3.12.58 \(\int \frac {1}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx\) [1158]
Optimal. Leaf size=174 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {c-i d} f}-\frac {\sqrt {c+d \tan (e+f x)}}{(i c+d) f \sqrt {a+i a \tan (e+f x)}}+\frac {2 d \sqrt {c+d \tan (e+f x)}}{\left (c^2+d^2\right ) f \sqrt {a+i a \tan (e+f x)}} \]
[Out]
-1/2*I*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2)/(a+I*a*tan(f*x+e))^(1/2))/f*2^(1/2)/a^(1/2
)/(c-I*d)^(1/2)-(c+d*tan(f*x+e))^(1/2)/(I*c+d)/f/(a+I*a*tan(f*x+e))^(1/2)+2*d*(c+d*tan(f*x+e))^(1/2)/(c^2+d^2)
/f/(a+I*a*tan(f*x+e))^(1/2)
________________________________________________________________________________________
Rubi [A]
time = 0.21, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3629, 3627,
3625, 214} \begin {gather*} \frac {2 d \sqrt {c+d \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {a+i a \tan (e+f x)}}-\frac {\sqrt {c+d \tan (e+f x)}}{f (d+i c) \sqrt {a+i a \tan (e+f x)}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} f \sqrt {c-i d}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[1/(Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]),x]
[Out]
((-I)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[2]
*Sqrt[a]*Sqrt[c - I*d]*f) - Sqrt[c + d*Tan[e + f*x]]/((I*c + d)*f*Sqrt[a + I*a*Tan[e + f*x]]) + (2*d*Sqrt[c +
d*Tan[e + f*x]])/((c^2 + d^2)*f*Sqrt[a + I*a*Tan[e + f*x]])
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 3625
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
-2*a*(b/f), Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*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]
Rule 3627
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*b*f*m)), x] - Dist[(a*c - b*d)/(2*b^2), Int[(a + b*Tan[e
+ f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), 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] && EqQ[m + n, 0] && LeQ[m, -2^(-1)]
Rule 3629
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-d)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*m*(c^2 + d^2))), x] + Dist[a/(a*c - b*d), Int[(
a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c -
a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n + 1, 0] && !LtQ[m, -1]
Rubi steps
\begin {align*} \int \frac {1}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx &=\frac {2 d \sqrt {c+d \tan (e+f x)}}{\left (c^2+d^2\right ) f \sqrt {a+i a \tan (e+f x)}}+\frac {\int \frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx}{c-i d}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{(i c+d) f \sqrt {a+i a \tan (e+f x)}}+\frac {2 d \sqrt {c+d \tan (e+f x)}}{\left (c^2+d^2\right ) f \sqrt {a+i a \tan (e+f x)}}+\frac {\int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 a}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{(i c+d) f \sqrt {a+i a \tan (e+f x)}}+\frac {2 d \sqrt {c+d \tan (e+f x)}}{\left (c^2+d^2\right ) f \sqrt {a+i a \tan (e+f x)}}-\frac {(i a) \text {Subst}\left (\int \frac {1}{a c-i a d-2 a^2 x^2} \, dx,x,\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}}\right )}{f}\\ &=-\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {c-i d} f}-\frac {\sqrt {c+d \tan (e+f x)}}{(i c+d) f \sqrt {a+i a \tan (e+f x)}}+\frac {2 d \sqrt {c+d \tan (e+f x)}}{\left (c^2+d^2\right ) f \sqrt {a+i a \tan (e+f x)}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A]
time = 3.42, size = 195, normalized size = 1.12 \begin {gather*} \frac {\frac {2 (-i c+d) e^{i (e+f x)} \log \left (2 \left (\sqrt {c-i d} e^{i (e+f x)}+\sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{\sqrt {1+e^{2 i (e+f x)}}}+2 i \sqrt {c-i d} \sqrt {c+d \tan (e+f x)}}{2 \sqrt {c-i d} (c+i d) f \sqrt {a+i a \tan (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[1/(Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]),x]
[Out]
((2*((-I)*c + d)*E^(I*(e + f*x))*Log[2*(Sqrt[c - I*d]*E^(I*(e + f*x)) + Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[c -
(I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))])])/Sqrt[1 + E^((2*I)*(e + f*x))] + (2*I)*Sqrt[c -
I*d]*Sqrt[c + d*Tan[e + f*x]])/(2*Sqrt[c - I*d]*(c + I*d)*f*Sqrt[a + I*a*Tan[e + f*x]])
________________________________________________________________________________________
Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 1737 vs. \(2 (143 ) = 286\).
