\(\int \frac {\cos ^5(c+d x)}{a+b \tan ^2(c+d x)} \, dx\) [455]

Optimal result

Integrand size = 23, antiderivative size = 126 \[ \int \frac {\cos ^5(c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^{7/2} d}+\frac {\left (a^2-3 a b+3 b^2\right ) \sin (c+d x)}{(a-b)^3 d}-\frac {(2 a-3 b) \sin ^3(c+d x)}{3 (a-b)^2 d}+\frac {\sin ^5(c+d x)}{5 (a-b) d} \]

[Out]

(a^2-3*a*b+3*b^2)*sin(d*x+c)/(a-b)^3/d-1/3*(2*a-3*b)*sin(d*x+c)^3/(a-b)^2/d+1/5*sin(d*x+c)^5/(a-b)/d-b^3*arcta
nh(sin(d*x+c)*(a-b)^(1/2)/a^(1/2))/(a-b)^(7/2)/d/a^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3757, 398, 214} \[ \int \frac {\cos ^5(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\left (a^2-3 a b+3 b^2\right ) \sin (c+d x)}{d (a-b)^3}-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} d (a-b)^{7/2}}+\frac {\sin ^5(c+d x)}{5 d (a-b)}-\frac {(2 a-3 b) \sin ^3(c+d x)}{3 d (a-b)^2} \]

[In]

Int[Cos[c + d*x]^5/(a + b*Tan[c + d*x]^2),x]

[Out]

-((b^3*ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)^(7/2)*d)) + ((a^2 - 3*a*b + 3*b^2)*Sin[c
+ d*x])/((a - b)^3*d) - ((2*a - 3*b)*Sin[c + d*x]^3)/(3*(a - b)^2*d) + Sin[c + d*x]^5/(5*(a - b)*d)

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 398

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 3757

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^3}{a-(a-b) x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a^2-3 a b+3 b^2}{(a-b)^3}-\frac {(2 a-3 b) x^2}{(a-b)^2}+\frac {x^4}{a-b}-\frac {b^3}{(a-b)^3 \left (a-(a-b) x^2\right )}\right ) \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {\left (a^2-3 a b+3 b^2\right ) \sin (c+d x)}{(a-b)^3 d}-\frac {(2 a-3 b) \sin ^3(c+d x)}{3 (a-b)^2 d}+\frac {\sin ^5(c+d x)}{5 (a-b) d}-\frac {b^3 \text {Subst}\left (\int \frac {1}{a-(a-b) x^2} \, dx,x,\sin (c+d x)\right )}{(a-b)^3 d} \\ & = -\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^{7/2} d}+\frac {\left (a^2-3 a b+3 b^2\right ) \sin (c+d x)}{(a-b)^3 d}-\frac {(2 a-3 b) \sin ^3(c+d x)}{3 (a-b)^2 d}+\frac {\sin ^5(c+d x)}{5 (a-b) d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.90 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.17 \[ \int \frac {\cos ^5(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\frac {120 b^3 \left (\log \left (\sqrt {a}-\sqrt {a-b} \sin (c+d x)\right )-\log \left (\sqrt {a}+\sqrt {a-b} \sin (c+d x)\right )\right )}{\sqrt {a} (a-b)^{7/2}}+\frac {30 \left (5 a^2-16 a b+19 b^2\right ) \sin (c+d x)}{(a-b)^3}+\frac {5 (5 a-9 b) \sin (3 (c+d x))}{(a-b)^2}+\frac {3 \sin (5 (c+d x))}{a-b}}{240 d} \]

[In]

Integrate[Cos[c + d*x]^5/(a + b*Tan[c + d*x]^2),x]

[Out]

((120*b^3*(Log[Sqrt[a] - Sqrt[a - b]*Sin[c + d*x]] - Log[Sqrt[a] + Sqrt[a - b]*Sin[c + d*x]]))/(Sqrt[a]*(a - b
)^(7/2)) + (30*(5*a^2 - 16*a*b + 19*b^2)*Sin[c + d*x])/(a - b)^3 + (5*(5*a - 9*b)*Sin[3*(c + d*x)])/(a - b)^2
+ (3*Sin[5*(c + d*x)])/(a - b))/(240*d)

Maple [A] (verified)

