3.177 \(\int \frac{\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

Optimal. Leaf size=183 \[ -\frac{63 \tanh ^{-1}(\sin (c+d x))}{2 a^8 d}-\frac{6 i \sec ^7(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{42 i \sec ^5(c+d x)}{5 a^2 d \left (a^2+i a^2 \tan (c+d x)\right )^3}+\frac{42 i \sec ^3(c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}-\frac{63 \tan (c+d x) \sec (c+d x)}{2 a^8 d}+\frac{2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7} \]

[Out]

(-63*ArcTanh[Sin[c + d*x]])/(2*a^8*d) - (63*Sec[c + d*x]*Tan[c + d*x])/(2*a^8*d) + (((2*I)/5)*Sec[c + d*x]^9)/
(a*d*(a + I*a*Tan[c + d*x])^7) - (((6*I)/5)*Sec[c + d*x]^7)/(a^3*d*(a + I*a*Tan[c + d*x])^5) + (((42*I)/5)*Sec
[c + d*x]^5)/(a^2*d*(a^2 + I*a^2*Tan[c + d*x])^3) + ((42*I)*Sec[c + d*x]^3)/(d*(a^8 + I*a^8*Tan[c + d*x]))

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Rubi [A]  time = 0.200439, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3500, 3768, 3770} \[ -\frac{63 \tanh ^{-1}(\sin (c+d x))}{2 a^8 d}-\frac{6 i \sec ^7(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{42 i \sec ^5(c+d x)}{5 a^2 d \left (a^2+i a^2 \tan (c+d x)\right )^3}+\frac{42 i \sec ^3(c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}-\frac{63 \tan (c+d x) \sec (c+d x)}{2 a^8 d}+\frac{2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(-63*ArcTanh[Sin[c + d*x]])/(2*a^8*d) - (63*Sec[c + d*x]*Tan[c + d*x])/(2*a^8*d) + (((2*I)/5)*Sec[c + d*x]^9)/
(a*d*(a + I*a*Tan[c + d*x])^7) - (((6*I)/5)*Sec[c + d*x]^7)/(a^3*d*(a + I*a*Tan[c + d*x])^5) + (((42*I)/5)*Sec
[c + d*x]^5)/(a^2*d*(a^2 + I*a^2*Tan[c + d*x])^3) + ((42*I)*Sec[c + d*x]^3)/(d*(a^8 + I*a^8*Tan[c + d*x]))

