3.2.66 \(\int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\) [166]

Optimal. Leaf size=134 \[ -\frac {192 x}{a^8}-\frac {192 i \log (\cos (c+d x))}{a^8 d}+\frac {129 \tan (c+d x)}{a^8 d}-\frac {36 i \tan ^2(c+d x)}{a^8 d}-\frac {10 \tan ^3(c+d x)}{a^8 d}+\frac {2 i \tan ^4(c+d x)}{a^8 d}+\frac {\tan ^5(c+d x)}{5 a^8 d}+\frac {64 i}{d \left (a^8+i a^8 \tan (c+d x)\right )} \]

[Out]

-192*x/a^8-192*I*ln(cos(d*x+c))/a^8/d+129*tan(d*x+c)/a^8/d-36*I*tan(d*x+c)^2/a^8/d-10*tan(d*x+c)^3/a^8/d+2*I*t
an(d*x+c)^4/a^8/d+1/5*tan(d*x+c)^5/a^8/d+64*I/d/(a^8+I*a^8*tan(d*x+c))

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Rubi [A]
time = 0.07, antiderivative size = 134, normalized size of antiderivative = 1.00, 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 = {3568, 45} \begin {gather*} \frac {\tan ^5(c+d x)}{5 a^8 d}+\frac {2 i \tan ^4(c+d x)}{a^8 d}-\frac {10 \tan ^3(c+d x)}{a^8 d}-\frac {36 i \tan ^2(c+d x)}{a^8 d}+\frac {129 \tan (c+d x)}{a^8 d}+\frac {64 i}{d \left (a^8+i a^8 \tan (c+d x)\right )}-\frac {192 i \log (\cos (c+d x))}{a^8 d}-\frac {192 x}{a^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^14/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(-192*x)/a^8 - ((192*I)*Log[Cos[c + d*x]])/(a^8*d) + (129*Tan[c + d*x])/(a^8*d) - ((36*I)*Tan[c + d*x]^2)/(a^8
*d) - (10*Tan[c + d*x]^3)/(a^8*d) + ((2*I)*Tan[c + d*x]^4)/(a^8*d) + Tan[c + d*x]^5/(5*a^8*d) + (64*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*} \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=-\frac {i \text {Subst}\left (\int \frac {(a-x)^6}{(a+x)^2} \, dx,x,i a \tan (c+d x)\right )}{a^{13} d}\\ &=-\frac {i \text {Subst}\left (\int \left (129 a^4-72 a^3 x+30 a^2 x^2-8 a x^3+x^4+\frac {64 a^6}{(a+x)^2}-\frac {192 a^5}{a+x}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^{13} d}\\ &=-\frac {192 x}{a^8}-\frac {192 i \log (\cos (c+d x))}{a^8 d}+\frac {129 \tan (c+d x)}{a^8 d}-\frac {36 i \tan ^2(c+d x)}{a^8 d}-\frac {10 \tan ^3(c+d x)}{a^8 d}+\frac {2 i \tan ^4(c+d x)}{a^8 d}+\frac {\tan ^5(c+d x)}{5 a^8 d}+\frac {64 i}{d \left (a^8+i a^8 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(599\) vs. \(2(134)=268\).
