\(\int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\) [1148]

Optimal result

Integrand size = 32, antiderivative size = 225 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=-\frac {i (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} f}+\frac {(i c+d) \sqrt {c+d \tan (e+f x)}}{4 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac {(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3628, 3627, 3625, 214} \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=-\frac {i (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} f}+\frac {(d+i c) \sqrt {c+d \tan (e+f x)}}{4 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {(c+d \tan (e+f x))^{5/2}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}+\frac {i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}} \]

[In]

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

[Out]

((-1/4*I)*(c - I*d)^(3/2)*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^(5/2)*f) + ((I*c + d)*Sqrt[c + d*Tan[e + f*x]])/(4*a^2*f*Sqrt[a + I*a*Tan[e + f*x]]) +
((I/6)*(c + d*Tan[e + f*x])^(3/2))/(a*f*(a + I*a*Tan[e + f*x])^(3/2)) - (c + d*Tan[e + f*x])^(5/2)/(5*(I*c - d
)*f*(a + I*a*Tan[e + f*x])^(5/2))

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 3627

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/(2*b*f*m)), x] - Dist[(a*c - b*d)/(2*b^2), Int[(a + b*Tan[e
+ f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), 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] && EqQ[m + n, 0] && LeQ[m, -2^(-1)]

Rule 3628

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac {\int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx}{2 a} \\ & = \frac {i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac {(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac {(c-i d) \int \frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx}{4 a^2} \\ & = \frac {(i c+d) \sqrt {c+d \tan (e+f x)}}{4 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac {(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac {(c-i d)^2 \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{8 a^3} \\ & = \frac {(i c+d) \sqrt {c+d \tan (e+f x)}}{4 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac {(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}-\frac {\left (i (c-i d)^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 )}{4 a f} \\ & = -\frac {i (c-i d)^{3/2} \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 )}{4 \sqrt {2} a^{5/2} f}+\frac {(i c+d) \sqrt {c+d \tan (e+f x)}}{4 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac {(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.31 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.25 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {24 i (c+i d) (c+d \tan (e+f x))^{5/2}-\frac {5 i (i c-d) (-i+\tan (e+f x)) \left (4 a d (-i+\tan (e+f x)) (c+d \tan (e+f x))^{3/2}-4 a (c+d \tan (e+f x))^{5/2}-3 i \sqrt {-a (c-i d)} (c+i d) (-i+\tan (e+f x)) \left (\sqrt {2} (c-i d) \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right ) \sqrt {a+i a \tan (e+f x)}-2 \sqrt {-a (c-i d)} \sqrt {c+d \tan (e+f x)}\right )\right )}{a}}{120 (c+i d)^2 f (a+i a \tan (e+f x))^{5/2}} \]

[In]

Integrate[(c + d*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^(5/2),x]

[Out]

((24*I)*(c + I*d)*(c + d*Tan[e + f*x])^(5/2) - ((5*I)*(I*c - d)*(-I + Tan[e + f*x])*(4*a*d*(-I + Tan[e + f*x])
*(c + d*Tan[e + f*x])^(3/2) - 4*a*(c + d*Tan[e + f*x])^(5/2) - (3*I)*Sqrt[-(a*(c - I*d))]*(c + I*d)*(-I + Tan[
e + f*x])*(Sqrt[2]*(c - I*d)*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*a*Sqrt[c + d*Ta
n[e + f*x]])]*Sqrt[a + I*a*Tan[e + f*x]] - 2*Sqrt[-(a*(c - I*d))]*Sqrt[c + d*Tan[e + f*x]])))/a)/(120*(c + I*d
)^2*f*(a + I*a*Tan[e + f*x])^(5/2))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1656 vs. \(2 (179 ) = 358\).