Rule 3500

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*d^2
*(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + 2*n)), x] - Dist[(d^2*(m - 2))/(b^2*(m + 2*n
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=\frac{2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac{9 \int \frac{\sec ^9(c+d x)}{(a+i a \tan (c+d x))^6} \, dx}{5 a^2}\\ &=\frac{2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac{6 i \sec ^7(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{21 \int \frac{\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx}{5 a^4}\\ &=\frac{2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac{6 i \sec ^7(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{42 i \sec ^5(c+d x)}{5 a^5 d (a+i a \tan (c+d x))^3}-\frac{21 \int \frac{\sec ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{a^6}\\ &=\frac{2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac{6 i \sec ^7(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{42 i \sec ^5(c+d x)}{5 a^5 d (a+i a \tan (c+d x))^3}+\frac{42 i \sec ^3(c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}-\frac{63 \int \sec ^3(c+d x) \, dx}{a^8}\\ &=-\frac{63 \sec (c+d x) \tan (c+d x)}{2 a^8 d}+\frac{2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac{6 i \sec ^7(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{42 i \sec ^5(c+d x)}{5 a^5 d (a+i a \tan (c+d x))^3}+\frac{42 i \sec ^3(c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}-\frac{63 \int \sec (c+d x) \, dx}{2 a^8}\\ &=-\frac{63 \tanh ^{-1}(\sin (c+d x))}{2 a^8 d}-\frac{63 \sec (c+d x) \tan (c+d x)}{2 a^8 d}+\frac{2 i \sec ^9(c+d x)}{5 a d (a+i a \tan (c+d x))^7}-\frac{6 i \sec ^7(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{42 i \sec ^5(c+d x)}{5 a^5 d (a+i a \tan (c+d x))^3}+\frac{42 i \sec ^3(c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 6.18385, size = 1244, normalized size = 6.8 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(63*Cos[8*c]*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^8*(Cos[d*x] + I*Sin[d*x])^8)/(2*d*(a +
I*a*Tan[c + d*x])^8) - (63*Cos[8*c]*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^8*(Cos[d*x] + I*
Sin[d*x])^8)/(2*d*(a + I*a*Tan[c + d*x])^8) + (Cos[5*d*x]*Sec[c + d*x]^8*(((8*I)/5)*Cos[3*c] - (8*Sin[3*c])/5)
*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a + I*a*Tan[c + d*x])^8) + (Cos[3*d*x]*Sec[c + d*x]^8*((-8*I)*Cos[5*c] + 8*Sin
[5*c])*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a + I*a*Tan[c + d*x])^8) + (Cos[d*x]*Sec[c + d*x]^8*((48*I)*Cos[7*c] - 4
8*Sin[7*c])*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a + I*a*Tan[c + d*x])^8) + (Sec[c]*Sec[c + d*x]^8*((8*I)*Cos[8*c] -
8*Sin[8*c])*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a + I*a*Tan[c + d*x])^8) + (((63*I)/2)*Log[Cos[c/2 + (d*x)/2] - Si
n[c/2 + (d*x)/2]]*Sec[c + d*x]^8*Sin[8*c]*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a + I*a*Tan[c + d*x])^8) - (((63*I)/2
)*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^8*Sin[8*c]*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a + I*a*
Tan[c + d*x])^8) + (Sec[c + d*x]^8*(48*Cos[7*c] + (48*I)*Sin[7*c])*(Cos[d*x] + I*Sin[d*x])^8*Sin[d*x])/(d*(a +
I*a*Tan[c + d*x])^8) + (Sec[c + d*x]^8*(-8*Cos[5*c] - (8*I)*Sin[5*c])*(Cos[d*x] + I*Sin[d*x])^8*Sin[3*d*x])/(
d*(a + I*a*Tan[c + d*x])^8) + (Sec[c + d*x]^8*((8*Cos[3*c])/5 + ((8*I)/5)*Sin[3*c])*(Cos[d*x] + I*Sin[d*x])^8*
Sin[5*d*x])/(d*(a + I*a*Tan[c + d*x])^8) + (Sec[c + d*x]^8*(Cos[8*c]/4 + (I/4)*Sin[8*c])*(Cos[d*x] + I*Sin[d*x
])^8)/(d*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2*(a + I*a*Tan[c + d*x])^8) + (Sec[c + d*x]^8*(-Cos[8*c]/4
- (I/4)*Sin[8*c])*(Cos[d*x] + I*Sin[d*x])^8)/(d*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2*(a + I*a*Tan[c + d
*x])^8) + (8*Sec[c + d*x]^8*(Cos[d*x] + I*Sin[d*x])^8*(Cos[8*c - (d*x)/2]/2 - Cos[8*c + (d*x)/2]/2 + (I/2)*Sin
[8*c - (d*x)/2] - (I/2)*Sin[8*c + (d*x)/2]))/(d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]
)*(a + I*a*Tan[c + d*x])^8) + (8*Sec[c + d*x]^8*(Cos[d*x] + I*Sin[d*x])^8*(-Cos[8*c - (d*x)/2]/2 + Cos[8*c + (
d*x)/2]/2 - (I/2)*Sin[8*c - (d*x)/2] + (I/2)*Sin[8*c + (d*x)/2]))/(d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2]
- Sin[c/2 + (d*x)/2])*(a + I*a*Tan[c + d*x])^8)

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Maple [A]  time = 0.12, size = 282, normalized size = 1.5 \begin{align*}{\frac{1}{2\,d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{8\,i}{d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2\,d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{63}{2\,d{a}^{8}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{32\,i}{d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -i \right ) ^{-2}}-{\frac{128\,i}{d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -i \right ) ^{-4}}+{\frac{256}{5\,d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -i \right ) ^{-5}}-64\,{\frac{1}{d{a}^{8} \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{3}}}+64\,{\frac{1}{d{a}^{8} \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) }}+{\frac{1}{2\,d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{8\,i}{d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{1}{2\,d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{63}{2\,d{a}^{8}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^8,x)

[Out]

1/2/d/a^8/(tan(1/2*d*x+1/2*c)+1)+8*I/d/a^8/(tan(1/2*d*x+1/2*c)+1)-1/2/d/a^8/(tan(1/2*d*x+1/2*c)+1)^2-63/2/d/a^
8*ln(tan(1/2*d*x+1/2*c)+1)-32*I/d/a^8/(tan(1/2*d*x+1/2*c)-I)^2-128*I/d/a^8/(tan(1/2*d*x+1/2*c)-I)^4+256/5/d/a^
8/(tan(1/2*d*x+1/2*c)-I)^5-64/d/a^8/(tan(1/2*d*x+1/2*c)-I)^3+64/d/a^8/(tan(1/2*d*x+1/2*c)-I)+1/2/d/a^8/(tan(1/
2*d*x+1/2*c)-1)-8*I/d/a^8/(tan(1/2*d*x+1/2*c)-1)+1/2/d/a^8/(tan(1/2*d*x+1/2*c)-1)^2+63/2/d/a^8*ln(tan(1/2*d*x+
1/2*c)-1)

