3.4.54 \(\int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [354]

Optimal. Leaf size=147 \[ \frac {256 i a^4 \sec ^{11}(c+d x)}{12155 d (a+i a \tan (c+d x))^{11/2}}+\frac {64 i a^3 \sec ^{11}(c+d x)}{1105 d (a+i a \tan (c+d x))^{9/2}}+\frac {8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

256/12155*I*a^4*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(11/2)+64/1105*I*a^3*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(9/
2)+8/85*I*a^2*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(7/2)+2/17*I*a*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(5/2)

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Rubi [A]
time = 0.19, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3575, 3574} \begin {gather*} \frac {256 i a^4 \sec ^{11}(c+d x)}{12155 d (a+i a \tan (c+d x))^{11/2}}+\frac {64 i a^3 \sec ^{11}(c+d x)}{1105 d (a+i a \tan (c+d x))^{9/2}}+\frac {8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((256*I)/12155)*a^4*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(11/2)) + (((64*I)/1105)*a^3*Sec[c + d*x]^11)/
(d*(a + I*a*Tan[c + d*x])^(9/2)) + (((8*I)/85)*a^2*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (((2*I)
/17)*a*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(5/2))

Rule 3574

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

Rule 3575

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

Rubi steps

\begin {align*} \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{17} (12 a) \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac {8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{85} \left (32 a^2\right ) \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\\ &=\frac {64 i a^3 \sec ^{11}(c+d x)}{1105 d (a+i a \tan (c+d x))^{9/2}}+\frac {8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}+\frac {\left (128 a^3\right ) \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{9/2}} \, dx}{1105}\\ &=\frac {256 i a^4 \sec ^{11}(c+d x)}{12155 d (a+i a \tan (c+d x))^{11/2}}+\frac {64 i a^3 \sec ^{11}(c+d x)}{1105 d (a+i a \tan (c+d x))^{9/2}}+\frac {8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 1.04, size = 108, normalized size = 0.73 \begin {gather*} \frac {2 \sec ^9(c+d x) (i \cos (4 (c+d x))+\sin (4 (c+d x))) (475 i-2242 i \cos (2 (c+d x))+1089 \sec (c+d x) \sin (3 (c+d x))+374 \tan (c+d x))}{12155 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(2*Sec[c + d*x]^9*(I*Cos[4*(c + d*x)] + Sin[4*(c + d*x)])*(475*I - (2242*I)*Cos[2*(c + d*x)] + 1089*Sec[c + d*
x]*Sin[3*(c + d*x)] + 374*Tan[c + d*x]))/(12155*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]
time = 3.52, size = 171, normalized size = 1.16

method	result	size
default	\(\frac {2 \left (4096 i \left (\cos ^{9}\left (d x +c \right )\right )+4096 \sin \left (d x +c \right ) \left (\cos ^{8}\left (d x +c \right )\right )-512 i \left (\cos ^{7}\left (d x +c \right )\right )+1536 \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )-160 i \left (\cos ^{5}\left (d x +c \right )\right )+1120 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )-84 i \left (\cos ^{3}\left (d x +c \right )\right )+924 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-1573 i \cos \left (d x +c \right )-715 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{12155 d \cos \left (d x +c \right )^{8} a^{2}}\)	\(171\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/12155/d*(4096*I*cos(d*x+c)^9+4096*sin(d*x+c)*cos(d*x+c)^8-512*I*cos(d*x+c)^7+1536*sin(d*x+c)*cos(d*x+c)^6-16
0*I*cos(d*x+c)^5+1120*sin(d*x+c)*cos(d*x+c)^4-84*I*cos(d*x+c)^3+924*cos(d*x+c)^2*sin(d*x+c)-1573*I*cos(d*x+c)-
715*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^8/a^2

