3.1169 \(\int \frac{\sqrt{a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=188 \[ -\frac{2 d (5 c+i d) \sqrt{a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 d \sqrt{a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{i \sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f (c-i d)^{5/2}} \]

[Out]

((-I)*Sqrt[2]*Sqrt[a]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f
*x]])])/((c - I*d)^(5/2)*f) - (2*d*Sqrt[a + I*a*Tan[e + f*x]])/(3*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) -
(2*(5*c + I*d)*d*Sqrt[a + I*a*Tan[e + f*x]])/(3*(c^2 + d^2)^2*f*Sqrt[c + d*Tan[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.471206, antiderivative size = 188, 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 = {3561, 3598, 12, 3544, 208} \[ -\frac{2 d (5 c+i d) \sqrt{a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 d \sqrt{a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{i \sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f (c-i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + I*a*Tan[e + f*x]]/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

((-I)*Sqrt[2]*Sqrt[a]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f
*x]])])/((c - I*d)^(5/2)*f) - (2*d*Sqrt[a + I*a*Tan[e + f*x]])/(3*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) -
(2*(5*c + I*d)*d*Sqrt[a + I*a*Tan[e + f*x]])/(3*(c^2 + d^2)^2*f*Sqrt[c + d*Tan[e + f*x]])

Rule 3561

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*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(c^2 + d^2)*
(n + 1)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*d*m - a*c*(n + 1) + a*d*(m + n + 1)*T
an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^
2 + d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || IntegersQ[2*m, 2*n])

Rule 3598

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*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]

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{a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac{2 d \sqrt{a+i a \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \int \frac{\sqrt{a+i a \tan (e+f x)} \left (\frac{1}{2} a (3 c+i d)-a d \tan (e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{3 a \left (c^2+d^2\right )}\\ &=-\frac{2 d \sqrt{a+i a \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 (5 c+i d) d \sqrt{a+i a \tan (e+f x)}}{3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{4 \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^2 \left (c^2+d^2\right )^2}\\ &=-\frac{2 d \sqrt{a+i a \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 (5 c+i d) d \sqrt{a+i a \tan (e+f x)}}{3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{\int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{(c-i d)^2}\\ &=-\frac{2 d \sqrt{a+i a \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 (5 c+i d) d \sqrt{a+i a \tan (e+f x)}}{3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{\left (2 i a^2\right ) \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 )}{(c-i d)^2 f}\\ &=-\frac{i \sqrt{2} \sqrt{a} \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 )}{(c-i d)^{5/2} f}-\frac{2 d \sqrt{a+i a \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 (5 c+i d) d \sqrt{a+i a \tan (e+f x)}}{3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 4.38484, size = 394, normalized size = 2.1 \[ \frac{\sqrt{2} \sqrt{e^{i f x}} \sqrt{a+i a \tan (e+f x)} \left (-\frac{4 d e^{i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}} \left (3 c^2 \left (1+e^{2 i (e+f x)}\right )-i c d \left (-3+2 e^{2 i (e+f x)}\right )+d^2 e^{2 i (e+f x)}\right )}{3 (c-i d)^2 (c+i d)^2 \left (c \left (1+e^{2 i (e+f x)}\right )-i d \left (-1+e^{2 i (e+f x)}\right )\right )^2}-\frac{i \log \left (2 \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{(c-i d)^{5/2}}\right )}{f \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \sqrt{\sec (e+f x)} \sqrt{\cos (f x)+i \sin (f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + I*a*Tan[e + f*x]]/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

(Sqrt[2]*Sqrt[E^(I*f*x)]*((-4*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)))]*(d^2*E^((2*I)*(e + f*x)) + 3*c^2*(1 + E^((2*I)*(e + f*x))) - I*c*d*(-3 +
2*E^((2*I)*(e + f*x)))))/(3*(c - I*d)^2*(c + I*d)^2*((-I)*d*(-1 + E^((2*I)*(e + f*x))) + c*(1 + E^((2*I)*(e +
f*x))))^2) - (I*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)))])])/(c - I*d)^(5/2))*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[E^(I*(e + f*
x))/(1 + E^((2*I)*(e + f*x)))]*Sqrt[1 + E^((2*I)*(e + f*x))]*f*Sqrt[Sec[e + f*x]]*Sqrt[Cos[f*x] + I*Sin[f*x]])

