Integrand size = 24, antiderivative size = 68 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {i \sec ^3(c+d x)}{5 d (a+i a \tan (c+d x))^4}+\frac {i \sec ^3(c+d x)}{15 a d (a+i a \tan (c+d x))^3} \] Output:
1/5*I*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^4+1/15*I*sec(d*x+c)^3/a/d/(a+I*a*t an(d*x+c))^3
Time = 0.32 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.59 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\sec ^3(c+d x) (-4 i+\tan (c+d x))}{15 a^4 d (-i+\tan (c+d x))^4} \] Input:
Integrate[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^4,x]
Output:
-1/15*(Sec[c + d*x]^3*(-4*I + Tan[c + d*x]))/(a^4*d*(-I + Tan[c + d*x])^4)
Time = 0.34 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {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.
\(\displaystyle \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^3}{(a+i a \tan (c+d x))^4}dx\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {\int \frac {\sec ^3(c+d x)}{(i \tan (c+d x) a+a)^3}dx}{5 a}+\frac {i \sec ^3(c+d x)}{5 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sec (c+d x)^3}{(i \tan (c+d x) a+a)^3}dx}{5 a}+\frac {i \sec ^3(c+d x)}{5 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3969 |
\(\displaystyle \frac {i \sec ^3(c+d x)}{15 a d (a+i a \tan (c+d x))^3}+\frac {i \sec ^3(c+d x)}{5 d (a+i a \tan (c+d x))^4}\) |
Input:
Int[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^4,x]
Output:
((I/5)*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^4) + ((I/15)*Sec[c + d*x] ^3)/(a*d*(a + I*a*Tan[c + d*x])^3)
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 = 0.96 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.56
method | result | size |
risch | \(\frac {i {\mathrm e}^{-3 i \left (d x +c \right )}}{6 a^{4} d}+\frac {i {\mathrm e}^{-5 i \left (d x +c \right )}}{10 a^{4} d}\) | \(38\) |
derivativedivides | \(\frac {\frac {2}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {28}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {8 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {16}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {6 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}}{a^{4} d}\) | \(90\) |
default | \(\frac {\frac {2}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {28}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {8 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {16}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {6 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}}{a^{4} d}\) | \(90\) |
Input:
int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/6*I/a^4/d*exp(-3*I*(d*x+c))+1/10*I/a^4/d*exp(-5*I*(d*x+c))
Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.44 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (5 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{30 \, a^{4} d} \] Input:
integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
Output:
1/30*(5*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-5*I*d*x - 5*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. 182 vs. \(2 (54) = 108\).
Time = 1.33 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.68 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} - \frac {\tan {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{15 a^{4} d \tan ^{4}{\left (c + d x \right )} - 60 i a^{4} d \tan ^{3}{\left (c + d x \right )} - 90 a^{4} d \tan ^{2}{\left (c + d x \right )} + 60 i a^{4} d \tan {\left (c + d x \right )} + 15 a^{4} d} + \frac {4 i \sec ^{3}{\left (c + d x \right )}}{15 a^{4} d \tan ^{4}{\left (c + d x \right )} - 60 i a^{4} d \tan ^{3}{\left (c + d x \right )} - 90 a^{4} d \tan ^{2}{\left (c + d x \right )} + 60 i a^{4} d \tan {\left (c + d x \right )} + 15 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \sec ^{3}{\left (c \right )}}{\left (i a \tan {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \] Input:
integrate(sec(d*x+c)**3/(a+I*a*tan(d*x+c))**4,x)
Output:
Piecewise((-tan(c + d*x)*sec(c + d*x)**3/(15*a**4*d*tan(c + d*x)**4 - 60*I *a**4*d*tan(c + d*x)**3 - 90*a**4*d*tan(c + d*x)**2 + 60*I*a**4*d*tan(c + d*x) + 15*a**4*d) + 4*I*sec(c + d*x)**3/(15*a**4*d*tan(c + d*x)**4 - 60*I* a**4*d*tan(c + d*x)**3 - 90*a**4*d*tan(c + d*x)**2 + 60*I*a**4*d*tan(c + d *x) + 15*a**4*d), Ne(d, 0)), (x*sec(c)**3/(I*a*tan(c) + a)**4, True))
Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {3 i \, \cos \left (5 \, d x + 5 \, c\right ) + 5 i \, \cos \left (3 \, d x + 3 \, c\right ) + 3 \, \sin \left (5 \, d x + 5 \, c\right ) + 5 \, \sin \left (3 \, d x + 3 \, c\right )}{30 \, a^{4} d} \] Input:
integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
Output:
1/30*(3*I*cos(5*d*x + 5*c) + 5*I*cos(3*d*x + 3*c) + 3*sin(5*d*x + 5*c) + 5 *sin(3*d*x + 3*c))/(a^4*d)
Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 15 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 25 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{15 \, a^{4} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{5}} \] Input:
integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
Output:
2/15*(15*tan(1/2*d*x + 1/2*c)^4 - 15*I*tan(1/2*d*x + 1/2*c)^3 - 25*tan(1/2 *d*x + 1/2*c)^2 + 5*I*tan(1/2*d*x + 1/2*c) + 4)/(a^4*d*(tan(1/2*d*x + 1/2* c) - I)^5)
Time = 0.75 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.96 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {2\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,15{}\mathrm {i}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,25{}\mathrm {i}-5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+4{}\mathrm {i}\right )}{15\,a^4\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,1{}\mathrm {i}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,10{}\mathrm {i}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,5{}\mathrm {i}+1\right )} \] Input:
int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^4),x)
Output:
(2*(15*tan(c/2 + (d*x)/2)^3 - tan(c/2 + (d*x)/2)^2*25i - 5*tan(c/2 + (d*x) /2) + tan(c/2 + (d*x)/2)^4*15i + 4i))/(15*a^4*d*(tan(c/2 + (d*x)/2)*5i - 1 0*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))
\[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^3/(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 - 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)*si n(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*t an((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(t an((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*x)/(2*a**4*d )