3.1116 \(\int \frac{(c+d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx\)
Optimal. Leaf size=185 \[ \frac{(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}-\frac{d (c+5 i d) \sqrt{c+d \tan (e+f x)}}{2 a f}-\frac{i (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{2 a f}+\frac{(c+i d)^{3/2} (4 d+i c) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{2 a f} \]
[Out]
((-I/2)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a*f) + ((c + I*d)^(3/2)*(I*c + 4*d)*
ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(2*a*f) - ((c + (5*I)*d)*d*Sqrt[c + d*Tan[e + f*x]])/(2*a*f)
+ ((I*c - d)*(c + d*Tan[e + f*x])^(3/2))/(2*f*(a + I*a*Tan[e + f*x]))
________________________________________________________________________________________
Rubi [A] time = 0.425263, antiderivative size = 185, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{3550, 3528, 3539, 3537, 63, 208} \[ \frac{(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}-\frac{d (c+5 i d) \sqrt{c+d \tan (e+f x)}}{2 a f}-\frac{i (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{2 a f}+\frac{(c+i d)^{3/2} (4 d+i c) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{2 a f} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x]),x]
[Out]
((-I/2)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a*f) + ((c + I*d)^(3/2)*(I*c + 4*d)*
ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(2*a*f) - ((c + (5*I)*d)*d*Sqrt[c + d*Tan[e + f*x]])/(2*a*f)
+ ((I*c - d)*(c + d*Tan[e + f*x])^(3/2))/(2*f*(a + I*a*Tan[e + f*x]))
Rule 3550
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b
*c - a*d)*(c + d*Tan[e + f*x])^(n - 1))/(2*a*f*(a + b*Tan[e + f*x])), x] + Dist[1/(2*a^2), Int[(c + d*Tan[e +
f*x])^(n - 2)*Simp[a*c^2 + a*d^2*(n - 1) - b*c*d*n - d*(a*c*(n - 2) + b*d*n)*Tan[e + f*x], x], x], x] /; FreeQ
[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1]
Rule 3528
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
Rule 3539
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Rule 3537
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{(c+d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx &=\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}+\frac{\int \sqrt{c+d \tan (e+f x)} \left (\frac{1}{2} a (2 c+i d) (c-3 i d)-\frac{1}{2} a (c+5 i d) d \tan (e+f x)\right ) \, dx}{2 a^2}\\ &=-\frac{(c+5 i d) d \sqrt{c+d \tan (e+f x)}}{2 a f}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}+\frac{\int \frac{\frac{1}{2} a \left (2 c^3-5 i c^2 d+4 c d^2+5 i d^3\right )+\frac{1}{2} a d \left (c^2-10 i c d+3 d^2\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 a^2}\\ &=-\frac{(c+5 i d) d \sqrt{c+d \tan (e+f x)}}{2 a f}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}+\frac{(c-i d)^3 \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{4 a}+\frac{\left ((c+i d)^2 (c-4 i d)\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{4 a}\\ &=-\frac{(c+5 i d) d \sqrt{c+d \tan (e+f x)}}{2 a f}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}-\frac{(i c+d)^3 \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{4 a f}-\frac{\left ((c+i d)^2 (i c+4 d)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{4 a f}\\ &=-\frac{(c+5 i d) d \sqrt{c+d \tan (e+f x)}}{2 a f}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}-\frac{(c-i d)^3 \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i c}{d}+\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{2 a d f}-\frac{\left ((c+i d)^2 (c-4 i d)\right ) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i c}{d}-\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{2 a d f}\\ &=-\frac{i (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{2 a f}+\frac{(c+i d)^{3/2} (i c+4 d) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{2 a f}-\frac{(c+5 i d) d \sqrt{c+d \tan (e+f x)}}{2 a f}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}\\ \end{align*}
