\(\int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx\) [361]

Optimal result

Integrand size = 25, antiderivative size = 81 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx=\frac {\arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a \sqrt {d} f}+\frac {\text {arctanh}\left (\frac {\sqrt {d} (1+\tan (e+f x))}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a \sqrt {d} f} \] Output:

arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a/d^(1/2)/f+1/2*arctanh(1/2*d^(1/2)*( 
1+tan(f*x+e))*2^(1/2)/(d*tan(f*x+e))^(1/2))*2^(1/2)/a/d^(1/2)/f
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(283\) vs. \(2(81)=162\).

Time = 0.26 (sec) , antiderivative size = 283, normalized size of antiderivative = 3.49 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx=\frac {8 d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+4 \left (-d^2\right )^{3/4} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt [4]{-d^2}}\right )-2 \sqrt {2} d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+2 \sqrt {2} d^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-4 \left (-d^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt [4]{-d^2}}\right )-\sqrt {2} d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )+\sqrt {2} d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{8 a d^2 f} \] Input:

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

Output:

(8*d^(3/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]] + 4*(-d^2)^(3/4)*ArcTan[Sq 
rt[d*Tan[e + f*x]]/(-d^2)^(1/4)] - 2*Sqrt[2]*d^(3/2)*ArcTan[1 - (Sqrt[2]*S 
qrt[d*Tan[e + f*x]])/Sqrt[d]] + 2*Sqrt[2]*d^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt 
[d*Tan[e + f*x]])/Sqrt[d]] - 4*(-d^2)^(3/4)*ArcTanh[Sqrt[d*Tan[e + f*x]]/( 
-d^2)^(1/4)] - Sqrt[2]*d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2 
]*Sqrt[d*Tan[e + f*x]]] + Sqrt[2]*d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f* 
x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(8*a*d^2*f)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4057, 3042, 4015, 221, 4117, 27, 73, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4057

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4015

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {\text {arctanh}\left (\frac {\sqrt {d} (a \tan (e+f x)+a)}{\sqrt {2} a \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a \sqrt {d} f}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {\int \frac {1}{a \sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{2 f}+\frac {\text {arctanh}\left (\frac {\sqrt {d} (a \tan (e+f x)+a)}{\sqrt {2} a \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a \sqrt {d} f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {1}{\sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{2 a f}+\frac {\text {arctanh}\left (\frac {\sqrt {d} (a \tan (e+f x)+a)}{\sqrt {2} a \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a \sqrt {d} f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\int \frac {1}{\tan (e+f x)+1}d\sqrt {d \tan (e+f x)}}{a d f}+\frac {\text {arctanh}\left (\frac {\sqrt {d} (a \tan (e+f x)+a)}{\sqrt {2} a \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a \sqrt {d} f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a \sqrt {d} f}+\frac {\text {arctanh}\left (\frac {\sqrt {d} (a \tan (e+f x)+a)}{\sqrt {2} a \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a \sqrt {d} f}\)

Input:

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

Output:

ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]]/(a*Sqrt[d]*f) + ArcTanh[(Sqrt[d]*(a + 
 a*Tan[e + f*x]))/(Sqrt[2]*a*Sqrt[d*Tan[e + f*x]])]/(Sqrt[2]*a*Sqrt[d]*f)
 

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 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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 4015

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

rule 4057

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[1/(c^2 + d^2)   Int[(a + b*Tan[e + f*x])^m 
*(c - d*Tan[e + f*x]), x], x] + Simp[d^2/(c^2 + d^2)   Int[(a + b*Tan[e + f 
*x])^m*((1 + Tan[e + f*x]^2)/(c + d*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 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(303\) vs. \(2(67)=134\).

Time = 1.42 (sec) , antiderivative size = 304, normalized size of antiderivative = 3.75

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 d^{\frac {5}{2}}}+\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{2 d^{2}}\right )}{f a}\)	\(304\)
default	\(\frac {2 d^{2} \left (\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 d^{\frac {5}{2}}}+\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{2 d^{2}}\right )}{f a}\)	\(304\)

Input:

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

Output:

2/f/a*d^2*(1/2/d^(5/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))+1/2/d^2*(1/8/d 
*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^ 
(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+ 
(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arcta 
n(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))-1/8/(d^2)^(1/4)*2^(1/2)*(l 
n((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*t 
an(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan( 
2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)* 
(d*tan(f*x+e))^(1/2)+1))))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.68 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx=\left [-\frac {\sqrt {2} \sqrt {-d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, d \tan \left (f x + e\right )}\right ) + \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right )}{2 \, a d f}, \frac {\sqrt {2} \sqrt {d} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) - 4 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right )}{4 \, a d f}\right ] \] Input:

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

Output:

[-1/2*(sqrt(2)*sqrt(-d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-d)*( 
tan(f*x + e) + 1)/(d*tan(f*x + e))) + sqrt(-d)*log((d*tan(f*x + e) - 2*sqr 
t(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1)))/(a*d*f), 1/4*(sqrt(2) 
*sqrt(d)*log((d*tan(f*x + e)^2 + 2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d)*(t 
an(f*x + e) + 1) + 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) - 4*sqrt(d) 
*arctan(sqrt(d*tan(f*x + e))/(sqrt(d)*tan(f*x + e))))/(a*d*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx=\frac {\frac {d {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{a} + \frac {4 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a}}{4 \, d f} \] Input:

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

Output:

1/4*(d*(sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) 
+ d)/sqrt(d) - sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*s 
qrt(d) + d)/sqrt(d))/a + 4*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/a) 
/(d*f)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 1.34 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{a\,\sqrt {d}\,f}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {12\,\sqrt {2}\,d^{9/2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{12\,d^5\,\mathrm {tan}\left (e+f\,x\right )+12\,d^5}\right )}{2\,a\,\sqrt {d}\,f} \] Input:

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

Output:

atan((d*tan(e + f*x))^(1/2)/d^(1/2))/(a*d^(1/2)*f) + (2^(1/2)*atanh((12*2^ 
(1/2)*d^(9/2)*(d*tan(e + f*x))^(1/2))/(12*d^5*tan(e + f*x) + 12*d^5)))/(2* 
a*d^(1/2)*f)
 

Reduce [F]

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

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

Output:

(sqrt(d)*int(sqrt(tan(e + f*x))/(tan(e + f*x)**2 + tan(e + f*x)),x))/(a*d)