Time = 1.20 (sec) , antiderivative size = 1657, normalized size of antiderivative = 7.36

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1657\)
default	\(\text {Expression too large to display}\)	\(1657\)

[In]

int((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/240/f/a^3*(-15*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^2-15*2^(1/2)*(-a*(I*d-c))^(1/2)*l
n((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^2-60*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^2*tan(f*x+e)+
220*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*d^2*tan(f*x+e)-52*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1
/2)*d^2*tan(f*x+e)^3+40*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c*d+60*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)
))^(1/2))/(tan(f*x+e)+I))*d^2*tan(f*x+e)^3-212*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*d^2*tan(f*x+e)^2+60
*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*d^2-220*c^2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*tan(f*x+e
)^2-15*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^2*tan(f*x+e)^4-15*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*(1+I*tan(f*x+e))*(c+d*tan(f
*x+e)))^(1/2))/(tan(f*x+e)+I))*d^2*tan(f*x+e)^4+90*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^
2*tan(f*x+e)^2+90*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*d^2*tan(f*x+e)^2+148*c^2*(a*(1+I*ta
n(f*x+e))*(c+d*tan(f*x+e)))^(1/2)+136*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c*d*tan(f*x+e)^2-40*(a*(1+
I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c*d*tan(f*x+e)^3+56*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c*d*tan(
f*x+e)-60*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*d^2*tan(f*x+e)-60*I*c^2*(a*(1+I*tan(f*x+e
))*(c+d*tan(f*x+e)))^(1/2)*tan(f*x+e)^3+60*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^2*tan(
f*x+e)^3+308*I*c^2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*tan(f*x+e))*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan
(f*x+e))^(1/2)/(-tan(f*x+e)+I)^4/(I*c-d)/(a*(1+I*tan(f*x+e))*(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. 571 vs. \(2 (171) = 342\).

Time = 0.28 (sec) , antiderivative size = 571, normalized size of antiderivative = 2.54 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {{\left (15 \, \sqrt {\frac {1}{2}} {\left (-i \, a^{3} c + a^{3} d\right )} f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{5} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{5} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} {\left ({\left (i \, c + 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}}}{i \, c + d}\right ) + 15 \, \sqrt {\frac {1}{2}} {\left (i \, a^{3} c - a^{3} d\right )} f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{5} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (-\frac {2 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{5} f^{2}}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} {\left ({\left (i \, c + 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}}}{i \, c + d}\right ) - \sqrt {2} {\left (3 \, c^{2} + 6 i \, c d - 3 \, d^{2} + {\left (23 \, c^{2} - 6 i \, c d + 17 \, d^{2}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, {\left (17 \, c^{2} + 2 i \, c d + 9 \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (7 \, c^{2} + 8 i \, c d - d^{2}\right )} e^{\left (2 i \, f x + 2 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}}\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{120 \, {\left (i \, a^{3} c - a^{3} d\right )} f} \]

[In]

integrate((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/120*(15*sqrt(1/2)*(-I*a^3*c + a^3*d)*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^5*f^2))*e^(5*I*f*x + 5*I
*e)*log((2*sqrt(1/2)*a^3*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^5*f^2))*e^(I*f*x + I*e) + sqrt(2)*((I*
c + 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)))/(I*c + d)) + 15*sqrt(1/2)*(I*a^3*c - a^3*d)*f*sqrt(-(c^3 - 3*I*c^2*d - 3*
c*d^2 + I*d^3)/(a^5*f^2))*e^(5*I*f*x + 5*I*e)*log(-(2*sqrt(1/2)*a^3*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3
)/(a^5*f^2))*e^(I*f*x + I*e) - sqrt(2)*((I*c + 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)))/(I*c + d)) - sqrt(2)*(3*c^2 +
6*I*c*d - 3*d^2 + (23*c^2 - 6*I*c*d + 17*d^2)*e^(6*I*f*x + 6*I*e) + 2*(17*c^2 + 2*I*c*d + 9*d^2)*e^(4*I*f*x +
4*I*e) + 2*(7*c^2 + 8*I*c*d - d^2)*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^(-5*I*f*x - 5*I*e)/((I*a^3*c - a^3*d)*f)

Sympy [F]

\[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((c+d*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e))**(5/2),x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Non regular value [0,0] was discarded and replaced randomly by 0=[-8,-38]Warning, replacing -8 by 29, a sub
stitution v

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\int \frac {{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

[In]

int((c + d*tan(e + f*x))^(3/2)/(a + a*tan(e + f*x)*1i)^(5/2),x)

[Out]

int((c + d*tan(e + f*x))^(3/2)/(a + a*tan(e + f*x)*1i)^(5/2), x)