Mathematica [A] time = 1.80597, size = 260, normalized size = 1.41 \[ \frac{\sec (e+f x) (\cos (f x)+i \sin (f x)) \left (2 (\sin (f x)+i \cos (f x)) \sqrt{c+d \tan (e+f x)} \left (\left (c^2+2 i c d-5 d^2\right ) \cos (e+f x)-4 i d^2 \sin (e+f x)\right )+\frac{2 (\cos (e)+i \sin (e)) \left (-\sqrt{-c-i d} (d+i c)^3 \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c+i d}}\right )-i \sqrt{-c+i d} (c+i d)^2 (c-4 i d) \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c-i d}}\right )\right )}{\sqrt{-c-i d} \sqrt{-c+i d}}\right )}{4 f (a+i a \tan (e+f x))} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x]),x]
[Out]
(Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])*((2*((-I)*Sqrt[-c + I*d]*(c + I*d)^2*(c - (4*I)*d)*ArcTan[Sqrt[c + d*Tan
[e + f*x]]/Sqrt[-c - I*d]] - Sqrt[-c - I*d]*(I*c + d)^3*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-c + I*d]])*(Cos[
e] + I*Sin[e]))/(Sqrt[-c - I*d]*Sqrt[-c + I*d]) + 2*(I*Cos[f*x] + Sin[f*x])*((c^2 + (2*I)*c*d - 5*d^2)*Cos[e +
f*x] - (4*I)*d^2*Sin[e + f*x])*Sqrt[c + d*Tan[e + f*x]]))/(4*f*(a + I*a*Tan[e + f*x]))
________________________________________________________________________________________
Maple [B] time = 0.056, size = 674, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e)),x)
[Out]
-2*I/f/a*d^2*(c+d*tan(f*x+e))^(1/2)+1/2*I/f/a/(I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))*c^3-3
/2*I/f/a*d^2/(I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))*c+3/2/f/a*d/(I*d-c)^(1/2)*arctan((c+d*
tan(f*x+e))^(1/2)/(I*d-c)^(1/2))*c^2-1/2/f/a*d^3/(I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))+3/
2*I/f/a*d^2/(c+I*d)*(c+d*tan(f*x+e))^(1/2)/(-I*d+d*tan(f*x+e))*c^2-1/2*I/f/a*d^4/(c+I*d)*(c+d*tan(f*x+e))^(1/2
)/(-I*d+d*tan(f*x+e))+1/2/f/a*d/(c+I*d)*(c+d*tan(f*x+e))^(1/2)/(-I*d+d*tan(f*x+e))*c^3-3/2/f/a*d^3/(c+I*d)*(c+
d*tan(f*x+e))^(1/2)/(-I*d+d*tan(f*x+e))*c-1/2*I/f/a/(c+I*d)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d
-c)^(1/2))*c^4-9/2*I/f/a*d^2/(c+I*d)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))*c^2+2*I/f/a*
d^4/(c+I*d)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))-1/2/f/a*d/(c+I*d)/(-I*d-c)^(1/2)*arct
an((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))*c^3+11/2/f/a*d^3/(c+I*d)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2
)/(-I*d-c)^(1/2))*c
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e)),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
________________________________________________________________________________________
Fricas [B] time = 4.70129, size = 2619, normalized size = 14.16 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e)),x, algorithm="fricas")
[Out]
-1/8*(a*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^2*f^2))*e^(2*I*f*x + 2*I*e)
*log(1/2*(4*c^3 - 8*I*c^2*d - 4*c*d^2 - (4*I*a*f*e^(2*I*f*x + 2*I*e) + 4*I*a*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2
*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d
^5)/(a^2*f^2)) + 2*(2*c^3 - 6*I*c^2*d - 6*c*d^2 + 2*I*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(c^2 - 2*
I*c*d - d^2)) - a*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^2*f^2))*e^(2*I*f*
x + 2*I*e)*log(1/2*(4*c^3 - 8*I*c^2*d - 4*c*d^2 - (-4*I*a*f*e^(2*I*f*x + 2*I*e) - 4*I*a*f)*sqrt(((c - I*d)*e^(
2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*
c*d^4 - I*d^5)/(a^2*f^2)) + 2*(2*c^3 - 6*I*c^2*d - 6*c*d^2 + 2*I*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e
