3.168 \(\int \frac{\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

Optimal. Leaf size=116 \[ \frac{\tan (c+d x)}{a^8 d}+\frac{24 i}{d \left (a^8+i a^8 \tan (c+d x)\right )}-\frac{16 i}{d \left (a^4+i a^4 \tan (c+d x)\right )^2}+\frac{16 i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac{8 i \log (\cos (c+d x))}{a^8 d}-\frac{8 x}{a^8} \]

[Out]

(-8*x)/a^8 - ((8*I)*Log[Cos[c + d*x]])/(a^8*d) + Tan[c + d*x]/(a^8*d) + ((16*I)/3)/(a^5*d*(a + I*a*Tan[c + d*x
])^3) - (16*I)/(d*(a^4 + I*a^4*Tan[c + d*x])^2) + (24*I)/(d*(a^8 + I*a^8*Tan[c + d*x]))

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Rubi [A]  time = 0.0682109, antiderivative size = 116, 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} \[ \frac{\tan (c+d x)}{a^8 d}+\frac{24 i}{d \left (a^8+i a^8 \tan (c+d x)\right )}-\frac{16 i}{d \left (a^4+i a^4 \tan (c+d x)\right )^2}+\frac{16 i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac{8 i \log (\cos (c+d x))}{a^8 d}-\frac{8 x}{a^8} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(-8*x)/a^8 - ((8*I)*Log[Cos[c + d*x]])/(a^8*d) + Tan[c + d*x]/(a^8*d) + ((16*I)/3)/(a^5*d*(a + I*a*Tan[c + d*x
])^3) - (16*I)/(d*(a^4 + I*a^4*Tan[c + d*x])^2) + (24*I)/(d*(a^8 + I*a^8*Tan[c + d*x]))

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 ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=-\frac{i \operatorname{Subst}\left (\int \frac{(a-x)^4}{(a+x)^4} \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (1+\frac{16 a^4}{(a+x)^4}-\frac{32 a^3}{(a+x)^3}+\frac{24 a^2}{(a+x)^2}-\frac{8 a}{a+x}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=-\frac{8 x}{a^8}-\frac{8 i \log (\cos (c+d x))}{a^8 d}+\frac{\tan (c+d x)}{a^8 d}+\frac{16 i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac{16 i}{d \left (a^4+i a^4 \tan (c+d x)\right )^2}+\frac{24 i}{d \left (a^8+i a^8 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.885939, size = 397, normalized size = 3.42 \[ \frac{\sec (c) \sec ^9(c+d x) (-\cos (5 (c+d x))-i \sin (5 (c+d x))) (12 i d x \sin (c+2 d x)+11 \sin (c+2 d x)+12 i d x \sin (3 c+2 d x)+14 \sin (3 c+2 d x)+12 i d x \sin (3 c+4 d x)-4 \sin (3 c+4 d x)+12 i d x \sin (5 c+4 d x)-\sin (5 c+4 d x)+12 d x \cos (3 c+2 d x)-10 i \cos (3 c+2 d x)+12 d x \cos (3 c+4 d x)+2 i \cos (3 c+4 d x)+12 d x \cos (5 c+4 d x)-i \cos (5 c+4 d x)+\cos (c+2 d x) (12 i \log (\cos (c+d x))+12 d x-7 i)+12 i \cos (3 c+2 d x) \log (\cos (c+d x))+12 i \cos (3 c+4 d x) \log (\cos (c+d x))+12 i \cos (5 c+4 d x) \log (\cos (c+d x))-12 \sin (c+2 d x) \log (\cos (c+d x))-12 \sin (3 c+2 d x) \log (\cos (c+d x))-12 \sin (3 c+4 d x) \log (\cos (c+d x))-12 \sin (5 c+4 d x) \log (\cos (c+d x))-12 i \cos (c))}{6 a^8 d (\tan (c+d x)-i)^8} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(Sec[c]*Sec[c + d*x]^9*(-Cos[5*(c + d*x)] - I*Sin[5*(c + d*x)])*((-12*I)*Cos[c] - (10*I)*Cos[3*c + 2*d*x] + 12
*d*x*Cos[3*c + 2*d*x] + (2*I)*Cos[3*c + 4*d*x] + 12*d*x*Cos[3*c + 4*d*x] - I*Cos[5*c + 4*d*x] + 12*d*x*Cos[5*c
+ 4*d*x] + Cos[c + 2*d*x]*(-7*I + 12*d*x + (12*I)*Log[Cos[c + d*x]]) + (12*I)*Cos[3*c + 2*d*x]*Log[Cos[c + d*
x]] + (12*I)*Cos[3*c + 4*d*x]*Log[Cos[c + d*x]] + (12*I)*Cos[5*c + 4*d*x]*Log[Cos[c + d*x]] + 11*Sin[c + 2*d*x
] + (12*I)*d*x*Sin[c + 2*d*x] - 12*Log[Cos[c + d*x]]*Sin[c + 2*d*x] + 14*Sin[3*c + 2*d*x] + (12*I)*d*x*Sin[3*c
+ 2*d*x] - 12*Log[Cos[c + d*x]]*Sin[3*c + 2*d*x] - 4*Sin[3*c + 4*d*x] + (12*I)*d*x*Sin[3*c + 4*d*x] - 12*Log[
Cos[c + d*x]]*Sin[3*c + 4*d*x] - Sin[5*c + 4*d*x] + (12*I)*d*x*Sin[5*c + 4*d*x] - 12*Log[Cos[c + d*x]]*Sin[5*c
+ 4*d*x]))/(6*a^8*d*(-I + Tan[c + d*x])^8)

