3.1142 \(\int \frac{\sqrt{c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=254 \[ -\frac{\left (20 c d-i \left (15 c^2+3 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{60 a^2 f (c+i d)^2 \sqrt{a+i a \tan (e+f x)}}-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{4 \sqrt{2} a^{5/2} f}+\frac{(-3 d+5 i c) \sqrt{c+d \tan (e+f x)}}{30 a f (c+i d) (a+i a \tan (e+f x))^{3/2}}+\frac{i \sqrt{c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}} \]
[Out]
((-I/4)*Sqrt[c - I*d]*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]*a^(5/2)*f) + ((I/5)*Sqrt[c + d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^(5/2)) + (((5*I)*c -
3*d)*Sqrt[c + d*Tan[e + f*x]])/(30*a*(c + I*d)*f*(a + I*a*Tan[e + f*x])^(3/2)) - ((20*c*d - I*(15*c^2 + 3*d^2)
)*Sqrt[c + d*Tan[e + f*x]])/(60*a^2*(c + I*d)^2*f*Sqrt[a + I*a*Tan[e + f*x]])
________________________________________________________________________________________
Rubi [A] time = 0.789802, antiderivative size = 254, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used =
{3557, 3596, 12, 3544, 208} \[ -\frac{\left (20 c d-i \left (15 c^2+3 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{60 a^2 f (c+i d)^2 \sqrt{a+i a \tan (e+f x)}}-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{4 \sqrt{2} a^{5/2} f}+\frac{(-3 d+5 i c) \sqrt{c+d \tan (e+f x)}}{30 a f (c+i d) (a+i a \tan (e+f x))^{3/2}}+\frac{i \sqrt{c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[c + d*Tan[e + f*x]]/(a + I*a*Tan[e + f*x])^(5/2),x]
[Out]
((-I/4)*Sqrt[c - I*d]*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]*a^(5/2)*f) + ((I/5)*Sqrt[c + d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^(5/2)) + (((5*I)*c -
3*d)*Sqrt[c + d*Tan[e + f*x]])/(30*a*(c + I*d)*f*(a + I*a*Tan[e + f*x])^(3/2)) - ((20*c*d - I*(15*c^2 + 3*d^2)
)*Sqrt[c + d*Tan[e + f*x]])/(60*a^2*(c + I*d)^2*f*Sqrt[a + I*a*Tan[e + f*x]])
Rule 3557
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Sim
p[(b*(a + b*Tan[e + f*x])^m*Sqrt[c + d*Tan[e + f*x]])/(2*a*f*m), x] + Dist[1/(4*a^2*m), Int[((a + b*Tan[e + f*
x])^(m + 1)*Simp[2*a*c*m + b*d + a*d*(2*m + 1)*Tan[e + f*x], x])/Sqrt[c + d*Tan[e + f*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] && LtQ[m, 0] && IntegersQ[2
*m]
Rule 3596
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n,
0]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 3544
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 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{\sqrt{c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx &=\frac{i \sqrt{c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac{\int \frac{-a (5 c-i d)-4 a d \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}} \, dx}{10 a^2}\\ &=\frac{i \sqrt{c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{(5 i c-3 d) \sqrt{c+d \tan (e+f x)}}{30 a (c+i d) f (a+i a \tan (e+f x))^{3/2}}+\frac{\int \frac{-\frac{1}{2} a^2 \left (10 c d-i \left (15 c^2+9 d^2\right )\right )+a^2 (5 i c-3 d) d \tan (e+f x)}{\sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{30 a^4 (i c-d)}\\ &=\frac{i \sqrt{c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{(5 i c-3 d) \sqrt{c+d \tan (e+f x)}}{30 a (c+i d) f (a+i a \tan (e+f x))^{3/2}}+\frac{\left (15 i c^2-20 c d+3 i d^2\right ) \sqrt{c+d \tan (e+f x)}}{60 a^2 (c+i d)^2 f \sqrt{a+i a \tan (e+f x)}}+\frac{\int \frac{15 a^3 (c-i d) (c+i d)^2 \sqrt{a+i a \tan (e+f x)}}{4 \sqrt{c+d \tan (e+f x)}} \, dx}{30 a^6 (c+i d)^2}\\ &=\frac{i \sqrt{c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{(5 i c-3 d) \sqrt{c+d \tan (e+f x)}}{30 a (c+i d) f (a+i a \tan (e+f x))^{3/2}}+\frac{\left (15 i c^2-20 c d+3 i d^2\right ) \sqrt{c+d \tan (e+f x)}}{60 a^2 (c+i d)^2 f \sqrt{a+i a \tan (e+f x)}}+\frac{(c-i d) \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{8 a^3}\\ &=\frac{i \sqrt{c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{(5 i c-3 d) \sqrt{c+d \tan (e+f x)}}{30 a (c+i d) f (a+i a \tan (e+f x))^{3/2}}+\frac{\left (15 i c^2-20 c d+3 i d^2\right ) \sqrt{c+d \tan (e+f x)}}{60 a^2 (c+i d)^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{(i c+d) \operatorname{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 )}{4 a f}\\ &=-\frac{i \sqrt{c-i d} \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 )}{4 \sqrt{2} a^{5/2} f}+\frac{i \sqrt{c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{(5 i c-3 d) \sqrt{c+d \tan (e+f x)}}{30 a (c+i d) f (a+i a \tan (e+f x))^{3/2}}+\frac{\left (15 i c^2-20 c d+3 i d^2\right ) \sqrt{c+d \tan (e+f x)}}{60 a^2 (c+i d)^2 f \sqrt{a+i a \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 5.50837, size = 302, normalized size = 1.19 \[ \frac{\sec ^{\frac{5}{2}}(e+f x) \left (\frac{2 i \sqrt{c+d \tan (e+f x)} \left (\left (26 c^2+40 i c d-6 d^2\right ) \cos (2 (e+f x))+11 c^2+4 i c (5 c+7 i d) \sin (2 (e+f x))+20 i c d-9 d^2\right )}{15 (c+i d)^2 \sqrt{\sec (e+f x)}}-i \sqrt{2} \sqrt{c-i d} e^{2 i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \log \left (2 \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )\right )}{8 f (a+i a \tan (e+f x))^{5/2}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sqrt[c + d*Tan[e + f*x]]/(a + I*a*Tan[e + f*x])^(5/2),x]
[Out]
(Sec[e + f*x]^(5/2)*((-I)*Sqrt[2]*Sqrt[c - I*d]*E^((2*I)*(e + f*x))*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*