time = 3.06, size = 599, normalized size = 4.47 \begin {gather*} \frac {\sec (c) \sec ^{13}(c+d x) (-\cos (7 (c+d x))-i \sin (7 (c+d x))) (-220 i \cos (3 c+2 d x)+900 d x \cos (3 c+2 d x)+238 i \cos (3 c+4 d x)+360 d x \cos (3 c+4 d x)-110 i \cos (5 c+4 d x)+360 d x \cos (5 c+4 d x)+77 i \cos (5 c+6 d x)+60 d x \cos (5 c+6 d x)-10 i \cos (7 c+6 d x)+60 d x \cos (7 c+6 d x)+10 \cos (c) (-7 i+120 d x+120 i \log (\cos (c+d x)))+5 \cos (c+2 d x) (43 i+180 d x+180 i \log (\cos (c+d x)))+900 i \cos (3 c+2 d x) \log (\cos (c+d x))+360 i \cos (3 c+4 d x) \log (\cos (c+d x))+360 i \cos (5 c+4 d x) \log (\cos (c+d x))+60 i \cos (5 c+6 d x) \log (\cos (c+d x))+60 i \cos (7 c+6 d x) \log (\cos (c+d x))+870 \sin (c)-985 \sin (c+2 d x)+300 i d x \sin (c+2 d x)-300 \log (\cos (c+d x)) \sin (c+2 d x)+320 \sin (3 c+2 d x)+300 i d x \sin (3 c+2 d x)-300 \log (\cos (c+d x)) \sin (3 c+2 d x)-512 \sin (3 c+4 d x)+240 i d x \sin (3 c+4 d x)-240 \log (\cos (c+d x)) \sin (3 c+4 d x)+10 \sin (5 c+4 d x)+240 i d x \sin (5 c+4 d x)-240 \log (\cos (c+d x)) \sin (5 c+4 d x)-97 \sin (5 c+6 d x)+60 i d x \sin (5 c+6 d x)-60 \log (\cos (c+d x)) \sin (5 c+6 d x)-10 \sin (7 c+6 d x)+60 i d x \sin (7 c+6 d x)-60 \log (\cos (c+d x)) \sin (7 c+6 d x))}{20 a^8 d (-i+\tan (c+d x))^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^14/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(Sec[c]*Sec[c + d*x]^13*(-Cos[7*(c + d*x)] - I*Sin[7*(c + d*x)])*((-220*I)*Cos[3*c + 2*d*x] + 900*d*x*Cos[3*c
+ 2*d*x] + (238*I)*Cos[3*c + 4*d*x] + 360*d*x*Cos[3*c + 4*d*x] - (110*I)*Cos[5*c + 4*d*x] + 360*d*x*Cos[5*c +
4*d*x] + (77*I)*Cos[5*c + 6*d*x] + 60*d*x*Cos[5*c + 6*d*x] - (10*I)*Cos[7*c + 6*d*x] + 60*d*x*Cos[7*c + 6*d*x]
+ 10*Cos[c]*(-7*I + 120*d*x + (120*I)*Log[Cos[c + d*x]]) + 5*Cos[c + 2*d*x]*(43*I + 180*d*x + (180*I)*Log[Cos
[c + d*x]]) + (900*I)*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]] + (360*I)*Cos[3*c + 4*d*x]*Log[Cos[c + d*x]] + (360*I
)*Cos[5*c + 4*d*x]*Log[Cos[c + d*x]] + (60*I)*Cos[5*c + 6*d*x]*Log[Cos[c + d*x]] + (60*I)*Cos[7*c + 6*d*x]*Log
[Cos[c + d*x]] + 870*Sin[c] - 985*Sin[c + 2*d*x] + (300*I)*d*x*Sin[c + 2*d*x] - 300*Log[Cos[c + d*x]]*Sin[c +
2*d*x] + 320*Sin[3*c + 2*d*x] + (300*I)*d*x*Sin[3*c + 2*d*x] - 300*Log[Cos[c + d*x]]*Sin[3*c + 2*d*x] - 512*Si
n[3*c + 4*d*x] + (240*I)*d*x*Sin[3*c + 4*d*x] - 240*Log[Cos[c + d*x]]*Sin[3*c + 4*d*x] + 10*Sin[5*c + 4*d*x] +
(240*I)*d*x*Sin[5*c + 4*d*x] - 240*Log[Cos[c + d*x]]*Sin[5*c + 4*d*x] - 97*Sin[5*c + 6*d*x] + (60*I)*d*x*Sin[
5*c + 6*d*x] - 60*Log[Cos[c + d*x]]*Sin[5*c + 6*d*x] - 10*Sin[7*c + 6*d*x] + (60*I)*d*x*Sin[7*c + 6*d*x] - 60*
Log[Cos[c + d*x]]*Sin[7*c + 6*d*x]))/(20*a^8*d*(-I + Tan[c + d*x])^8)

