\(\int \frac {\sqrt {e \tan (c+d x)}}{(a+a \sec (c+d x))^2} \, dx\) [133]

Optimal result

Integrand size = 25, antiderivative size = 305 \[ \int \frac {\sqrt {e \tan (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=-\frac {\sqrt {e} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}+\frac {\sqrt {e} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}+\sqrt {e} \tan (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {4 e^3}{5 a^2 d (e \tan (c+d x))^{5/2}}+\frac {4 e^3 \sec (c+d x)}{5 a^2 d (e \tan (c+d x))^{5/2}}+\frac {2 e}{a^2 d \sqrt {e \tan (c+d x)}}-\frac {12 e \cos (c+d x)}{5 a^2 d \sqrt {e \tan (c+d x)}}-\frac {12 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{5 a^2 d \sqrt {\sin (2 c+2 d x)}} \] Output:

-1/2*e^(1/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/a^2/d+ 
1/2*e^(1/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/a^2/d-1 
/2*e^(1/2)*arctanh(2^(1/2)*(e*tan(d*x+c))^(1/2)/(e^(1/2)+e^(1/2)*tan(d*x+c 
)))*2^(1/2)/a^2/d-4/5*e^3/a^2/d/(e*tan(d*x+c))^(5/2)+4/5*e^3*sec(d*x+c)/a^ 
2/d/(e*tan(d*x+c))^(5/2)+2*e/a^2/d/(e*tan(d*x+c))^(1/2)-12/5*e*cos(d*x+c)/ 
a^2/d/(e*tan(d*x+c))^(1/2)+12/5*cos(d*x+c)*EllipticE(cos(c+1/4*Pi+d*x),2^( 
1/2))*(e*tan(d*x+c))^(1/2)/a^2/d/sin(2*d*x+2*c)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 8.58 (sec) , antiderivative size = 2792, normalized size of antiderivative = 9.15 \[ \int \frac {\sqrt {e \tan (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\text {Result too large to show} \] Input:

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

Output:

(Cos[c/2 + (d*x)/2]^4*Sec[c + d*x]^2*((-24*Cos[c/2]*Cos[d*x]*Sec[2*c]*(4*S 
in[c/2] + Sin[(3*c)/2] + Sin[(5*c)/2]))/(5*d*(1 + 2*Cos[c])) - (56*Sec[c/2 
]*Sec[c/2 + (d*x)/2]*Sin[(d*x)/2])/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^ 
3*Sin[(d*x)/2])/(5*d) - (12*(-2 - 5*Cos[c] - 6*Cos[2*c] + Cos[3*c])*Sec[2* 
c]*Sin[d*x])/(5*d*(1 + 2*Cos[c])) - (56*Tan[c/2])/(5*d) + (4*Sec[c/2 + (d* 
x)/2]^2*Tan[c/2])/(5*d))*Sqrt[e*Tan[c + d*x]])/(a + a*Sec[c + d*x])^2 + (( 
E^((2*I)*c)*Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + 
d*x))]] - 2*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*A 
rcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])*Cos[c/ 
2 + (d*x)/2]^4*Sec[2*c]*Sec[c + d*x]^2*Sqrt[e*Tan[c + d*x]])/(d*E^(I*c)*Sq 
rt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2 
*I)*(c + d*x)))*(1 + 2*Cos[c])*(a + a*Sec[c + d*x])^2*Sqrt[Tan[c + d*x]]) 
- ((-(E^((4*I)*c)*Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I) 
*(c + d*x))]]) + 2*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d 
*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]) 
*Cos[c/2 + (d*x)/2]^4*Sec[2*c]*Sec[c + d*x]^2*Sqrt[e*Tan[c + d*x]])/(d*E^( 
(2*I)*c)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))] 
*(1 + E^((2*I)*(c + d*x)))*(1 + 2*Cos[c])*(a + a*Sec[c + d*x])^2*Sqrt[Tan[ 
c + d*x]]) - ((-(E^((6*I)*c)*Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 
 + E^((4*I)*(c + d*x))]]) + 2*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E...
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.24, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4376, 3042, 4374, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4376

\(\displaystyle \frac {e^4 \int \frac {(a-a \sec (c+d x))^2}{(e \tan (c+d x))^{7/2}}dx}{a^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^4 \int \frac {\left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}{\left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx}{a^4}\)

