Integrand size = 22, antiderivative size = 132 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {i \sec (c+d x)}{7 d (a+i a \tan (c+d x))^4}+\frac {3 i \sec (c+d x)}{35 a d (a+i a \tan (c+d x))^3}+\frac {2 i \sec (c+d x)}{35 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {2 i \sec (c+d x)}{35 d \left (a^4+i a^4 \tan (c+d x)\right )} \] Output:
1/7*I*sec(d*x+c)/d/(a+I*a*tan(d*x+c))^4+3/35*I*sec(d*x+c)/a/d/(a+I*a*tan(d *x+c))^3+2/35*I*sec(d*x+c)/d/(a^2+I*a^2*tan(d*x+c))^2+2/35*I*sec(d*x+c)/d/ (a^4+I*a^4*tan(d*x+c))
Time = 0.34 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.55 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {i \sec ^4(c+d x) (28 \cos (c+d x)+20 \cos (3 (c+d x))+7 i \sin (c+d x)+15 i \sin (3 (c+d x)))}{140 a^4 d (-i+\tan (c+d x))^4} \] Input:
Integrate[Sec[c + d*x]/(a + I*a*Tan[c + d*x])^4,x]
Output:
((I/140)*Sec[c + d*x]^4*(28*Cos[c + d*x] + 20*Cos[3*(c + d*x)] + (7*I)*Sin [c + d*x] + (15*I)*Sin[3*(c + d*x)]))/(a^4*d*(-I + Tan[c + d*x])^4)
Time = 0.55 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3983, 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.
| \(\displaystyle \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^4}dx\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {3 \int \frac {\sec (c+d x)}{(i \tan (c+d x) a+a)^3}dx}{7 a}+\frac {i \sec (c+d x)}{7 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \int \frac {\sec (c+d x)}{(i \tan (c+d x) a+a)^3}dx}{7 a}+\frac {i \sec (c+d x)}{7 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {3 \left (\frac {2 \int \frac {\sec (c+d x)}{(i \tan (c+d x) a+a)^2}dx}{5 a}+\frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}\right )}{7 a}+\frac {i \sec (c+d x)}{7 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {2 \int \frac {\sec (c+d x)}{(i \tan (c+d x) a+a)^2}dx}{5 a}+\frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}\right )}{7 a}+\frac {i \sec (c+d x)}{7 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {3 \left (\frac {2 \left (\frac {\int \frac {\sec (c+d x)}{i \tan (c+d x) a+a}dx}{3 a}+\frac {i \sec (c+d x)}{3 d (a+i a \tan (c+d x))^2}\right )}{5 a}+\frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}\right )}{7 a}+\frac {i \sec (c+d x)}{7 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {2 \left (\frac {\int \frac {\sec (c+d x)}{i \tan (c+d x) a+a}dx}{3 a}+\frac {i \sec (c+d x)}{3 d (a+i a \tan (c+d x))^2}\right )}{5 a}+\frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}\right )}{7 a}+\frac {i \sec (c+d x)}{7 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3969 |
| \(\displaystyle \frac {i \sec (c+d x)}{7 d (a+i a \tan (c+d x))^4}+\frac {3 \left (\frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}+\frac {2 \left (\frac {i \sec (c+d x)}{3 a d (a+i a \tan (c+d x))}+\frac {i \sec (c+d x)}{3 d (a+i a \tan (c+d x))^2}\right )}{5 a}\right )}{7 a}\) |
Input:
Int[Sec[c + d*x]/(a + I*a*Tan[c + d*x])^4,x]
Output:
((I/7)*Sec[c + d*x])/(d*(a + I*a*Tan[c + d*x])^4) + (3*(((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)) )/(7*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 = 0.87 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.56
| method | result | size |
| risch | \(\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{8 a^{4} d}+\frac {i {\mathrm e}^{-3 i \left (d x +c \right )}}{8 a^{4} d}+\frac {3 i {\mathrm e}^{-5 i \left (d x +c \right )}}{40 a^{4} d}+\frac {i {\mathrm e}^{-7 i \left (d x +c \right )}}{56 a^{4} d}\) | \(74\) |
| derivativedivides | \(\frac {\frac {2}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {12}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {16 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {16}{7 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {72}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {8 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {6 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}}{a^{4} d}\) | \(123\) |
| default | \(\frac {\frac {2}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {12}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {16 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {16}{7 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {72}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {8 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {6 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}}{a^{4} d}\) | \(123\) |
Input:
int(sec(d*x+c)/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/8*I/a^4/d*exp(-I*(d*x+c))+1/8*I/a^4/d*exp(-3*I*(d*x+c))+3/40*I/a^4/d*exp (-5*I*(d*x+c))+1/56*I/a^4/d*exp(-7*I*(d*x+c))
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.39 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (35 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 35 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 21 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{280 \, a^{4} d} \] Input:
integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
Output:
1/280*(35*I*e^(6*I*d*x + 6*I*c) + 35*I*e^(4*I*d*x + 4*I*c) + 21*I*e^(2*I*d *x + 2*I*c) + 5*I)*e^(-7*I*d*x - 7*I*c)/(a^4*d)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (112) = 224\).
