^(3/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+ 
(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))-1/4/f/d/(c^2 
+d^2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e) 
-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2-1/4/f*d/(c^2+d^2)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2796 vs. \(2 (104) = 208\).

Time = 0.23 (sec) , antiderivative size = 2796, normalized size of antiderivative = 20.87 \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(d*f*sqrt(-((c^2 + d^2)*f^2*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2*b^6)* 
c^2 - 8*(a^7*b - 7*a^5*b^3 + 7*a^3*b^5 - a*b^7)*c*d + (a^8 - 12*a^6*b^2 + 
38*a^4*b^4 - 12*a^2*b^6 + b^8)*d^2)/((c^4 + 2*c^2*d^2 + d^4)*f^4)) + (a^4 
- 6*a^2*b^2 + b^4)*c + 4*(a^3*b - a*b^3)*d)/((c^2 + d^2)*f^2))*log(-(4*(a^ 
7*b + a^5*b^3 - a^3*b^5 - a*b^7)*c - (a^8 - 4*a^6*b^2 - 10*a^4*b^4 - 4*a^2 
*b^6 + b^8)*d)*sqrt(d*tan(f*x + e) + c) + ((2*a*b*c^2*d + 2*a*b*d^3 + (a^2 
 - b^2)*c^3 + (a^2 - b^2)*c*d^2)*f^3*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2* 
b^6)*c^2 - 8*(a^7*b - 7*a^5*b^3 + 7*a^3*b^5 - a*b^7)*c*d + (a^8 - 12*a^6*b 
^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)*d^2)/((c^4 + 2*c^2*d^2 + d^4)*f^4)) + 
(8*(a^4*b^2 - a^2*b^4)*c^2 - 2*(3*a^5*b - 10*a^3*b^3 + 3*a*b^5)*c*d + (a^6 
 - 7*a^4*b^2 + 7*a^2*b^4 - b^6)*d^2)*f)*sqrt(-((c^2 + d^2)*f^2*sqrt(-(16*( 
a^6*b^2 - 2*a^4*b^4 + a^2*b^6)*c^2 - 8*(a^7*b - 7*a^5*b^3 + 7*a^3*b^5 - a* 
b^7)*c*d + (a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)*d^2)/((c^4 + 
 2*c^2*d^2 + d^4)*f^4)) + (a^4 - 6*a^2*b^2 + b^4)*c + 4*(a^3*b - a*b^3)*d) 
/((c^2 + d^2)*f^2))) - d*f*sqrt(-((c^2 + d^2)*f^2*sqrt(-(16*(a^6*b^2 - 2*a 
^4*b^4 + a^2*b^6)*c^2 - 8*(a^7*b - 7*a^5*b^3 + 7*a^3*b^5 - a*b^7)*c*d + (a 
^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)*d^2)/((c^4 + 2*c^2*d^2 + 
d^4)*f^4)) + (a^4 - 6*a^2*b^2 + b^4)*c + 4*(a^3*b - a*b^3)*d)/((c^2 + d^2) 
*f^2))*log(-(4*(a^7*b + a^5*b^3 - a^3*b^5 - a*b^7)*c - (a^8 - 4*a^6*b^2 - 
10*a^4*b^4 - 4*a^2*b^6 + b^8)*d)*sqrt(d*tan(f*x + e) + c) - ((2*a*b*c^2...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

integrate((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,[0,9,3]%%%}+%%%{4,[0,7,3]%%%}+%%%{6,[0,5,3]%%%}+%%%{ 
4,[0,3,3]
 

Mupad [B] (verification not implemented)

Time = 4.30 (sec) , antiderivative size = 2287, normalized size of antiderivative = 17.07 \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*b^2*(c + d*tan(e + f*x))^(1/2))/(d*f) - atan(((((16*(2*b^2*d^3*f^2 - 2* 
a^2*d^3*f^2 + 4*a*b*c*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)* 
(-(a*b^3*4i - a^3*b*4i + a^4 + b^4 - 6*a^2*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1 
/2))*(-(a*b^3*4i - a^3*b*4i + a^4 + b^4 - 6*a^2*b^2)/(4*(c*f^2 - d*f^2*1i) 
))^(1/2) - (16*(c + d*tan(e + f*x))^(1/2)*(a^4*d^2 + b^4*d^2 - 6*a^2*b^2*d 
^2))/f^2)*(-(a*b^3*4i - a^3*b*4i + a^4 + b^4 - 6*a^2*b^2)/(4*(c*f^2 - d*f^ 
2*1i)))^(1/2)*1i - (((16*(2*b^2*d^3*f^2 - 2*a^2*d^3*f^2 + 4*a*b*c*d^2*f^2) 
)/f^3 + 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(a*b^3*4i - a^3*b*4i + a^4 + 
 b^4 - 6*a^2*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2))*(-(a*b^3*4i - a^3*b*4i + 
a^4 + b^4 - 6*a^2*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2) + (16*(c + d*tan(e + 
f*x))^(1/2)*(a^4*d^2 + b^4*d^2 - 6*a^2*b^2*d^2))/f^2)*(-(a*b^3*4i - a^3*b* 
4i + a^4 + b^4 - 6*a^2*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2)*1i)/((((16*(2*b^ 
2*d^3*f^2 - 2*a^2*d^3*f^2 + 4*a*b*c*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e 
+ f*x))^(1/2)*(-(a*b^3*4i - a^3*b*4i + a^4 + b^4 - 6*a^2*b^2)/(4*(c*f^2 - 
d*f^2*1i)))^(1/2))*(-(a*b^3*4i - a^3*b*4i + a^4 + b^4 - 6*a^2*b^2)/(4*(c*f 
^2 - d*f^2*1i)))^(1/2) - (16*(c + d*tan(e + f*x))^(1/2)*(a^4*d^2 + b^4*d^2 
 - 6*a^2*b^2*d^2))/f^2)*(-(a*b^3*4i - a^3*b*4i + a^4 + b^4 - 6*a^2*b^2)/(4 
*(c*f^2 - d*f^2*1i)))^(1/2) + (((16*(2*b^2*d^3*f^2 - 2*a^2*d^3*f^2 + 4*a*b 
*c*d^2*f^2))/f^3 + 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(a*b^3*4i - a^3*b 
*4i + a^4 + b^4 - 6*a^2*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2))*(-(a*b^3*4i...
 

Reduce [F]

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

int((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)*a**2 - int((sqrt(tan(e + f*x)*d + c)*tan(e + f 
*x)**2)/(tan(e + f*x)*d + c),x)*a**2*d*f + int((sqrt(tan(e + f*x)*d + c)*t 
an(e + f*x)**2)/(tan(e + f*x)*d + c),x)*b**2*d*f + 2*int((sqrt(tan(e + f*x 
)*d + c)*tan(e + f*x))/(tan(e + f*x)*d + c),x)*a*b*d*f)/(d*f)