\(\int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx\) [259]

Optimal result

Integrand size = 28, antiderivative size = 132 \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {2 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{9 a d (a+i a \tan (c+d x))^3}-\frac {4 i e^4}{15 d \sqrt {e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )} \] Output:

-2/15*e^4*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^4/d/cos(d*x+c)^(1/2)/(e* 
sec(d*x+c))^(1/2)+4/9*I*e^2*(e*sec(d*x+c))^(3/2)/a/d/(a+I*a*tan(d*x+c))^3- 
4/15*I*e^4/d/(e*sec(d*x+c))^(1/2)/(a^4+I*a^4*tan(d*x+c))
 

Mathematica [C] (verified)

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

Time = 1.66 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.13 \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {e^3 e^{-i d x} \sec ^4(c+d x) \sqrt {e \sec (c+d x)} \left (-7-7 \cos (2 (c+d x))+6 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+3 i \sin (2 (c+d x))\right ) (-i \cos (c+2 d x)+\sin (c+2 d x))}{45 a^4 d (-i+\tan (c+d x))^4} \] Input:

Integrate[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

(e^3*Sec[c + d*x]^4*Sqrt[e*Sec[c + d*x]]*(-7 - 7*Cos[2*(c + d*x)] + 6*E^(( 
2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 
 3/4, -E^((2*I)*(c + d*x))] + (3*I)*Sin[2*(c + d*x)])*((-I)*Cos[c + 2*d*x] 
 + Sin[c + 2*d*x]))/(45*a^4*d*E^(I*d*x)*(-I + Tan[c + d*x])^4)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3981, 3042, 3981, 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.

\(\displaystyle \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4}dx\)

\(\Big \downarrow \) 3981

\(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{3/2}}{9 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \int \frac {(e \sec (c+d x))^{3/2}}{(i \tan (c+d x) a+a)^2}dx}{3 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{3/2}}{9 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \int \frac {(e \sec (c+d x))^{3/2}}{(i \tan (c+d x) a+a)^2}dx}{3 a^2}\)

\(\Big \downarrow \) 3981

\(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{3/2}}{9 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \left (\frac {e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx}{5 a^2}+\frac {4 i e^2}{5 d \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}\right )}{3 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{3/2}}{9 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \left (\frac {e^2 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 a^2}+\frac {4 i e^2}{5 d \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}\right )}{3 a^2}\)

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{3/2}}{9 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \left (\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{5 d \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}\right )}{3 a^2}\)

Input:

Int[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

(((4*I)/9)*e^2*(e*Sec[c + d*x])^(3/2))/(a*d*(a + I*a*Tan[c + d*x])^3) - (e 
^2*((2*e^2*EllipticE[(c + d*x)/2, 2])/(5*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[e*S 
ec[c + d*x]]) + (((4*I)/5)*e^2)/(d*Sqrt[e*Sec[c + d*x]]*(a^2 + I*a^2*Tan[c 
 + d*x]))))/(3*a^2)
 

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 3981

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

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. 500 vs. \(2 (116 ) = 232\).

Time = 3.73 (sec) , antiderivative size = 501, normalized size of antiderivative = 3.80

method	result	size
default	\(-\frac {2 \left (-6 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\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 )+3 i \left (-2 \cos \left (d x +c \right )^{4}-4 \cos \left (d x +c \right )^{3}-\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\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 )-6 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\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 )+3 i \left (2 \cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\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 )-\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-10 \cos \left (d x +c \right )^{2}-16 \cos \left (d x +c \right )-3\right )+2 i \cos \left (d x +c \right )^{2} \left (-3+5 \cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )\right )\right ) \sqrt {e \sec \left (d x +c \right )}\, e^{3}}{45 a^{4} d \left (2 i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-\cos \left (d x +c \right )-1\right )-2 \cos \left (d x +c \right )^{3}-2 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+1\right )}\)	\(501\)

Input:

int((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
 

Output:

-2/45/a^4/d*(-6*sin(d*x+c)*cos(d*x+c)*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(1/(c 
os(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x 
+c)-cot(d*x+c)),I)+3*I*(-2*cos(d*x+c)^4-4*cos(d*x+c)^3-cos(d*x+c)^2+2*cos( 
d*x+c)+1)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ellip 
ticE(I*(csc(d*x+c)-cot(d*x+c)),I)-6*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)^2+2* 
cos(d*x+c)+1)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*E 
llipticF(I*(csc(d*x+c)-cot(d*x+c)),I)+3*I*(2*cos(d*x+c)^4+4*cos(d*x+c)^3+c 
os(d*x+c)^2-2*cos(d*x+c)-1)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)-sin(d*x+c)*cos(d*x+c)* 
(-10*cos(d*x+c)^2-16*cos(d*x+c)-3)+2*I*cos(d*x+c)^2*(-3+5*cos(d*x+c)^2+2*c 
os(d*x+c)))*(e*sec(d*x+c))^(1/2)/(2*I*sin(d*x+c)*cos(d*x+c)*(-cos(d*x+c)-1 
)-2*cos(d*x+c)^3-2*cos(d*x+c)^2+cos(d*x+c)+1)*e^3
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.97 \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (-6 i \, \sqrt {2} e^{\frac {7}{2}} e^{\left (5 i \, d x + 5 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (-6 i \, e^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 4 i \, e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 7 i \, e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{45 \, a^{4} d} \] Input:

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
 

Output:

1/45*(-6*I*sqrt(2)*e^(7/2)*e^(5*I*d*x + 5*I*c)*weierstrassZeta(-4, 0, weie 
rstrassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(2)*(-6*I*e^3*e^(6*I*d*x + 
6*I*c) - 4*I*e^3*e^(4*I*d*x + 4*I*c) + 7*I*e^3*e^(2*I*d*x + 2*I*c) + 5*I*e 
^3)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-5*I*d*x 
 - 5*I*c)/(a^4*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Timed out} \] Input:

integrate((e*sec(d*x+c))**(7/2)/(a+I*a*tan(d*x+c))**4,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

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

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
 

Output:

integrate((e*sec(d*x + c))^(7/2)/(I*a*tan(d*x + c) + a)^4, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}}{\tan \left (d x +c \right )^{4}-4 \tan \left (d x +c \right )^{3} i -6 \tan \left (d x +c \right )^{2}+4 \tan \left (d x +c \right ) i +1}d x \right ) e^{3}}{a^{4}} \] Input:

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

Output:

(sqrt(e)*int((sqrt(sec(c + d*x))*sec(c + d*x)**3)/(tan(c + d*x)**4 - 4*tan 
(c + d*x)**3*i - 6*tan(c + d*x)**2 + 4*tan(c + d*x)*i + 1),x)*e**3)/a**4