\(\int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx\) [1165]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F(-2)]
   Giac [F(-1)]
   Mupad [F(-1)]

Optimal result

Integrand size = 32, antiderivative size = 269 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} (c-i d)^{3/2} f}-\frac {1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}+\frac {3 i c-11 d}{6 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {(3 c-5 i d) (c+5 i d) d \sqrt {a+i a \tan (e+f x)}}{6 a^2 (c-i d) (c+i d)^3 f \sqrt {c+d \tan (e+f x)}} \]

[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)/(c-I*d)^
(3/2)/f*2^(1/2)+1/6*(3*I*c-11*d)/a/(c+I*d)^2/f/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(1/2)+1/6*(3*c-5*I*d)
*(c+5*I*d)*d*(a+I*a*tan(f*x+e))^(1/2)/a^2/(c-I*d)/(c+I*d)^3/f/(c+d*tan(f*x+e))^(1/2)-1/3/(I*c-d)/f/(c+d*tan(f*
x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3640, 3677, 3679, 12, 3625, 214} \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f (c-i d)^{3/2}}+\frac {d (3 c-5 i d) (c+5 i d) \sqrt {a+i a \tan (e+f x)}}{6 a^2 f (c-i d) (c+i d)^3 \sqrt {c+d \tan (e+f x)}}+\frac {-11 d+3 i c}{6 a f (c+i d)^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \]

[In]

Int[1/((a + I*a*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(3/2)),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)*(c - I*d)^(3/2)*f) - 1/(3*(I*c - d)*f*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]]) + ((
3*I)*c - 11*d)/(6*a*(c + I*d)^2*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]) + ((3*c - (5*I)*d)*(c +
 (5*I)*d)*d*Sqrt[a + I*a*Tan[e + f*x]])/(6*a^2*(c - I*d)*(c + I*d)^3*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 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]

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*} \text {integral}& = -\frac {1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {\int \frac {-\frac {1}{2} a (3 i c-7 d)-2 i a d \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx}{3 a^2 (i c-d)} \\ & = -\frac {1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}+\frac {3 i c-11 d}{6 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {\int \frac {\sqrt {a+i a \tan (e+f x)} \left (-\frac {1}{4} a^2 \left (3 c^2+12 i c d-25 d^2\right )-\frac {1}{2} a^2 (3 c+11 i d) d \tan (e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{3 a^4 (c+i d)^2} \\ & = -\frac {1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}+\frac {3 i c-11 d}{6 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {(3 c-5 i d) (c+5 i d) d \sqrt {a+i a \tan (e+f x)}}{6 a^2 (c-i d) (c+i d)^3 f \sqrt {c+d \tan (e+f x)}}-\frac {2 \int -\frac {3 a^3 (c+i d)^3 \sqrt {a+i a \tan (e+f x)}}{8 \sqrt {c+d \tan (e+f x)}} \, dx}{3 a^5 (c+i d)^2 \left (c^2+d^2\right )} \\ & = -\frac {1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}+\frac {3 i c-11 d}{6 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {(3 c-5 i d) (c+5 i d) d \sqrt {a+i a \tan (e+f x)}}{6 a^2 (c-i d) (c+i d)^3 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}{4 a^2 (c-i d)} \\ & = -\frac {1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}+\frac {3 i c-11 d}{6 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {(3 c-5 i d) (c+5 i d) d \sqrt {a+i a \tan (e+f x)}}{6 a^2 (c-i d) (c+i d)^3 f \sqrt {c+d \tan (e+f x)}}+\frac {\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 (i c+d) f} \\ & = -\frac {i \text {arctanh}\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} (c-i d)^{3/2} f}-\frac {1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}+\frac {3 i c-11 d}{6 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {(3 c-5 i d) (c+5 i d) d \sqrt {a+i a \tan (e+f x)}}{6 a^2 (c-i d) (c+i d)^3 f \sqrt {c+d \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.02 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\frac {\frac {4 i (c+i d)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}+\frac {6 i c-22 d}{a \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {\frac {3 i \sqrt {2} a (c+i d)^3 \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a (c-i d)}}-\frac {2 d \left (3 c^2+10 i c d+25 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{a^2 \left (c^2+d^2\right )}}{12 (c+i d)^2 f} \]

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(3/2)),x]

[Out]

(((4*I)*(c + I*d))/((a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]]) + ((6*I)*c - 22*d)/(a*Sqrt[a + I*a*
Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]) - (((3*I)*Sqrt[2]*a*(c + I*d)^3*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[a +
I*a*Tan[e + f*x]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[-(a*(c - I*d))] - (2*d*(3*c^2 + (10*I)*c*d + 25
*d^2)*Sqrt[a + I*a*Tan[e + f*x]])/Sqrt[c + d*Tan[e + f*x]])/(a^2*(c^2 + d^2)))/(12*(c + I*d)^2*f)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 4834 vs. \(2 (219 ) = 438\).

