\(\int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx\) [422]

Optimal result

Integrand size = 30, antiderivative size = 206 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i}{9 d (e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {32 i}{105 d e^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {256 i \sqrt {e \sec (c+d x)}}{315 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}-\frac {128 i \sqrt {a+i a \tan (c+d x)}}{315 a d e^2 (e \sec (c+d x))^{3/2}} \] Output:

2/9*I/d/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^(1/2)+32/105*I/d/e^2/(e*se 
c(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2)+256/315*I*(e*sec(d*x+c))^(1/2)/d/ 
e^4/(a+I*a*tan(d*x+c))^(1/2)-16/63*I*(a+I*a*tan(d*x+c))^(1/2)/a/d/(e*sec(d 
*x+c))^(7/2)-128/315*I*(a+I*a*tan(d*x+c))^(1/2)/a/d/e^2/(e*sec(d*x+c))^(3/ 
2)
 

Mathematica [A] (verified)

Time = 1.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.42 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {e \sec (c+d x)} (945 i-84 i \cos (2 (c+d x))-5 i \cos (4 (c+d x))+336 \sin (2 (c+d x))+40 \sin (4 (c+d x)))}{1260 d e^4 \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[1/((e*Sec[c + d*x])^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]
 

Output:

(Sqrt[e*Sec[c + d*x]]*(945*I - (84*I)*Cos[2*(c + d*x)] - (5*I)*Cos[4*(c + 
d*x)] + 336*Sin[2*(c + d*x)] + 40*Sin[4*(c + d*x)]))/(1260*d*e^4*Sqrt[a + 
I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3983, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3983

\(\displaystyle \frac {8 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{7/2}}dx}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{7/2}}dx}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3978

\(\displaystyle \frac {8 \left (\frac {6 a \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {i \tan (c+d x) a+a}}dx}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8 \left (\frac {6 a \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {i \tan (c+d x) a+a}}dx}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {8 \left (\frac {6 a \left (\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{3/2}}dx}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8 \left (\frac {6 a \left (\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{3/2}}dx}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3978

\(\displaystyle \frac {8 \left (\frac {6 a \left (\frac {4 \left (\frac {2 a \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}dx}{3 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d (e \sec (c+d x))^{3/2}}\right )}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8 \left (\frac {6 a \left (\frac {4 \left (\frac {2 a \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}dx}{3 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d (e \sec (c+d x))^{3/2}}\right )}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3969

\(\displaystyle \frac {8 \left (\frac {6 a \left (\frac {4 \left (\frac {4 i a \sqrt {e \sec (c+d x)}}{3 d e^2 \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d (e \sec (c+d x))^{3/2}}\right )}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\)

Input:

Int[1/((e*Sec[c + d*x])^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]
 

Output:

((2*I)/9)/(d*(e*Sec[c + d*x])^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (8*((((- 
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) 
))/(9*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 = 6.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.43

method	result	size
default	\(\frac {\frac {16 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}{63}-\frac {2 i \cos \left (d x +c \right )^{3}}{63}+\frac {128 \sin \left (d x +c \right )}{315}-\frac {32 i \cos \left (d x +c \right )}{315}+\frac {256 i \sec \left (d x +c \right )}{315}}{d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {e \sec \left (d x +c \right )}\, e^{3}}\)	\(88\)

Input:

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

Output:

2/315/d/(a*(1+I*tan(d*x+c)))^(1/2)/(e*sec(d*x+c))^(1/2)/e^3*(40*sin(d*x+c) 
*cos(d*x+c)^2-5*I*cos(d*x+c)^3+64*sin(d*x+c)-16*I*cos(d*x+c)+128*I*sec(d*x 
+c))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.54 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-45 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 465 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1470 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 2142 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 287 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 35 i\right )} e^{\left (-\frac {9}{2} i \, d x - \frac {9}{2} i \, c\right )}}{2520 \, a d e^{4}} \] Input:

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

Output:

1/2520*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1)) 
*(-45*I*e^(10*I*d*x + 10*I*c) - 465*I*e^(8*I*d*x + 8*I*c) + 1470*I*e^(6*I* 
d*x + 6*I*c) + 2142*I*e^(4*I*d*x + 4*I*c) + 287*I*e^(2*I*d*x + 2*I*c) + 35 
*I)*e^(-9/2*I*d*x - 9/2*I*c)/(a*d*e^4)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {35 i \, \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) - 45 i \, \cos \left (\frac {7}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 252 i \, \cos \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) - 420 i \, \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 1890 i \, \cos \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 35 \, \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 45 \, \sin \left (\frac {7}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 252 \, \sin \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 420 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 1890 \, \sin \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right )}{2520 \, \sqrt {a} d e^{\frac {7}{2}}} \] Input:

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

Output:

1/2520*(35*I*cos(9/2*d*x + 9/2*c) - 45*I*cos(7/9*arctan2(sin(9/2*d*x + 9/2 
*c), cos(9/2*d*x + 9/2*c))) + 252*I*cos(5/9*arctan2(sin(9/2*d*x + 9/2*c), 
cos(9/2*d*x + 9/2*c))) - 420*I*cos(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9 
/2*d*x + 9/2*c))) + 1890*I*cos(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d 
*x + 9/2*c))) + 35*sin(9/2*d*x + 9/2*c) + 45*sin(7/9*arctan2(sin(9/2*d*x + 
 9/2*c), cos(9/2*d*x + 9/2*c))) + 252*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c) 
, cos(9/2*d*x + 9/2*c))) + 420*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9 
/2*d*x + 9/2*c))) + 1890*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x 
 + 9/2*c))))/(sqrt(a)*d*e^(7/2))
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(e*sec(d*x+c))^(7/2)/(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 = 1.51 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (336\,\sin \left (2\,c+2\,d\,x\right )-\cos \left (4\,c+4\,d\,x\right )\,5{}\mathrm {i}-\cos \left (2\,c+2\,d\,x\right )\,84{}\mathrm {i}+40\,\sin \left (4\,c+4\,d\,x\right )+945{}\mathrm {i}\right )}{1260\,d\,e^4\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \] Input:

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

Output:

((e/cos(c + d*x))^(1/2)*(336*sin(2*c + 2*d*x) - cos(4*c + 4*d*x)*5i - cos( 
2*c + 2*d*x)*84i + 40*sin(4*c + 4*d*x) + 945i))/(1260*d*e^4*((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]

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

int(1/(e*sec(d*x+c))^(7/2)/(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)**4*tan(c + d*x)**2 + sec(c + d*x)**4),x)*i + int(( 
sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1))/(sec(c + d*x)**4*tan(c + d*x) 
**2 + sec(c + d*x)**4),x)))/(a*e**4)