\(\int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x)) \, dx\) [185]

Optimal result

Integrand size = 26, antiderivative size = 123 \[ \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x)) \, dx=-\frac {6 a e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i a (e \sec (c+d x))^{7/2}}{7 d}+\frac {6 a e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a e (e \sec (c+d x))^{5/2} \sin (c+d x)}{5 d} \] Output:

-6/5*a*e^4*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(e*sec 
(d*x+c))^(1/2)+2/7*I*a*(e*sec(d*x+c))^(7/2)/d+6/5*a*e^3*(e*sec(d*x+c))^(1/ 
2)*sin(d*x+c)/d+2/5*a*e*(e*sec(d*x+c))^(5/2)*sin(d*x+c)/d
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.97 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.27 \[ \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x)) \, dx=\frac {a e e^{-i d x} (e \sec (c+d x))^{5/2} (\cos (d x)-i \sin (d x)) (\cos (c+3 d x)+i \sin (c+3 d x)) \left (-36 i-28 i \cos (2 (c+d x))+7 i e^{-2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+7 \sec (c+d x) \sin (3 (c+d x))+27 \tan (c+d x)\right )}{70 d} \] Input:

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

Output:

(a*e*(e*Sec[c + d*x])^(5/2)*(Cos[d*x] - I*Sin[d*x])*(Cos[c + 3*d*x] + I*Si 
n[c + 3*d*x])*(-36*I - (28*I)*Cos[2*(c + d*x)] + ((7*I)*(1 + E^((2*I)*(c + 
 d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(( 
2*I)*(c + d*x)) + 7*Sec[c + d*x]*Sin[3*(c + d*x)] + 27*Tan[c + d*x]))/(70* 
d*E^(I*d*x))
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 3967, 3042, 4255, 3042, 4255, 3042, 4258, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3967

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

\(\Big \downarrow \) 3042

\(\displaystyle a \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx+\frac {2 i a (e \sec (c+d x))^{7/2}}{7 d}\)

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {3}{5} e^2 \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 i a (e \sec (c+d x))^{7/2}}{7 d}\)

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 i a (e \sec (c+d x))^{7/2}}{7 d}\)

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle a \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 i a (e \sec (c+d x))^{7/2}}{7 d}\)

Input:

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

Output:

(((2*I)/7)*a*(e*Sec[c + d*x])^(7/2))/d + a*((2*e*(e*Sec[c + d*x])^(5/2)*Si 
n[c + d*x])/(5*d) + (3*e^2*((-2*e^2*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos 
[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*e*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x]) 
/d))/5)
 

Defintions of rubi rules used

rule 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3967

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

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (106 ) = 212\).

Time = 33.58 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.86

method	result	size
default	\(\frac {2 a \sqrt {e \sec \left (d x +c \right )}\, \left (21 \sin \left (d x +c \right )+7 \tan \left (d x +c \right )+7 \sec \left (d x +c \right ) \tan \left (d x +c \right )+5 i \left (\sec \left (d x +c \right )^{3}+\sec \left (d x +c \right )^{2}\right )+21 i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+21 i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\right ) e^{3}}{35 d \left (\cos \left (d x +c \right )+1\right )}\)	\(229\)
parts	\(-\frac {2 a \sqrt {e \sec \left (d x +c \right )}\, e^{3} \left (-3 \sin \left (d x +c \right )-\tan \left (d x +c \right )-\sec \left (d x +c \right ) \tan \left (d x +c \right )+i \left (3 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+3\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+i \left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 i a \left (e \sec \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d}\)	\(229\)

Input:

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

Output:

2/35*a/d*(e*sec(d*x+c))^(1/2)/(cos(d*x+c)+1)*(21*sin(d*x+c)+7*tan(d*x+c)+7 
*sec(d*x+c)*tan(d*x+c)+5*I*(sec(d*x+c)^3+sec(d*x+c)^2)+21*I*(-cos(d*x+c)^2 
-2*cos(d*x+c)-1)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cot(d*x+c) 
-csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)+21*I*(cos(d*x+c)^2+2*cos(d*x+c)+1 
)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I) 
*(1/(cos(d*x+c)+1))^(1/2))*e^3
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.69 \[ \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x)) \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (21 i \, a e^{3} e^{\left (7 i \, d x + 7 i \, c\right )} + 77 i \, a e^{3} e^{\left (5 i \, d x + 5 i \, c\right )} + 23 i \, a e^{3} e^{\left (3 i \, d x + 3 i \, c\right )} + 7 i \, a e^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 21 \, \sqrt {2} {\left (i \, a e^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, a e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, a e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a e^{3}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{35 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

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

Output:

-2/35*(sqrt(2)*(21*I*a*e^3*e^(7*I*d*x + 7*I*c) + 77*I*a*e^3*e^(5*I*d*x + 5 
*I*c) + 23*I*a*e^3*e^(3*I*d*x + 3*I*c) + 7*I*a*e^3*e^(I*d*x + I*c))*sqrt(e 
/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 21*sqrt(2)*(I*a*e^3* 
e^(6*I*d*x + 6*I*c) + 3*I*a*e^3*e^(4*I*d*x + 4*I*c) + 3*I*a*e^3*e^(2*I*d*x 
 + 2*I*c) + I*a*e^3)*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4 
, 0, e^(I*d*x + I*c))))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 
 3*d*e^(2*I*d*x + 2*I*c) + d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x)) \, dx=\frac {\sqrt {e}\, a \,e^{3} \left (2 \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3} i +7 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) d \right )}{7 d} \] Input:

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

Output:

(sqrt(e)*a*e**3*(2*sqrt(sec(c + d*x))*sec(c + d*x)**3*i + 7*int(sqrt(sec(c 
 + d*x))*sec(c + d*x)**3,x)*d))/(7*d)