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 764 vs. \(2 (115) = 230\).
time = 0.53, size = 764, normalized size = 5.20 \begin {gather*} -\frac {2 \, {\left (-1767 i \, \sqrt {a} - \frac {6854 \, \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2088 i \, \sqrt {a} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {16438 \, \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {5661 i \, \sqrt {a} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {56984 \, \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {13328 i \, \sqrt {a} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {129336 \, \sqrt {a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {7514 i \, \sqrt {a} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {156468 \, \sqrt {a} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {156468 \, \sqrt {a} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {7514 i \, \sqrt {a} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} - \frac {129336 \, \sqrt {a} \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} + \frac {13328 i \, \sqrt {a} \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}} - \frac {56984 \, \sqrt {a} \sin \left (d x + c\right )^{15}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{15}} + \frac {5661 i \, \sqrt {a} \sin \left (d x + c\right )^{16}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{16}} - \frac {16438 \, \sqrt {a} \sin \left (d x + c\right )^{17}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{17}} - \frac {2088 i \, \sqrt {a} \sin \left (d x + c\right )^{18}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{18}} - \frac {6854 \, \sqrt {a} \sin \left (d x + c\right )^{19}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{19}} + \frac {1767 i \, \sqrt {a} \sin \left (d x + c\right )^{20}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{20}}\right )} {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {3}{2}} {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}^{\frac {3}{2}}}{12155 \, {\left (a^{2} - \frac {10 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {45 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {120 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {210 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {252 \, a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {210 \, a^{2} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} - \frac {120 \, a^{2} \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}} + \frac {45 \, a^{2} \sin \left (d x + c\right )^{16}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{16}} - \frac {10 \, a^{2} \sin \left (d x + c\right )^{18}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{18}} + \frac {a^{2} \sin \left (d x + c\right )^{20}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{20}}\right )} d {\left (-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-2/12155*(-1767*I*sqrt(a) - 6854*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) + 2088*I*sqrt(a)*sin(d*x + c)^2/(cos(
d*x + c) + 1)^2 - 16438*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 5661*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4 - 56984*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 13328*I*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c)
+ 1)^6 - 129336*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 7514*I*sqrt(a)*sin(d*x + c)^8/(cos(d*x + c) + 1)
^8 - 156468*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 156468*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11
- 7514*I*sqrt(a)*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 129336*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13
+ 13328*I*sqrt(a)*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 - 56984*sqrt(a)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15
+ 5661*I*sqrt(a)*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 16438*sqrt(a)*sin(d*x + c)^17/(cos(d*x + c) + 1)^17
- 2088*I*sqrt(a)*sin(d*x + c)^18/(cos(d*x + c) + 1)^18 - 6854*sqrt(a)*sin(d*x + c)^19/(cos(d*x + c) + 1)^19 +
1767*I*sqrt(a)*sin(d*x + c)^20/(cos(d*x + c) + 1)^20)*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)*(sin(d*x + c
)/(cos(d*x + c) + 1) - 1)^(3/2)/((a^2 - 10*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 45*a^2*sin(d*x + c)^4/(co
s(d*x + c) + 1)^4 - 120*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 210*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8
- 252*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 210*a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 120*a^2*sin(
d*x + c)^14/(cos(d*x + c) + 1)^14 + 45*a^2*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 10*a^2*sin(d*x + c)^18/(cos
(d*x + c) + 1)^18 + a^2*sin(d*x + c)^20/(cos(d*x + c) + 1)^20)*d*(-2*I*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d
*x + c)^2/(cos(d*x + c) + 1)^2 - 1)^(3/2))

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Fricas [A]
time = 0.44, size = 184, normalized size = 1.25 \begin {gather*} -\frac {512 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-1105 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 510 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 136 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i\right )}}{12155 \, {\left (a^{2} d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, a^{2} d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, a^{2} d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-512/12155*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-1105*I*e^(6*I*d*x + 6*I*c) - 510*I*e^(4*I*d*x + 4*I*c)
- 136*I*e^(2*I*d*x + 2*I*c) - 16*I)/(a^2*d*e^(16*I*d*x + 16*I*c) + 8*a^2*d*e^(14*I*d*x + 14*I*c) + 28*a^2*d*e^
(12*I*d*x + 12*I*c) + 56*a^2*d*e^(10*I*d*x + 10*I*c) + 70*a^2*d*e^(8*I*d*x + 8*I*c) + 56*a^2*d*e^(6*I*d*x + 6*
I*c) + 28*a^2*d*e^(4*I*d*x + 4*I*c) + 8*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**11/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^11/(I*a*tan(d*x + c) + a)^(3/2), x)

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Mupad [B]
time = 9.86, size = 301, normalized size = 2.05 \begin {gather*} \frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{11\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1536{}\mathrm {i}}{13\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}+\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{5\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{17\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^11*(a + a*tan(c + d*x)*1i)^(3/2)),x)

[Out]

(exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*512i)/(11*a^2*d
*(exp(c*2i + d*x*2i) + 1)^5) - (exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*
2i) + 1))^(1/2)*1536i)/(13*a^2*d*(exp(c*2i + d*x*2i) + 1)^6) + (exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2
i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*512i)/(5*a^2*d*(exp(c*2i + d*x*2i) + 1)^7) - (exp(- c*1i - d*x
*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*512i)/(17*a^2*d*(exp(c*2i + d*x*
2i) + 1)^8)

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