\(\int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx\) [181]

Optimal result

Integrand size = 29, antiderivative size = 61 \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {x}{a^3}+\frac {2}{a^3 d (i+\cot (c+d x))}-\frac {i \log (\sin (c+d x))}{a^3 d}+\frac {i \log (\tan (c+d x))}{a^3 d} \]

[Out]

-x/a^3+2/a^3/d/(I+cot(d*x+c))-I*ln(sin(d*x+c))/a^3/d+I*ln(tan(d*x+c))/a^3/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3167, 862, 78} \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {2}{a^3 d (\cot (c+d x)+i)}-\frac {i \log (\sin (c+d x))}{a^3 d}+\frac {i \log (\tan (c+d x))}{a^3 d}-\frac {x}{a^3} \]

[In]

Int[Sec[c + d*x]/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^3,x]

[Out]

-(x/a^3) + 2/(a^3*d*(I + Cot[c + d*x])) - (I*Log[Sin[c + d*x]])/(a^3*d) + (I*Log[Tan[c + d*x]])/(a^3*d)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 862

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c/e)*x)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 3167

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[-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])

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1+x^2}{x (i a+a x)^3} \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \frac {-\frac {i}{a}+\frac {x}{a}}{x (i a+a x)^2} \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {i}{a^3 x}+\frac {2}{a^3 (i+x)^2}-\frac {i}{a^3 (i+x)}\right ) \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {x}{a^3}+\frac {2}{a^3 d (i+\cot (c+d x))}-\frac {i \log (\sin (c+d x))}{a^3 d}+\frac {i \log (\tan (c+d x))}{a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.66 \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {i \left (\log (i-\tan (c+d x))-\frac {2 i}{-i+\tan (c+d x)}\right )}{a^3 d} \]

[In]

Integrate[Sec[c + d*x]/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^3,x]

[Out]

(I*(Log[I - Tan[c + d*x]] - (2*I)/(-I + Tan[c + d*x])))/(a^3*d)

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.57

method	result	size
derivativedivides	\(\frac {\frac {2}{\tan \left (d x +c \right )-i}+i \ln \left (\tan \left (d x +c \right )-i\right )}{d \,a^{3}}\)	\(35\)
default	\(\frac {\frac {2}{\tan \left (d x +c \right )-i}+i \ln \left (\tan \left (d x +c \right )-i\right )}{d \,a^{3}}\)	\(35\)
risch	\(\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{d \,a^{3}}-\frac {2 x}{a^{3}}-\frac {2 c}{d \,a^{3}}-\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d \,a^{3}}\)	\(56\)
norman	\(\frac {-\frac {x}{a}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}-\frac {2 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {8 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} a^{2}}+\frac {i \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{3}}-\frac {i \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}-\frac {i \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}\)	\(183\)

[In]

int(sec(d*x+c)/(cos(d*x+c)*a+I*a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d/a^3*(2/(tan(d*x+c)-I)+I*ln(tan(d*x+c)-I))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {{\left (2 \, d x e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - i\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a^{3} d} \]

[In]

integrate(sec(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-(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)

Sympy [F]

\[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {\int \frac {\sec {\left (c + d x \right )}}{- i \sin ^{3}{\left (c + d x \right )} - 3 \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )} + 3 i \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} + \cos ^{3}{\left (c + d x \right )}}\, dx}{a^{3}} \]

[In]

integrate(sec(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)/(-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

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.62 \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {4 \, d x + 4 \, c - 2 \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right ) - 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i \, \log \left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right ) - 2 \, \sin \left (2 \, d x + 2 \, c\right )}{2 \, a^{3} d} \]

[In]

integrate(sec(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(4*d*x + 4*c - 2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1) - 2*I*cos(2*d*x + 2*c) + I*log(cos(2*d*x
+ 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) - 2*sin(2*d*x + 2*c))/(a^3*d)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.64 \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {\frac {i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3}} - \frac {2 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a^{3}} + \frac {i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{3}} + \frac {3 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 i}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{2}}}{d} \]

[In]

integrate(sec(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-(I*log(tan(1/2*d*x + 1/2*c) + 1)/a^3 - 2*I*log(tan(1/2*d*x + 1/2*c) - I)/a^3 + I*log(tan(1/2*d*x + 1/2*c) - 1
)/a^3 + (3*I*tan(1/2*d*x + 1/2*c)^2 + 10*tan(1/2*d*x + 1/2*c) - 3*I)/(a^3*(tan(1/2*d*x + 1/2*c) - I)^2))/d

Mupad [B] (verification not implemented)

Time = 23.36 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.66 \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,4{}\mathrm {i}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,1{}\mathrm {i}+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a^3\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )\,2{}\mathrm {i}}{a^3\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,1{}\mathrm {i}}{a^3\,d} \]

[In]

int(1/(cos(c + d*x)*(a*cos(c + d*x) + a*sin(c + d*x)*1i)^3),x)

[Out]

(log(tan(c/2 + (d*x)/2) - 1i)*2i)/(a^3*d) - (tan(c/2 + (d*x)/2)*4i)/(d*(a^3*tan(c/2 + (d*x)/2)^2*1i - a^3*1i +
2*a^3*tan(c/2 + (d*x)/2))) - (log(tan(c/2 + (d*x)/2)^2 - 1)*1i)/(a^3*d)