\(\int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx\) [421]

Optimal result

Integrand size = 30, antiderivative size = 165 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}+\frac {16 i}{35 d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {12 i \sqrt {a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}-\frac {32 i \sqrt {a+i a \tan (c+d x)}}{35 a d e^2 \sqrt {e \sec (c+d x)}} \] Output:

2/7*I/d/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2)+16/35*I/d/e^2/(e*sec 
(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-12/35*I*(a+I*a*tan(d*x+c))^(1/2)/a 
/d/(e*sec(d*x+c))^(5/2)-32/35*I*(a+I*a*tan(d*x+c))^(1/2)/a/d/e^2/(e*sec(d* 
x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.00 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.48 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i (17+\cos (2 (c+d x))+3 i \sec (c+d x) \sin (3 (c+d x))+35 i \tan (c+d x))}{35 d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[1/((e*Sec[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]
 

Output:

((-1/35*I)*(17 + Cos[2*(c + d*x)] + (3*I)*Sec[c + d*x]*Sin[3*(c + d*x)] + 
(35*I)*Tan[c + d*x]))/(d*e^2*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x 
]])
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3983, 3042, 3978, 3042, 3983, 3042, 3969}

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.

\(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {6 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{5/2}}dx}{7 a}+\frac {2 i}{7 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{5/2}}dx}{7 a}+\frac {2 i}{7 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3978

\(\displaystyle \frac {6 \left (\frac {4 a \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )}{7 a}+\frac {2 i}{7 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6 \left (\frac {4 a \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )}{7 a}+\frac {2 i}{7 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {6 \left (\frac {4 a \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\right )}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )}{7 a}+\frac {2 i}{7 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6 \left (\frac {4 a \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\right )}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )}{7 a}+\frac {2 i}{7 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3969

\(\displaystyle \frac {6 \left (\frac {4 a \left (\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {e \sec (c+d x)}}\right )}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )}{7 a}+\frac {2 i}{7 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}}\)

Input:

Int[1/((e*Sec[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]
 

Output:

((2*I)/7)/(d*(e*Sec[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (6*((((- 
2*I)/5)*Sqrt[a + I*a*Tan[c + d*x]])/(d*(e*Sec[c + d*x])^(5/2)) + (4*a*(((2 
*I)/3)/(d*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((4*I)/3)*Sq 
rt[a + I*a*Tan[c + d*x]])/(a*d*Sqrt[e*Sec[c + d*x]])))/(5*e^2)))/(7*a)
 

Defintions of rubi rules used

rule 3042

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

rule 3969

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ 
(a*f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && EqQ 
[Simplify[m + n], 0]
 

rule 3978

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/( 
a*f*m)), x] + Simp[a*((m + n)/(m*d^2))   Int[(d*Sec[e + f*x])^(m + 2)*(a + 
b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b 
^2, 0] && GtQ[n, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
 

rule 3983

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[a*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ 
(b*f*(m + 2*n))), x] + Simp[Simplify[m + n]/(a*(m + 2*n))   Int[(d*Sec[e + 
f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x 
] && EqQ[a^2 + b^2, 0] && LtQ[n, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2* 
n]
 

Maple [A] (verified)

Time = 6.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.42

method	result	size
default	\(-\frac {2 \left (-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )+i \cos \left (d x +c \right )^{2}-16 \tan \left (d x +c \right )+8 i\right )}{35 d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {e \sec \left (d x +c \right )}\, e^{2}}\)	\(70\)

Input:

int(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOS 
E)
 

Output:

-2/35/d/(a*(1+I*tan(d*x+c)))^(1/2)/(e*sec(d*x+c))^(1/2)/e^2*(-6*cos(d*x+c) 
*sin(d*x+c)+I*cos(d*x+c)^2-16*tan(d*x+c)+8*I)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-7 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 112 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 40 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-\frac {7}{2} i \, d x - \frac {7}{2} i \, c\right )}}{140 \, a d e^{3}} \] Input:

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fr 
icas")
 

Output:

1/140*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))* 
(-7*I*e^(8*I*d*x + 8*I*c) - 112*I*e^(6*I*d*x + 6*I*c) - 70*I*e^(4*I*d*x + 
4*I*c) + 40*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(-7/2*I*d*x - 7/2*I*c)/(a*d*e^3 
)
 

Sympy [F]

\[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \] Input:

integrate(1/(e*sec(d*x+c))**(5/2)/(a+I*a*tan(d*x+c))**(1/2),x)
 

Output:

Integral(1/((e*sec(c + d*x))**(5/2)*sqrt(I*a*(tan(c + d*x) - I))), x)
 

Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {5 i \, \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) - 7 i \, \cos \left (\frac {5}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 35 i \, \cos \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) - 105 i \, \cos \left (\frac {1}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 5 \, \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, \sin \left (\frac {5}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 35 \, \sin \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 105 \, \sin \left (\frac {1}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right )}{140 \, \sqrt {a} d e^{\frac {5}{2}}} \] Input:

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="ma 
xima")
 

Output:

1/140*(5*I*cos(7/2*d*x + 7/2*c) - 7*I*cos(5/7*arctan2(sin(7/2*d*x + 7/2*c) 
, cos(7/2*d*x + 7/2*c))) + 35*I*cos(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos( 
7/2*d*x + 7/2*c))) - 105*I*cos(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d 
*x + 7/2*c))) + 5*sin(7/2*d*x + 7/2*c) + 7*sin(5/7*arctan2(sin(7/2*d*x + 7 
/2*c), cos(7/2*d*x + 7/2*c))) + 35*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), c 
os(7/2*d*x + 7/2*c))) + 105*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2* 
d*x + 7/2*c))))/(sqrt(a)*d*e^(5/2))
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="gi 
ac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 1.36 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (-\sin \left (c+d\,x\right )-\frac {3\,\sin \left (3\,c+3\,d\,x\right )}{35}+\frac {\cos \left (c+d\,x\right )\,1{}\mathrm {i}}{2}+\frac {\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{70}\right )}{d\,e^3\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \] Input:

int(1/((e/cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)^(1/2)),x)
 

Output:

-((e/cos(c + d*x))^(1/2)*((cos(c + d*x)*1i)/2 - sin(c + d*x) + (cos(3*c + 
3*d*x)*1i)/70 - (3*sin(3*c + 3*d*x))/35))/(d*e^3*((a*(cos(2*c + 2*d*x) + s 
in(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2))
 

Reduce [F]

\[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {e}\, \sqrt {a}\, \left (-\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{3} \tan \left (d x +c \right )^{2}+\sec \left (d x +c \right )^{3}}d x \right ) i +\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\sec \left (d x +c \right )^{3} \tan \left (d x +c \right )^{2}+\sec \left (d x +c \right )^{3}}d x \right )}{a \,e^{3}} \] Input:

int(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x)
 

Output:

(sqrt(e)*sqrt(a)*( - int((sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan( 
c + d*x))/(sec(c + d*x)**3*tan(c + d*x)**2 + sec(c + d*x)**3),x)*i + int(( 
sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1))/(sec(c + d*x)**3*tan(c + d*x) 
**2 + sec(c + d*x)**3),x)))/(a*e**3)