\(\int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\) [178]

Optimal result

Integrand size = 24, antiderivative size = 156 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a^8 d}+\frac {2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7}-\frac {2 i \sec ^5(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac {2 i \sec ^3(c+d x)}{3 a^2 d \left (a^2+i a^2 \tan (c+d x)\right )^3}-\frac {2 i \sec (c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )} \]

[Out]

arctanh(sin(d*x+c))/a^8/d+2/7*I*sec(d*x+c)^7/a/d/(a+I*a*tan(d*x+c))^7-2/5*I*sec(d*x+c)^5/a^3/d/(a+I*a*tan(d*x+
c))^5+2/3*I*sec(d*x+c)^3/a^2/d/(a^2+I*a^2*tan(d*x+c))^3-2*I*sec(d*x+c)/d/(a^8+I*a^8*tan(d*x+c))

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3581, 3855} \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a^8 d}-\frac {2 i \sec (c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}-\frac {2 i \sec ^5(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac {2 i \sec ^3(c+d x)}{3 a^2 d \left (a^2+i a^2 \tan (c+d x)\right )^3}+\frac {2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7} \]

[In]

Int[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^8,x]

[Out]

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

Rule 3581

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 3855

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7}-\frac {\int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^6} \, dx}{a^2} \\ & = \frac {2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7}-\frac {2 i \sec ^5(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac {\int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx}{a^4} \\ & = \frac {2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7}-\frac {2 i \sec ^5(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac {2 i \sec ^3(c+d x)}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{a^6} \\ & = \frac {2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7}-\frac {2 i \sec ^5(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac {2 i \sec ^3(c+d x)}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac {2 i \sec (c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}+\frac {\int \sec (c+d x) \, dx}{a^8} \\ & = \frac {\text {arctanh}(\sin (c+d x))}{a^8 d}+\frac {2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7}-\frac {2 i \sec ^5(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac {2 i \sec ^3(c+d x)}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac {2 i \sec (c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.49 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.95 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\sec ^8(c+d x) \left (70 i \cos \left (\frac {1}{2} (c+d x)\right )-42 i \cos \left (\frac {3}{2} (c+d x)\right )-210 i \cos \left (\frac {5}{2} (c+d x)\right )+30 i \cos \left (\frac {7}{2} (c+d x)\right )-105 \cos \left (\frac {7}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+105 \cos \left (\frac {7}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-70 \sin \left (\frac {1}{2} (c+d x)\right )-42 \sin \left (\frac {3}{2} (c+d x)\right )+210 \sin \left (\frac {5}{2} (c+d x)\right )+30 \sin \left (\frac {7}{2} (c+d x)\right )-105 i \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin \left (\frac {7}{2} (c+d x)\right )+105 i \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin \left (\frac {7}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {9}{2} (c+d x)\right )+i \sin \left (\frac {9}{2} (c+d x)\right )\right )}{105 a^8 d (-i+\tan (c+d x))^8} \]

[In]

Integrate[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(Sec[c + d*x]^8*((70*I)*Cos[(c + d*x)/2] - (42*I)*Cos[(3*(c + d*x))/2] - (210*I)*Cos[(5*(c + d*x))/2] + (30*I)
*Cos[(7*(c + d*x))/2] - 105*Cos[(7*(c + d*x))/2]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 105*Cos[(7*(c + d*
x))/2]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 70*Sin[(c + d*x)/2] - 42*Sin[(3*(c + d*x))/2] + 210*Sin[(5*(
c + d*x))/2] + 30*Sin[(7*(c + d*x))/2] - (105*I)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[(7*(c + d*x))/2]
+ (105*I)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[(7*(c + d*x))/2])*(Cos[(9*(c + d*x))/2] + I*Sin[(9*(c
+ d*x))/2]))/(105*a^8*d*(-I + Tan[c + d*x])^8)

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {2 i {\mathrm e}^{-i \left (d x +c \right )}}{a^{8} d}+\frac {2 i {\mathrm e}^{-3 i \left (d x +c \right )}}{3 a^{8} d}-\frac {2 i {\mathrm e}^{-5 i \left (d x +c \right )}}{5 a^{8} d}+\frac {2 i {\mathrm e}^{-7 i \left (d x +c \right )}}{7 a^{8} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{8} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{8} d}\)	\(115\)
derivativedivides	\(\frac {-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {128 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {16 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {128 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {256}{7 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {896}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {160}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}}{a^{8} d}\)	\(137\)
default	\(\frac {-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {128 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {16 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {128 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {256}{7 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {896}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {160}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}}{a^{8} d}\)	\(137\)

[In]

int(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^8,x,method=_RETURNVERBOSE)

[Out]

-2*I/a^8/d*exp(-I*(d*x+c))+2/3*I/a^8/d*exp(-3*I*(d*x+c))-2/5*I/a^8/d*exp(-5*I*(d*x+c))+2/7*I/a^8/d*exp(-7*I*(d
*x+c))-1/a^8/d*ln(exp(I*(d*x+c))-I)+1/a^8/d*ln(exp(I*(d*x+c))+I)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.63 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {{\left (105 \, e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 105 \, e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 210 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 42 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 30 i\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{105 \, a^{8} d} \]

