Integrand size = 31, antiderivative size = 75 \[ \int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {4 x}{a^3}+\frac {4 i \log (\sin (c+d x))}{a^3 d}-\frac {4 i \log (\tan (c+d x))}{a^3 d}-\frac {3 \tan (c+d x)}{a^3 d}+\frac {i \tan ^2(c+d x)}{2 a^3 d} \]
4*x/a^3+4*I*ln(sin(d*x+c))/a^3/d-4*I*ln(tan(d*x+c))/a^3/d-3*tan(d*x+c)/a^3 /d+1/2*I*tan(d*x+c)^2/a^3/d
Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.64 \[ \int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {i \left (-4 \log (i-\tan (c+d x))+3 i \tan (c+d x)+\frac {1}{2} \tan ^2(c+d x)\right )}{a^3 d} \]
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.76, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 3567, 27, 516, 99, 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.
\(\displaystyle \int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (c+d x)^3 (a \cos (c+d x)+i a \sin (c+d x))^3}dx\) |
\(\Big \downarrow \) 3567 |
\(\displaystyle -\frac {\int \frac {\left (\cot ^2(c+d x)+1\right )^2 \tan ^3(c+d x)}{a^3 (\cot (c+d x)+i)^3}d\cot (c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {\left (\cot ^2(c+d x)+1\right )^2 \tan ^3(c+d x)}{(\cot (c+d x)+i)^3}d\cot (c+d x)}{a^3 d}\) |
\(\Big \downarrow \) 516 |
\(\displaystyle -\frac {\int \frac {(\cot (c+d x)-i)^2 \tan ^3(c+d x)}{\cot (c+d x)+i}d\cot (c+d x)}{a^3 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (i \tan ^3(c+d x)-3 \tan ^2(c+d x)-4 i \tan (c+d x)+\frac {4 i}{\cot (c+d x)+i}\right )d\cot (c+d x)}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{2} i \tan ^2(c+d x)+3 \tan (c+d x)-4 i \log (\cot (c+d x))+4 i \log (\cot (c+d x)+i)}{a^3 d}\) |
-(((-4*I)*Log[Cot[c + d*x]] + (4*I)*Log[I + Cot[c + d*x]] + 3*Tan[c + d*x] - (I/2)*Tan[c + d*x]^2)/(a^3*d))
3.2.83.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[(e*x)^m*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; Free Q[{a, b, c, d, e, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !IntegerQ[n]))
Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*si n[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[x^m*((b + a*x)^n/(1 + x^2)^((m + n + 2)/2)), x], x, Cot[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] && !(GtQ[ n, 0] && GtQ[m, 1])
Time = 0.76 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.55
method | result | size |
derivativedivides | \(\frac {-3 \tan \left (d x +c \right )+\frac {i \tan \left (d x +c \right )^{2}}{2}-4 i \ln \left (\tan \left (d x +c \right )-i\right )}{d \,a^{3}}\) | \(41\) |
default | \(\frac {-3 \tan \left (d x +c \right )+\frac {i \tan \left (d x +c \right )^{2}}{2}-4 i \ln \left (\tan \left (d x +c \right )-i\right )}{d \,a^{3}}\) | \(41\) |
risch | \(\frac {8 x}{a^{3}}+\frac {8 c}{d \,a^{3}}-\frac {2 i \left (2 \,{\mathrm e}^{2 i \left (d x +c \right )}+3\right )}{a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {4 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d \,a^{3}}\) | \(73\) |
norman | \(\frac {\frac {4 x}{a}-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}-\frac {8 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {4 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}+\frac {2 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4 i \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}+\frac {4 i \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}-\frac {4 i \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{3}}\) | \(194\) |
Time = 0.25 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.51 \[ \int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (4 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d x + 2 \, {\left (4 \, d x - i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, {\left (-i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 3 i\right )}}{a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d} \]
2*(4*d*x*e^(4*I*d*x + 4*I*c) + 4*d*x + 2*(4*d*x - I)*e^(2*I*d*x + 2*I*c) - 2*(-I*e^(4*I*d*x + 4*I*c) - 2*I*e^(2*I*d*x + 2*I*c) - I)*log(e^(2*I*d*x + 2*I*c) + 1) - 3*I)/(a^3*d*e^(4*I*d*x + 4*I*c) + 2*a^3*d*e^(2*I*d*x + 2*I* c) + a^3*d)
\[ \int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {\int \frac {\sec ^{3}{\left (c + d x \right )}}{- i \sin ^{3}{\left (c + d x \right )} - 3 \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )} + 3 i \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} + \cos ^{3}{\left (c + d x \right )}}\, dx}{a^{3}} \]
Integral(sec(c + d*x)**3/(-I*sin(c + d*x)**3 - 3*sin(c + d*x)**2*cos(c + d *x) + 3*I*sin(c + d*x)*cos(c + d*x)**2 + cos(c + d*x)**3), x)/a**3
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (67) = 134\).
