Integrand size = 22, antiderivative size = 98 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}+\frac {2 i \sec (c+d x)}{15 a d (a+i a \tan (c+d x))^2}+\frac {2 i \sec (c+d x)}{15 d \left (a^3+i a^3 \tan (c+d x)\right )} \] Output:
1/5*I*sec(d*x+c)/d/(a+I*a*tan(d*x+c))^3+2/15*I*sec(d*x+c)/a/d/(a+I*a*tan(d *x+c))^2+2/15*I*sec(d*x+c)/d/(a^3+I*a^3*tan(d*x+c))
Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.55 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\sec ^3(c+d x) (5+9 \cos (2 (c+d x))+6 i \sin (2 (c+d x)))}{30 a^3 d (-i+\tan (c+d x))^3} \] Input:
Integrate[Sec[c + d*x]/(a + I*a*Tan[c + d*x])^3,x]
Output:
-1/30*(Sec[c + d*x]^3*(5 + 9*Cos[2*(c + d*x)] + (6*I)*Sin[2*(c + d*x)]))/( a^3*d*(-I + Tan[c + d*x])^3)
Time = 0.43 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 3983, 3042, 3983, 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.
\(\displaystyle \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3}dx\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {2 \int \frac {\sec (c+d x)}{(i \tan (c+d x) a+a)^2}dx}{5 a}+\frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \frac {\sec (c+d x)}{(i \tan (c+d x) a+a)^2}dx}{5 a}+\frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {\sec (c+d x)}{i \tan (c+d x) a+a}dx}{3 a}+\frac {i \sec (c+d x)}{3 d (a+i a \tan (c+d x))^2}\right )}{5 a}+\frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {\sec (c+d x)}{i \tan (c+d x) a+a}dx}{3 a}+\frac {i \sec (c+d x)}{3 d (a+i a \tan (c+d x))^2}\right )}{5 a}+\frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3969 |
\(\displaystyle \frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}+\frac {2 \left (\frac {i \sec (c+d x)}{3 a d (a+i a \tan (c+d x))}+\frac {i \sec (c+d x)}{3 d (a+i a \tan (c+d x))^2}\right )}{5 a}\) |
Input:
Int[Sec[c + d*x]/(a + I*a*Tan[c + d*x])^3,x]
Output:
((I/5)*Sec[c + d*x])/(d*(a + I*a*Tan[c + d*x])^3) + (2*(((I/3)*Sec[c + d*x ])/(d*(a + I*a*Tan[c + d*x])^2) + ((I/3)*Sec[c + d*x])/(a*d*(a + I*a*Tan[c + d*x]))))/(5*a)
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]
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]
Time = 0.78 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.57
method | result | size |
risch | \(\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{4 a^{3} d}+\frac {i {\mathrm e}^{-3 i \left (d x +c \right )}}{6 a^{3} d}+\frac {i {\mathrm e}^{-5 i \left (d x +c \right )}}{20 a^{3} d}\) | \(56\) |
derivativedivides | \(\frac {-\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {8}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {16}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{a^{3} d}\) | \(90\) |
default | \(\frac {-\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {8}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {16}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{a^{3} d}\) | \(90\) |
Input:
int(sec(d*x+c)/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/4*I/a^3/d*exp(-I*(d*x+c))+1/6*I/a^3/d*exp(-3*I*(d*x+c))+1/20*I/a^3/d*exp (-5*I*(d*x+c))
Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.42 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 10 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{60 \, a^{3} d} \] Input:
integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/60*(15*I*e^(4*I*d*x + 4*I*c) + 10*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-5*I*d *x - 5*I*c)/(a^3*d)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (82) = 164\).
Time = 0.86 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.23 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {2 \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{15 a^{3} d \tan ^{3}{\left (c + d x \right )} - 45 i a^{3} d \tan ^{2}{\left (c + d x \right )} - 45 a^{3} d \tan {\left (c + d x \right )} + 15 i a^{3} d} - \frac {6 i \tan {\left (c + d x \right )} \sec {\left (c + d x \right )}}{15 a^{3} d \tan ^{3}{\left (c + d x \right )} - 45 i a^{3} d \tan ^{2}{\left (c + d x \right )} - 45 a^{3} d \tan {\left (c + d x \right )} + 15 i a^{3} d} - \frac {7 \sec {\left (c + d x \right )}}{15 a^{3} d \tan ^{3}{\left (c + d x \right )} - 45 i a^{3} d \tan ^{2}{\left (c + d x \right )} - 45 a^{3} d \tan {\left (c + d x \right )} + 15 i a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sec {\left (c \right )}}{\left (i a \tan {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))**3,x)