________________________________________________________________________________________

Maple [B]  time = 0.132, size = 2448, normalized size = 13. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x)

[Out]

1/6/f*2^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(-12*2^(1/2)*tan(f*x+e)^2*c*d^3*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e)
)*(1+I*tan(f*x+e)))^(1/2)-3*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)*a*c^5-9*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))*a*c
^4*d+10*I*2^(1/2)*tan(f*x+e)^2*c^2*d^2*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+12*I*2^(
1/2)*tan(f*x+e)*c^3*d*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+12*I*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)*c*d^3-4*2^(1/2)*tan(f*x+e)*c^2*d^2*(-a*(I*d-c))^
(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-9*I*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*a*c^2*d^3-2*I
*2^(1/2)*tan(f*x+e)^2*d^4*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-15*I*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)^2*a*c^3*d^2-3*I*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*a*c*d^4-3*I*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)*a*c^4*d-15*I*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)*a*c^2*d^3+14*I*2^(1/2)*(-a*(I
*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^2*d^2-4*2^(1/2)*tan(f*x+e)*d^4*(-a*(I*d-c))^(1/2)*(
a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-3*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*a*c^3*d^2+9*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*a*c*d^4-6*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*a*c^4*d+9*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*a*c^2*d^3-9*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)*a*c^3*d^2+6*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)*a*c*d^4+12*2^(1/2)*c^3*d*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/
2)+3*I*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*a*d^5+2*I*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+
I*tan(f*x+e)))^(1/2)*d^4-9*I*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))*a*c^3*d^2+3*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*ta
n(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))*a*c^2*d^3
+3*I*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))*a*c^5+3*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*a*d^5)/(c+d*tan(f*x+e))^
(3/2)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(c^2+d^2)^2/(I*c-d)/(-tan(f*x+e)+I)/(-a*(I*d-c))^(1/2)

________________________________________________________________________________________

Maxima [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+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [B]  time = 1.67414, size = 2491, normalized size = 13.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-(sqrt(2)*(24*c^2*d + 24*I*c*d^2 + (24*c^2*d - 16*I*c*d^2 + 8*d^3)*e^(4*I*f*x + 4*I*e) + (48*c^2*d + 8*I*c*d^2
+ 8*d^3)*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) - ((3*c^6 - 6*I*c^5*d + 3*c^4*d^2 - 12*I*c^3*d^3 - 3*c^2*d^4 - 6*
I*c*d^5 - 3*d^6)*f*e^(4*I*f*x + 4*I*e) + 6*(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6)*f*e^(2*I*f*x + 2*I*e) + (3*c^6
+ 6*I*c^5*d + 3*c^4*d^2 + 12*I*c^3*d^3 - 3*c^2*d^4 + 6*I*c*d^5 - 3*d^6)*f)*sqrt(-2*I*a/((I*c^5 + 5*c^4*d - 10*
I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(((I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*f*sqrt(-2*I*a/((I*c^5
+ 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^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^(-2*I*f*x - 2*I*e)) + ((3*c^6 - 6*I*c^5*d + 3*c^4*d^2 - 12*I*c^3*d^3 - 3*c^2*d^4 -
6*I*c*d^5 - 3*d^6)*f*e^(4*I*f*x + 4*I*e) + 6*(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6)*f*e^(2*I*f*x + 2*I*e) + (3*c^
6 + 6*I*c^5*d + 3*c^4*d^2 + 12*I*c^3*d^3 - 3*c^2*d^4 + 6*I*c*d^5 - 3*d^6)*f)*sqrt(-2*I*a/((I*c^5 + 5*c^4*d - 1
0*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f*sqrt(-2*I*a/((I*
c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^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^(-2*I*f*x - 2*I*e)))/((6*c^6 - 12*I*c^5*d + 6*c^4*d^2 - 24*I*c^3*d^3 - 6*c^2*d^4
- 12*I*c*d^5 - 6*d^6)*f*e^(4*I*f*x + 4*I*e) + 12*(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6)*f*e^(2*I*f*x + 2*I*e) +
(6*c^6 + 12*I*c^5*d + 6*c^4*d^2 + 24*I*c^3*d^3 - 6*c^2*d^4 + 12*I*c*d^5 - 6*d^6)*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+I*a*tan(f*x+e))**(1/2)/(c+d*tan(f*x+e))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out