Time = 0.32 (sec) , antiderivative size = 301, normalized size of antiderivative = 4.01 \[ \int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (4 i \, d x + 2 \, {\left (-i \, \cos \left (4 \, d x + 4 \, c\right ) - 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + \sin \left (4 \, d x + 4 \, c\right ) + 2 \, \sin \left (2 \, d x + 2 \, c\right ) - i\right )} \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right ) + 4 \, {\left (i \, d x + i \, c\right )} \cos \left (4 \, d x + 4 \, c\right ) + 2 \, {\left (4 i \, d x + 4 i \, c + 1\right )} \cos \left (2 \, d x + 2 \, c\right ) - {\left (\cos \left (4 \, d x + 4 \, c\right ) + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + i \, \sin \left (4 \, d x + 4 \, c\right ) + 2 i \, \sin \left (2 \, d x + 2 \, c\right ) + 1\right )} \log \left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right ) - 4 \, {\left (d x + c\right )} \sin \left (4 \, d x + 4 \, c\right ) - 2 \, {\left (4 \, d x + 4 \, c - i\right )} \sin \left (2 \, d x + 2 \, c\right ) + 4 i \, c + 3\right )}}{{\left (-i \, a^{3} \cos \left (4 \, d x + 4 \, c\right ) - 2 i \, a^{3} \cos \left (2 \, d x + 2 \, c\right ) + a^{3} \sin \left (4 \, d x + 4 \, c\right ) + 2 \, a^{3} \sin \left (2 \, d x + 2 \, c\right ) - i \, a^{3}\right )} d} \]
-2*(4*I*d*x + 2*(-I*cos(4*d*x + 4*c) - 2*I*cos(2*d*x + 2*c) + sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c) - I)*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1) + 4*(I*d*x + I*c)*cos(4*d*x + 4*c) + 2*(4*I*d*x + 4*I*c + 1)*cos(2*d* x + 2*c) - (cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + I*sin(4*d*x + 4*c) + 2 *I*sin(2*d*x + 2*c) + 1)*log(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*c os(2*d*x + 2*c) + 1) - 4*(d*x + c)*sin(4*d*x + 4*c) - 2*(4*d*x + 4*c - I)* sin(2*d*x + 2*c) + 4*I*c + 3)/((-I*a^3*cos(4*d*x + 4*c) - 2*I*a^3*cos(2*d* x + 2*c) + a^3*sin(4*d*x + 4*c) + 2*a^3*sin(2*d*x + 2*c) - I*a^3)*d)
Time = 0.34 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.71 \[ \int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (\frac {2 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3}} - \frac {4 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a^{3}} + \frac {2 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{3}} + \frac {-3 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 i}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}}\right )}}{d} \]
2*(2*I*log(tan(1/2*d*x + 1/2*c) + 1)/a^3 - 4*I*log(tan(1/2*d*x + 1/2*c) - I)/a^3 + 2*I*log(tan(1/2*d*x + 1/2*c) - 1)/a^3 + (-3*I*tan(1/2*d*x + 1/2*c )^4 + 3*tan(1/2*d*x + 1/2*c)^3 + 7*I*tan(1/2*d*x + 1/2*c)^2 - 3*tan(1/2*d* x + 1/2*c) - 3*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3))/d
Time = 23.22 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )\,8{}\mathrm {i}-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,4{}\mathrm {i}}{a^3\,d}+\frac {6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,2{}\mathrm {i}-6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}^2} \]