time = 0.91, size = 1738, normalized size = 9.99
| | |
| method |
result |
size |
| | |
| derivativedivides |
\(\text {Expression too large to display}\) |
\(1738\) |
| default |
\(\text {Expression too large to display}\) |
\(1738\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
[Out]
1/4/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)/a*(-I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e
)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c
))^(1/2)*d^3-4*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^3+2*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*ta
n(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a
*(I*d-c))^(1/2)*c^3*tan(f*x+e)-4*I*c*d^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)-ln((3*a*c+I*a*
tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/
(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^3*tan(f*x+e)^2+3*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e
)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c
))^(1/2)*c*d^2*tan(f*x+e)^2-3*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)
*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*d*tan(f*x+e)^2+3*
I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(
f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*d-6*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(
f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(
I*d-c))^(1/2)*c^2*d*tan(f*x+e)+2*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2
)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*d^3*tan(f*x+e)-4*I*(
a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^2*d-6*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/
2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*
c*d^2*tan(f*x+e)+ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*
x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^3-3*ln((3*a*c+I*a*tan(f*x+e)*c-I*a
*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*
2^(1/2)*(-a*(I*d-c))^(1/2)*c*d^2-4*I*c^3*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)+I*ln((3*a*c+I*
a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)
)/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*d^3*tan(f*x+e)^2+4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^
2*d*tan(f*x+e)+4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^3*tan(f*x+e)-4*c^3*(a*(c+d*tan(f*x+e))*(1+I*tan
(f*x+e)))^(1/2)-4*c*d^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/
2)/(I*d-c)/(c+I*d)^2/(I*c-d)/(-tan(f*x+e)+I)^2
________________________________________________________________________________________
Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i
________________________________________________________________________________________
Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 398 vs. \(2 (142) = 284\).
time = 1.48, size = 398, normalized size = 2.29 \begin {gather*} -\frac {{\left ({\left (-i \, a c + a d\right )} f \sqrt {-\frac {2 i}{{\left (i \, a c + a d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left ({\left (i \, a c + a d\right )} f \sqrt {-\frac {2 i}{{\left (i \, a c + a d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) + {\left (i \, a c - a d\right )} f \sqrt {-\frac {2 i}{{\left (i \, a c + a d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left ({\left (-i \, a c - a d\right )} f \sqrt {-\frac {2 i}{{\left (i \, a c + a d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) + 2 \, \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, {\left (i \, a c - a d\right )} f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
-1/4*((-I*a*c + a*d)*f*sqrt(-2*I/((I*a*c + a*d)*f^2))*e^(I*f*x + I*e)*log((I*a*c + a*d)*f*sqrt(-2*I/((I*a*c +
a*d)*f^2))*e^(I*f*x + I*e) + sqrt(2)*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(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) + 1)) + (I*a*c - a*d)*f*sqrt(-2*I/((I*a*c + a*d)*f^2))
*e^(I*f*x + I*e)*log((-I*a*c - a*d)*f*sqrt(-2*I/((I*a*c + a*d)*f^2))*e^(I*f*x + I*e) + sqrt(2)*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(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I