Time = 1.39 (sec) , antiderivative size = 4835, normalized size of antiderivative = 17.97

method result size
derivativedivides \(\text {Expression too large to display}\) \(4835\)
default \(\text {Expression too large to display}\) \(4835\)

[In]

int(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/24/f*(-12*I*c^6*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*tan(f*x+e)^2-256*I*(a*(1+I*tan(f*x+e))*(c+d*tan(
f*x+e)))^(1/2)*d^6*tan(f*x+e)^2+40*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^4*d^2+68*I*(a*(1+I*tan(f*x+
e))*(c+d*tan(f*x+e)))^(1/2)*c^2*d^4-3*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^6*tan(f*x+e)^
3-9*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*(1+I*tan(f*x+e))*(c+d*ta
n(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*d^6*tan(f*x+e)^3+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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*
2^(1/2)*(-a*(I*d-c))^(1/2)*c^6*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*d^6*tan(f
*x+e)+15*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*(1+I*tan(f*x+e))*(c
+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^5*d+20*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+
e)))^(1/2)*c^6+48*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*d^6-30*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(
-a*(I*d-c))^(1/2)*c^3*d^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*
(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c*d^5-204*(a*(1+I*tan(f*x
+e))*(c+d*tan(f*x+e)))^(1/2)*d^6*tan(f*x+e)-60*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^5*d-72*(a*(1+I*ta
n(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^3*d^3+152*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^2*d^4*tan(f*x+e)^3
+20*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^5*d*tan(f*x+e)^2-104*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(
1/2)*c^3*d^3*tan(f*x+e)^2-124*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c*d^5*tan(f*x+e)^2-124*(a*(1+I*tan(f
*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^4*d^2*tan(f*x+e)-296*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^2*d^4*tan(
f*x+e)+52*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^4*d^2*tan(f*x+e)^3-12*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x
+e)))^(1/2)*c*d^5-32*c^6*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*tan(f*x+e)+100*(a*(1+I*tan(f*x+e))*(c+d*t
an(f*x+e)))^(1/2)*d^6*tan(f*x+e)^3-12*I*c^5*d*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*tan(f*x+e)^3-72*I*(a
*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^3*d^3*tan(f*x+e)^3-60*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2
)*c*d^5*tan(f*x+e)^3-3*I*2^(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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^6*(-a*(I*d-c))^(1/2)-136*I*(a*(1+I*tan(f*x+e)
)*(c+d*tan(f*x+e)))^(1/2)*c^4*d^2*tan(f*x+e)^2-380*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^2*d^4*tan(f
*x+e)^2-92*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^5*d*tan(f*x+e)-40*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*
x+e)))^(1/2)*c^3*d^3*tan(f*x+e)+52*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c*d^5*tan(f*x+e)+75*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1
/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*d^4*tan(f*x+e)^3-36*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*
(-a*(I*d-c))^(1/2)*c^5*d*tan(f*x+e)^2+36*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c*d^5*tan(f*
x+e)^2-75*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*(1+I*tan(f*x+e))*(
c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^4*d^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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x
+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^5*d*tan(f*x+e)^4+30*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1
/2)*c^3*d^3*tan(f*x+e)^4-15*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*
(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c*d^5*tan(f*x+e)^4-15*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e
)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^4*d^2*tan(f*x+e)^3+15*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^
(1/2)*(-a*(I*d-c))^(1/2)*c^2*d^4*tan(f*x+e)-3*I*2^(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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*d^6*(-a*(I*d-c))^(1/2)*t
an(f*x+e)^4+9*I*2^(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*(1+I
*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^6*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2+9*I*2^(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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(
1/2))/(tan(f*x+e)+I))*d^6*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2+30*I*2^(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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^4*d^2*(
-a*(I*d-c))^(1/2)-15*I*2^(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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^2*d^4*(-a*(I*d-c))^(1/2)-45*I*2^(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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1
/2))/(tan(f*x+e)+I))*c^2*d^4*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2-6*I*2^(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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c*d^5*(
-a*(I*d-c))^(1/2)*tan(f*x+e)+42*I*2^(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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^5*d*(-a*(I*d-c))^(1/2)*tan(f*x+e)-60
*I*2^(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*(1+I*tan(f*x+e))*
(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^3*d^3*(-a*(I*d-c))^(1/2)*tan(f*x+e)-15*I*2^(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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan
(f*x+e)+I))*c^4*d^2*(-a*(I*d-c))^(1/2)*tan(f*x+e)^4+30*I*2^(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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^2*d^4*(-a*(I*
d-c))^(1/2)*tan(f*x+e)^4-6*I*2^(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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^5*d*(-a*(I*d-c))^(1/2)*tan(f*x+e)^3-60*I*
2^(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*(1+I*tan(f*x+e))*(c+
d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^3*d^3*(-a*(I*d-c))^(1/2)*tan(f*x+e)^3+42*I*2^(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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(
f*x+e)+I))*c*d^5*(-a*(I*d-c))^(1/2)*tan(f*x+e)^3-45*I*2^(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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^4*d^2*(-a*(I*d-c
))^(1/2)*tan(f*x+e)^2)/a^2*(a*(1+I*tan(f*x+e)))^(1/2)/(-tan(f*x+e)+I)^3/(I*c-d)/(c+I*d)^4/(I*d-c)^2/(a*(1+I*ta
n(f*x+e))*(c+d*tan(f*x+e)))^(1/2)/(c+d*tan(f*x+e))^(1/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 982 vs. \(2 (205) = 410\).