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Maple [A]  time = 0.094, size = 92, normalized size = 0.8 \begin{align*}{\frac{\tan \left ( dx+c \right ) }{d{a}^{8}}}-{\frac{16}{3\,d{a}^{8} \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}+{\frac{8\,i\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{d{a}^{8}}}+24\,{\frac{1}{d{a}^{8} \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{16\,i}{d{a}^{8} \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^8,x)

[Out]

tan(d*x+c)/a^8/d-16/3/d/a^8/(tan(d*x+c)-I)^3+8*I/d/a^8*ln(tan(d*x+c)-I)+24/d/a^8/(tan(d*x+c)-I)+16*I/d/a^8/(ta
n(d*x+c)-I)^2

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Maxima [A]  time = 1.25337, size = 255, normalized size = 2.2 \begin{align*} \frac{\frac{2520 \, \tan \left (d x + c\right )^{6} - 13440 i \, \tan \left (d x + c\right )^{5} - 29960 \, \tan \left (d x + c\right )^{4} + 35840 i \, \tan \left (d x + c\right )^{3} + 24360 \, \tan \left (d x + c\right )^{2} - 8960 i \, \tan \left (d x + c\right ) - 1400}{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}} + \frac{8 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{8}} + \frac{\tan \left (d x + c\right )}{a^{8}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

((2520*tan(d*x + c)^6 - 13440*I*tan(d*x + c)^5 - 29960*tan(d*x + c)^4 + 35840*I*tan(d*x + c)^3 + 24360*tan(d*x
+ c)^2 - 8960*I*tan(d*x + c) - 1400)/(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) + 8*I*log(I*tan(d*x + c) + 1)/a^8 + tan(d*x + c)/a^8)/d

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Fricas [A]  time = 2.85278, size = 370, normalized size = 3.19 \begin{align*} -\frac{48 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (48 \, d x - 24 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} -{\left (-24 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 24 i \, e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 12 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i}{3 \,{\left (a^{8} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{8} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

-1/3*(48*d*x*e^(8*I*d*x + 8*I*c) + (48*d*x - 24*I)*e^(6*I*d*x + 6*I*c) - (-24*I*e^(8*I*d*x + 8*I*c) - 24*I*e^(
6*I*d*x + 6*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) - 12*I*e^(4*I*d*x + 4*I*c) + 4*I*e^(2*I*d*x + 2*I*c) - 2*I)/(a^
8*d*e^(8*I*d*x + 8*I*c) + a^8*d*e^(6*I*d*x + 6*I*c))

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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)**10/(a+I*a*tan(d*x+c))**8,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.21933, size = 270, normalized size = 2.33 \begin{align*} -\frac{2 \,{\left (-\frac{120 i \, \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}{a^{8}} + \frac{60 i \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{8}} + \frac{60 i \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{8}} - \frac{15 \,{\left (4 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 i\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a^{8}} + \frac{294 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 1884 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 4890 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 6920 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4890 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1884 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 294 i}{a^{8}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{6}}\right )}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

-2/15*(-120*I*log(tan(1/2*d*x + 1/2*c) - I)/a^8 + 60*I*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^8 + 60*I*log(abs(t
an(1/2*d*x + 1/2*c) - 1))/a^8 - 15*(4*I*tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c) - 4*I)/((tan(1/2*d*x + 1
/2*c)^2 - 1)*a^8) + (294*I*tan(1/2*d*x + 1/2*c)^6 + 1884*tan(1/2*d*x + 1/2*c)^5 - 4890*I*tan(1/2*d*x + 1/2*c)^
4 - 6920*tan(1/2*d*x + 1/2*c)^3 + 4890*I*tan(1/2*d*x + 1/2*c)^2 + 1884*tan(1/2*d*x + 1/2*c) - 294*I)/(a^8*(tan
(1/2*d*x + 1/2*c) - I)^6))/d