\(\int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx\) [435]

Optimal result

Integrand size = 30, antiderivative size = 162 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 i}{11 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {12 i}{77 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {16 i}{77 a^2 d \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {32 i \sqrt {a+i a \tan (c+d x)}}{77 a^3 d \sqrt {e \sec (c+d x)}} \] Output:

2/11*I/d/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2)+12/77*I/a/d/(e*sec( 
d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2)+16/77*I/a^2/d/(e*sec(d*x+c))^(1/2)/ 
(a+I*a*tan(d*x+c))^(1/2)-32/77*I*(a+I*a*tan(d*x+c))^(1/2)/a^3/d/(e*sec(d*x 
+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \sec ^3(c+d x) (-55 \cos (c+d x)+35 \cos (3 (c+d x))-22 i \sin (c+d x)+42 i \sin (3 (c+d x)))}{154 a^2 d \sqrt {e \sec (c+d x)} (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

((I/154)*Sec[c + d*x]^3*(-55*Cos[c + d*x] + 35*Cos[3*(c + d*x)] - (22*I)*S 
in[c + d*x] + (42*I)*Sin[3*(c + d*x)]))/(a^2*d*Sqrt[e*Sec[c + d*x]]*(-I + 
Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.06, 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, 3983, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3983

\(\displaystyle \frac {6 \int \frac {1}{\sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6 \int \frac {1}{\sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {6 \left (\frac {4 \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{7 a}+\frac {2 i}{7 d (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}}\right )}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6 \left (\frac {4 \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{7 a}+\frac {2 i}{7 d (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}}\right )}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {6 \left (\frac {4 \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\right )}{7 a}+\frac {2 i}{7 d (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}}\right )}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6 \left (\frac {4 \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\right )}{7 a}+\frac {2 i}{7 d (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}}\right )}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3969

\(\displaystyle \frac {6 \left (\frac {4 \left (\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {e \sec (c+d x)}}\right )}{7 a}+\frac {2 i}{7 d (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}}\right )}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\)

Input:

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

Output:

((2*I)/11)/(d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + (6*(((2 
*I)/7)/(d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) + (4*(((2*I)/ 
3)/(d*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((4*I)/3)*Sqrt[a 
 + I*a*Tan[c + d*x]])/(a*d*Sqrt[e*Sec[c + d*x]])))/(7*a)))/(11*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 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 = 6.63 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {\frac {4 i \cos \left (d x +c \right )^{4}}{11}+\frac {4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{11}-\frac {2 i \cos \left (d x +c \right )^{2}}{77}+\frac {12 \cos \left (d x +c \right ) \sin \left (d x +c \right )}{77}-\frac {16 i}{77}+\frac {32 \tan \left (d x +c \right )}{77}}{d \sqrt {e \sec \left (d x +c \right )}\, a^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(97\)

Input:

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

Output:

2/77/d/(e*sec(d*x+c))^(1/2)/a^2/(a*(1+I*tan(d*x+c)))^(1/2)*(14*I*cos(d*x+c 
)^4+14*sin(d*x+c)*cos(d*x+c)^3-I*cos(d*x+c)^2+6*cos(d*x+c)*sin(d*x+c)-8*I+ 
16*tan(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-77 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 110 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 40 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i\right )} e^{\left (-\frac {11}{2} i \, d x - \frac {11}{2} i \, c\right )}}{308 \, a^{3} d e} \] Input:

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

Output:

1/308*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))* 
(-77*I*e^(8*I*d*x + 8*I*c) + 110*I*e^(4*I*d*x + 4*I*c) + 40*I*e^(2*I*d*x + 
 2*I*c) + 7*I)*e^(-11/2*I*d*x - 11/2*I*c)/(a^3*d*e)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {7 i \, \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 33 i \, \cos \left (\frac {7}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 77 i \, \cos \left (\frac {3}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) - 77 i \, \cos \left (\frac {1}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 7 \, \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 33 \, \sin \left (\frac {7}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 77 \, \sin \left (\frac {3}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 77 \, \sin \left (\frac {1}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right )}{308 \, a^{\frac {5}{2}} d \sqrt {e}} \] Input:

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

Output:

1/308*(7*I*cos(11/2*d*x + 11/2*c) + 33*I*cos(7/11*arctan2(sin(11/2*d*x + 1 
1/2*c), cos(11/2*d*x + 11/2*c))) + 77*I*cos(3/11*arctan2(sin(11/2*d*x + 11 
/2*c), cos(11/2*d*x + 11/2*c))) - 77*I*cos(1/11*arctan2(sin(11/2*d*x + 11/ 
2*c), cos(11/2*d*x + 11/2*c))) + 7*sin(11/2*d*x + 11/2*c) + 33*sin(7/11*ar 
ctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 77*sin(3/11*arcta 
n2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 77*sin(1/11*arctan2( 
sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))))/(a^(5/2)*d*sqrt(e))
 

Giac [F(-2)]

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

integrate(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),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 = 1.69 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (154\,\sin \left (c+d\,x\right )+33\,\sin \left (3\,c+3\,d\,x\right )+7\,\sin \left (5\,c+5\,d\,x\right )+\cos \left (3\,c+3\,d\,x\right )\,33{}\mathrm {i}+\cos \left (5\,c+5\,d\,x\right )\,7{}\mathrm {i}\right )}{308\,a^2\,d\,e\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \] Input:

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

Output:

((e/cos(c + d*x))^(1/2)*(154*sin(c + d*x) + cos(3*c + 3*d*x)*33i + cos(5*c 
 + 5*d*x)*7i + 33*sin(3*c + 3*d*x) + 7*sin(5*c + 5*d*x)))/(308*a^2*d*e*((a 
*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/ 
2))
 

Reduce [F]

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

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

Output:

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