Integrand size = 24, antiderivative size = 48 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {x}{a^3}-\frac {i \log (\cos (c+d x))}{a^3 d}+\frac {2 i}{a^3 d (1+i \tan (c+d x))} \] Output:
-x/a^3-I*ln(cos(d*x+c))/a^3/d+2*I/a^3/d/(1+I*tan(d*x+c))
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.92 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {i \left (-\log (i-\tan (c+d x))-\frac {2 a}{a+i a \tan (c+d x)}\right )}{a^3 d} \] Input:
Integrate[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^3,x]
Output:
((-I)*(-Log[I - Tan[c + d*x]] - (2*a)/(a + I*a*Tan[c + d*x])))/(a^3*d)
Time = 0.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3968, 49, 2009}
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.
\(\displaystyle \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^4}{(a+i a \tan (c+d x))^3}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i \int \frac {a-i a \tan (c+d x)}{(i \tan (c+d x) a+a)^2}d(i a \tan (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i \int \left (\frac {2 a}{(i \tan (c+d x) a+a)^2}+\frac {1}{-i \tan (c+d x) a-a}\right )d(i a \tan (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (-\frac {2 a}{a+i a \tan (c+d x)}-\log (a+i a \tan (c+d x))\right )}{a^3 d}\) |
Input:
Int[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^3,x]
Output:
((-I)*(-Log[a + I*a*Tan[c + d*x]] - (2*a)/(a + I*a*Tan[c + d*x])))/(a^3*d)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Time = 0.56 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.17
method | result | size |
derivativedivides | \(\frac {2}{a^{3} d \left (-i+\tan \left (d x +c \right )\right )}+\frac {i \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 a^{3} d}-\frac {\arctan \left (\tan \left (d x +c \right )\right )}{a^{3} d}\) | \(56\) |
default | \(\frac {2}{a^{3} d \left (-i+\tan \left (d x +c \right )\right )}+\frac {i \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 a^{3} d}-\frac {\arctan \left (\tan \left (d x +c \right )\right )}{a^{3} d}\) | \(56\) |
risch | \(\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{a^{3} d}-\frac {2 x}{a^{3}}-\frac {2 c}{a^{3} d}-\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{3} d}\) | \(56\) |
Input:
int(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
2/a^3/d/(-I+tan(d*x+c))+1/2*I/a^3/d*ln(1+tan(d*x+c)^2)-1/a^3/d*arctan(tan( d*x+c))
Time = 0.10 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.15 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (2 \, d x e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - i\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a^{3} d} \] Input:
integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
-(2*d*x*e^(2*I*d*x + 2*I*c) + I*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c ) + 1) - I)*e^(-2*I*d*x - 2*I*c)/(a^3*d)
\[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {\sec ^{4}{\left (c + d x \right )}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \] Input:
integrate(sec(d*x+c)**4/(a+I*a*tan(d*x+c))**3,x)
Output:
I*Integral(sec(c + d*x)**4/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3*tan( c + d*x) + I), x)/a**3
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.38 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {4 \, {\left (-i \, \tan \left (d x + c\right ) - 1\right )}}{2 i \, a^{3} \tan \left (d x + c\right )^{2} + 4 \, a^{3} \tan \left (d x + c\right ) - 2 i \, a^{3}} - \frac {i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{3}}}{d} \] Input:
integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
-(4*(-I*tan(d*x + c) - 1)/(2*I*a^3*tan(d*x + c)^2 + 4*a^3*tan(d*x + c) - 2 *I*a^3) - I*log(I*tan(d*x + c) + 1)/a^3)/d
Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.75 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3} d} + \frac {2}{a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}} \] Input:
integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
I*log(tan(d*x + c) - I)/(a^3*d) + 2/(a^3*d*(tan(d*x + c) - I))
Time = 0.43 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.88 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{a^3\,d}+\frac {2{}\mathrm {i}}{a^3\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \] Input:
int(1/(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^3),x)
Output:
(log(tan(c + d*x) - 1i)*1i)/(a^3*d) + 2i/(a^3*d*(tan(c + d*x)*1i + 1))
\[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\text {too large to display} \] Input:
int(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x)
Output:
( - 408*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 + 432*i nt(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 + 24*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 - 1368*int(sin(c + d*x)**4/(4*cos(c + d*x)*s in(c + d*x)**3*i - 3*cos(c + d*x)*sin(c + d*x)*i - 4*sin(c + d*x)**4 + 5*s in(c + d*x)**2 - 1),x)*d + 1440*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 + 1632*int(sin(c + d*x)**3/(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*i + 1632*int(sin(c + d*x)**3/(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 + 168*int(sin(c + d*x)**3/(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 + 1836*int(sin (c + d*x)**2/(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 - 1944*int(sin(c + d*x)**2/(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 - 1224*int(sin(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...