\(\int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\) [168]

Optimal result

Integrand size = 24, antiderivative size = 116 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {8 x}{a^8}-\frac {8 i \log (\cos (c+d x))}{a^8 d}+\frac {\tan (c+d x)}{a^8 d}+\frac {16 i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac {16 i}{d \left (a^4+i a^4 \tan (c+d x)\right )^2}+\frac {24 i}{d \left (a^8+i a^8 \tan (c+d x)\right )} \]

[Out]

-8*x/a^8-8*I*ln(cos(d*x+c))/a^8/d+tan(d*x+c)/a^8/d+16/3*I/a^5/d/(a+I*a*tan(d*x+c))^3-16*I/d/(a^4+I*a^4*tan(d*x
+c))^2+24*I/d/(a^8+I*a^8*tan(d*x+c))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\tan (c+d x)}{a^8 d}+\frac {24 i}{d \left (a^8+i a^8 \tan (c+d x)\right )}-\frac {8 i \log (\cos (c+d x))}{a^8 d}-\frac {8 x}{a^8}+\frac {16 i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac {16 i}{d \left (a^4+i a^4 \tan (c+d x)\right )^2} \]

[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 45

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])

Rule 3568

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int \frac {(a-x)^4}{(a+x)^4} \, dx,x,i a \tan (c+d x)\right )}{a^9 d} \\ & = -\frac {i \text {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 [A] (verified)

Time = 0.80 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.68 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {i \left (-8 a \log (i-\tan (c+d x))+i a \tan (c+d x)+\frac {8 a \left (-5 i+12 \tan (c+d x)+9 i \tan ^2(c+d x)\right )}{3 (-i+\tan (c+d x))^3}\right )}{a^9 d} \]

[In]

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

[Out]

((-I)*(-8*a*Log[I - Tan[c + d*x]] + I*a*Tan[c + d*x] + (8*a*(-5*I + 12*Tan[c + d*x] + (9*I)*Tan[c + d*x]^2))/(
3*(-I + Tan[c + d*x])^3)))/(a^9*d)

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {\tan \left (d x +c \right )}{a^{8} d}+\frac {16 i}{a^{8} d \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {16}{3 a^{8} d \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {8 \arctan \left (\tan \left (d x +c \right )\right )}{a^{8} d}+\frac {4 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{a^{8} d}+\frac {24}{a^{8} d \left (\tan \left (d x +c \right )-i\right )}\)	\(108\)
default	\(\frac {\tan \left (d x +c \right )}{a^{8} d}+\frac {16 i}{a^{8} d \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {16}{3 a^{8} d \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {8 \arctan \left (\tan \left (d x +c \right )\right )}{a^{8} d}+\frac {4 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{a^{8} d}+\frac {24}{a^{8} d \left (\tan \left (d x +c \right )-i\right )}\)	\(108\)
risch	\(\frac {6 i {\mathrm e}^{-2 i \left (d x +c \right )}}{a^{8} d}-\frac {2 i {\mathrm e}^{-4 i \left (d x +c \right )}}{a^{8} d}+\frac {2 i {\mathrm e}^{-6 i \left (d x +c \right )}}{3 a^{8} d}-\frac {16 x}{a^{8}}-\frac {16 c}{a^{8} d}+\frac {2 i}{d \,a^{8} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {8 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{8} d}\)	\(114\)

[In]

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

[Out]

tan(d*x+c)/a^8/d+16*I/a^8/d/(tan(d*x+c)-I)^2-16/3/a^8/d/(tan(d*x+c)-I)^3-8/a^8/d*arctan(tan(d*x+c))+4*I/a^8/d*
ln(1+tan(d*x+c)^2)+24/a^8/d/(tan(d*x+c)-I)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {2 \, {\left (24 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} + 12 \, {\left (2 \, d x - i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 12 \, {\left (i \, e^{\left (8 i \, d x + 8 i \, c\right )} + i \, e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 6 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{3 \, {\left (a^{8} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{8} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\int \frac {\sec ^{10}{\left (c + d x \right )}}{\tan ^{8}{\left (c + d x \right )} - 8 i \tan ^{7}{\left (c + d x \right )} - 28 \tan ^{6}{\left (c + d x \right )} + 56 i \tan ^{5}{\left (c + d x \right )} + 70 \tan ^{4}{\left (c + d x \right )} - 56 i \tan ^{3}{\left (c + d x \right )} - 28 \tan ^{2}{\left (c + d x \right )} + 8 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{8}} \]