Output:
Piecewise((2*tan(c + d*x)**2*sec(c + d*x)/(15*a**3*d*tan(c + d*x)**3 - 45* I*a**3*d*tan(c + d*x)**2 - 45*a**3*d*tan(c + d*x) + 15*I*a**3*d) - 6*I*tan (c + d*x)*sec(c + d*x)/(15*a**3*d*tan(c + d*x)**3 - 45*I*a**3*d*tan(c + d* x)**2 - 45*a**3*d*tan(c + d*x) + 15*I*a**3*d) - 7*sec(c + d*x)/(15*a**3*d* tan(c + d*x)**3 - 45*I*a**3*d*tan(c + d*x)**2 - 45*a**3*d*tan(c + d*x) + 1 5*I*a**3*d), Ne(d, 0)), (x*sec(c)/(I*a*tan(c) + a)**3, True))
Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.70 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {3 i \, \cos \left (5 \, d x + 5 \, c\right ) + 10 i \, \cos \left (3 \, d x + 3 \, c\right ) + 15 i \, \cos \left (d x + c\right ) + 3 \, \sin \left (5 \, d x + 5 \, c\right ) + 10 \, \sin \left (3 \, d x + 3 \, c\right ) + 15 \, \sin \left (d x + c\right )}{60 \, a^{3} d} \] Input:
integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
1/60*(3*I*cos(5*d*x + 5*c) + 10*I*cos(3*d*x + 3*c) + 15*I*cos(d*x + c) + 3 *sin(5*d*x + 5*c) + 10*sin(3*d*x + 3*c) + 15*sin(d*x + c))/(a^3*d)
Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.74 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7\right )}}{15 \, a^{3} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{5}} \] Input:
integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
2/15*(15*tan(1/2*d*x + 1/2*c)^4 - 30*I*tan(1/2*d*x + 1/2*c)^3 - 40*tan(1/2 *d*x + 1/2*c)^2 + 20*I*tan(1/2*d*x + 1/2*c) + 7)/(a^3*d*(tan(1/2*d*x + 1/2 *c) - I)^5)
Time = 0.73 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.36 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {2\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,15{}\mathrm {i}+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,40{}\mathrm {i}-20\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+7{}\mathrm {i}\right )}{15\,a^3\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,1{}\mathrm {i}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,10{}\mathrm {i}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,5{}\mathrm {i}+1\right )} \] Input:
int(1/(cos(c + d*x)*(a + a*tan(c + d*x)*1i)^3),x)
Output:
(2*(30*tan(c/2 + (d*x)/2)^3 - tan(c/2 + (d*x)/2)^2*40i - 20*tan(c/2 + (d*x )/2) + tan(c/2 + (d*x)/2)^4*15i + 7i))/(15*a^3*d*(tan(c/2 + (d*x)/2)*5i - 10*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^3*10i + 5*tan(c/2 + (d*x)/2)^ 4 + tan(c/2 + (d*x)/2)^5*1i + 1))
\[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)/(a+I*a*tan(d*x+c))^3,x)
Output:
( - 3*int(cos(c + d*x)/(4*cos(c + d*x)*sin(c + d*x)**2*i - cos(c + d*x)*i - 4*sin(c + d*x)**3 + 3*sin(c + d*x)),x)*d*i - 24*int(sin(c + d*x)**3/(4*c os(c + d*x)*sin(c + d*x)**2*i - cos(c + d*x)*i - 4*sin(c + d*x)**3 + 3*sin (c + d*x)),x)*d + 3*int(sin(c + d*x)**2/(4*cos(c + d*x)*sin(c + d*x)**2*i - cos(c + d*x)*i - 4*sin(c + d*x)**3 + 3*sin(c + d*x)),x)*d*i + 12*int(( - sin(c + d*x)**3)/(4*cos(c + d*x)*sin(c + d*x)**2 - cos(c + d*x) + 4*sin(c + d*x)**3*i - 3*sin(c + d*x)*i),x)*d*i - 9*int(( - sin(c + d*x))/(4*cos(c + d*x)*sin(c + d*x)**2 - cos(c + d*x) + 4*sin(c + d*x)**3*i - 3*sin(c + d *x)*i),x)*d*i + 12*int(( - cos(c + d*x)*sin(c + d*x)**2)/(4*cos(c + d*x)*s in(c + d*x)**2 - cos(c + d*x) + 4*sin(c + d*x)**3*i - 3*sin(c + d*x)*i),x) *d + 18*int(sin(c + d*x)/(4*cos(c + d*x)*sin(c + d*x)**2*i - cos(c + d*x)* i - 4*sin(c + d*x)**3 + 3*sin(c + d*x)),x)*d + 12*int((cos(c + d*x)*sin(c + d*x)**2)/(4*cos(c + d*x)*sin(c + d*x)**2*i - cos(c + d*x)*i - 4*sin(c + d*x)**3 + 3*sin(c + d*x)),x)*d*i + 12*int((cos(c + d*x)*sin(c + d*x)**2)/( 4*cos(c + d*x)*sin(c + d*x)**2 - cos(c + d*x) + 4*sin(c + d*x)**3*i - 3*si n(c + d*x)*i),x)*d - 3*int(1/(4*cos(c + d*x)*sin(c + d*x)**2*i - cos(c + d *x)*i - 4*sin(c + d*x)**3 + 3*sin(c + d*x)),x)*d*i + 2*log(tan((c + d*x)/2 )**6 - 6*tan((c + d*x)/2)**5*i - 15*tan((c + d*x)/2)**4 + 20*tan((c + d*x) /2)**3*i + 15*tan((c + d*x)/2)**2 - 6*tan((c + d*x)/2)*i - 1)*i - 12*log(t an((c + d*x)/2)**2 + 1)*i + 2*log(tan((c + d*x)/2)**6*i + 6*tan((c + d*...