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

1/105*(105*e^(7*I*d*x + 7*I*c)*log(e^(I*d*x + I*c) + I) - 105*e^(7*I*d*x + 7*I*c)*log(e^(I*d*x + I*c) - I) - 2
10*I*e^(6*I*d*x + 6*I*c) + 70*I*e^(4*I*d*x + 4*I*c) - 42*I*e^(2*I*d*x + 2*I*c) + 30*I)*e^(-7*I*d*x - 7*I*c)/(a
^8*d)

Sympy [F]

\[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\int \frac {\sec ^{9}{\left (c + d x \right )}}{\tan ^{8}{\left (c + d x \right )} - 8 i \tan ^{7}{\left (c + d x \right )} - 28 \tan ^{6}{\left (c + d x \right )} + 56 i \tan ^{5}{\left (c + d x \right )} + 70 \tan ^{4}{\left (c + d x \right )} - 56 i \tan ^{3}{\left (c + d x \right )} - 28 \tan ^{2}{\left (c + d x \right )} + 8 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{8}} \]

[In]

integrate(sec(d*x+c)**9/(a+I*a*tan(d*x+c))**8,x)

[Out]

Integral(sec(c + d*x)**9/(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 = 0.32 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.19 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {-210 i \, \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) - 210 i \, \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) + 60 i \, \cos \left (7 \, d x + 7 \, c\right ) - 84 i \, \cos \left (5 \, d x + 5 \, c\right ) + 140 i \, \cos \left (3 \, d x + 3 \, c\right ) - 420 i \, \cos \left (d x + c\right ) + 105 \, \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - 105 \, \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right ) + 60 \, \sin \left (7 \, d x + 7 \, c\right ) - 84 \, \sin \left (5 \, d x + 5 \, c\right ) + 140 \, \sin \left (3 \, d x + 3 \, c\right ) - 420 \, \sin \left (d x + c\right )}{210 \, a^{8} d} \]

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

1/210*(-210*I*arctan2(cos(d*x + c), sin(d*x + c) + 1) - 210*I*arctan2(cos(d*x + c), -sin(d*x + c) + 1) + 60*I*
cos(7*d*x + 7*c) - 84*I*cos(5*d*x + 5*c) + 140*I*cos(3*d*x + 3*c) - 420*I*cos(d*x + c) + 105*log(cos(d*x + c)^
2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - 105*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) + 60*
sin(7*d*x + 7*c) - 84*sin(5*d*x + 5*c) + 140*sin(3*d*x + 3*c) - 420*sin(d*x + c))/(a^8*d)

Giac [A] (verification not implemented)

none

Time = 1.64 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.79 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\frac {105 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{8}} - \frac {105 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{8}} - \frac {16 \, {\left (-105 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 175 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 490 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 294 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 133 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 19\right )}}{a^{8} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{7}}}{105 \, d} \]

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

1/105*(105*log(tan(1/2*d*x + 1/2*c) + 1)/a^8 - 105*log(tan(1/2*d*x + 1/2*c) - 1)/a^8 - 16*(-105*I*tan(1/2*d*x
+ 1/2*c)^5 - 175*tan(1/2*d*x + 1/2*c)^4 + 490*I*tan(1/2*d*x + 1/2*c)^3 + 294*tan(1/2*d*x + 1/2*c)^2 - 133*I*ta
n(1/2*d*x + 1/2*c) - 19)/(a^8*(tan(1/2*d*x + 1/2*c) - I)^7))/d

Mupad [B] (verification not implemented)

Time = 8.00 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.33 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^8\,d}+\frac {\frac {16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a^8}-\frac {224\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,a^8}+\frac {304\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15\,a^8}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,224{}\mathrm {i}}{5\,a^8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,80{}\mathrm {i}}{3\,a^8}-\frac {304{}\mathrm {i}}{105\,a^8}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,1{}\mathrm {i}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,21{}\mathrm {i}+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,35{}\mathrm {i}-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,7{}\mathrm {i}+1\right )} \]

[In]

int(1/(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^8),x)

[Out]

(2*atanh(tan(c/2 + (d*x)/2)))/(a^8*d) + ((tan(c/2 + (d*x)/2)^2*224i)/(5*a^8) - (224*tan(c/2 + (d*x)/2)^3)/(3*a
^8) - (tan(c/2 + (d*x)/2)^4*80i)/(3*a^8) + (16*tan(c/2 + (d*x)/2)^5)/a^8 - 304i/(105*a^8) + (304*tan(c/2 + (d*
x)/2))/(15*a^8))/(d*(tan(c/2 + (d*x)/2)*7i - 21*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^3*35i + 35*tan(c/2 +
(d*x)/2)^4 + tan(c/2 + (d*x)/2)^5*21i - 7*tan(c/2 + (d*x)/2)^6 - tan(c/2 + (d*x)/2)^7*1i + 1))