x)))]*Sqrt[1 + E^((2*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)))])] + (((2*I)/15)*(11*c^2 + (20*I)*c*d - 9*d^2 +
(26*c^2 + (40*I)*c*d - 6*d^2)*Cos[2*(e + f*x)] + (4*I)*c*(5*c + (7*I)*d)*Sin[2*(e + f*x)])*Sqrt[c + d*Tan[e +
f*x]])/((c + I*d)^2*Sqrt[Sec[e + f*x]])))/(8*f*(a + I*a*Tan[e + f*x])^(5/2))
________________________________________________________________________________________
Maple [B] time = 0.082, size = 2226, normalized size = 8.8 \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))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x)
[Out]
1/240/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^3*(-30*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*ta
n(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+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))/(t
an(f*x+e)+I))*tan(f*x+e)^4*c*d+120*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e
)*d+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))*tan(f*x+e)^3*c*d
-120*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+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))*tan(f*x+e)*c*d-60*2^(1/2)*(-a*(I*d-c))^(1/2)*l
n((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+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))*tan(f*x+e)*c^2+60*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a
*tan(f*x+e)*d+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))*tan(f*
x+e)*d^2-30*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+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))*c*d-80*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x
+e)))^(1/2)*tan(f*x+e)^3*c*d+15*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d
+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))*c^2-15*I*2^(1/2)*(-
a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+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))*d^2+464*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e
)*c*d+60*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+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))*tan(f*x+e)^3*c^2-60*2^(1/2)*(-a*(I*d-c))^(1/
2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+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))*tan(f*x+e)^3*d^2+240*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d-12*(a*(c
+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^3*d^2-148*I*c^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+
60*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^2+308*tan(f*x+e)*c^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^
(1/2)-60*tan(f*x+e)^3*c^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-60*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))
^(1/2)*tan(f*x+e)*d^2+220*I*tan(f*x+e)^2*c^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+12*I*(a*(c+d*tan(f*x+
e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*d^2-304*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*c*d+
180*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+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))*tan(f*x+e)^2*c*d+15*I*2^(1/2)*(-a*(I*d-c))^(1/2)*
ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+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))*tan(f*x+e)^4*c^2-15*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*
d+3*a*tan(f*x+e)*d+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))*t
an(f*x+e)^4*d^2-90*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+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))*tan(f*x+e)^2*c^2+90*I*2^(1/2)*(-
a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+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))*tan(f*x+e)^2*d^2)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(
I*c-d)^2/(-tan(f*x+e)+I)^4
________________________________________________________________________________________
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))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
________________________________________________________________________________________
Fricas [B] time = 1.95099, size = 1494, normalized size = 5.88 \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))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
1/2*(sqrt(1/2)*(30*a^3*c^2 + 60*I*a^3*c*d - 30*a^3*d^2)*f*sqrt(-(c - I*d)/(a^5*f^2))*e^(6*I*f*x + 6*I*e)*log((
2*I*sqrt(1/2)*a^3*f*sqrt(-(c - I*d)/(a^5*f^2))*e^(2*I*f*x + 2*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)*e^(I*f*x
+ I*e))*e^(-I*f*x - I*e)) - sqrt(1/2)*(30*a^3*c^2 + 60*I*a^3*c*d - 30*a^3*d^2)*f*sqrt(-(c - I*d)/(a^5*f^2))*e^
(6*I*f*x + 6*I*e)*log((-2*I*sqrt(1/2)*a^3*f*sqrt(-(c - I*d)/(a^5*f^2))*e^(2*I*f*x + 2*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)*e^(I*f*x + I*e))*e^(-I*f*x - I*e)) - 2*sqrt(2)*(-3*I*c^2 + 6*c*d + 3*I*d^2 + (-23*I*c^2 + 34*c*
d + 3*I*d^2)*e^(6*I*f*x + 6*I*e) + (-34*I*c^2 + 54*c*d + 12*I*d^2)*e^(4*I*f*x + 4*I*e) + (-14*I*c^2 + 26*c*d +
12*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))*sqrt
(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e))*e^(-6*I*f*x - 6*I*e)/((120*a^3*c^2 + 240*I*a^3*c*d - 120*a^3*d^
2)*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((c+d*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**(5/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="giac")
[Out]
Exception raised: TypeError