Time = 1.41 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.68 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {2 \tan ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{35 a^{4} d \tan ^{4}{\left (c + d x \right )} - 140 i a^{4} d \tan ^{3}{\left (c + d x \right )} - 210 a^{4} d \tan ^{2}{\left (c + d x \right )} + 140 i a^{4} d \tan {\left (c + d x \right )} + 35 a^{4} d} - \frac {8 i \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{35 a^{4} d \tan ^{4}{\left (c + d x \right )} - 140 i a^{4} d \tan ^{3}{\left (c + d x \right )} - 210 a^{4} d \tan ^{2}{\left (c + d x \right )} + 140 i a^{4} d \tan {\left (c + d x \right )} + 35 a^{4} d} - \frac {13 \tan {\left (c + d x \right )} \sec {\left (c + d x \right )}}{35 a^{4} d \tan ^{4}{\left (c + d x \right )} - 140 i a^{4} d \tan ^{3}{\left (c + d x \right )} - 210 a^{4} d \tan ^{2}{\left (c + d x \right )} + 140 i a^{4} d \tan {\left (c + d x \right )} + 35 a^{4} d} + \frac {12 i \sec {\left (c + d x \right )}}{35 a^{4} d \tan ^{4}{\left (c + d x \right )} - 140 i a^{4} d \tan ^{3}{\left (c + d x \right )} - 210 a^{4} d \tan ^{2}{\left (c + d x \right )} + 140 i a^{4} d \tan {\left (c + d x \right )} + 35 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \sec {\left (c \right )}}{\left (i a \tan {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \] Input:
integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))**4,x)
Output:
Piecewise((2*tan(c + d*x)**3*sec(c + d*x)/(35*a**4*d*tan(c + d*x)**4 - 140 *I*a**4*d*tan(c + d*x)**3 - 210*a**4*d*tan(c + d*x)**2 + 140*I*a**4*d*tan( c + d*x) + 35*a**4*d) - 8*I*tan(c + d*x)**2*sec(c + d*x)/(35*a**4*d*tan(c + d*x)**4 - 140*I*a**4*d*tan(c + d*x)**3 - 210*a**4*d*tan(c + d*x)**2 + 14 0*I*a**4*d*tan(c + d*x) + 35*a**4*d) - 13*tan(c + d*x)*sec(c + d*x)/(35*a* *4*d*tan(c + d*x)**4 - 140*I*a**4*d*tan(c + d*x)**3 - 210*a**4*d*tan(c + d *x)**2 + 140*I*a**4*d*tan(c + d*x) + 35*a**4*d) + 12*I*sec(c + d*x)/(35*a* *4*d*tan(c + d*x)**4 - 140*I*a**4*d*tan(c + d*x)**3 - 210*a**4*d*tan(c + d *x)**2 + 140*I*a**4*d*tan(c + d*x) + 35*a**4*d), Ne(d, 0)), (x*sec(c)/(I*a *tan(c) + a)**4, True))
Time = 0.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.69 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {5 i \, \cos \left (7 \, d x + 7 \, c\right ) + 21 i \, \cos \left (5 \, d x + 5 \, c\right ) + 35 i \, \cos \left (3 \, d x + 3 \, c\right ) + 35 i \, \cos \left (d x + c\right ) + 5 \, \sin \left (7 \, d x + 7 \, c\right ) + 21 \, \sin \left (5 \, d x + 5 \, c\right ) + 35 \, \sin \left (3 \, d x + 3 \, c\right ) + 35 \, \sin \left (d x + c\right )}{280 \, a^{4} d} \] Input:
integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
Output:
1/280*(5*I*cos(7*d*x + 7*c) + 21*I*cos(5*d*x + 5*c) + 35*I*cos(3*d*x + 3*c ) + 35*I*cos(d*x + c) + 5*sin(7*d*x + 7*c) + 21*sin(5*d*x + 5*c) + 35*sin( 3*d*x + 3*c) + 35*sin(d*x + c))/(a^4*d)
Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.75 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {2 \, {\left (35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 105 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 210 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 210 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 147 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 49 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12\right )}}{35 \, a^{4} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{7}} \] Input:
integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
Output:
2/35*(35*tan(1/2*d*x + 1/2*c)^6 - 105*I*tan(1/2*d*x + 1/2*c)^5 - 210*tan(1 /2*d*x + 1/2*c)^4 + 210*I*tan(1/2*d*x + 1/2*c)^3 + 147*tan(1/2*d*x + 1/2*c )^2 - 49*I*tan(1/2*d*x + 1/2*c) - 12)/(a^4*d*(tan(1/2*d*x + 1/2*c) - I)^7)
Time = 0.85 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.48 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{8}+\frac {{\mathrm {e}}^{-c\,3{}\mathrm {i}-d\,x\,3{}\mathrm {i}}\,1{}\mathrm {i}}{8}+\frac {{\mathrm {e}}^{-c\,5{}\mathrm {i}-d\,x\,5{}\mathrm {i}}\,3{}\mathrm {i}}{40}+\frac {{\mathrm {e}}^{-c\,7{}\mathrm {i}-d\,x\,7{}\mathrm {i}}\,1{}\mathrm {i}}{56}}{a^4\,d} \] Input:
int(1/(cos(c + d*x)*(a + a*tan(c + d*x)*1i)^4),x)
Output:
((exp(- c*1i - d*x*1i)*1i)/8 + (exp(- c*3i - d*x*3i)*1i)/8 + (exp(- c*5i - d*x*5i)*3i)/40 + (exp(- c*7i - d*x*7i)*1i)/56)/(a^4*d)
\[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^4} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)/(a+I*a*tan(d*x+c))^4,x)
Output:
( - 2*int(cos(c + d*x)/(8*cos(c + d*x)*sin(c + d*x)**3*i - 4*cos(c + d*x)* sin(c + d*x)*i - 8*sin(c + d*x)**4 + 8*sin(c + d*x)**2 - 1),x)*d - 16*int( sin(c + d*x)**4/(8*cos(c + d*x)*sin(c + d*x)**3*i - 4*cos(c + d*x)*sin(c + d*x)*i - 8*sin(c + d*x)**4 + 8*sin(c + d*x)**2 - 1),x)*d - 16*int(sin(c + d*x)**2/(8*cos(c + d*x)*sin(c + d*x)**3 - 4*cos(c + d*x)*sin(c + d*x) + 8 *sin(c + d*x)**4*i - 8*sin(c + d*x)**2*i + i),x)*d*i + 16*int((cos(c + d*x )*sin(c + d*x)**3)/(8*cos(c + d*x)*sin(c + d*x)**3 - 4*cos(c + d*x)*sin(c + d*x) + 8*sin(c + d*x)**4*i - 8*sin(c + d*x)**2*i + i),x)*d - 2*int((cos( c + d*x)*sin(c + d*x)**2)/(8*cos(c + d*x)*sin(c + d*x)**3 - 4*cos(c + d*x) *sin(c + d*x) + 8*sin(c + d*x)**4*i - 8*sin(c + d*x)**2*i + i),x)*d*i - 8* int((cos(c + d*x)*sin(c + d*x))/(8*cos(c + d*x)*sin(c + d*x)**3 - 4*cos(c + d*x)*sin(c + d*x) + 8*sin(c + d*x)**4*i - 8*sin(c + d*x)**2*i + i),x)*d - 2*int(1/(8*cos(c + d*x)*sin(c + d*x)**3*i - 4*cos(c + d*x)*sin(c + d*x)* i - 8*sin(c + d*x)**4 + 8*sin(c + d*x)**2 - 1),x)*d + log(tan((c + d*x)/2) **8 - 8*tan((c + d*x)/2)**7*i - 28*tan((c + d*x)/2)**6 + 56*tan((c + d*x)/ 2)**5*i + 70*tan((c + d*x)/2)**4 - 56*tan((c + d*x)/2)**3*i - 28*tan((c + d*x)/2)**2 + 8*tan((c + d*x)/2)*i + 1)*i - 8*log(tan((c + d*x)/2)**2 + 1)* i + log(tan((c + d*x)/2)**8*i + 8*tan((c + d*x)/2)**7 - 28*tan((c + d*x)/2 )**6*i - 56*tan((c + d*x)/2)**5 + 70*tan((c + d*x)/2)**4*i + 56*tan((c + d *x)/2)**3 - 28*tan((c + d*x)/2)**2*i - 8*tan((c + d*x)/2) + i)*i + 6*d*...