3.1159 \(\int \frac{1}{(a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}} \, dx\)
Optimal. Leaf size=193 \[ -\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f \sqrt{c-i d}}+\frac{(-7 d+3 i c) \sqrt{c+d \tan (e+f x)}}{6 a f (c+i d)^2 \sqrt{a+i a \tan (e+f x)}}-\frac{\sqrt{c+d \tan (e+f x)}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}} \]
[Out]
((-I/2)*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^(3/2)*Sqrt[c - I*d]*f) - Sqrt[c + d*Tan[e + f*x]]/(3*(I*c - d)*f*(a + I*a*Tan[e + f*x])^(3/2)) + (((3*I)*
c - 7*d)*Sqrt[c + d*Tan[e + f*x]])/(6*a*(c + I*d)^2*f*Sqrt[a + I*a*Tan[e + f*x]])
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Rubi [A] time = 0.467614, antiderivative size = 193, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used =
{3559, 3596, 12, 3544, 208} \[ -\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f \sqrt{c-i d}}+\frac{(-7 d+3 i c) \sqrt{c+d \tan (e+f x)}}{6 a f (c+i d)^2 \sqrt{a+i a \tan (e+f x)}}-\frac{\sqrt{c+d \tan (e+f x)}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[1/((a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]]),x]
[Out]
((-I/2)*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^(3/2)*Sqrt[c - I*d]*f) - Sqrt[c + d*Tan[e + f*x]]/(3*(I*c - d)*f*(a + I*a*Tan[e + f*x])^(3/2)) + (((3*I)*
c - 7*d)*Sqrt[c + d*Tan[e + f*x]])/(6*a*(c + I*d)^2*f*Sqrt[a + I*a*Tan[e + f*x]])
Rule 3559
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 + 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*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan
[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2
+ d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
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{1}{(a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}} \, dx &=-\frac{\sqrt{c+d \tan (e+f x)}}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2}}-\frac{\int \frac{-\frac{1}{2} a (3 i c-5 d)-i a d \tan (e+f x)}{\sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{3 a^2 (i c-d)}\\ &=-\frac{\sqrt{c+d \tan (e+f x)}}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2}}+\frac{(3 i c-7 d) \sqrt{c+d \tan (e+f x)}}{6 a (c+i d)^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{\int -\frac{3 a^2 (c+i d)^2 \sqrt{a+i a \tan (e+f x)}}{4 \sqrt{c+d \tan (e+f x)}} \, dx}{3 a^4 (c+i d)^2}\\ &=-\frac{\sqrt{c+d \tan (e+f x)}}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2}}+\frac{(3 i c-7 d) \sqrt{c+d \tan (e+f x)}}{6 a (c+i d)^2 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}{4 a^2}\\ &=-\frac{\sqrt{c+d \tan (e+f x)}}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2}}+\frac{(3 i c-7 d) \sqrt{c+d \tan (e+f x)}}{6 a (c+i d)^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{i \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 )}{2 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 )}{2 \sqrt{2} a^{3/2} \sqrt{c-i d} f}-\frac{\sqrt{c+d \tan (e+f x)}}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2}}+\frac{(3 i c-7 d) \sqrt{c+d \tan (e+f x)}}{6 a (c+i d)^2 f \sqrt{a+i a \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 3.57417, size = 249, normalized size = 1.29 \[ \frac{\sec ^{\frac{3}{2}}(e+f x) \left (-\frac{i \sqrt{2} \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{3/2} \left (1+e^{2 i (e+f x)}\right )^{3/2} \log \left (2 \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{\sqrt{c-i d}}-\frac{2 ((3 c+7 i d) \tan (e+f x)-5 i c+9 d) \sqrt{c+d \tan (e+f x)}}{3 (c+i d)^2 \sec ^{\frac{3}{2}}(e+f x)}\right )}{4 f (a+i a \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/((a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]]),x]
[Out]
(Sec[e + f*x]^(3/2)*(((-I)*Sqrt[2]*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(3/2)*(1 + E^((2*I)*(e + f*x)))
^(3/2)*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[c - I*d] - (2*((-5*I)*c + 9*d + (3*c + (7*I)*d)*Tan[e + f*x])*Sqrt[
c + d*Tan[e + f*x]])/(3*(c + I*d)^2*Sec[e + f*x]^(3/2))))/(4*f*(a + I*a*Tan[e + f*x])^(3/2))
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Maple [B] time = 0.095, size = 2946, normalized size = 15.3 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2),x)
[Out]
1/24/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^2*(-18*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))/(ta
n(f*x+e)+I))*tan(f*x+e)^3*c^2*d^2+16*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^2*d^2+12*I*tan(f*x+e)^2*c
^4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-28*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*d
^4-12*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^3*d+12*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^3+3*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^4+3*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^4-16*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*c^2
