3.831 \(\int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{\sqrt{a+i a \tan (e+f x)}} \, dx\)
Optimal. Leaf size=169 \[ \frac{2 c^{3/2} (-2 B+i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{c (-2 B+i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{(-B+i A) (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}} \]
[Out]
(2*(I*A - 2*B)*c^(3/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(Sqr
t[a]*f) + ((I*A - 2*B)*c*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(a*f) + ((I*A - B)*(c - I*c*Ta
n[e + f*x])^(3/2))/(f*Sqrt[a + I*a*Tan[e + f*x]])
________________________________________________________________________________________
Rubi [A] time = 0.26395, antiderivative size = 169, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used =
{3588, 78, 50, 63, 217, 203} \[ \frac{2 c^{3/2} (-2 B+i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{c (-2 B+i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{(-B+i A) (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2))/Sqrt[a + I*a*Tan[e + f*x]],x]
[Out]
(2*(I*A - 2*B)*c^(3/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(Sqr
t[a]*f) + ((I*A - 2*B)*c*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(a*f) + ((I*A - B)*(c - I*c*Ta
n[e + f*x])^(3/2))/(f*Sqrt[a + I*a*Tan[e + f*x]])
Rule 3588
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(a*c)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
Rule 78
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ
[p, n]))))
Rule 50
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]
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 217
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 0]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{\sqrt{a+i a \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) \sqrt{c-i c x}}{(a+i a x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(i A-B) (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{((A+2 i B) c) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{\sqrt{a+i a x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(i A-2 B) c \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{(i A-B) (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\left ((A+2 i B) c^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(i A-2 B) c \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{(i A-B) (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}+\frac{\left (2 (i A-2 B) c^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 c-\frac{c x^2}{a}}} \, dx,x,\sqrt{a+i a \tan (e+f x)}\right )}{a f}\\ &=\frac{(i A-2 B) c \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{(i A-B) (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}+\frac{\left (2 (i A-2 B) c^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{c x^2}{a}} \, dx,x,\frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c-i c \tan (e+f x)}}\right )}{a f}\\ &=\frac{2 (i A-2 B) c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{(i A-2 B) c \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{(i A-B) (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 6.35624, size = 161, normalized size = 0.95 \[ \frac{c^2 (\cos (f x)+i \sin (f x)) (\sin (f x)+i \cos (f x)) (A+B \tan (e+f x)) \left (\cos (e+f x) (\tan (e+f x)+i) (-2 i A+i B \tan (e+f x)+3 B)+2 (A+2 i B) \tan ^{-1}(\cos (e+f x)+i \sin (e+f x))\right )}{f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)} (A \cos (e+f x)+B \sin (e+f x))} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2))/Sqrt[a + I*a*Tan[e + f*x]],x]
[Out]
(c^2*(Cos[f*x] + I*Sin[f*x])*(I*Cos[f*x] + Sin[f*x])*(A + B*Tan[e + f*x])*(2*(A + (2*I)*B)*ArcTan[Cos[e + f*x]
+ I*Sin[e + f*x]] + Cos[e + f*x]*(I + Tan[e + f*x])*((-2*I)*A + 3*B + I*B*Tan[e + f*x])))/(f*(A*Cos[e + f*x]
+ B*Sin[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])
________________________________________________________________________________________
Maple [B] time = 0.219, size = 499, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(1/2),x)
[Out]
1/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a*c*(-2*I*B*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^
2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*tan(f*x+e)^2*a*c+2*I*A*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*
c)^(1/2))/(a*c)^(1/2))*tan(f*x+e)*a*c-A*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/
2))*tan(f*x+e)^2*a*c+2*I*B*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c+4*I*B
*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)-4*B*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)
^(1/2))/(a*c)^(1/2))*tan(f*x+e)*a*c-B*tan(f*x+e)^2*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-2*I*A*(a*c*(1+tan(
f*x+e)^2))^(1/2)*(a*c)^(1/2)+A*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c+2
*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+3*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c*(1+t
an(f*x+e)^2))^(1/2)/(-tan(f*x+e)+I)^2/(a*c)^(1/2)
