\(\int \frac {1}{(a+a \sec (c+d x))^2 \sqrt {e \tan (c+d x)}} \, dx\) [134]

Optimal result

Integrand size = 25, antiderivative size = 306 \[ \int \frac {1}{(a+a \sec (c+d x))^2 \sqrt {e \tan (c+d x)}} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d \sqrt {e}}+\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d \sqrt {e}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}+\sqrt {e} \tan (c+d x)}\right )}{\sqrt {2} a^2 d \sqrt {e}}-\frac {4 e^3}{7 a^2 d (e \tan (c+d x))^{7/2}}+\frac {4 e^3 \sec (c+d x)}{7 a^2 d (e \tan (c+d x))^{7/2}}+\frac {2 e}{3 a^2 d (e \tan (c+d x))^{3/2}}-\frac {20 e \sec (c+d x)}{21 a^2 d (e \tan (c+d x))^{3/2}}-\frac {10 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{21 a^2 d \sqrt {e \tan (c+d x)}} \] Output:

-1/2*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/a^2/d/e^(1/2)+ 
1/2*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/a^2/d/e^(1/2)+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/e^(1/2)-4/7*e^3/a^2/d/(e*tan(d*x+c))^(7/2)+4/7*e^3*sec(d*x+c)/a^ 
2/d/(e*tan(d*x+c))^(7/2)+2/3*e/a^2/d/(e*tan(d*x+c))^(3/2)-20/21*e*sec(d*x+ 
c)/a^2/d/(e*tan(d*x+c))^(3/2)-10/21*InverseJacobiAM(c-1/4*Pi+d*x,2^(1/2))* 
sec(d*x+c)*sin(2*d*x+2*c)^(1/2)/a^2/d/(e*tan(d*x+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 = 8.52 (sec) , antiderivative size = 1281, normalized size of antiderivative = 4.19 \[ \int \frac {1}{(a+a \sec (c+d x))^2 \sqrt {e \tan (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

(40*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(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[Ta 
n[c + d*x]])/(21*d*E^(I*(c + d*x))*(a + a*Sec[c + d*x])^2*Sqrt[e*Tan[c + d 
*x]]) + (Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(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[Tan[c + d*x]])/(d*E^((2*I)*c 
)*(-1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])^2*Sqrt[e*Tan[c + d*x]]) 
+ (Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(Sqrt 
[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d*x))]] + 2*E^(( 
4*I)*c)*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTa 
nh[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[Tan[c + d*x]])/(d*E^((2*I)*c)*(-1 
+ E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])^2*Sqrt[e*Tan[c + d*x]]) - (2*S 
qrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*Cos[c/2 + 
 (d*x)/2]^4*(3*(-1 + E^((4*I)*(c + d*x))) + E^((4*I)*(c + d*x))*(-1 + E^(( 
2*I)*c))*Sqrt[1 - E^((4*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, E^ 
((4*I)*(c + d*x))])*Sec[2*c]*Sec[c + d*x]^2*Sqrt[Tan[c + d*x]])/(3*d*E^(I* 
(2*c + d*x))*(-1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])^2*Sqrt[e*T...
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 379, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4376

\(\displaystyle \frac {e^4 \int \frac {(a-a \sec (c+d x))^2}{(e \tan (c+d x))^{9/2}}dx}{a^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^4 \int \frac {\left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}{\left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{9/2}}dx}{a^4}\)

\(\Big \downarrow \) 4374

\(\displaystyle \frac {e^4 \int \left (\frac {\sec ^2(c+d x) a^2}{(e \tan (c+d x))^{9/2}}-\frac {2 \sec (c+d x) a^2}{(e \tan (c+d x))^{9/2}}+\frac {a^2}{(e \tan (c+d x))^{9/2}}\right )dx}{a^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^4 \left (-\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{9/2}}+\frac {a^2 \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d e^{9/2}}-\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{9/2}}+\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{9/2}}-\frac {10 a^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{21 d e^4 \sqrt {e \tan (c+d x)}}+\frac {2 a^2}{3 d e^3 (e \tan (c+d x))^{3/2}}-\frac {20 a^2 \sec (c+d x)}{21 d e^3 (e \tan (c+d x))^{3/2}}-\frac {4 a^2}{7 d e (e \tan (c+d x))^{7/2}}+\frac {4 a^2 \sec (c+d x)}{7 d e (e \tan (c+d x))^{7/2}}\right )}{a^4}\)

Input:

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

Output:

(e^4*(-((a^2*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]* 
d*e^(9/2))) + (a^2*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sq 
rt[2]*d*e^(9/2)) - (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[ 
e*Tan[c + d*x]]])/(2*Sqrt[2]*d*e^(9/2)) + (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c 
 + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*d*e^(9/2)) - (4*a^2)/( 
7*d*e*(e*Tan[c + d*x])^(7/2)) + (4*a^2*Sec[c + d*x])/(7*d*e*(e*Tan[c + d*x 
])^(7/2)) + (2*a^2)/(3*d*e^3*(e*Tan[c + d*x])^(3/2)) - (20*a^2*Sec[c + d*x 
])/(21*d*e^3*(e*Tan[c + d*x])^(3/2)) - (10*a^2*EllipticF[c - Pi/4 + d*x, 2 
]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]])/(21*d*e^4*Sqrt[e*Tan[c + 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 = 1.48 (sec) , antiderivative size = 533, normalized size of antiderivative = 1.74

method	result	size
default	\(-\frac {\left (1-\cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \left (-21 i \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 )+21 i \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 )+3 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+62 \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 )-21 \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 )-21 \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 )-26 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+23 \csc \left (d x +c \right )-23 \cot \left (d x +c \right )\right ) \sec \left (d x +c \right ) \csc \left (d x +c \right )^{2}}{42 a^{2} d \sqrt {e \tan \left (d x +c \right )}}\)	\(533\)

Input:

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

Output:

-1/42/a^2/d*(1-cos(d*x+c))*(1+cos(d*x+c))^2/(e*tan(d*x+c))^(1/2)*(-21*I*(- 
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))+21*I*(-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)+cs 
c(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))+3*(1-cos(d*x+c))^5*csc(d*x+c)^5+6 
2*(-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(-c 
sc(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))-21*(-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))-21*(-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))-26*(1-cos(d*x+c))^3* 
csc(d*x+c)^3+23*csc(d*x+c)-23*cot(d*x+c))*sec(d*x+c)*csc(d*x+c)^2
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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