[In]

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

[Out]

Integral(sec(c + d*x)**10/(tan(c + d*x)**8 - 8*I*tan(c + d*x)**7 - 28*tan(c + d*x)**6 + 56*I*tan(c + d*x)**5 +
70*tan(c + d*x)**4 - 56*I*tan(c + d*x)**3 - 28*tan(c + d*x)**2 + 8*I*tan(c + d*x) + 1), x)/a**8

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.65 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\frac {8 \, {\left (9 \, \tan \left (d x + c\right )^{6} - 48 i \, \tan \left (d x + c\right )^{5} - 107 \, \tan \left (d x + c\right )^{4} + 128 i \, \tan \left (d x + c\right )^{3} + 87 \, \tan \left (d x + c\right )^{2} - 32 i \, \tan \left (d x + c\right ) - 5\right )}}{a^{8} \tan \left (d x + c\right )^{7} - 7 i \, a^{8} \tan \left (d x + c\right )^{6} - 21 \, a^{8} \tan \left (d x + c\right )^{5} + 35 i \, a^{8} \tan \left (d x + c\right )^{4} + 35 \, a^{8} \tan \left (d x + c\right )^{3} - 21 i \, a^{8} \tan \left (d x + c\right )^{2} - 7 \, a^{8} \tan \left (d x + c\right ) + i \, a^{8}} + \frac {24 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{8}} + \frac {3 \, \tan \left (d x + c\right )}{a^{8}}}{3 \, d} \]

[In]

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

[Out]

1/3*(8*(9*tan(d*x + c)^6 - 48*I*tan(d*x + c)^5 - 107*tan(d*x + c)^4 + 128*I*tan(d*x + c)^3 + 87*tan(d*x + c)^2
- 32*I*tan(d*x + c) - 5)/(a^8*tan(d*x + c)^7 - 7*I*a^8*tan(d*x + c)^6 - 21*a^8*tan(d*x + c)^5 + 35*I*a^8*tan(
d*x + c)^4 + 35*a^8*tan(d*x + c)^3 - 21*I*a^8*tan(d*x + c)^2 - 7*a^8*tan(d*x + c) + I*a^8) + 24*I*log(I*tan(d*
x + c) + 1)/a^8 + 3*tan(d*x + c)/a^8)/d

Giac [A] (verification not implemented)

none

Time = 1.79 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.72 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {2 \, {\left (\frac {60 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{8}} - \frac {120 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a^{8}} + \frac {60 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{8}} - \frac {15 \, {\left (4 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{8}} + \frac {2 \, {\left (147 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 942 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2445 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3460 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2445 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 942 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 147 i\right )}}{a^{8} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{6}}\right )}}{15 \, d} \]

[In]

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

[Out]

-2/15*(60*I*log(tan(1/2*d*x + 1/2*c) + 1)/a^8 - 120*I*log(tan(1/2*d*x + 1/2*c) - I)/a^8 + 60*I*log(tan(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) + 2*(147*I*tan(1/2*d*x + 1/2*c)^6 + 942*tan(1/2*d*x + 1/2*c)^5 - 2445*I*tan(1/2*d*x + 1/2*c)^4 - 3460*t
an(1/2*d*x + 1/2*c)^3 + 2445*I*tan(1/2*d*x + 1/2*c)^2 + 942*tan(1/2*d*x + 1/2*c) - 147*I)/(a^8*(tan(1/2*d*x +
1/2*c) - I)^6))/d

Mupad [B] (verification not implemented)

Time = 4.53 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )}{a^8\,d}-\frac {\frac {32\,\mathrm {tan}\left (c+d\,x\right )}{a^8}-\frac {40{}\mathrm {i}}{3\,a^8}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,24{}\mathrm {i}}{a^8}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,8{}\mathrm {i}}{a^8\,d} \]

[In]

int(1/(cos(c + d*x)^10*(a + a*tan(c + d*x)*1i)^8),x)

[Out]

(log(tan(c + d*x) - 1i)*8i)/(a^8*d) - ((32*tan(c + d*x))/a^8 - 40i/(3*a^8) + (tan(c + d*x)^2*24i)/a^8)/(d*(tan
(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) + tan(c + d*x)/(a^8*d)