3.12.69 \(\int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx\) [1169]

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

[Out]

-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))*2^(1/2)*a^(1/2)/(c-I
*d)^(5/2)/f-2/3*(5*c+I*d)*d*(a+I*a*tan(f*x+e))^(1/2)/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^(1/2)-2/3*d*(a+I*a*tan(f*x
+e))^(1/2)/(c^2+d^2)/f/(c+d*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.31, antiderivative size = 188, normalized size of antiderivative = 1.00, 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 = {3642, 3679, 12, 3625, 214} \begin {gather*} -\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}} \end {gather*}

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 3642

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 3679

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]

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 ) \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 )}{(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*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(394\) vs. \(2(188)=376\).
time = 4.97, size = 394, normalized size = 2.10 \begin {gather*} \frac {\sqrt {2} \sqrt {e^{i 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 (d^2 e^{2 i (e+f x)}+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 )\right )}{3 (c-i d)^2 (c+i d)^2 \left (-i d \left (-1+e^{2 i (e+f x)}\right )+c \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 ) \sqrt {a+i a \tan (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)}} f \sqrt {\sec (e+f x)} \sqrt {\cos (f x)+i \sin (f x)}} \end {gather*}

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]])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2447 vs. \(2 (153 ) = 306\).
time = 0.73, size = 2448, normalized size = 13.02

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2448\)
default	\(\text {Expression too large to display}\)	\(2448\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 992 vs. \(2 (152) = 304\).
time = 1.05, size = 992, normalized size = 5.28 \begin {gather*} -\frac {8 \, \sqrt {2} {\left ({\left (3 \, c^{2} d - 2 i \, c d^{2} + d^{3}\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (6 \, c^{2} d + i \, c d^{2} + d^{3}\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + 3 \, {\left (c^{2} d + i \, c d^{2}\right )} e^{\left (i \, f x + i \, e\right )}\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}} - 3 \, {\left ({\left (c^{6} - 2 i \, c^{5} d + c^{4} d^{2} - 4 i \, c^{3} d^{3} - c^{2} d^{4} - 2 i \, c d^{5} - d^{6}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{6} + 2 i \, c^{5} d + c^{4} d^{2} + 4 i \, c^{3} d^{3} - c^{2} d^{4} + 2 i \, c d^{5} - d^{6}\right )} f\right )} \sqrt {-\frac {2 i \, a}{{\left (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}\right )} f^{2}}} \log \left ({\left ({\left (i \, c^{3} + 3 \, c^{2} d - 3 i \, c d^{2} - d^{3}\right )} f \sqrt {-\frac {2 i \, a}{{\left (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}\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 )} e^{\left (-i \, f x - i \, e\right )}\right ) + 3 \, {\left ({\left (c^{6} - 2 i \, c^{5} d + c^{4} d^{2} - 4 i \, c^{3} d^{3} - c^{2} d^{4} - 2 i \, c d^{5} - d^{6}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{6} + 2 i \, c^{5} d + c^{4} d^{2} + 4 i \, c^{3} d^{3} - c^{2} d^{4} + 2 i \, c d^{5} - d^{6}\right )} f\right )} \sqrt {-\frac {2 i \, a}{{\left (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}\right )} f^{2}}} \log \left ({\left ({\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f \sqrt {-\frac {2 i \, a}{{\left (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}\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 )} e^{\left (-i \, f x - i \, e\right )}\right )}{6 \, {\left ({\left (c^{6} - 2 i \, c^{5} d + c^{4} d^{2} - 4 i \, c^{3} d^{3} - c^{2} d^{4} - 2 i \, c d^{5} - d^{6}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{6} + 2 i \, c^{5} d + c^{4} d^{2} + 4 i \, c^{3} d^{3} - c^{2} d^{4} + 2 i \, c d^{5} - d^{6}\right )} f\right )}} \end {gather*}

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]

-1/6*(8*sqrt(2)*((3*c^2*d - 2*I*c*d^2 + d^3)*e^(5*I*f*x + 5*I*e) + (6*c^2*d + I*c*d^2 + d^3)*e^(3*I*f*x + 3*I*
e) + 3*(c^2*d + I*c*d^2)*e^(I*f*x + 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)) - 3*((c^6 - 2*I*c^5*d + c^4*d^2 - 4*I*c^3*d^3 - c^2*d^4 - 2*I*c*d^5 -
d^6)*f*e^(4*I*f*x + 4*I*e) + 2*(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6)*f*e^(2*I*f*x + 2*I*e) + (c^6 + 2*I*c^5*d +
c^4*d^2 + 4*I*c^3*d^3 - c^2*d^4 + 2*I*c*d^5 - 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^(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))*e^(-I*f*x - I*e)
) + 3*((c^6 - 2*I*c^5*d + c^4*d^2 - 4*I*c^3*d^3 - c^2*d^4 - 2*I*c*d^5 - d^6)*f*e^(4*I*f*x + 4*I*e) + 2*(c^6 +
3*c^4*d^2 + 3*c^2*d^4 + d^6)*f*e^(2*I*f*x + 2*I*e) + (c^6 + 2*I*c^5*d + c^4*d^2 + 4*I*c^3*d^3 - c^2*d^4 + 2*I*
c*d^5 - 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^(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))*e^(-I*f*x - I*e)))/((c^6 - 2*I*c^5*d + c^4*d^2 - 4*I*
c^3*d^3 - c^2*d^4 - 2*I*c*d^5 - d^6)*f*e^(4*I*f*x + 4*I*e) + 2*(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6)*f*e^(2*I*f*
x + 2*I*e) + (c^6 + 2*I*c^5*d + c^4*d^2 + 4*I*c^3*d^3 - c^2*d^4 + 2*I*c*d^5 - d^6)*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

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]

Integral(sqrt(I*a*(tan(e + f*x) - I))/(c + d*tan(e + f*x))**(5/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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