\(\Big \downarrow \) 4374

\(\displaystyle \frac {e^4 \int \left (\frac {\sec ^2(c+d x) a^2}{(e \tan (c+d x))^{7/2}}-\frac {2 \sec (c+d x) a^2}{(e \tan (c+d x))^{7/2}}+\frac {a^2}{(e \tan (c+d x))^{7/2}}\right )dx}{a^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^4 \left (-\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{7/2}}+\frac {a^2 \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d e^{7/2}}+\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{7/2}}-\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{7/2}}-\frac {12 a^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{5 d e^4 \sqrt {\sin (2 c+2 d x)}}+\frac {2 a^2}{d e^3 \sqrt {e \tan (c+d x)}}-\frac {12 a^2 \cos (c+d x)}{5 d e^3 \sqrt {e \tan (c+d x)}}-\frac {4 a^2}{5 d e (e \tan (c+d x))^{5/2}}+\frac {4 a^2 \sec (c+d x)}{5 d e (e \tan (c+d x))^{5/2}}\right )}{a^4}\)

Input:

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

Output:

(e^4*(-((a^2*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]* 
d*e^(7/2))) + (a^2*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sq 
rt[2]*d*e^(7/2)) + (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[ 
e*Tan[c + d*x]]])/(2*Sqrt[2]*d*e^(7/2)) - (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c 
 + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*d*e^(7/2)) - (4*a^2)/( 
5*d*e*(e*Tan[c + d*x])^(5/2)) + (4*a^2*Sec[c + d*x])/(5*d*e*(e*Tan[c + d*x 
])^(5/2)) + (2*a^2)/(d*e^3*Sqrt[e*Tan[c + d*x]]) - (12*a^2*Cos[c + d*x])/( 
5*d*e^3*Sqrt[e*Tan[c + d*x]]) - (12*a^2*Cos[c + d*x]*EllipticE[c - Pi/4 + 
d*x, 2]*Sqrt[e*Tan[c + d*x]])/(5*d*e^4*Sqrt[Sin[2*c + 2*d*x]])))/a^4
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4374

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_))^(n_), x_Symbol] :> Int[ExpandIntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[ 
c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]
 

rule 4376

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_))^(n_), x_Symbol] :> Simp[a^(2*n)/e^(2*n)   Int[(e*Cot[c + d*x])^(m + 2* 
n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[a 
^2 - b^2, 0] && ILtQ[n, 0]
 

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.90 (sec) , antiderivative size = 669, normalized size of antiderivative = 2.19

method	result	size
default	\(\frac {\left (i \left (5 \cos \left (d x +c \right )^{2}+10 \cos \left (d x +c \right )+5\right ) \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+i \left (-5 \cos \left (d x +c \right )^{2}-10 \cos \left (d x +c \right )-5\right ) \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\left (-5 \cos \left (d x +c \right )^{2}-10 \cos \left (d x +c \right )-5\right ) \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\left (-5 \cos \left (d x +c \right )^{2}-10 \cos \left (d x +c \right )-5\right ) \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\left (24 \cos \left (d x +c \right )^{2}+48 \cos \left (d x +c \right )+24\right ) \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+\left (-12 \cos \left (d x +c \right )^{2}-24 \cos \left (d x +c \right )-12\right ) \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+\cos \left (d x +c \right ) \left (4 \cos \left (d x +c \right )-4\right )\right ) \sqrt {e \tan \left (d x +c \right )}\, \csc \left (d x +c \right )}{10 a^{2} d \left (1+\cos \left (d x +c \right )\right )}\)	\(669\)

Input:

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

Output:

1/10/a^2/d*(I*(5*cos(d*x+c)^2+10*cos(d*x+c)+5)*(-cot(d*x+c)+csc(d*x+c)+1)^ 
(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2)*E 
llipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))+I*(-5*co 
s(d*x+c)^2-10*cos(d*x+c)-5)*(-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*(2*cot(d*x+c) 
-2*csc(d*x+c)+2)^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2)*EllipticPi((-cot(d*x 
+c)+csc(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))+(-5*cos(d*x+c)^2-10*cos(d*x 
+c)-5)*(-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2 
)*(-csc(d*x+c)+cot(d*x+c))^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/ 
2),1/2-1/2*I,1/2*2^(1/2))+(-5*cos(d*x+c)^2-10*cos(d*x+c)-5)*(-cot(d*x+c)+c 
sc(d*x+c)+1)^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(-csc(d*x+c)+cot(d* 
x+c))^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1 
/2))+(24*cos(d*x+c)^2+48*cos(d*x+c)+24)*(-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*( 
2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2)*Elliptic 
E((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2*2^(1/2))+(-12*cos(d*x+c)^2-24*cos(d 
*x+c)-12)*(-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^( 
1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2)*EllipticF((-cot(d*x+c)+csc(d*x+c)+1)^( 
1/2),1/2*2^(1/2))+cos(d*x+c)*(4*cos(d*x+c)-4))*(e*tan(d*x+c))^(1/2)/(1+cos 
(d*x+c))*csc(d*x+c)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt {e \tan (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {e \tan (c+d x)}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(e)*int(sqrt(tan(c + d*x))/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1),x)) 
/a**2