\(\int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [142]

Optimal result

Integrand size = 24, antiderivative size = 65 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {3 \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {3 i \sec (c+d x)}{a^3 d}+\frac {2 i \sec ^3(c+d x)}{a d (a+i a \tan (c+d x))^2} \] Output:

-3*arctanh(sin(d*x+c))/a^3/d+3*I*sec(d*x+c)/a^3/d+2*I*sec(d*x+c)^3/a/d/(a+ 
I*a*tan(d*x+c))^2
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.66 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\sec ^3(c+d x) (i \cos (d x)-\sin (d x))^3 \left (6 \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {d x}{2}\right )\right ) (\cos (3 c)+i \sin (3 c))+(\cos (2 c-d x)+i \sin (2 c-d x)) (-5 i+\tan (c+d x))\right )}{a^3 d (-i+\tan (c+d x))^3} \] Input:

Integrate[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^3,x]
 

Output:

(Sec[c + d*x]^3*(I*Cos[d*x] - Sin[d*x])^3*(6*ArcTanh[Sin[c] + Cos[c]*Tan[( 
d*x)/2]]*(Cos[3*c] + I*Sin[3*c]) + (Cos[2*c - d*x] + I*Sin[2*c - d*x])*(-5 
*I + Tan[c + d*x])))/(a^3*d*(-I + Tan[c + d*x])^3)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3981, 3042, 3982, 3042, 4257}

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.

\(\displaystyle \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3982

\(\displaystyle \frac {2 i \sec ^3(c+d x)}{a d (a+i a \tan (c+d x))^2}-\frac {3 \left (\frac {\int \sec (c+d x)dx}{a}-\frac {i \sec (c+d x)}{a d}\right )}{a^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

\(\displaystyle \frac {2 i \sec ^3(c+d x)}{a d (a+i a \tan (c+d x))^2}-\frac {3 \left (\frac {\text {arctanh}(\sin (c+d x))}{a d}-\frac {i \sec (c+d x)}{a d}\right )}{a^2}\)

Input:

Int[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^3,x]
 

Output:

(-3*(ArcTanh[Sin[c + d*x]]/(a*d) - (I*Sec[c + d*x])/(a*d)))/a^2 + ((2*I)*S 
ec[c + d*x]^3)/(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 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 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {\frac {8}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {i}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 i}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2}-3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}\)	\(86\)
default	\(\frac {\frac {8}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {i}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 i}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2}-3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}\)	\(86\)
risch	\(\frac {4 i {\mathrm e}^{-i \left (d x +c \right )}}{a^{3} d}+\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}\)	\(93\)

Input:

int(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

2/d/a^3*(4/(-I+tan(1/2*d*x+1/2*c))-1/2*I/(tan(1/2*d*x+1/2*c)-1)+3/2*ln(tan 
(1/2*d*x+1/2*c)-1)+1/2*I/(tan(1/2*d*x+1/2*c)+1)-3/2*ln(tan(1/2*d*x+1/2*c)+ 
1))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.72 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {3 \, {\left (e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3 \, {\left (e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i}{a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{3} d e^{\left (i \, d x + i \, c\right )}} \] Input:

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

-(3*(e^(3*I*d*x + 3*I*c) + e^(I*d*x + I*c))*log(e^(I*d*x + I*c) + I) - 3*( 
e^(3*I*d*x + 3*I*c) + e^(I*d*x + I*c))*log(e^(I*d*x + I*c) - I) - 6*I*e^(2 
*I*d*x + 2*I*c) - 4*I)/(a^3*d*e^(3*I*d*x + 3*I*c) + a^3*d*e^(I*d*x + I*c))
 

Sympy [F]

\[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {\sec ^{5}{\left (c + d x \right )}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \] Input:

integrate(sec(d*x+c)**5/(a+I*a*tan(d*x+c))**3,x)
 

Output:

I*Integral(sec(c + d*x)**5/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3*tan( 
c + d*x) + I), x)/a**3
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (59) = 118\).

Time = 0.14 (sec) , antiderivative size = 319, normalized size of antiderivative = 4.91 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {6 \, {\left (\cos \left (3 \, d x + 3 \, c\right ) + \cos \left (d x + c\right ) + i \, \sin \left (3 \, d x + 3 \, c\right ) + i \, \sin \left (d x + c\right )\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) + 6 \, {\left (\cos \left (3 \, d x + 3 \, c\right ) + \cos \left (d x + c\right ) + i \, \sin \left (3 \, d x + 3 \, c\right ) + i \, \sin \left (d x + c\right )\right )} \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (i \, \cos \left (3 \, d x + 3 \, c\right ) + i \, \cos \left (d x + c\right ) - \sin \left (3 \, d x + 3 \, c\right ) - \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (-i \, \cos \left (3 \, d x + 3 \, c\right ) - i \, \cos \left (d x + c\right ) + \sin \left (3 \, d x + 3 \, c\right ) + \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right ) + 12 \, \cos \left (2 \, d x + 2 \, c\right ) + 12 i \, \sin \left (2 \, d x + 2 \, c\right ) + 8}{-2 \, {\left (i \, a^{3} \cos \left (3 \, d x + 3 \, c\right ) + i \, a^{3} \cos \left (d x + c\right ) - a^{3} \sin \left (3 \, d x + 3 \, c\right ) - a^{3} \sin \left (d x + c\right )\right )} d} \] Input:

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

(6*(cos(3*d*x + 3*c) + cos(d*x + c) + I*sin(3*d*x + 3*c) + I*sin(d*x + c)) 
*arctan2(cos(d*x + c), sin(d*x + c) + 1) + 6*(cos(3*d*x + 3*c) + cos(d*x + 
 c) + I*sin(3*d*x + 3*c) + I*sin(d*x + c))*arctan2(cos(d*x + c), -sin(d*x 
+ c) + 1) + 3*(I*cos(3*d*x + 3*c) + I*cos(d*x + c) - sin(3*d*x + 3*c) - si 
n(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) + 3* 
(-I*cos(3*d*x + 3*c) - I*cos(d*x + c) + sin(3*d*x + 3*c) + sin(d*x + c))*l 
og(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) + 12*cos(2*d*x + 
2*c) + 12*I*sin(2*d*x + 2*c) + 8)/((-2*I*a^3*cos(3*d*x + 3*c) - 2*I*a^3*co 
s(d*x + c) + 2*a^3*sin(3*d*x + 3*c) + 2*a^3*sin(d*x + c))*d)
 

Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.69 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {3 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3}} - \frac {3 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{3}} - \frac {2 \, {\left (4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )} a^{3}}}{d} \] Input:

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

-(3*log(tan(1/2*d*x + 1/2*c) + 1)/a^3 - 3*log(tan(1/2*d*x + 1/2*c) - 1)/a^ 
3 - 2*(4*tan(1/2*d*x + 1/2*c)^2 - I*tan(1/2*d*x + 1/2*c) - 5)/((tan(1/2*d* 
x + 1/2*c)^3 - I*tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c) + I)*a^3))/ 
d
 

Mupad [B] (verification not implemented)

Time = 0.89 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.62 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {6\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,8{}\mathrm {i}}{a^3}-\frac {10{}\mathrm {i}}{a^3}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,1{}\mathrm {i}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}+1\right )} \] Input:

int(1/(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^3),x)
 

Output:

- (6*atanh(tan(c/2 + (d*x)/2)))/(a^3*d) - ((tan(c/2 + (d*x)/2)^2*8i)/a^3 - 
 10i/a^3 + (2*tan(c/2 + (d*x)/2))/a^3)/(d*(tan(c/2 + (d*x)/2)*1i - tan(c/2 
 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^3*1i + 1))
 

Reduce [F]

\[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\text {too large to display} \] Input:

int(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^3,x)
 

Output:

(8160*cos(c + d*x)*int(cos(c + d*x)/(4*cos(c + d*x)*sin(c + d*x)**4*i - 5* 
cos(c + d*x)*sin(c + d*x)**2*i + cos(c + d*x)*i - 4*sin(c + d*x)**5 + 7*si 
n(c + d*x)**3 - 3*sin(c + d*x)),x)*d*i + 696*cos(c + d*x)*int(cos(c + d*x) 
/(4*cos(c + d*x)*sin(c + d*x)**3*i - 3*cos(c + d*x)*sin(c + d*x)*i - 4*sin 
(c + d*x)**4 + 5*sin(c + d*x)**2 - 1),x)*d - 720*cos(c + d*x)*int(cos(c + 
d*x)/(4*cos(c + d*x)*sin(c + d*x)**3 - 3*cos(c + d*x)*sin(c + d*x) + 4*sin 
(c + d*x)**4*i - 5*sin(c + d*x)**2*i + i),x)*d*i - 12048*cos(c + d*x)*int( 
sin(c + d*x)**5/(4*cos(c + d*x)*sin(c + d*x)**4*i - 5*cos(c + d*x)*sin(c + 
 d*x)**2*i + cos(c + d*x)*i - 4*sin(c + d*x)**5 + 7*sin(c + d*x)**3 - 3*si 
n(c + d*x)),x)*d + 27360*cos(c + d*x)*int(sin(c + d*x)**4/(4*cos(c + d*x)* 
sin(c + d*x)**4*i - 5*cos(c + d*x)*sin(c + d*x)**2*i + cos(c + d*x)*i - 4* 
sin(c + d*x)**5 + 7*sin(c + d*x)**3 - 3*sin(c + d*x)),x)*d*i + 2352*cos(c 
+ d*x)*int(sin(c + d*x)**4/(4*cos(c + d*x)*sin(c + d*x)**3*i - 3*cos(c + d 
*x)*sin(c + d*x)*i - 4*sin(c + d*x)**4 + 5*sin(c + d*x)**2 - 1),x)*d - 240 
0*cos(c + d*x)*int(sin(c + d*x)**4/(4*cos(c + d*x)*sin(c + d*x)**3 - 3*cos 
(c + d*x)*sin(c + d*x) + 4*sin(c + d*x)**4*i - 5*sin(c + d*x)**2*i + i),x) 
*d*i + 38940*cos(c + d*x)*int(sin(c + d*x)**3/(4*cos(c + d*x)*sin(c + d*x) 
**4*i - 5*cos(c + d*x)*sin(c + d*x)**2*i + cos(c + d*x)*i - 4*sin(c + d*x) 
**5 + 7*sin(c + d*x)**3 - 3*sin(c + d*x)),x)*d - 2784*cos(c + d*x)*int(sin 
(c + d*x)**3/(4*cos(c + d*x)*sin(c + d*x)**3*i - 3*cos(c + d*x)*sin(c +...