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Maxima [B]  time = 1.90993, size = 730, normalized size = 3.99 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

((630*cos(9*d*x + 9*c) + 1260*cos(7*d*x + 7*c) + 630*cos(5*d*x + 5*c) + 630*I*sin(9*d*x + 9*c) + 1260*I*sin(7*
d*x + 7*c) + 630*I*sin(5*d*x + 5*c))*arctan2(cos(d*x + c), sin(d*x + c) + 1) + (630*cos(9*d*x + 9*c) + 1260*co
s(7*d*x + 7*c) + 630*cos(5*d*x + 5*c) + 630*I*sin(9*d*x + 9*c) + 1260*I*sin(7*d*x + 7*c) + 630*I*sin(5*d*x + 5
*c))*arctan2(cos(d*x + c), -sin(d*x + c) + 1) - (-315*I*cos(9*d*x + 9*c) - 630*I*cos(7*d*x + 7*c) - 315*I*cos(
5*d*x + 5*c) + 315*sin(9*d*x + 9*c) + 630*sin(7*d*x + 7*c) + 315*sin(5*d*x + 5*c))*log(cos(d*x + c)^2 + sin(d*
x + c)^2 + 2*sin(d*x + c) + 1) - (315*I*cos(9*d*x + 9*c) + 630*I*cos(7*d*x + 7*c) + 315*I*cos(5*d*x + 5*c) - 3
15*sin(9*d*x + 9*c) - 630*sin(7*d*x + 7*c) - 315*sin(5*d*x + 5*c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin
(d*x + c) + 1) + 1260*cos(8*d*x + 8*c) + 2100*cos(6*d*x + 6*c) + 672*cos(4*d*x + 4*c) - 96*cos(2*d*x + 2*c) +
1260*I*sin(8*d*x + 8*c) + 2100*I*sin(6*d*x + 6*c) + 672*I*sin(4*d*x + 4*c) - 96*I*sin(2*d*x + 2*c) + 32)/((-20
*I*a^8*cos(9*d*x + 9*c) - 40*I*a^8*cos(7*d*x + 7*c) - 20*I*a^8*cos(5*d*x + 5*c) + 20*a^8*sin(9*d*x + 9*c) + 40
*a^8*sin(7*d*x + 7*c) + 20*a^8*sin(5*d*x + 5*c))*d)

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Fricas [A]  time = 3.02538, size = 554, normalized size = 3.03 \begin{align*} -\frac{315 \,{\left (e^{\left (9 i \, d x + 9 i \, c\right )} + 2 \, e^{\left (7 i \, d x + 7 i \, c\right )} + e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 315 \,{\left (e^{\left (9 i \, d x + 9 i \, c\right )} + 2 \, e^{\left (7 i \, d x + 7 i \, c\right )} + e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 630 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 1050 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 336 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 48 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i}{10 \,{\left (a^{8} d e^{\left (9 i \, d x + 9 i \, c\right )} + 2 \, a^{8} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{8} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

-1/10*(315*(e^(9*I*d*x + 9*I*c) + 2*e^(7*I*d*x + 7*I*c) + e^(5*I*d*x + 5*I*c))*log(e^(I*d*x + I*c) + I) - 315*
(e^(9*I*d*x + 9*I*c) + 2*e^(7*I*d*x + 7*I*c) + e^(5*I*d*x + 5*I*c))*log(e^(I*d*x + I*c) - I) - 630*I*e^(8*I*d*
x + 8*I*c) - 1050*I*e^(6*I*d*x + 6*I*c) - 336*I*e^(4*I*d*x + 4*I*c) + 48*I*e^(2*I*d*x + 2*I*c) - 16*I)/(a^8*d*
e^(9*I*d*x + 9*I*c) + 2*a^8*d*e^(7*I*d*x + 7*I*c) + a^8*d*e^(5*I*d*x + 5*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)**11/(a+I*a*tan(d*x+c))**8,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.25661, size = 225, normalized size = 1.23 \begin{align*} -\frac{\frac{315 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{8}} - \frac{315 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{8}} - \frac{10 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 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 ) + 16 i\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} a^{8}} - \frac{4 \,{\left (160 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 720 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 1360 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 880 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 208\right )}}{a^{8}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{5}}}{10 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

-1/10*(315*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^8 - 315*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^8 - 10*(tan(1/2*d
*x + 1/2*c)^3 - 16*I*tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) + 16*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^8
) - 4*(160*tan(1/2*d*x + 1/2*c)^4 - 720*I*tan(1/2*d*x + 1/2*c)^3 - 1360*tan(1/2*d*x + 1/2*c)^2 + 880*I*tan(1/2
*d*x + 1/2*c) + 208)/(a^8*(tan(1/2*d*x + 1/2*c) - I)^5))/d