\(\int \frac {\sec ^6(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx\) [174]

Optimal result

Integrand size = 31, antiderivative size = 70 \[ \int \frac {\sec ^6(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {\tan (c+d x)}{a^2 d}-\frac {i \tan ^2(c+d x)}{a^2 d}-\frac {i \tan ^4(c+d x)}{2 a^2 d}-\frac {\tan ^5(c+d x)}{5 a^2 d} \]

[Out]

tan(d*x+c)/a^2/d-I*tan(d*x+c)^2/a^2/d-1/2*I*tan(d*x+c)^4/a^2/d-1/5*tan(d*x+c)^5/a^2/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3167, 862, 76} \[ \int \frac {\sec ^6(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {\tan ^5(c+d x)}{5 a^2 d}-\frac {i \tan ^4(c+d x)}{2 a^2 d}-\frac {i \tan ^2(c+d x)}{a^2 d}+\frac {\tan (c+d x)}{a^2 d} \]

[In]

Int[Sec[c + d*x]^6/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

Tan[c + d*x]/(a^2*d) - (I*Tan[c + d*x]^2)/(a^2*d) - ((I/2)*Tan[c + d*x]^4)/(a^2*d) - Tan[c + d*x]^5/(5*a^2*d)

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

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 {\left (1+x^2\right )^3}{x^6 (i a+a x)^2} \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \frac {\left (-\frac {i}{a}+\frac {x}{a}\right )^3 (i a+a x)}{x^6} \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \left (-\frac {1}{a^2 x^6}-\frac {2 i}{a^2 x^5}-\frac {2 i}{a^2 x^3}+\frac {1}{a^2 x^2}\right ) \, dx,x,\cot (c+d x)\right )}{d} \\ & = \frac {\tan (c+d x)}{a^2 d}-\frac {i \tan ^2(c+d x)}{a^2 d}-\frac {i \tan ^4(c+d x)}{2 a^2 d}-\frac {\tan ^5(c+d x)}{5 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.71 \[ \int \frac {\sec ^6(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {\tan (c+d x) \left (-10+10 i \tan (c+d x)+5 i \tan ^3(c+d x)+2 \tan ^4(c+d x)\right )}{10 a^2 d} \]

[In]

Integrate[Sec[c + d*x]^6/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

-1/10*(Tan[c + d*x]*(-10 + (10*I)*Tan[c + d*x] + (5*I)*Tan[c + d*x]^3 + 2*Tan[c + d*x]^4))/(a^2*d)

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.51

method	result	size
risch	\(\frac {8 i \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{5 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\)	\(36\)
derivativedivides	\(\frac {\tan \left (d x +c \right )-\frac {\tan \left (d x +c \right )^{5}}{5}-\frac {i \tan \left (d x +c \right )^{4}}{2}-i \tan \left (d x +c \right )^{2}}{d \,a^{2}}\)	\(47\)
default	\(\frac {\tan \left (d x +c \right )-\frac {\tan \left (d x +c \right )^{5}}{5}-\frac {i \tan \left (d x +c \right )^{4}}{2}-i \tan \left (d x +c \right )^{2}}{d \,a^{2}}\)	\(47\)
norman	\(\frac {\frac {4 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a d}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}-\frac {28 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 a d}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a d}-\frac {4 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a d}+\frac {4 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a d}-\frac {4 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(205\)

[In]

int(sec(d*x+c)^6/(cos(d*x+c)*a+I*a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

8/5*I*(5*exp(2*I*(d*x+c))+1)/d/a^2/(exp(2*I*(d*x+c))+1)^5

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^6(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {8 \, {\left (-5 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{5 \, {\left (a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \]

[In]

integrate(sec(d*x+c)^6/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-8/5*(-5*I*e^(2*I*d*x + 2*I*c) - I)/(a^2*d*e^(10*I*d*x + 10*I*c) + 5*a^2*d*e^(8*I*d*x + 8*I*c) + 10*a^2*d*e^(6
*I*d*x + 6*I*c) + 10*a^2*d*e^(4*I*d*x + 4*I*c) + 5*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)

Sympy [F]

\[ \int \frac {\sec ^6(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sec ^{6}{\left (c + d x \right )}}{- \sin ^{2}{\left (c + d x \right )} + 2 i \sin {\left (c + d x \right )} \cos {\left (c + d x \right )} + \cos ^{2}{\left (c + d x \right )}}\, dx}{a^{2}} \]

[In]

integrate(sec(d*x+c)**6/(a*cos(d*x+c)+I*a*sin(d*x+c))**2,x)

[Out]

Integral(sec(c + d*x)**6/(-sin(c + d*x)**2 + 2*I*sin(c + d*x)*cos(c + d*x) + cos(c + d*x)**2), x)/a**2

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.67 \[ \int \frac {\sec ^6(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {2 \, \tan \left (d x + c\right )^{5} + 5 i \, \tan \left (d x + c\right )^{4} + 10 i \, \tan \left (d x + c\right )^{2} - 10 \, \tan \left (d x + c\right )}{10 \, a^{2} d} \]

[In]

integrate(sec(d*x+c)^6/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/10*(2*tan(d*x + c)^5 + 5*I*tan(d*x + c)^4 + 10*I*tan(d*x + c)^2 - 10*tan(d*x + c))/(a^2*d)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.67 \[ \int \frac {\sec ^6(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {2 \, \tan \left (d x + c\right )^{5} + 5 i \, \tan \left (d x + c\right )^{4} + 10 i \, \tan \left (d x + c\right )^{2} - 10 \, \tan \left (d x + c\right )}{10 \, a^{2} d} \]

[In]

integrate(sec(d*x+c)^6/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/10*(2*tan(d*x + c)^5 + 5*I*tan(d*x + c)^4 + 10*I*tan(d*x + c)^2 - 10*tan(d*x + c))/(a^2*d)

Mupad [B] (verification not implemented)

Time = 23.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^6(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {\sin \left (c+d\,x\right )\,\left (-4\,{\cos \left (c+d\,x\right )}^4+\frac {5{}\mathrm {i}\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^3}{2}-2\,{\cos \left (c+d\,x\right )}^2+\frac {5{}\mathrm {i}\,\sin \left (c+d\,x\right )\,\cos \left (c+d\,x\right )}{2}+1\right )}{5\,a^2\,d\,{\cos \left (c+d\,x\right )}^5} \]

[In]

int(1/(cos(c + d*x)^6*(a*cos(c + d*x) + a*sin(c + d*x)*1i)^2),x)

[Out]

-(sin(c + d*x)*((cos(c + d*x)*sin(c + d*x)*5i)/2 + (cos(c + d*x)^3*sin(c + d*x)*5i)/2 - 2*cos(c + d*x)^2 - 4*c
os(c + d*x)^4 + 1))/(5*a^2*d*cos(c + d*x)^5)