\(\int \frac {\sqrt {e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx\) [21]

Optimal result

Integrand size = 25, antiderivative size = 204 \[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx=-\frac {2 \sqrt {2} \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (-\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right ),-1\right ) \sqrt {e \tan (c+d x)}}{\sqrt {-a+b} \sqrt {a+b} d \sqrt {\sin (c+d x)}}+\frac {2 \sqrt {2} \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right ),-1\right ) \sqrt {e \tan (c+d x)}}{\sqrt {-a+b} \sqrt {a+b} d \sqrt {\sin (c+d x)}} \] Output:

-2*2^(1/2)*cos(d*x+c)^(1/2)*EllipticPi(sin(d*x+c)^(1/2)/(1+cos(d*x+c))^(1/ 
2),-(-a+b)^(1/2)/(a+b)^(1/2),I)*(e*tan(d*x+c))^(1/2)/(-a+b)^(1/2)/(a+b)^(1 
/2)/d/sin(d*x+c)^(1/2)+2*2^(1/2)*cos(d*x+c)^(1/2)*EllipticPi(sin(d*x+c)^(1 
/2)/(1+cos(d*x+c))^(1/2),(-a+b)^(1/2)/(a+b)^(1/2),I)*(e*tan(d*x+c))^(1/2)/ 
(-a+b)^(1/2)/(a+b)^(1/2)/d/sin(d*x+c)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 19.18 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.78 \[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx=\frac {2 \left (b+a \sqrt {\sec ^2(c+d x)}\right ) \sqrt {e \tan (c+d x)} \left (\frac {-2 \arctan \left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+a \tan (c+d x)\right )-\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+a \tan (c+d x)\right )}{4 \sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2}}+\frac {b \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},-\tan ^2(c+d x),-\frac {a^2 \tan ^2(c+d x)}{a^2-b^2}\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 \left (-a^2+b^2\right )}\right )}{d (a+b \cos (c+d x)) \sqrt {\sec ^2(c+d x)} \sqrt {\tan (c+d x)}} \] Input:

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

Output:

(2*(b + a*Sqrt[Sec[c + d*x]^2])*Sqrt[e*Tan[c + d*x]]*((-2*ArcTan[1 - (Sqrt 
[2]*Sqrt[a]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2] 
*Sqrt[a]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)] + Log[Sqrt[a^2 - b^2] - Sq 
rt[2]*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Tan[c + d*x]] + a*Tan[c + d*x]] - Log 
[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Tan[c + d*x]] + 
a*Tan[c + d*x]])/(4*Sqrt[2]*Sqrt[a]*(a^2 - b^2)^(1/4)) + (b*AppellF1[3/4, 
1/2, 1, 7/4, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))]*Tan[c + 
 d*x]^(3/2))/(3*(-a^2 + b^2))))/(d*(a + b*Cos[c + d*x])*Sqrt[Sec[c + d*x]^ 
2]*Sqrt[Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.83, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3212, 3042, 3209, 3042, 3385, 3042, 3384, 993, 1537, 412}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )}}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 3212

\(\displaystyle \sqrt {e \tan (c+d x)} \sqrt {e \cot (c+d x)} \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \cot (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {e \tan (c+d x)} \sqrt {e \cot (c+d x)} \int \frac {1}{\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right ) \sqrt {-e \tan \left (c+d x-\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3209

\(\displaystyle \frac {\sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)} \int \frac {\sqrt {\sin (c+d x)}}{\sqrt {-\cos (c+d x)} (a+b \cos (c+d x))}dx}{\sqrt {\sin (c+d x)}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3385

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \int \frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{\sqrt {\sin (c+d x)}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3384

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

\(\Big \downarrow \) 993

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

\(\Big \downarrow \) 1537

\(\displaystyle \frac {4 \sqrt {2} \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \left (\frac {\int \frac {1}{\sqrt {1-\frac {\sin (c+d x)}{\cos (c+d x)+1}} \sqrt {\frac {\sin (c+d x)}{\cos (c+d x)+1}+1} \left (\sqrt {a+b}-\frac {\sqrt {b-a} \sin (c+d x)}{\cos (c+d x)+1}\right )}d\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}}{2 \sqrt {b-a}}-\frac {\int \frac {1}{\sqrt {1-\frac {\sin (c+d x)}{\cos (c+d x)+1}} \sqrt {\frac {\sin (c+d x)}{\cos (c+d x)+1}+1} \left (\frac {\sqrt {b-a} \sin (c+d x)}{\cos (c+d x)+1}+\sqrt {a+b}\right )}d\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}}{2 \sqrt {b-a}}\right )}{d \sqrt {\sin (c+d x)}}\)

\(\Big \downarrow \) 412

\(\displaystyle \frac {4 \sqrt {2} \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \left (\frac {\operatorname {EllipticPi}\left (\frac {\sqrt {b-a}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right ),-1\right )}{2 \sqrt {b-a} \sqrt {a+b}}-\frac {\operatorname {EllipticPi}\left (-\frac {\sqrt {b-a}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right ),-1\right )}{2 \sqrt {b-a} \sqrt {a+b}}\right )}{d \sqrt {\sin (c+d x)}}\)