Time = 0.30 (sec) , antiderivative size = 982, normalized size of antiderivative = 3.65 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/12*(sqrt(2)*(c^3 + I*c^2*d + c*d^2 + I*d^3 + 2*(2*c^3 + 3*I*c^2*d + 12*c*d^2 - 19*I*d^3)*e^(6*I*f*x + 6*I*e
) + (9*c^3 + 19*I*c^2*d + 29*c*d^2 - 25*I*d^3)*e^(4*I*f*x + 4*I*e) + 2*(3*c^3 + 7*I*c^2*d + 3*c*d^2 + 7*I*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)) - 3*((-I*a^2*c^5 + a^2*c^4*d - 2*I*a^2*c^3*d^2 + 2*a^2*c^2*d^3 - I*a^2*c*d^4 + a^2*d^5)*f*
e^(5*I*f*x + 5*I*e) + (-I*a^2*c^5 + 3*a^2*c^4*d + 2*I*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + 3*I*a^2*c*d^4 - a^2*d^5)*f
*e^(3*I*f*x + 3*I*e))*sqrt(-1/2*I/((I*a^3*c^3 + 3*a^3*c^2*d - 3*I*a^3*c*d^2 - a^3*d^3)*f^2))*log(-2*(I*a^2*c^2
 + 2*a^2*c*d - I*a^2*d^2)*f*sqrt(-1/2*I/((I*a^3*c^3 + 3*a^3*c^2*d - 3*I*a^3*c*d^2 - a^3*d^3)*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*((I*a^2*c^5 - a^2*c^4*d + 2*I*a^2*c^3*d^2 - 2*a^2*c^2*d^3 + I*a^2*
c*d^4 - a^2*d^5)*f*e^(5*I*f*x + 5*I*e) + (I*a^2*c^5 - 3*a^2*c^4*d - 2*I*a^2*c^3*d^2 - 2*a^2*c^2*d^3 - 3*I*a^2*
c*d^4 + a^2*d^5)*f*e^(3*I*f*x + 3*I*e))*sqrt(-1/2*I/((I*a^3*c^3 + 3*a^3*c^2*d - 3*I*a^3*c*d^2 - a^3*d^3)*f^2))
*log(-2*(-I*a^2*c^2 - 2*a^2*c*d + I*a^2*d^2)*f*sqrt(-1/2*I/((I*a^3*c^3 + 3*a^3*c^2*d - 3*I*a^3*c*d^2 - a^3*d^3
)*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))*sq
rt(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) + 1)))/((I*a^2*c^5 - a^2*c^4*d + 2*I*a^2*c^3*d^2 - 2*a^2*
c^2*d^3 + I*a^2*c*d^4 - a^2*d^5)*f*e^(5*I*f*x + 5*I*e) + (I*a^2*c^5 - 3*a^2*c^4*d - 2*I*a^2*c^3*d^2 - 2*a^2*c^
2*d^3 - 3*I*a^2*c*d^4 + a^2*d^5)*f*e^(3*I*f*x + 3*I*e))

Sympy [F]

\[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(a+I*a*tan(f*x+e))**(3/2)/(c+d*tan(f*x+e))**(3/2),x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F(-1)]

Timed out. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]

[In]

int(1/((a + a*tan(e + f*x)*1i)^(3/2)*(c + d*tan(e + f*x))^(3/2)),x)

[Out]

int(1/((a + a*tan(e + f*x)*1i)^(3/2)*(c + d*tan(e + f*x))^(3/2)), x)