3.12.59 \(\int \frac {1}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx\) [1159]
Optimal. Leaf size=193 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} \sqrt {c-i d} f}-\frac {\sqrt {c+d \tan (e+f x)}}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2}}+\frac {(3 i c-7 d) \sqrt {c+d \tan (e+f x)}}{6 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)}} \]
[Out]
-1/4*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))/a^(3/2)/f*2^(1/2
)/(c-I*d)^(1/2)+1/6*(3*I*c-7*d)*(c+d*tan(f*x+e))^(1/2)/a/(c+I*d)^2/f/(a+I*a*tan(f*x+e))^(1/2)-1/3*(c+d*tan(f*x
+e))^(1/2)/(I*c-d)/f/(a+I*a*tan(f*x+e))^(3/2)
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Rubi [A]
time = 0.31, antiderivative size = 193, 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 = {3640, 3677, 12,
3625, 214} \begin {gather*} -\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}} \end {gather*}
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]
((-1/2*I)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqr
t[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 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 3640
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 3677
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]
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 \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 )}{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*}
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Mathematica [A]
time = 4.01, size = 249, normalized size = 1.29 \begin {gather*} \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 (-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 ^{\frac {3}{2}}(e+f x)}\right )}{4 f (a+i a \tan (e+f x))^{3/2}} \end {gather*}
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] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 2945 vs. \(2 (154 ) = 308\).
time = 0.82, size = 2946, normalized size = 15.26
| | |
| method |
result |
size |
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| derivativedivides |
\(\text {Expression too large to display}\) |
\(2946\) |
| default |
\(\text {Expression too large to display}\) |
\(2946\) |
| | |
|
|
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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,method=_RETURNVERBOSE)
[Out]
-1/24/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^2*(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*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))*d^4*tan(f*x+e)+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*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))*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*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))*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*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))*
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*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))*d^4+16*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e))
)^(1/2)*c^2*d^2*tan(f*x+e)^2-96*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^3*d*tan(f*x+e)-96*I*(a*(c+d*ta
n(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d^3*tan(f*x+e)-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*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))*
c^4*tan(f*x+e)^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*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))*d^4*tan(f*x+e)^3-12*I*c^4*(a*(c+d*
tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2+28*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^4*tan(f*x+
e)^2-16*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^2*d^2+40*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c
^3*d*tan(f*x+e)^2+40*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d^3*tan(f*x+e)^2+32*(a*(c+d*tan(f*x+e))*(1+
I*tan(f*x+e)))^(1/2)*c^2*d^2*tan(f*x+e)+64*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^4*tan(f*x+e)-56*(a*(c
+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^3*d-56*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d^3+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^4-32*c^4*(a*(c+d*t
an(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)-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*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))*c^
3*d*tan(f*x+e)^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*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))*c*d^3*tan(f*x+e)^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*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))*c^2*d^2*tan(f*x+e)^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*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))*c^3*d*tan(f*x+e)-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*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))*c*d^3*ta
n(f*x+e)+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*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))*c^4*tan(f*x+e)^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*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))*c^2*d^2*tan(f*x+e)^3-36*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*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))*c^3*d*tan(f*x+e)^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*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))*c*d^3*tan(f*x+e)^2-54*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*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))*c^2*d^2*tan(f*x+e)+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*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))*d^4*tan(f*x+e)^2+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*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))*c^2*d
^2+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*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))*c^4*tan(f*x+e))/(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
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 >> ECL says: THROW: The catch RAT-ERR is undefined.
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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. 506 vs. \(2 (151) = 302\).
time = 1.35, size = 506, normalized size = 2.62 \begin {gather*} -\frac {{\left (3 \, {\left (a^{2} c^{2} + 2 i \, a^{2} c d - a^{2} d^{2}\right )} f \sqrt {\frac {i}{2 \, {\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \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 (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 ) - 3 \, {\left (a^{2} c^{2} + 2 i \, a^{2} c d - a^{2} d^{2}\right )} f \sqrt {\frac {i}{2 \, {\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \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 (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 ) - \sqrt {2} {\left (4 \, {\left (i \, c - 2 \, 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}}\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{12 \, {\left (a^{2} c^{2} + 2 i \, a^{2} c d - a^{2} d^{2}\right )} f} \end {gather*}
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]
-1/12*(3*(a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f*sqrt(1/2*I/((-I*a^3*c - a^3*d)*f^2))*e^(3*I*f*x + 3*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^(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)) - 3
*(a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f*sqrt(1/2*I/((-I*a^3*c - a^3*d)*f^2))*e^(3*I*f*x + 3*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^(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)) - sqrt(2)*
(4*(I*c - 2*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^(-3*I*f*x - 3*I*e)/((a^2*c
^2 + 2*I*a^2*c*d - a^2*d^2)*f)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \end {gather*}
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/((I*a*(tan(e + f*x) - I))**(3/2)*sqrt(c + d*tan(e + f*x))), x)
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
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]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regu
lar value [
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/((a + a*tan(e + f*x)*1i)^(3/2)*(c + d*tan(e + f*x))^(1/2)),x)
[Out]
int(1/((a + a*tan(e + f*x)*1i)^(3/2)*(c + d*tan(e + f*x))^(1/2)), x)
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