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Maple [A]
time = 0.37, size = 85, normalized size = 0.63

method	result	size
derivativedivides	\(\frac {129 \tan \left (d x +c \right )+\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}+2 i \left (\tan ^{4}\left (d x +c \right )\right )-10 \left (\tan ^{3}\left (d x +c \right )\right )-36 i \left (\tan ^{2}\left (d x +c \right )\right )+192 i \ln \left (\tan \left (d x +c \right )-i\right )+\frac {64}{\tan \left (d x +c \right )-i}}{d \,a^{8}}\)	\(85\)
default	\(\frac {129 \tan \left (d x +c \right )+\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}+2 i \left (\tan ^{4}\left (d x +c \right )\right )-10 \left (\tan ^{3}\left (d x +c \right )\right )-36 i \left (\tan ^{2}\left (d x +c \right )\right )+192 i \ln \left (\tan \left (d x +c \right )-i\right )+\frac {64}{\tan \left (d x +c \right )-i}}{d \,a^{8}}\)	\(85\)
risch	\(\frac {32 i {\mathrm e}^{-2 i \left (d x +c \right )}}{a^{8} d}-\frac {384 x}{a^{8}}-\frac {384 c}{a^{8} d}+\frac {16 i \left (50 \,{\mathrm e}^{8 i \left (d x +c \right )}+220 \,{\mathrm e}^{6 i \left (d x +c \right )}+370 \,{\mathrm e}^{4 i \left (d x +c \right )}+285 \,{\mathrm e}^{2 i \left (d x +c \right )}+87\right )}{5 d \,a^{8} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {192 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{8} d}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^8,x,method=_RETURNVERBOSE)

[Out]

1/d/a^8*(129*tan(d*x+c)+1/5*tan(d*x+c)^5+2*I*tan(d*x+c)^4-10*tan(d*x+c)^3-36*I*tan(d*x+c)^2+192*I*ln(tan(d*x+c
)-I)+64/(tan(d*x+c)-I))

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Maxima [A]
time = 0.31, size = 229, normalized size = 1.71 \begin {gather*} \frac {\frac {320 \, {\left (\tan \left (d x + c\right )^{6} - 6 i \, \tan \left (d x + c\right )^{5} - 15 \, \tan \left (d x + c\right )^{4} + 20 i \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )^{2} - 6 i \, \tan \left (d x + c\right ) - 1\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 {\tan \left (d x + c\right )^{5} + 10 i \, \tan \left (d x + c\right )^{4} - 50 \, \tan \left (d x + c\right )^{3} - 180 i \, \tan \left (d x + c\right )^{2} + 645 \, \tan \left (d x + c\right )}{a^{8}} + \frac {960 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{8}}}{5 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

1/5*(320*(tan(d*x + c)^6 - 6*I*tan(d*x + c)^5 - 15*tan(d*x + c)^4 + 20*I*tan(d*x + c)^3 + 15*tan(d*x + c)^2 -
6*I*tan(d*x + c) - 1)/(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) + (tan(d*x + c)^5 + 10*
I*tan(d*x + c)^4 - 50*tan(d*x + c)^3 - 180*I*tan(d*x + c)^2 + 645*tan(d*x + c))/a^8 + 960*I*log(I*tan(d*x + c)
+ 1)/a^8)/d

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (122) = 244\).
time = 0.48, size = 273, normalized size = 2.04 \begin {gather*} -\frac {16 \, {\left (120 \, d x e^{\left (12 i \, d x + 12 i \, c\right )} + 60 \, {\left (10 \, d x - i\right )} e^{\left (10 i \, d x + 10 i \, c\right )} + 30 \, {\left (40 \, d x - 9 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, {\left (120 \, d x - 47 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 5 \, {\left (120 \, d x - 77 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (120 \, d x - 137 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 60 \, {\left (i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 5 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 10 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 10 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 5 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 10 i\right )}}{5 \, {\left (a^{8} d e^{\left (12 i \, d x + 12 i \, c\right )} + 5 \, a^{8} d e^{\left (10 i \, d x + 10 i \, c\right )} + 10 \, a^{8} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{8} d e^{\left (6 i \, d x + 6 i \, c\right )} + 5 \, a^{8} d e^{\left (4 i \, d x + 4 i \, c\right )} + a^{8} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