Input:

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

Output:

(4*Sqrt[2]*Sqrt[Cos[c + d*x]]*(-1/2*EllipticPi[-(Sqrt[-a + b]/Sqrt[a + b]) 
, ArcSin[Sqrt[Sin[c + d*x]]/Sqrt[1 + Cos[c + d*x]]], -1]/(Sqrt[-a + b]*Sqr 
t[a + b]) + EllipticPi[Sqrt[-a + b]/Sqrt[a + b], ArcSin[Sqrt[Sin[c + d*x]] 
/Sqrt[1 + Cos[c + d*x]]], -1]/(2*Sqrt[-a + b]*Sqrt[a + b]))*Sqrt[e*Tan[c + 
 d*x]])/(d*Sqrt[Sin[c + d*x]])
 

Defintions of rubi rules used

rule 412

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x 
_)^2]), x_Symbol] :> Simp[(1/(a*Sqrt[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b* 
(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, 
 f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && S 
implerSqrtQ[-f/e, -d/c])
 

rule 993

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> 
With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2* 
b)   Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Simp[s/(2*b)   Int[1/((r 
 - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
a*d, 0]
 

rule 1537

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[ 
{q = Rt[(-a)*c, 2]}, Simp[Sqrt[-c]   Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqr 
t[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] & 
& GtQ[a, 0] && LtQ[c, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3209

Int[1/(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(g_)*tan[(e_.) + (f_.)*( 
x_)]]), x_Symbol] :> Simp[Sqrt[Sin[e + f*x]]/(Sqrt[Cos[e + f*x]]*Sqrt[g*Tan 
[e + f*x]])   Int[Sqrt[Cos[e + f*x]]/(Sqrt[Sin[e + f*x]]*(a + b*Sin[e + f*x 
])), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]
 

rule 3212

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.), x_Symbol] :> Simp[g^(2*IntPart[p])*(g*Cot[e + f*x])^FracPart[p] 
*(g*Tan[e + f*x])^FracPart[p]   Int[(a + b*Sin[e + f*x])^m/(g*Tan[e + f*x]) 
^p, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&  !IntegerQ[p]
 

rule 3384

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[sin[(e_.) + (f_.)*(x_)]]*((a_ 
) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[-4*Sqrt[2]*(g/f)   S 
ubst[Int[x^2/(((a + b)*g^2 + (a - b)*x^4)*Sqrt[1 - x^4/g^2]), x], x, Sqrt[g 
*Cos[e + f*x]]/Sqrt[1 + Sin[e + f*x]]], x] /; FreeQ[{a, b, e, f, g}, x] && 
NeQ[a^2 - b^2, 0]
 

rule 3385

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[(d_)*sin[(e_.) + (f_.)*(x_)]] 
*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[Sqrt[Sin[e + f* 
x]]/Sqrt[d*Sin[e + f*x]]   Int[Sqrt[g*Cos[e + f*x]]/(Sqrt[Sin[e + f*x]]*(a 
+ b*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2 
, 0]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(452\) vs. \(2(164)=328\).

Time = 3.14 (sec) , antiderivative size = 453, normalized size of antiderivative = 2.22

method	result	size
default	\(-\frac {\left (\sqrt {-a^{2}+b^{2}}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )-\sqrt {-a^{2}+b^{2}}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )-a \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )+\operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b -a \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )+\operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b \right ) \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {e \tan \left (d x +c \right )}\, \left (\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{d \left (\sqrt {-a^{2}+b^{2}}-a +b \right ) \left (\sqrt {-a^{2}+b^{2}}+a -b \right )}\)	\(453\)

Input:

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

Output:

-1/d*((-a^2+b^2)^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),(a-b)/( 
a-b+(-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))-(-a^2+b^2)^(1/2)*EllipticPi((-cot(d 
*x+c)+csc(d*x+c)+1)^(1/2),-(a-b)/(-a+b+(-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))- 
a*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),(a-b)/(a-b+(-(a-b)*(a+b))^(1 
/2)),1/2*2^(1/2))+EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),(a-b)/(a-b+( 
-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b-a*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1 
)^(1/2),-(a-b)/(-a+b+(-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))+EllipticPi((-cot(d 
*x+c)+csc(d*x+c)+1)^(1/2),-(a-b)/(-a+b+(-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))* 
b)*(cot(d*x+c)-csc(d*x+c))^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(-cot 
(d*x+c)+csc(d*x+c)+1)^(1/2)*(e*tan(d*x+c))^(1/2)*(cot(d*x+c)+csc(d*x+c))/( 
(-a^2+b^2)^(1/2)-a+b)/((-a^2+b^2)^(1/2)+a-b)
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(sqrt(e*tan(d*x + c))/(b*cos(d*x + c) + a), x)
 

Giac [F]

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

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

Output:

integrate(sqrt(e*tan(d*x + c))/(b*cos(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx=\int \frac {\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}}{a+b\,\cos \left (c+d\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(e)*int(sqrt(tan(c + d*x))/(cos(c + d*x)*b + a),x)