________________________________________________________________________________________
Maxima [B] time = 2.75616, size = 1229, normalized size = 7.27 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
((4*A + 8*I*B)*c*cos(2*f*x + 2*e) + 4*(I*A - 2*B)*c*sin(2*f*x + 2*e) + (4*A + 4*I*B)*c + ((2*A + 4*I*B)*c*cos(
3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (2*A + 4*I*B)*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) + 2*(I*A - 2*B)*c*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*(I*A - 2*B)*c*sin(1/2*arc
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1
/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + ((2*A + 4*I*B)*c*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(
2*f*x + 2*e))) + (2*A + 4*I*B)*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*(I*A - 2*B)*c*sin(3/
2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*(I*A - 2*B)*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x +
2*e))))*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(sin(2*f*x + 2*e), cos(
2*f*x + 2*e))) + 1) - ((-I*A + 2*B)*c*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (-I*A + 2*B)*c*co
s(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (A + 2*I*B)*c*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) + (A + 2*I*B)*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*log(cos(1/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - ((I*A - 2*B)*c*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
(I*A - 2*B)*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (A + 2*I*B)*c*sin(3/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e))) - (A + 2*I*B)*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*log(cos(1/2*ar
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(
1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1))*sqrt(a)*sqrt(c)/((-2*I*a*cos(3/2*arctan2(sin(2*f*x + 2*
e), cos(2*f*x + 2*e))) - 2*I*a*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*a*sin(3/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*a*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*f)
________________________________________________________________________________________
Fricas [B] time = 1.68285, size = 1308, normalized size = 7.74 \begin{align*} -\frac{{\left (a \sqrt{\frac{{\left (4 \, A^{2} + 16 i \, A B - 16 \, B^{2}\right )} c^{3}}{a f^{2}}} f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac{2 \,{\left ({\left (2 i \, A - 4 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (2 i \, A - 4 \, B\right )} c\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} +{\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} - a f\right )} \sqrt{\frac{{\left (4 \, A^{2} + 16 i \, A B - 16 \, B^{2}\right )} c^{3}}{a f^{2}}}}{{\left (-4 i \, A + 8 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-4 i \, A + 8 \, B\right )} c}\right ) - a \sqrt{\frac{{\left (4 \, A^{2} + 16 i \, A B - 16 \, B^{2}\right )} c^{3}}{a f^{2}}} f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac{2 \,{\left ({\left (2 i \, A - 4 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (2 i \, A - 4 \, B\right )} c\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} -{\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} - a f\right )} \sqrt{\frac{{\left (4 \, A^{2} + 16 i \, A B - 16 \, B^{2}\right )} c^{3}}{a f^{2}}}}{{\left (-4 i \, A + 8 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-4 i \, A + 8 \, B\right )} c}\right ) - 2 \,{\left ({\left (-4 i \, A + 6 \, B\right )} c e^{\left (3 i \, f x + 3 i \, e\right )} +{\left (4 i \, A - 8 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-4 i \, A + 6 \, B\right )} c e^{\left (i \, f x + i \, e\right )} +{\left (4 i \, A - 4 \, B\right )} c\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
-1/4*(a*sqrt((4*A^2 + 16*I*A*B - 16*B^2)*c^3/(a*f^2))*f*e^(2*I*f*x + 2*I*e)*log((2*((2*I*A - 4*B)*c*e^(2*I*f*x
+ 2*I*e) + (2*I*A - 4*B)*c)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*
e) + (a*f*e^(2*I*f*x + 2*I*e) - a*f)*sqrt((4*A^2 + 16*I*A*B - 16*B^2)*c^3/(a*f^2)))/((-4*I*A + 8*B)*c*e^(2*I*f
*x + 2*I*e) + (-4*I*A + 8*B)*c)) - a*sqrt((4*A^2 + 16*I*A*B - 16*B^2)*c^3/(a*f^2))*f*e^(2*I*f*x + 2*I*e)*log((
2*((2*I*A - 4*B)*c*e^(2*I*f*x + 2*I*e) + (2*I*A - 4*B)*c)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x
+ 2*I*e) + 1))*e^(I*f*x + I*e) - (a*f*e^(2*I*f*x + 2*I*e) - a*f)*sqrt((4*A^2 + 16*I*A*B - 16*B^2)*c^3/(a*f^2)
))/((-4*I*A + 8*B)*c*e^(2*I*f*x + 2*I*e) + (-4*I*A + 8*B)*c)) - 2*((-4*I*A + 6*B)*c*e^(3*I*f*x + 3*I*e) + (4*I
*A - 8*B)*c*e^(2*I*f*x + 2*I*e) + (-4*I*A + 6*B)*c*e^(I*f*x + I*e) + (4*I*A - 4*B)*c)*sqrt(a/(e^(2*I*f*x + 2*I
*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e))*e^(-2*I*f*x - 2*I*e)/(a*f)
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e))**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (f x + e\right ) + A\right )}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{\sqrt{i \, a \tan \left (f x + e\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate((B*tan(f*x + e) + A)*(-I*c*tan(f*x + e) + c)^(3/2)/sqrt(I*a*tan(f*x + e) + a), x)