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

Optimal. Leaf size=70 \[ -\frac{\tan ^5(c+d x)}{5 a^2 d}-\frac{i \tan ^4(c+d x)}{2 a^2 d}-\frac{i \tan ^2(c+d x)}{a^2 d}+\frac{\tan (c+d x)}{a^2 d} \]

[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)

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Rubi [A]  time = 0.0784553, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {3088, 848, 75} \[ -\frac{\tan ^5(c+d x)}{5 a^2 d}-\frac{i \tan ^4(c+d x)}{2 a^2 d}-\frac{i \tan ^2(c+d x)}{a^2 d}+\frac{\tan (c+d x)}{a^2 d} \]

Antiderivative was successfully verified.

[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 3088

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])

Rule 848

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*x)/e)^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 75

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])

Rubi steps

\begin{align*} \int \frac{\sec ^6(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx &=-\frac{\operatorname{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{\operatorname{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{\operatorname{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]  time = 0.407311, size = 77, normalized size = 1.1 \[ \frac{\sec (c) \sec ^5(c+d x) (-5 \sin (2 c+d x)+5 \sin (2 c+3 d x)+\sin (4 c+5 d x)-5 i \cos (2 c+d x)+5 \sin (d x)-5 i \cos (d x))}{20 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c]*Sec[c + d*x]^5*((-5*I)*Cos[d*x] - (5*I)*Cos[2*c + d*x] + 5*Sin[d*x] - 5*Sin[2*c + d*x] + 5*Sin[2*c + 3
*d*x] + Sin[4*c + 5*d*x]))/(20*a^2*d)

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Maple [A]  time = 0.175, size = 47, normalized size = 0.7 \begin{align*}{\frac{1}{d{a}^{2}} \left ( \tan \left ( dx+c \right ) -{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5}}-{\frac{i}{2}} \left ( \tan \left ( dx+c \right ) \right ) ^{4}-i \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.998652, size = 63, normalized size = 0.9 \begin{align*} -\frac{6 \, \tan \left (d x + c\right )^{5} + 15 i \, \tan \left (d x + c\right )^{4} + 30 i \, \tan \left (d x + c\right )^{2} - 30 \, \tan \left (d x + c\right )}{30 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(6*tan(d*x + c)^5 + 15*I*tan(d*x + c)^4 + 30*I*tan(d*x + c)^2 - 30*tan(d*x + c))/(a^2*d)

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Fricas [A]  time = 0.456679, size = 267, normalized size = 3.81 \begin{align*} \frac{40 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i}{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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(40*I*e^(2*I*d*x + 2*I*c) + 8*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)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.13239, size = 63, normalized size = 0.9 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)