Integrand size = 22, antiderivative size = 54 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{2 d}+\frac {i a \sec ^3(c+d x)}{3 d}+\frac {a \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
1/2*a*arctanh(sin(d*x+c))/d+1/3*I*a*sec(d*x+c)^3/d+1/2*a*sec(d*x+c)*tan(d* x+c)/d
Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{2 d}+\frac {i a \sec ^3(c+d x)}{3 d}+\frac {a \sec (c+d x) \tan (c+d x)}{2 d} \] Input:
Integrate[Sec[c + d*x]^3*(a + I*a*Tan[c + d*x]),x]
Output:
(a*ArcTanh[Sin[c + d*x]])/(2*d) + ((I/3)*a*Sec[c + d*x]^3)/d + (a*Sec[c + d*x]*Tan[c + d*x])/(2*d)
Time = 0.34 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 3967, 3042, 4255, 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.
\(\displaystyle \int \sec ^3(c+d x) (a+i a \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^3 (a+i a \tan (c+d x))dx\) |
\(\Big \downarrow \) 3967 |
\(\displaystyle a \int \sec ^3(c+d x)dx+\frac {i a \sec ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {i a \sec ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle a \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {i a \sec ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {i a \sec ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle a \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {i a \sec ^3(c+d x)}{3 d}\) |
Input:
Int[Sec[c + d*x]^3*(a + I*a*Tan[c + d*x]),x]
Output:
((I/3)*a*Sec[c + d*x]^3)/d + a*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x ]*Tan[c + d*x])/(2*d))
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a Int[(d *Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] || NeQ[a^2 + b^2, 0])
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.63 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {\frac {i a}{3 \cos \left (d x +c \right )^{3}}+a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(51\) |
default | \(\frac {\frac {i a}{3 \cos \left (d x +c \right )^{3}}+a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(51\) |
risch | \(-\frac {i a \left (3 \,{\mathrm e}^{5 i \left (d x +c \right )}-8 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(94\) |
Input:
int(sec(d*x+c)^3*(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/3*I*a/cos(d*x+c)^3+a*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+t an(d*x+c))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (46) = 92\).
Time = 0.11 (sec) , antiderivative size = 180, normalized size of antiderivative = 3.33 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {-6 i \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 16 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} + 6 i \, a e^{\left (i \, d x + i \, c\right )} + 3 \, {\left (a e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3 \, {\left (a e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{6 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="fricas")
Output:
1/6*(-6*I*a*e^(5*I*d*x + 5*I*c) + 16*I*a*e^(3*I*d*x + 3*I*c) + 6*I*a*e^(I* d*x + I*c) + 3*(a*e^(6*I*d*x + 6*I*c) + 3*a*e^(4*I*d*x + 4*I*c) + 3*a*e^(2 *I*d*x + 2*I*c) + a)*log(e^(I*d*x + I*c) + I) - 3*(a*e^(6*I*d*x + 6*I*c) + 3*a*e^(4*I*d*x + 4*I*c) + 3*a*e^(2*I*d*x + 2*I*c) + a)*log(e^(I*d*x + I*c ) - I))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \sec ^3(c+d x) (a+i a \tan (c+d x)) \, dx=i a \left (\int \left (- i \sec ^{3}{\left (c + d x \right )}\right )\, dx + \int \tan {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)**3*(a+I*a*tan(d*x+c)),x)
Output:
I*a*(Integral(-I*sec(c + d*x)**3, x) + Integral(tan(c + d*x)*sec(c + d*x)* *3, x))
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.13 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {3 \, a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {4 i \, a}{\cos \left (d x + c\right )^{3}}}{12 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="maxima")
Output:
-1/12*(3*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 4*I*a/cos(d*x + c)^3)/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (46) = 92\).
Time = 0.16 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.80 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {3 \, a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 3 \, a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + \frac {2 \, {\left (3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 i \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="giac")
Output:
1/6*(3*a*log(tan(1/2*d*x + 1/2*c) + 1) - 3*a*log(tan(1/2*d*x + 1/2*c) - 1) + 2*(3*a*tan(1/2*d*x + 1/2*c)^5 - 6*I*a*tan(1/2*d*x + 1/2*c)^4 - 3*a*tan( 1/2*d*x + 1/2*c) - 2*I*a)/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 2.32 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.98 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2{}\mathrm {i}\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a\,2{}\mathrm {i}}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int((a + a*tan(c + d*x)*1i)/cos(c + d*x)^3,x)
Output:
(a*atanh(tan(c/2 + (d*x)/2)))/d - ((a*2i)/3 + a*tan(c/2 + (d*x)/2) + a*tan (c/2 + (d*x)/2)^4*2i - a*tan(c/2 + (d*x)/2)^5)/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))
Time = 0.18 (sec) , antiderivative size = 166, normalized size of antiderivative = 3.07 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \left (-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} i -3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+2 \cos \left (d x +c \right ) i -2 i \right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^3*(a+I*a*tan(d*x+c)),x)
Output:
(a*( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1) + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1 )*sin(c + d*x)**2 - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1) - 2*cos(c + d *x)*sin(c + d*x)**2*i - 3*cos(c + d*x)*sin(c + d*x) + 2*cos(c + d*x)*i - 2 *i))/(6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))