Time = 10.57 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.31

method	result	size
derivativedivides	\(\frac {\frac {\frac {a^{2} \sin \left (d x +c \right )^{5}}{5}-\frac {2 a b \sin \left (d x +c \right )^{5}}{5}+\frac {b^{2} \sin \left (d x +c \right )^{5}}{5}-\frac {2 a^{2} \sin \left (d x +c \right )^{3}}{3}+\frac {5 a b \sin \left (d x +c \right )^{3}}{3}-b^{2} \sin \left (d x +c \right )^{3}+a^{2} \sin \left (d x +c \right )-3 a b \sin \left (d x +c \right )+3 b^{2} \sin \left (d x +c \right )}{\left (a -b \right )^{3}}-\frac {b^{3} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{\left (a -b \right )^{3} \sqrt {a \left (a -b \right )}}}{d}\)	\(165\)
default	\(\frac {\frac {\frac {a^{2} \sin \left (d x +c \right )^{5}}{5}-\frac {2 a b \sin \left (d x +c \right )^{5}}{5}+\frac {b^{2} \sin \left (d x +c \right )^{5}}{5}-\frac {2 a^{2} \sin \left (d x +c \right )^{3}}{3}+\frac {5 a b \sin \left (d x +c \right )^{3}}{3}-b^{2} \sin \left (d x +c \right )^{3}+a^{2} \sin \left (d x +c \right )-3 a b \sin \left (d x +c \right )+3 b^{2} \sin \left (d x +c \right )}{\left (a -b \right )^{3}}-\frac {b^{3} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{\left (a -b \right )^{3} \sqrt {a \left (a -b \right )}}}{d}\)	\(165\)
risch	\(-\frac {5 i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{16 \left (a -b \right )^{3} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} a b}{\left (a -b \right )^{3} d}-\frac {19 i {\mathrm e}^{i \left (d x +c \right )} b^{2}}{16 \left (a -b \right )^{3} d}+\frac {5 i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{16 \left (a -b \right ) \left (a^{2}-2 a b +b^{2}\right ) d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} a b}{\left (a -b \right ) \left (a^{2}-2 a b +b^{2}\right ) d}+\frac {19 i {\mathrm e}^{-i \left (d x +c \right )} b^{2}}{16 \left (a -b \right ) \left (a^{2}-2 a b +b^{2}\right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{\sqrt {a^{2}-a b}}-1\right )}{2 \sqrt {a^{2}-a b}\, \left (a -b \right )^{3} d}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{\sqrt {a^{2}-a b}}-1\right )}{2 \sqrt {a^{2}-a b}\, \left (a -b \right )^{3} d}+\frac {\sin \left (5 d x +5 c \right )}{80 \left (a -b \right ) d}+\frac {5 \sin \left (3 d x +3 c \right ) a}{48 \left (a -b \right )^{2} d}-\frac {3 \sin \left (3 d x +3 c \right ) b}{16 \left (a -b \right )^{2} d}\)	\(374\)

[In]

int(cos(d*x+c)^5/(a+b*tan(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/(a-b)^3*(1/5*a^2*sin(d*x+c)^5-2/5*a*b*sin(d*x+c)^5+1/5*b^2*sin(d*x+c)^5-2/3*a^2*sin(d*x+c)^3+5/3*a*b*si
n(d*x+c)^3-b^2*sin(d*x+c)^3+a^2*sin(d*x+c)-3*a*b*sin(d*x+c)+3*b^2*sin(d*x+c))-b^3/(a-b)^3/(a*(a-b))^(1/2)*arct
anh((a-b)*sin(d*x+c)/(a*(a-b))^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 395, normalized size of antiderivative = 3.13 \[ \int \frac {\cos ^5(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\left [-\frac {15 \, \sqrt {a^{2} - a b} b^{3} \log \left (-\frac {{\left (a - b\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - a b} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) - 2 \, {\left (3 \, {\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} - 34 \, a^{3} b + 59 \, a^{2} b^{2} - 33 \, a b^{3} + {\left (4 \, a^{4} - 17 \, a^{3} b + 22 \, a^{2} b^{2} - 9 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} d}, \frac {15 \, \sqrt {-a^{2} + a b} b^{3} \arctan \left (\frac {\sqrt {-a^{2} + a b} \sin \left (d x + c\right )}{a}\right ) + {\left (3 \, {\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} - 34 \, a^{3} b + 59 \, a^{2} b^{2} - 33 \, a b^{3} + {\left (4 \, a^{4} - 17 \, a^{3} b + 22 \, a^{2} b^{2} - 9 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} d}\right ] \]

[In]

integrate(cos(d*x+c)^5/(a+b*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

