3.170 \(\int \frac{\sec ^6(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

Optimal. Leaf size=81 \[ \frac{i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac{i}{d \left (a^2+i a^2 \tan (c+d x)\right )^4}+\frac{4 i}{5 a^3 d (a+i a \tan (c+d x))^5} \]

[Out]

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

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Rubi [A]  time = 0.0556106, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3487, 43} \[ \frac{i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac{i}{d \left (a^2+i a^2 \tan (c+d x)\right )^4}+\frac{4 i}{5 a^3 d (a+i a \tan (c+d x))^5} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^8,x]

[Out]

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

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^6(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=-\frac{i \operatorname{Subst}\left (\int \frac{(a-x)^2}{(a+x)^6} \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (\frac{4 a^2}{(a+x)^6}-\frac{4 a}{(a+x)^5}+\frac{1}{(a+x)^4}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=\frac{4 i}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac{i}{d \left (a^2+i a^2 \tan (c+d x)\right )^4}\\ \end{align*}

Mathematica [A]  time = 0.153923, size = 56, normalized size = 0.69 \[ \frac{i \sec ^8(c+d x) (4 i \sin (2 (c+d x))+16 \cos (2 (c+d x))+15)}{240 a^8 d (\tan (c+d x)-i)^8} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^8,x]

[Out]

((I/240)*Sec[c + d*x]^8*(15 + 16*Cos[2*(c + d*x)] + (4*I)*Sin[2*(c + d*x)]))/(a^8*d*(-I + Tan[c + d*x])^8)

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Maple [A]  time = 0.102, size = 49, normalized size = 0.6 \begin{align*}{\frac{1}{d{a}^{8}} \left ({\frac{-i}{ \left ( \tan \left ( dx+c \right ) -i \right ) ^{4}}}+{\frac{4}{5\, \left ( \tan \left ( dx+c \right ) -i \right ) ^{5}}}-{\frac{1}{3\, \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}} \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))^8,x)

[Out]

1/d/a^8*(-I/(tan(d*x+c)-I)^4+4/5/(tan(d*x+c)-I)^5-1/3/(tan(d*x+c)-I)^3)

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Maxima [B]  time = 1.14954, size = 192, normalized size = 2.37 \begin{align*} -\frac{35 \, \tan \left (d x + c\right )^{4} - 35 i \, \tan \left (d x + c\right )^{3} + 21 \, \tan \left (d x + c\right )^{2} - 7 i \, \tan \left (d x + c\right ) + 14}{{\left (105 \, a^{8} \tan \left (d x + c\right )^{7} - 735 i \, a^{8} \tan \left (d x + c\right )^{6} - 2205 \, a^{8} \tan \left (d x + c\right )^{5} + 3675 i \, a^{8} \tan \left (d x + c\right )^{4} + 3675 \, a^{8} \tan \left (d x + c\right )^{3} - 2205 i \, a^{8} \tan \left (d x + c\right )^{2} - 735 \, a^{8} \tan \left (d x + c\right ) + 105 i \, a^{8}\right )} 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))^8,x, algorithm="maxima")

[Out]

-(35*tan(d*x + c)^4 - 35*I*tan(d*x + c)^3 + 21*tan(d*x + c)^2 - 7*I*tan(d*x + c) + 14)/((105*a^8*tan(d*x + c)^
7 - 735*I*a^8*tan(d*x + c)^6 - 2205*a^8*tan(d*x + c)^5 + 3675*I*a^8*tan(d*x + c)^4 + 3675*a^8*tan(d*x + c)^3 -
2205*I*a^8*tan(d*x + c)^2 - 735*a^8*tan(d*x + c) + 105*I*a^8)*d)

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Fricas [A]  time = 2.46611, size = 132, normalized size = 1.63 \begin{align*} \frac{{\left (10 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 15 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i\right )} e^{\left (-10 i \, d x - 10 i \, c\right )}}{240 \, a^{8} 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))^8,x, algorithm="fricas")

[Out]

1/240*(10*I*e^(4*I*d*x + 4*I*c) + 15*I*e^(2*I*d*x + 2*I*c) + 6*I)*e^(-10*I*d*x - 10*I*c)/(a^8*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))**8,x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.18484, size = 185, normalized size = 2.28 \begin{align*} -\frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 30 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 140 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 170 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 282 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 170 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 140 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{15 \, a^{8} d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{10}} \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))^8,x, algorithm="giac")

[Out]

-2/15*(15*tan(1/2*d*x + 1/2*c)^9 - 30*I*tan(1/2*d*x + 1/2*c)^8 - 140*tan(1/2*d*x + 1/2*c)^7 + 170*I*tan(1/2*d*
x + 1/2*c)^6 + 282*tan(1/2*d*x + 1/2*c)^5 - 170*I*tan(1/2*d*x + 1/2*c)^4 - 140*tan(1/2*d*x + 1/2*c)^3 + 30*I*t
an(1/2*d*x + 1/2*c)^2 + 15*tan(1/2*d*x + 1/2*c))/(a^8*d*(tan(1/2*d*x + 1/2*c) - I)^10)