*d^2+96*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)*c^3*d-20*I*c^4*(a*(c+d*tan(f*x+e))*(1+I*tan(f
*x+e)))^(1/2)+36*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^4-40*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1
/2)*tan(f*x+e)^2*c^3*d-40*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*c*d^3-32*(a*(c+d*tan(f*x+e)
)*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)*c^2*d^2+56*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d^3+32*tan(f*x+e
)*c^4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-64*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)*d^
4+56*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^3*d+96*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*
x+e)*c*d^3+3*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^4+3*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^4-9*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^4-9*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^4+54*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^2*d^2+36*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^3*d-36*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^3-9*
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^4-9*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^4-18*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*
d^2+36*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^3+12*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*t
an(f*x+e)))^(1/2))/(tan(f*x+e)+I))*tan(f*x+e)^3*c^3*d-12*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^3+54*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*d^2-36
*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^3*d)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x
+e)))^(1/2)/(c+I*d)^3/(I*c-d)/(-tan(f*x+e)+I)^3/(I*d-c)
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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(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
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Fricas [B] time = 1.82954, size = 1368, normalized size = 7.09 \begin{align*} -\frac{{\left ({\left (3 \, a^{2} c^{2} + 6 i \, a^{2} c d - 3 \, a^{2} d^{2}\right )} f \sqrt{\frac{i}{2 \,{\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-{\left (2 \,{\left (i \, a^{2} c + a^{2} d\right )} f \sqrt{\frac{i}{2 \,{\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (2 i \, f x + 2 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 )} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}\right ) -{\left (3 \, a^{2} c^{2} + 6 i \, a^{2} c d - 3 \, a^{2} d^{2}\right )} f \sqrt{\frac{i}{2 \,{\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-{\left (2 \,{\left (-i \, a^{2} c - a^{2} d\right )} f \sqrt{\frac{i}{2 \,{\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (2 i \, f x + 2 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 )} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}\right ) + \sqrt{2}{\left ({\left (-4 i \, c + 8 \, d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-5 i \, c + 9 \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c + d\right )} \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}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{{\left (12 \, a^{2} c^{2} + 24 i \, a^{2} c d - 12 \, a^{2} d^{2}\right )} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
-((3*a^2*c^2 + 6*I*a^2*c*d - 3*a^2*d^2)*f*sqrt(1/2*I/((-I*a^3*c - a^3*d)*f^2))*e^(4*I*f*x + 4*I*e)*log(-(2*(I*
a^2*c + a^2*d)*f*sqrt(1/2*I/((-I*a^3*c - a^3*d)*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)) - (3*a^2*c^2 + 6*I*a^2*c*d - 3*a^2*d^2)*f*sqrt(1/2*I/((-I*a^3*c - a^3*d)*f^2)
)*e^(4*I*f*x + 4*I*e)*log(-(2*(-I*a^2*c - a^2*d)*f*sqrt(1/2*I/((-I*a^3*c - a^3*d)*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(2)*((-4*I*c + 8*d)*e^(4*I*f*x + 4*I*
e) + (-5*I*c + 9*d)*e^(2*I*f*x + 2*I*e) - I*c + d)*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^(-4*I*f*x - 4*I*e)/((12*a^2*c^2 + 24*I*a^2
*c*d - 12*a^2*d^2)*f)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \left (i \tan{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}} \sqrt{c + d \tan{\left (e + f x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(c+d*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**(3/2),x)
[Out]
Integral(1/((a*(I*tan(e + f*x) + 1))**(3/2)*sqrt(c + d*tan(e + f*x))), x)
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Giac [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(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="giac")
[Out]
Timed out