[-1/30*(15*sqrt(a^2 - a*b)*b^3*log(-((a - b)*cos(d*x + c)^2 - 2*sqrt(a^2 - a*b)*sin(d*x + c) - 2*a + b)/((a -
b)*cos(d*x + c)^2 + b)) - 2*(3*(a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*cos(d*x + c)^4 + 8*a^4 - 34*a^3*b + 59*a^2*
b^2 - 33*a*b^3 + (4*a^4 - 17*a^3*b + 22*a^2*b^2 - 9*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/((a^5 - 4*a^4*b + 6*a
^3*b^2 - 4*a^2*b^3 + a*b^4)*d), 1/15*(15*sqrt(-a^2 + a*b)*b^3*arctan(sqrt(-a^2 + a*b)*sin(d*x + c)/a) + (3*(a^
4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*cos(d*x + c)^4 + 8*a^4 - 34*a^3*b + 59*a^2*b^2 - 33*a*b^3 + (4*a^4 - 17*a^3*b
+ 22*a^2*b^2 - 9*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5/(a+b*tan(d*x+c)**2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^5(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(cos(d*x+c)^5/(a+b*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (114) = 228\).

Time = 0.51 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.53 \[ \int \frac {\cos ^5(c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {\frac {15 \, b^{3} \arctan \left (-\frac {a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt {-a^{2} + a b}}\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {-a^{2} + a b}} - \frac {3 \, a^{4} \sin \left (d x + c\right )^{5} - 12 \, a^{3} b \sin \left (d x + c\right )^{5} + 18 \, a^{2} b^{2} \sin \left (d x + c\right )^{5} - 12 \, a b^{3} \sin \left (d x + c\right )^{5} + 3 \, b^{4} \sin \left (d x + c\right )^{5} - 10 \, a^{4} \sin \left (d x + c\right )^{3} + 45 \, a^{3} b \sin \left (d x + c\right )^{3} - 75 \, a^{2} b^{2} \sin \left (d x + c\right )^{3} + 55 \, a b^{3} \sin \left (d x + c\right )^{3} - 15 \, b^{4} \sin \left (d x + c\right )^{3} + 15 \, a^{4} \sin \left (d x + c\right ) - 75 \, a^{3} b \sin \left (d x + c\right ) + 150 \, a^{2} b^{2} \sin \left (d x + c\right ) - 135 \, a b^{3} \sin \left (d x + c\right ) + 45 \, b^{4} \sin \left (d x + c\right )}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}}}{15 \, d} \]

[In]

integrate(cos(d*x+c)^5/(a+b*tan(d*x+c)^2),x, algorithm="giac")

[Out]

-1/15*(15*b^3*arctan(-(a*sin(d*x + c) - b*sin(d*x + c))/sqrt(-a^2 + a*b))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqr
t(-a^2 + a*b)) - (3*a^4*sin(d*x + c)^5 - 12*a^3*b*sin(d*x + c)^5 + 18*a^2*b^2*sin(d*x + c)^5 - 12*a*b^3*sin(d*
x + c)^5 + 3*b^4*sin(d*x + c)^5 - 10*a^4*sin(d*x + c)^3 + 45*a^3*b*sin(d*x + c)^3 - 75*a^2*b^2*sin(d*x + c)^3
+ 55*a*b^3*sin(d*x + c)^3 - 15*b^4*sin(d*x + c)^3 + 15*a^4*sin(d*x + c) - 75*a^3*b*sin(d*x + c) + 150*a^2*b^2*
sin(d*x + c) - 135*a*b^3*sin(d*x + c) + 45*b^4*sin(d*x + c))/(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^
4 - b^5))/d

Mupad [B] (verification not implemented)

Time = 15.39 (sec) , antiderivative size = 1493, normalized size of antiderivative = 11.85 \[ \int \frac {\cos ^5(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\text {Too large to display} \]

[In]

int(cos(c + d*x)^5/(a + b*tan(c + d*x)^2),x)

[Out]

