\(\int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^3} \, dx\) [249]

Optimal result

Integrand size = 28, antiderivative size = 116 \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^3} \, dx=-\frac {6 i e^4}{5 a^3 d \sqrt {e \sec (c+d x)}}-\frac {6 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{5 a d (a+i a \tan (c+d x))^2} \] Output:

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

Mathematica [C] (verified)

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

Time = 1.55 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.01 \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^3} \, dx=\frac {2 e e^{-i d x} \left (-2+\frac {6 e^{2 i (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}\right ) (e \sec (c+d x))^{5/2} (\cos (c+2 d x)+i \sin (c+2 d x))}{5 a^3 d (-i+\tan (c+d x))^3} \] Input:

Integrate[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^3,x]
 

Output:

(2*e*(-2 + (6*E^((2*I)*(c + d*x))*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2 
*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))])*(e*Sec[c + d*x])^(5/2)*(Co 
s[c + 2*d*x] + I*Sin[c + 2*d*x]))/(5*a^3*d*E^(I*d*x)*(-I + Tan[c + d*x])^3 
)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3981, 3042, 3982, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3982

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

Int[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^3,x]
 

Output:

(-3*e^2*(((2*I)*e^2)/(a*d*Sqrt[e*Sec[c + d*x]]) + (2*e^2*EllipticE[(c + d* 
x)/2, 2])/(a*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]])))/(5*a^2) + (((4*I 
)/5)*e^2*(e*Sec[c + d*x])^(3/2))/(a*d*(a + I*a*Tan[c + d*x])^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 3982

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + 
f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Simp[d^2*((m - 2)/(a*(m + n - 1))) 
 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] && LtQ[n, 0] && GtQ[m, 1] &&  !IL 
tQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 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. 445 vs. \(2 (101 ) = 202\).

Time = 3.63 (sec) , antiderivative size = 446, normalized size of antiderivative = 3.84

method	result	size
default	\(-\frac {2 \left (i \left (3 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+3\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )+\sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \left (3 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )\right )+i \left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )+\sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \left (-3 \cos \left (d x +c \right )^{3}-6 \cos \left (d x +c \right )^{2}-3 \cos \left (d x +c \right )\right )+i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-2 \cos \left (d x +c \right )-5\right )+\cos \left (d x +c \right ) \left (2 \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )-3\right )\right ) e^{3} \sqrt {e \sec \left (d x +c \right )}}{5 a^{3} d \left (-\cos \left (d x +c \right ) \sin \left (d x +c \right )-\sin \left (d x +c \right )+i \cos \left (d x +c \right )^{2}+i \cos \left (d x +c \right )\right )}\)	\(446\)

Input:

int((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

-2/5/a^3/d*(I*(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*sin(d*x+c)*(1/(cos(d*x+c)+1) 
)^(1/2)*(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)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Elliptic 
E(I*(cot(d*x+c)-csc(d*x+c)),I)*(3*cos(d*x+c)^3+6*cos(d*x+c)^2+3*cos(d*x+c) 
)+I*(-3*cos(d*x+c)^2-6*cos(d*x+c)-3)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*( 
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)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cot( 
d*x+c)-csc(d*x+c)),I)*(-3*cos(d*x+c)^3-6*cos(d*x+c)^2-3*cos(d*x+c))+I*sin( 
d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)-5)+cos(d*x+c)*(2*cos(d*x+c)^2-cos(d*x+c)- 
3))*e^3*(e*sec(d*x+c))^(1/2)/(-cos(d*x+c)*sin(d*x+c)-sin(d*x+c)+I*cos(d*x+ 
c)^2+I*cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98 \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^3} \, dx=-\frac {2 \, {\left (3 i \, \sqrt {2} e^{\frac {7}{2}} e^{\left (3 i \, d x + 3 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (3 i \, e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, e^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{5 \, a^{3} d} \] Input:

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

-2/5*(3*I*sqrt(2)*e^(7/2)*e^(3*I*d*x + 3*I*c)*weierstrassZeta(-4, 0, weier 
strassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(2)*(3*I*e^3*e^(4*I*d*x + 4* 
I*c) + 2*I*e^3*e^(2*I*d*x + 2*I*c) - 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^(-3*I*d*x - 3*I*c)/(a^3*d)
 

Sympy [F(-1)]

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

integrate((e*sec(d*x+c))**(7/2)/(a+I*a*tan(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

integrate((e*sec(d*x + c))^(7/2)/(I*a*tan(d*x + c) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^3} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \] Input:

int((e/cos(c + d*x))^(7/2)/(a + a*tan(c + d*x)*1i)^3,x)
 

Output:

int((e/cos(c + d*x))^(7/2)/(a + a*tan(c + d*x)*1i)^3, x)
 

Reduce [F]

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

int((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^3,x)
 

Output:

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