\(\int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx\) [399]

Optimal result

Integrand size = 30, antiderivative size = 164 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {12 i a}{35 d e^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {32 i a \sqrt {e \sec (c+d x)}}{35 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{35 d e^2 (e \sec (c+d x))^{3/2}} \] Output:

12/35*I*a/d/e^2/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2)+32/35*I*a*(e 
*sec(d*x+c))^(1/2)/d/e^4/(a+I*a*tan(d*x+c))^(1/2)-2/7*I*(a+I*a*tan(d*x+c)) 
^(1/2)/d/(e*sec(d*x+c))^(7/2)-16/35*I*(a+I*a*tan(d*x+c))^(1/2)/d/e^2/(e*se 
c(d*x+c))^(3/2)
 

Mathematica [A] (verified)

Time = 1.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {(35 i \cos (c+d x)+i \cos (3 (c+d x))+70 \sin (c+d x)+6 \sin (3 (c+d x))) \sqrt {a+i a \tan (c+d x)}}{70 d e^3 \sqrt {e \sec (c+d x)}} \] Input:

Integrate[Sqrt[a + I*a*Tan[c + d*x]]/(e*Sec[c + d*x])^(7/2),x]
 

Output:

(((35*I)*Cos[c + d*x] + I*Cos[3*(c + d*x)] + 70*Sin[c + d*x] + 6*Sin[3*(c 
+ d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/(70*d*e^3*Sqrt[e*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 174, 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, 3978, 3042, 3983, 3042, 3978, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3978

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3983

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3978

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3969

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

Input:

Int[Sqrt[a + I*a*Tan[c + d*x]]/(e*Sec[c + d*x])^(7/2),x]
 

Output:

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

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 = 6.57 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.48

method	result	size
default	\(\frac {2 \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (i \cos \left (d x +c \right )^{3}+6 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}+8 i \cos \left (d x +c \right )+16 \sin \left (d x +c \right )\right )}{35 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}\)	\(79\)

Input:

int((a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

2/35/d*(a*(1+I*tan(d*x+c)))^(1/2)*(I*cos(d*x+c)^3+6*sin(d*x+c)*cos(d*x+c)^ 
2+8*I*cos(d*x+c)+16*sin(d*x+c))/(e*sec(d*x+c))^(1/2)/e^3
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.59 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-5 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 40 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 112 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i\right )} e^{\left (-\frac {5}{2} i \, d x - \frac {5}{2} i \, c\right )}}{140 \, d e^{4}} \] Input:

integrate((a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(7/2),x, algorithm="fric 
as")
 

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(d*x+c))**(1/2)/(e*sec(d*x+c))**(7/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {a} {\left (7 i \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 5 i \, \cos \left (\frac {7}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) - 35 i \, \cos \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 105 i \, \cos \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 7 \, \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sin \left (\frac {7}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 35 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 105 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )}}{140 \, d e^{\frac {7}{2}}} \] Input:

integrate((a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(7/2),x, algorithm="maxi 
ma")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(7/2),x, algorithm="giac 
")
 

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 = 2.37 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,36{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}+76\,\sin \left (2\,c+2\,d\,x\right )+6\,\sin \left (4\,c+4\,d\,x\right )+35{}\mathrm {i}\right )}{140\,d\,e^4} \] Input:

int((a + a*tan(c + d*x)*1i)^(1/2)/(e/cos(c + d*x))^(7/2),x)
 

Output:

((e/cos(c + d*x))^(1/2)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/ 
(cos(2*c + 2*d*x) + 1))^(1/2)*(cos(2*c + 2*d*x)*36i + cos(4*c + 4*d*x)*1i 
+ 76*sin(2*c + 2*d*x) + 6*sin(4*c + 4*d*x) + 35i))/(140*d*e^4)
 

Reduce [F]

\[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {e}\, \sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\sec \left (d x +c \right )^{4}}d x \right )}{e^{4}} \] Input:

int((a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(7/2),x)
 

Output:

(sqrt(e)*sqrt(a)*int((sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1))/sec(c + 
 d*x)**4,x))/e**4