((2*tan(c/2 + (d*x)/2)*(a^2 - 3*a*b + 3*b^2))/(3*a*b^2 - 3*a^2*b + a^3 - b^3) + (tan(c/2 + (d*x)/2)^9*(2*a^2 -
6*a*b + 6*b^2))/(3*a*b^2 - 3*a^2*b + a^3 - b^3) + (tan(c/2 + (d*x)/2)^3*((8*a^2)/3 - (32*a*b)/3 + 16*b^2))/(3
*a*b^2 - 3*a^2*b + a^3 - b^3) + (tan(c/2 + (d*x)/2)^7*((8*a^2)/3 - (32*a*b)/3 + 16*b^2))/(3*a*b^2 - 3*a^2*b +
a^3 - b^3) + (tan(c/2 + (d*x)/2)^5*((116*a^2)/15 - (332*a*b)/15 + (132*b^2)/5))/(3*a*b^2 - 3*a^2*b + a^3 - b^3
))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + t
an(c/2 + (d*x)/2)^10 + 1)) - (b^3*atan(((b^3*((tan(c/2 + (d*x)/2)*(16*a*b^10 - 96*a^2*b^9 + 240*a^3*b^8 - 320*
a^4*b^7 + 240*a^5*b^6 - 96*a^6*b^5 + 16*a^7*b^4))/2 + (b^3*(tan(c/2 + (d*x)/2)^2*(4*a^12 - 44*a^11*b + 8*a^2*b
^10 - 76*a^3*b^9 + 324*a^4*b^8 - 816*a^5*b^7 + 1344*a^6*b^6 - 1512*a^7*b^5 + 1176*a^8*b^4 - 624*a^9*b^3 + 216*
a^10*b^2) + 36*a^11*b - 4*a^12 + 4*a^3*b^9 - 36*a^4*b^8 + 144*a^5*b^7 - 336*a^6*b^6 + 504*a^7*b^5 - 504*a^8*b^
4 + 336*a^9*b^3 - 144*a^10*b^2))/(a^(1/2)*(a - b)^(7/2)))*1i)/(a^(1/2)*(a - b)^(7/2)) + (b^3*((tan(c/2 + (d*x)
/2)*(16*a*b^10 - 96*a^2*b^9 + 240*a^3*b^8 - 320*a^4*b^7 + 240*a^5*b^6 - 96*a^6*b^5 + 16*a^7*b^4))/2 - (b^3*(ta
n(c/2 + (d*x)/2)^2*(4*a^12 - 44*a^11*b + 8*a^2*b^10 - 76*a^3*b^9 + 324*a^4*b^8 - 816*a^5*b^7 + 1344*a^6*b^6 -
1512*a^7*b^5 + 1176*a^8*b^4 - 624*a^9*b^3 + 216*a^10*b^2) + 36*a^11*b - 4*a^12 + 4*a^3*b^9 - 36*a^4*b^8 + 144*
a^5*b^7 - 336*a^6*b^6 + 504*a^7*b^5 - 504*a^8*b^4 + 336*a^9*b^3 - 144*a^10*b^2))/(a^(1/2)*(a - b)^(7/2)))*1i)/
(a^(1/2)*(a - b)^(7/2)))/(tan(c/2 + (d*x)/2)^2*(8*a*b^9 - 24*a^2*b^8 + 24*a^3*b^7 - 8*a^4*b^6) - 8*a*b^9 + 24*
a^2*b^8 - 24*a^3*b^7 + 8*a^4*b^6 + (b^3*((tan(c/2 + (d*x)/2)*(16*a*b^10 - 96*a^2*b^9 + 240*a^3*b^8 - 320*a^4*b
^7 + 240*a^5*b^6 - 96*a^6*b^5 + 16*a^7*b^4))/2 + (b^3*(tan(c/2 + (d*x)/2)^2*(4*a^12 - 44*a^11*b + 8*a^2*b^10 -
76*a^3*b^9 + 324*a^4*b^8 - 816*a^5*b^7 + 1344*a^6*b^6 - 1512*a^7*b^5 + 1176*a^8*b^4 - 624*a^9*b^3 + 216*a^10*
b^2) + 36*a^11*b - 4*a^12 + 4*a^3*b^9 - 36*a^4*b^8 + 144*a^5*b^7 - 336*a^6*b^6 + 504*a^7*b^5 - 504*a^8*b^4 + 3
36*a^9*b^3 - 144*a^10*b^2))/(a^(1/2)*(a - b)^(7/2))))/(a^(1/2)*(a - b)^(7/2)) - (b^3*((tan(c/2 + (d*x)/2)*(16*
a*b^10 - 96*a^2*b^9 + 240*a^3*b^8 - 320*a^4*b^7 + 240*a^5*b^6 - 96*a^6*b^5 + 16*a^7*b^4))/2 - (b^3*(tan(c/2 +
(d*x)/2)^2*(4*a^12 - 44*a^11*b + 8*a^2*b^10 - 76*a^3*b^9 + 324*a^4*b^8 - 816*a^5*b^7 + 1344*a^6*b^6 - 1512*a^7
*b^5 + 1176*a^8*b^4 - 624*a^9*b^3 + 216*a^10*b^2) + 36*a^11*b - 4*a^12 + 4*a^3*b^9 - 36*a^4*b^8 + 144*a^5*b^7
- 336*a^6*b^6 + 504*a^7*b^5 - 504*a^8*b^4 + 336*a^9*b^3 - 144*a^10*b^2))/(a^(1/2)*(a - b)^(7/2))))/(a^(1/2)*(a
- b)^(7/2))))*1i)/(a^(1/2)*d*(a - b)^(7/2))