3.116 \(\int \frac{\sec ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=27 \[ \frac{i (a-i a \tan (c+d x))^3}{3 a^5 d} \]

[Out]

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

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Rubi [A]  time = 0.0444558, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3487, 32} \[ \frac{i (a-i a \tan (c+d x))^3}{3 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [B]  time = 0.209109, size = 68, normalized size = 2.52 \[ \frac{\sec (c) \sec ^3(c+d x) (-3 \sin (2 c+d x)+2 \sin (2 c+3 d x)-3 i \cos (2 c+d x)+3 \sin (d x)-3 i \cos (d x))}{6 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c]*Sec[c + d*x]^3*((-3*I)*Cos[d*x] - (3*I)*Cos[2*c + d*x] + 3*Sin[d*x] - 3*Sin[2*c + d*x] + 2*Sin[2*c + 3
*d*x]))/(6*a^2*d)

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Maple [A]  time = 0.068, size = 36, normalized size = 1.3 \begin{align*}{\frac{1}{{a}^{2}d} \left ( \tan \left ( dx+c \right ) -{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3}}-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+I*a*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.09483, size = 47, normalized size = 1.74 \begin{align*} -\frac{\tan \left (d x + c\right )^{3} + 3 i \, \tan \left (d x + c\right )^{2} - 3 \, \tan \left (d x + c\right )}{3 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.9394, size = 139, normalized size = 5.15 \begin{align*} \frac{8 i}{3 \,{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

8/3*I/(a^2*d*e^(6*I*d*x + 6*I*c) + 3*a^2*d*e^(4*I*d*x + 4*I*c) + 3*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+I*a*tan(d*x+c))**2,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.18927, size = 47, normalized size = 1.74 \begin{align*} -\frac{\tan \left (d x + c\right )^{3} + 3 i \, \tan \left (d x + c\right )^{2} - 3 \, \tan \left (d x + c\right )}{3 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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