-16/5*(120*d*x*e^(12*I*d*x + 12*I*c) + 60*(10*d*x - I)*e^(10*I*d*x + 10*I*c) + 30*(40*d*x - 9*I)*e^(8*I*d*x +
8*I*c) + 10*(120*d*x - 47*I)*e^(6*I*d*x + 6*I*c) + 5*(120*d*x - 77*I)*e^(4*I*d*x + 4*I*c) + (120*d*x - 137*I)*
e^(2*I*d*x + 2*I*c) + 60*(I*e^(12*I*d*x + 12*I*c) + 5*I*e^(10*I*d*x + 10*I*c) + 10*I*e^(8*I*d*x + 8*I*c) + 10*
I*e^(6*I*d*x + 6*I*c) + 5*I*e^(4*I*d*x + 4*I*c) + I*e^(2*I*d*x + 2*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) - 10*I)/
(a^8*d*e^(12*I*d*x + 12*I*c) + 5*a^8*d*e^(10*I*d*x + 10*I*c) + 10*a^8*d*e^(8*I*d*x + 8*I*c) + 10*a^8*d*e^(6*I*
d*x + 6*I*c) + 5*a^8*d*e^(4*I*d*x + 4*I*c) + a^8*d*e^(2*I*d*x + 2*I*c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sec ^{14}{\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**14/(a+I*a*tan(d*x+c))**8,x)

[Out]

Integral(sec(c + d*x)**14/(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

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (122) = 244\).
time = 1.16, size = 250, normalized size = 1.87 \begin {gather*} -\frac {2 \, {\left (\frac {480 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{8}} - \frac {960 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a^{8}} + \frac {480 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{8}} + \frac {160 \, {\left (9 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 i\right )}}{a^{8} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{2}} + \frac {-1096 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 645 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 5840 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 2780 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12120 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 4286 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12120 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2780 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5840 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 645 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1096 i}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5} a^{8}}\right )}}{5 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

-2/5*(480*I*log(tan(1/2*d*x + 1/2*c) + 1)/a^8 - 960*I*log(tan(1/2*d*x + 1/2*c) - I)/a^8 + 480*I*log(tan(1/2*d*
x + 1/2*c) - 1)/a^8 + 160*(9*I*tan(1/2*d*x + 1/2*c)^2 + 20*tan(1/2*d*x + 1/2*c) - 9*I)/(a^8*(tan(1/2*d*x + 1/2
*c) - I)^2) + (-1096*I*tan(1/2*d*x + 1/2*c)^10 + 645*tan(1/2*d*x + 1/2*c)^9 + 5840*I*tan(1/2*d*x + 1/2*c)^8 -
2780*tan(1/2*d*x + 1/2*c)^7 - 12120*I*tan(1/2*d*x + 1/2*c)^6 + 4286*tan(1/2*d*x + 1/2*c)^5 + 12120*I*tan(1/2*d
*x + 1/2*c)^4 - 2780*tan(1/2*d*x + 1/2*c)^3 - 5840*I*tan(1/2*d*x + 1/2*c)^2 + 645*tan(1/2*d*x + 1/2*c) + 1096*
I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^5*a^8))/d

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Mupad [B]
time = 3.45, size = 105, normalized size = 0.78 \begin {gather*} \frac {\frac {129\,\mathrm {tan}\left (c+d\,x\right )}{a^8}-\frac {10\,{\mathrm {tan}\left (c+d\,x\right )}^3}{a^8}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,a^8}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,192{}\mathrm {i}}{a^8}+\frac {64{}\mathrm {i}}{a^8\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,36{}\mathrm {i}}{a^8}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,2{}\mathrm {i}}{a^8}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^14*(a + a*tan(c + d*x)*1i)^8),x)

[Out]

((log(tan(c + d*x) - 1i)*192i)/a^8 + (129*tan(c + d*x))/a^8 + 64i/(a^8*(tan(c + d*x)*1i + 1)) - (tan(c + d*x)^
2*36i)/a^8 - (10*tan(c + d*x)^3)/a^8 + (tan(c + d*x)^4*2i)/a^8 + tan(c + d*x)^5/(5*a^8))/d

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