3.171 \(\int \frac{\sec ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

Optimal. Leaf size=55 \[ \frac{i}{3 a^2 d (a+i a \tan (c+d x))^6}-\frac{i}{5 a^3 d (a+i a \tan (c+d x))^5} \]

[Out]

(I/3)/(a^2*d*(a + I*a*Tan[c + d*x])^6) - (I/5)/(a^3*d*(a + I*a*Tan[c + d*x])^5)

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Rubi [A]  time = 0.0479161, antiderivative size = 55, 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} \[ \frac{i}{3 a^2 d (a+i a \tan (c+d x))^6}-\frac{i}{5 a^3 d (a+i a \tan (c+d x))^5} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(I/3)/(a^2*d*(a + I*a*Tan[c + d*x])^6) - (I/5)/(a^3*d*(a + I*a*Tan[c + d*x])^5)

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 ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=-\frac{i \operatorname{Subst}\left (\int \frac{a-x}{(a+x)^7} \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (\frac{2 a}{(a+x)^7}-\frac{1}{(a+x)^6}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=\frac{i}{3 a^2 d (a+i a \tan (c+d x))^6}-\frac{i}{5 a^3 d (a+i a \tan (c+d x))^5}\\ \end{align*}

Mathematica [A]  time = 0.20986, size = 78, normalized size = 1.42 \[ \frac{i \sec ^8(c+d x) (16 i \sin (2 (c+d x))+10 i \sin (4 (c+d x))+64 \cos (2 (c+d x))+20 \cos (4 (c+d x))+45)}{960 a^8 d (\tan (c+d x)-i)^8} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^8,x]

[Out]

((I/960)*Sec[c + d*x]^8*(45 + 64*Cos[2*(c + d*x)] + 20*Cos[4*(c + d*x)] + (16*I)*Sin[2*(c + d*x)] + (10*I)*Sin
[4*(c + d*x)]))/(a^8*d*(-I + Tan[c + d*x])^8)

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Maple [A]  time = 0.12, size = 36, normalized size = 0.7 \begin{align*}{\frac{1}{d{a}^{8}} \left ( -{\frac{1}{5\, \left ( \tan \left ( dx+c \right ) -i \right ) ^{5}}}-{\frac{{\frac{i}{3}}}{ \left ( \tan \left ( dx+c \right ) -i \right ) ^{6}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^8,x)

[Out]

1/d/a^8*(-1/5/(tan(d*x+c)-I)^5-1/3*I/(tan(d*x+c)-I)^6)

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Maxima [B]  time = 1.1587, size = 165, normalized size = 3. \begin{align*} -\frac{7 \,{\left (3 \, \tan \left (d x + c\right )^{2} - i \, \tan \left (d x + c\right ) + 2\right )}}{{\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)^4/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

-7*(3*tan(d*x + c)^2 - I*tan(d*x + c) + 2)/((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 = 2.32729, size = 205, normalized size = 3.73 \begin{align*} \frac{{\left (15 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 40 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 45 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 24 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-12 i \, d x - 12 i \, c\right )}}{960 \, a^{8} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

1/960*(15*I*e^(8*I*d*x + 8*I*c) + 40*I*e^(6*I*d*x + 6*I*c) + 45*I*e^(4*I*d*x + 4*I*c) + 24*I*e^(2*I*d*x + 2*I*
c) + 5*I)*e^(-12*I*d*x - 12*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)**4/(a+I*a*tan(d*x+c))**8,x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.17525, size = 220, normalized size = 4. \begin{align*} -\frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 60 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} - 235 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 480 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 822 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 904 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 822 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 480 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 235 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 60 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 )}^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

-2/15*(15*tan(1/2*d*x + 1/2*c)^11 - 60*I*tan(1/2*d*x + 1/2*c)^10 - 235*tan(1/2*d*x + 1/2*c)^9 + 480*I*tan(1/2*
d*x + 1/2*c)^8 + 822*tan(1/2*d*x + 1/2*c)^7 - 904*I*tan(1/2*d*x + 1/2*c)^6 - 822*tan(1/2*d*x + 1/2*c)^5 + 480*
I*tan(1/2*d*x + 1/2*c)^4 + 235*tan(1/2*d*x + 1/2*c)^3 - 60*I*tan(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)^12)