*e) + 1)) + 2*sqrt(2)*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(a/(e^(2*I
*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) + 1))*e^(-I*f*x - I*e)/((I*a*c - a*d)*f)
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+I*a*tan(f*x+e))**(1/2)/(c+d*tan(f*x+e))**(1/2),x)
[Out]
Integral(1/(sqrt(I*a*(tan(e + f*x) - I))*sqrt(c + d*tan(e + f*x))), x)
________________________________________________________________________________________
Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regu
lar value [
________________________________________________________________________________________
Mupad [B]
time = 19.62, size = 1508, normalized size = 8.67 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/((a + a*tan(e + f*x)*1i)^(1/2)*(c + d*tan(e + f*x))^(1/2)),x)
[Out]
(2*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(d*f*((c + d*tan(e + f*x))^(1/2) - c^(1/2))*((a*1i)/d + ((a + a*
tan(e + f*x)*1i)^(1/2) - a^(1/2))^2/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2 - (a^(1/2)*c^(1/2)*((a + a*tan(e
+ f*x)*1i)^(1/2) - a^(1/2))*2i)/(d*((c + d*tan(e + f*x))^(1/2) - c^(1/2))))) - (2^(1/2)*atan(((2^(1/2)*((4*d^7
*f*(a*c - a*d*1i)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/((c + d*tan(e + f*x))^(1/2) - c^(1/2)) - 4*a^(3/2
)*c^(1/2)*d^7*f + (2^(1/2)*(d^7*(a^2*c*f^2*1i - a^2*d*f^2) + (d^8*(5*a*c*f^2 - a*d*f^2*3i)*((a + a*tan(e + f*x
)*1i)^(1/2) - a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2 - (d^7*f*(a^(3/2)*c^(3/2)*f*2i + 6*a^(3/2)*
c^(1/2)*d*f)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/((c + d*tan(e + f*x))^(1/2) - c^(1/2))))/(a^(1/2)*f*(d
*1i - c)^(1/2)) + (a^(1/2)*c^(1/2)*d^8*f*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2*4i)/((c + d*tan(e + f*x))
^(1/2) - c^(1/2))^2)*1i)/(a^(1/2)*f*(d*1i - c)^(1/2)) - (2^(1/2)*(4*a^(3/2)*c^(1/2)*d^7*f - (4*d^7*f*(a*c - a*
d*1i)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/((c + d*tan(e + f*x))^(1/2) - c^(1/2)) + (2^(1/2)*(d^7*(a^2*c
*f^2*1i - a^2*d*f^2) + (d^8*(5*a*c*f^2 - a*d*f^2*3i)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2)/((c + d*tan(
e + f*x))^(1/2) - c^(1/2))^2 - (d^7*f*(a^(3/2)*c^(3/2)*f*2i + 6*a^(3/2)*c^(1/2)*d*f)*((a + a*tan(e + f*x)*1i)^
(1/2) - a^(1/2)))/((c + d*tan(e + f*x))^(1/2) - c^(1/2))))/(a^(1/2)*f*(d*1i - c)^(1/2)) - (a^(1/2)*c^(1/2)*d^8
*f*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2*4i)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2)*1i)/(a^(1/2)*f*(d
*1i - c)^(1/2)))/((8*d^8*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2
- a*d^7*8i + (2^(1/2)*((4*d^7*f*(a*c - a*d*1i)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/((c + d*tan(e + f*x
))^(1/2) - c^(1/2)) - 4*a^(3/2)*c^(1/2)*d^7*f + (2^(1/2)*(d^7*(a^2*c*f^2*1i - a^2*d*f^2) + (d^8*(5*a*c*f^2 - a
*d*f^2*3i)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2 - (d^7*f*(a^(
3/2)*c^(3/2)*f*2i + 6*a^(3/2)*c^(1/2)*d*f)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/((c + d*tan(e + f*x))^(1
/2) - c^(1/2))))/(a^(1/2)*f*(d*1i - c)^(1/2)) + (a^(1/2)*c^(1/2)*d^8*f*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2
))^2*4i)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2))/(a^(1/2)*f*(d*1i - c)^(1/2)) + (2^(1/2)*(4*a^(3/2)*c^(1/2)
*d^7*f - (4*d^7*f*(a*c - a*d*1i)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/((c + d*tan(e + f*x))^(1/2) - c^(1
/2)) + (2^(1/2)*(d^7*(a^2*c*f^2*1i - a^2*d*f^2) + (d^8*(5*a*c*f^2 - a*d*f^2*3i)*((a + a*tan(e + f*x)*1i)^(1/2)
- a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2 - (d^7*f*(a^(3/2)*c^(3/2)*f*2i + 6*a^(3/2)*c^(1/2)*d*f
)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/((c + d*tan(e + f*x))^(1/2) - c^(1/2))))/(a^(1/2)*f*(d*1i - c)^(1
/2)) - (a^(1/2)*c^(1/2)*d^8*f*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2*4i)/((c + d*tan(e + f*x))^(1/2) - c^
(1/2))^2))/(a^(1/2)*f*(d*1i - c)^(1/2))))*1i)/(2*a^(1/2)*f*(d*1i - c)^(1/2))
________________________________________________________________________________________