\(\int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx\) [1153]

Optimal result

Integrand size = 32, antiderivative size = 257 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {2 \sqrt [4]{-1} d^{5/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{a^{3/2} f}-\frac {i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {(c+i d) (i c+3 d) \sqrt {c+d \tan (e+f x)}}{2 a f \sqrt {a+i a \tan (e+f x)}}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {3639, 3676, 3682, 3625, 214, 3680, 65, 223, 212} \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {2 \sqrt [4]{-1} d^{5/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{a^{3/2} f}-\frac {i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {(c+i d) (3 d+i c) \sqrt {c+d \tan (e+f x)}}{2 a f \sqrt {a+i a \tan (e+f x)}} \]

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Dist[1/(2*a^2*m), Int[(
a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)
) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], x], 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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m
, 2*n])

Rule 3676

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

Rule 3680

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b*(B/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]

Rule 3682

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), x], x] /; FreeQ[{a, b
, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}-\frac {\int \frac {\sqrt {c+d \tan (e+f x)} \left (-\frac {3}{2} a \left (c^2-2 i c d+d^2\right )+3 i a d^2 \tan (e+f x)\right )}{\sqrt {a+i a \tan (e+f x)}} \, dx}{3 a^2} \\ & = \frac {(c+i d) (i c+3 d) \sqrt {c+d \tan (e+f x)}}{2 a f \sqrt {a+i a \tan (e+f x)}}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {\int \frac {\sqrt {a+i a \tan (e+f x)} \left (\frac {3}{4} a^2 \left (c^3-3 i c^2 d-3 c d^2-3 i d^3\right )-3 a^2 d^3 \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx}{3 a^4} \\ & = \frac {(c+i d) (i c+3 d) \sqrt {c+d \tan (e+f x)}}{2 a f \sqrt {a+i a \tan (e+f x)}}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {(c-i d)^3 \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{4 a^2}-\frac {\left (i d^3\right ) \int \frac {(a-i a \tan (e+f x)) \sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{a^3} \\ & = \frac {(c+i d) (i c+3 d) \sqrt {c+d \tan (e+f x)}}{2 a f \sqrt {a+i a \tan (e+f x)}}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}-\frac {\left (i d^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{a f}+\frac {(i c+d)^3 \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 (c-i d)^{5/2} \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 )}{2 \sqrt {2} a^{3/2} f}+\frac {(c+i d) (i c+3 d) \sqrt {c+d \tan (e+f x)}}{2 a f \sqrt {a+i a \tan (e+f x)}}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}-\frac {\left (2 d^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+i d-\frac {i d x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{a^2 f} \\ & = -\frac {i (c-i d)^{5/2} \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 )}{2 \sqrt {2} a^{3/2} f}+\frac {(c+i d) (i c+3 d) \sqrt {c+d \tan (e+f x)}}{2 a f \sqrt {a+i a \tan (e+f x)}}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}-\frac {\left (2 d^3\right ) \text {Subst}\left (\int \frac {1}{1+\frac {i d x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{a^2 f} \\ & = \frac {2 \sqrt [4]{-1} d^{5/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{a^{3/2} f}-\frac {i (c-i d)^{5/2} \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 )}{2 \sqrt {2} a^{3/2} f}+\frac {(c+i d) (i c+3 d) \sqrt {c+d \tan (e+f x)}}{2 a f \sqrt {a+i a \tan (e+f x)}}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.01 (sec) , antiderivative size = 400, normalized size of antiderivative = 1.56 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {3 a (c-i d)^2 \sqrt {-2 c+2 i d} \arctan \left (\frac {\sqrt {-c+i d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}\right ) (1+i \tan (e+f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}+24 \sqrt [4]{-1} \sqrt {i a} \sqrt {i a (c+i d)} d^{5/2} \text {arcsinh}\left (\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {i a} \sqrt {i a (c+i d)}}\right ) (1+i \tan (e+f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}+2 a^{3/2} (c+i d) \left (c (5 c-9 i d)+\left (3 i c^2+16 c d-9 i d^2\right ) \tan (e+f x)+d (3 i c+11 d) \tan ^2(e+f x)\right )}{12 a^{5/2} f (-i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \]

[In]

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

[Out]

(3*a*(c - I*d)^2*Sqrt[-2*c + (2*I)*d]*ArcTan[(Sqrt[-c + I*d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*Sqrt[a]*Sqrt
[c + d*Tan[e + f*x]])]*(1 + I*Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]] + 24*(-1)^(1/4
)*Sqrt[I*a]*Sqrt[I*a*(c + I*d)]*d^(5/2)*ArcSinh[((-1)^(1/4)*Sqrt[a]*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[
I*a]*Sqrt[I*a*(c + I*d)])]*(1 + I*Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[(c + d*Tan[e + f*x])/(c + I*d)
] + 2*a^(3/2)*(c + I*d)*(c*(5*c - (9*I)*d) + ((3*I)*c^2 + 16*c*d - (9*I)*d^2)*Tan[e + f*x] + d*((3*I)*c + 11*d
)*Tan[e + f*x]^2))/(12*a^(5/2)*f*(-I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3160 vs. \(2 (201 ) = 402\).