)/(c^2 - 2*I*c*d - d^2)) - a*f*sqrt(-(c^5 - 5*I*c^4*d + 5*c^3*d^2 - 25*I*c^2*d^3 + 40*c*d^4 + 16*I*d^5)/(a^2*f
^2))*e^(2*I*f*x + 2*I*e)*log(1/2*(I*c^3 + 2*c^2*d + 7*I*c*d^2 - 4*d^3 + (a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt((
(c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(c^5 - 5*I*c^4*d + 5*c^3*d^2 - 25*I*
c^2*d^3 + 40*c*d^4 + 16*I*d^5)/(a^2*f^2)) + (I*c^3 + 3*c^2*d + 4*I*c*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2
*I*e)/(a*f)) + a*f*sqrt(-(c^5 - 5*I*c^4*d + 5*c^3*d^2 - 25*I*c^2*d^3 + 40*c*d^4 + 16*I*d^5)/(a^2*f^2))*e^(2*I*
f*x + 2*I*e)*log(1/2*(I*c^3 + 2*c^2*d + 7*I*c*d^2 - 4*d^3 - (a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt(((c - I*d)*e^
(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(c^5 - 5*I*c^4*d + 5*c^3*d^2 - 25*I*c^2*d^3 + 40
*c*d^4 + 16*I*d^5)/(a^2*f^2)) + (I*c^3 + 3*c^2*d + 4*I*c*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a*f))
- 2*(I*c^2 - 2*c*d - I*d^2 + (I*c^2 - 2*c*d - 9*I*d^2)*e^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*
e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*f)
________________________________________________________________________________________
Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e)),x)
[Out]
Exception raised: AttributeError
________________________________________________________________________________________
Giac [B] time = 1.603, size = 636, normalized size = 3.44 \begin{align*} -\frac{1}{2} \, d^{2}{\left (\frac{4 i \, \sqrt{d \tan \left (f x + e\right ) + c}}{a f} - \frac{\sqrt{d \tan \left (f x + e\right ) + c} c^{2} + 2 i \, \sqrt{d \tan \left (f x + e\right ) + c} c d - \sqrt{d \tan \left (f x + e\right ) + c} d^{2}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )} a d f} - \frac{\sqrt{2}{\left (c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}\right )} \arctan \left (\frac{-16 i \, \sqrt{d \tan \left (f x + e\right ) + c} c - 16 i \, \sqrt{c^{2} + d^{2}} \sqrt{d \tan \left (f x + e\right ) + c}}{8 \, \sqrt{2} \sqrt{c + \sqrt{c^{2} + d^{2}}} c - 8 i \, \sqrt{2} \sqrt{c + \sqrt{c^{2} + d^{2}}} d + 8 \, \sqrt{2} \sqrt{c^{2} + d^{2}} \sqrt{c + \sqrt{c^{2} + d^{2}}}}\right )}{a \sqrt{c + \sqrt{c^{2} + d^{2}}} d^{2} f{\left (-\frac{i \, d}{c + \sqrt{c^{2} + d^{2}}} + 1\right )}} + \frac{4 \,{\left (i \, c^{3} + 2 \, c^{2} d + 7 i \, c d^{2} - 4 \, d^{3}\right )} \arctan \left (\frac{4 \,{\left (\sqrt{d \tan \left (f x + e\right ) + c} c - \sqrt{c^{2} + d^{2}} \sqrt{d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} + i \, \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d - \sqrt{c^{2} + d^{2}} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}}\right )}{a \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d^{2} f{\left (\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e)),x, algorithm="giac")
[Out]
-1/2*d^2*(4*I*sqrt(d*tan(f*x + e) + c)/(a*f) - (sqrt(d*tan(f*x + e) + c)*c^2 + 2*I*sqrt(d*tan(f*x + e) + c)*c*
d - sqrt(d*tan(f*x + e) + c)*d^2)/((d*tan(f*x + e) - I*d)*a*d*f) - sqrt(2)*(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)
*arctan((-16*I*sqrt(d*tan(f*x + e) + c)*c - 16*I*sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(8*sqrt(2)*sqrt(c +
sqrt(c^2 + d^2))*c - 8*I*sqrt(2)*sqrt(c + sqrt(c^2 + d^2))*d + 8*sqrt(2)*sqrt(c^2 + d^2)*sqrt(c + sqrt(c^2 +
d^2))))/(a*sqrt(c + sqrt(c^2 + d^2))*d^2*f*(-I*d/(c + sqrt(c^2 + d^2)) + 1)) + 4*(I*c^3 + 2*c^2*d + 7*I*c*d^2
- 4*d^3)*arctan(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-8*c + 8*sqr
t(c^2 + d^2)) + I*sqrt(-8*c + 8*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-8*c + 8*sqrt(c^2 + d^2))))/(a*sqrt(
-8*c + 8*sqrt(c^2 + d^2))*d^2*f*(I*d/(c - sqrt(c^2 + d^2)) + 1)))