Time = 0.88 (sec) , antiderivative size = 3161, normalized size of antiderivative = 12.30

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(3161\)
default	\(\text {Expression too large to display}\)	\(3161\)

[In]

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

[Out]

1/24*I/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^2*(3*I*(I*a*d)^(1/2)*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))*c*d^2*tan(f*x+e)^3-9*I*(I*a*d)^(1/2)*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))/(ta
n(f*x+e)+I))*c^2*d*tan(f*x+e)^2-9*I*(I*a*d)^(1/2)*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*d
^2*tan(f*x+e)-24*I*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2
)+a*d)/(I*a*d)^(1/2))*a*d^4+36*I*(I*a*d)^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*d^3-12*c^3*(I*a*d)^
(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*tan(f*x+e)^2-24*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*t
an(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*d^4*tan(f*x+e)^3+72*ln(1/2*(2*I*a*d*tan
(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*d^4*tan(f*x+e)
-24*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^
(1/2))*a*c*d^3+52*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*c*d^2-80*(a*(1+I*tan(f*x+e))*(c+d*
tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*d^3*tan(f*x+e)+20*c^3*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/
2)+3*(I*a*d)^(1/2)*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^3*tan(f*x+e)^3+24*I*ln(1/2*(2*I*
a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d^3*t
an(f*x+e)^3+3*I*(I*a*d)^(1/2)*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^3-72*I*ln(1/2*(2*I*a*
d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d^3*tan
(f*x+e)+20*I*(I*a*d)^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^2*d*tan(f*x+e)^2+128*I*(I*a*d)^(1/2)*
(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c*d^2*tan(f*x+e)-44*I*(I*a*d)^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f
*x+e)))^(1/2)*d^3*tan(f*x+e)^2+32*I*c^3*(I*a*d)^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*tan(f*x+e)+4
*I*(I*a*d)^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^2*d-3*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(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^3-76*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*c*d^2*tan(f*x+e
)^2+16*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*c^2*d*tan(f*x+e)+72*ln(1/2*(2*I*a*d*tan(f*x+e
)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d^3*tan(f*x+e)^2+7
2*I*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^
(1/2))*a*d^4*tan(f*x+e)^2-9*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*ta
n(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*tan(f
*x+e)+9*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(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^2*tan(f*x+e)^2+9*2^(1/2)*(-
a*(I*d-c))^(1/2)*(I*a*d)^(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*tan(f*x+e)^2-9*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*
a*d)^(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^3*tan(f*x+e)-3*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(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^2-9*I*(I*a*d)^(1/2)*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*t
an(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*tan(f*
x+e)+3*I*(I*a*d)^(1/2)*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*d+3*(I*a*d)^(1/2)*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*d*tan(f*x+e)^3+3*I*(I*a*d)^(1/2)*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^3*tan(f*x+e)^3-9*I*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*t
an(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^3*tan(f*x+e)^2)/(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)/(I*a*d)^(1/2)/(I*c-d)/(-tan(f*x+e
)+I)^3

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. 1071 vs. \(2 (189) = 378\).

Time = 0.35 (sec) , antiderivative size = 1071, normalized size of antiderivative = 4.17 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/12*(3*sqrt(1/2)*a^2*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^3*f^2))*e^(3
*I*f*x + 3*I*e)*log(-(2*I*sqrt(1/2)*a^2*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5
)/(a^3*f^2))*e^(I*f*x + I*e) - sqrt(2)*(c^2 - 2*I*c*d - d^2 + (c^2 - 2*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)))/(c^2 -
2*I*c*d - d^2)) - 3*sqrt(1/2)*a^2*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^
3*f^2))*e^(3*I*f*x + 3*I*e)*log(-(-2*I*sqrt(1/2)*a^2*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*
c*d^4 - I*d^5)/(a^3*f^2))*e^(I*f*x + I*e) - sqrt(2)*(c^2 - 2*I*c*d - d^2 + (c^2 - 2*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)))/(c^2 - 2*I*c*d - d^2)) + 3*a^2*f*sqrt(4*I*d^5/(a^3*f^2))*e^(3*I*f*x + 3*I*e)*log(2*(4*sqrt(2)*(d^3*e^(3
*I*f*x + 3*I*e) + d^3*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)) + ((a^2*c - 3*I*a^2*d)*f*e^(2*I*f*x + 2*I*e) + (a^2*c + I*a^2*d)*f)*sqrt(
4*I*d^5/(a^3*f^2)))/(I*c^3 + c^2*d + I*c*d^2 + d^3 + (I*c^3 + c^2*d + I*c*d^2 + d^3)*e^(2*I*f*x + 2*I*e))) - 3
*a^2*f*sqrt(4*I*d^5/(a^3*f^2))*e^(3*I*f*x + 3*I*e)*log(2*(4*sqrt(2)*(d^3*e^(3*I*f*x + 3*I*e) + d^3*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)) - ((a^2*c - 3*I*a^2*d)*f*e^(2*I*f*x + 2*I*e) + (a^2*c + I*a^2*d)*f)*sqrt(4*I*d^5/(a^3*f^2)))/(I*c^3 + c^2*
d + I*c*d^2 + d^3 + (I*c^3 + c^2*d + I*c*d^2 + d^3)*e^(2*I*f*x + 2*I*e))) - sqrt(2)*(I*c^2 - 2*c*d - I*d^2 - 2
*(-2*I*c^2 - 3*c*d - 5*I*d^2)*e^(4*I*f*x + 4*I*e) + (5*I*c^2 + 4*c*d + 9*I*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^(-3*I*f*
x - 3*I*e)/(a^2*f)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((c+d*tan(f*x+e))^(5/2)/(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.

Giac [F(-2)]

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

[In]

integrate((c+d*tan(f*x+e))^(5/2)/(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:Non regular value [0,0] was discarded and replaced randomly by 0=[36,86]Warning, replacing 36 by -68, a sub
stitution v

Mupad [F(-1)]

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

[In]

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

[Out]

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