\(\int \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx\) [1151]

Optimal result

Integrand size = 32, antiderivative size = 257 \[ \int \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx=-\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {d} \left (15 c^2-10 i c d-7 d^2\right ) \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{4 f}-\frac {i \sqrt {2} \sqrt {a} (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}+\frac {(7 c-i d) d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {d \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f} \]

[Out]

-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))*2^(1/2
)*a^(1/2)/f-1/4*(-1)^(1/4)*(15*c^2-10*I*c*d-7*d^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^(1/2)*d^(1/2)/f+1/4*(7*c-I*d)*d*(a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/f+1
/2*d*(a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2)/f

Rubi [A] (verified)

Time = 1.38 (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 = {3641, 3678, 3682, 3625, 214, 3680, 65, 223, 212} \[ \int \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx=-\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {d} \left (15 c^2-10 i c d-7 d^2\right ) \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{4 f}-\frac {i \sqrt {2} \sqrt {a} (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}+\frac {d \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac {d (7 c-i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f} \]

[In]

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

[Out]

-1/4*((-1)^(1/4)*Sqrt[a]*Sqrt[d]*(15*c^2 - (10*I)*c*d - 7*d^2)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e
+ f*x]])/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])])/f - (I*Sqrt[2]*Sqrt[a]*(c - I*d)^(5/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*
Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/f + ((7*c - I*d)*d*Sqrt[a + I*a*Tan[e +
f*x]]*Sqrt[c + d*Tan[e + f*x]])/(4*f) + (d*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(2*f)

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 3641

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*(m + n - 1))), x] - Dist[1/(a*(m + n - 1)), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 2)*Simp[d*(b*c*m + a*d*(-1 + n)) - a*c^2*(m + n - 1) + d*(b*d*m
- a*c*(m + 2*n - 2))*Tan[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] && GtQ[n, 1] && NeQ[m + n - 1, 0] && (IntegerQ[n] || IntegersQ[2*m, 2*n])

Rule 3678

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*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] && 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 {d \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}-\frac {\int \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (-\frac {1}{2} a \left (4 c^2-i c d-3 d^2\right )-\frac {1}{2} a (7 c-i d) d \tan (e+f x)\right ) \, dx}{2 a} \\ & = \frac {(7 c-i d) d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {d \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}-\frac {\int \frac {\sqrt {a+i a \tan (e+f x)} \left (-\frac {1}{4} a^2 \left (8 c^3-9 i c^2 d-14 c d^2+i d^3\right )-\frac {1}{4} a^2 d \left (15 c^2-10 i c d-7 d^2\right ) \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 a^2} \\ & = \frac {(7 c-i d) d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {d \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+(c-i d)^3 \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx+\frac {\left (d \left (15 i c^2+10 c d-7 i d^2\right )\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}{8 a} \\ & = \frac {(7 c-i d) d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {d \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac {\left (2 a^2 (i c+d)^3\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 )}{f}+\frac {\left (a d \left (15 i c^2+10 c d-7 i d^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{8 f} \\ & = -\frac {i \sqrt {2} \sqrt {a} (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 )}{f}+\frac {(7 c-i d) d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {d \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac {\left (d \left (15 c^2-10 i c d-7 d^2\right )\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 )}{4 f} \\ & = -\frac {i \sqrt {2} \sqrt {a} (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 )}{f}+\frac {(7 c-i d) d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {d \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac {\left (d \left (15 c^2-10 i c d-7 d^2\right )\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 )}{4 f} \\ & = -\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {d} \left (15 c^2-10 i c d-7 d^2\right ) \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 )}{4 f}-\frac {i \sqrt {2} \sqrt {a} (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 )}{f}+\frac {(7 c-i d) d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {d \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.24 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.27 \[ \int \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx=\frac {4 i \sqrt {2} (c-i d)^2 \sqrt {-a (c-i d)} \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )+(7 c-i d) d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}+2 d \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}+\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {i a (c+i d)} \sqrt {d} \left (15 c^2-10 i c d-7 d^2\right ) \text {arcsinh}\left (\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {i a} \sqrt {i a (c+i d)}}\right ) \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}}{\sqrt {i a} \sqrt {c+d \tan (e+f x)}}}{4 f} \]

[In]

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

[Out]

((4*I)*Sqrt[2]*(c - I*d)^2*Sqrt[-(a*(c - I*d))]*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt
[2]*a*Sqrt[c + d*Tan[e + f*x]])] + (7*c - I*d)*d*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]] + 2*d*Sqr
t[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2) + ((-1)^(1/4)*Sqrt[a]*Sqrt[I*a*(c + I*d)]*Sqrt[d]*(15*c^2 -
(10*I)*c*d - 7*d^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)])]*Sqrt[(c + d*Tan[e + f*x])/(c + I*d)])/(Sqrt[I*a]*Sqrt[c + d*Tan[e + f*x]]))/(4*f)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2129 vs. \(2 (202 ) = 404\).

Time = 1.00 (sec) , antiderivative size = 2130, normalized size of antiderivative = 8.29

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2130\)
default	\(\text {Expression too large to display}\)	\(2130\)

[In]

int((a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/16/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*(18*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I
*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*d*tan(f*x+e)+8*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*(I*a*d)^(1/2)*a*d^4+
6*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*d^3*tan(f*x+e)+4*I*(a
*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c*d^2*tan(f*x+e)^2-16*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e
)))^(1/2))/(tan(f*x+e)+I))*(I*a*d)^(1/2)*a*c*d^3*tan(f*x+e)+16*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(
I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c*d^2-5*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c^2*d^2*tan(f*x+e)+15*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))*2
^(1/2)*(-a*(I*d-c))^(1/2)*a*c^3*d+7*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*d^4*tan(f*x+e)-15*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))*2^(1/2)*(-a*(I*d-c)
)^(1/2)*a*c^3*d*tan(f*x+e)-3*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c*d^3*tan(f*x+e)-4*(a*(1+I*tan(f*x+e))*(c+d*tan(f
*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*d^3*tan(f*x+e)^2-16*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*(I*
a*d)^(1/2)*a*c^3*d*tan(f*x+e)-8*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*(I*a*d)^(1/2)*a*c^4+3*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))*2^(1/2)*(-a*(I*d-c)
)^(1/2)*a*c*d^3-5*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c^2*d^2+7*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*d^4-12*(a*(1+I*tan
(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c*d^2*tan(f*x+e)+8*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))*(I*a*d)^(1/2)*a*c^4*tan(f*x+e)-8*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))*(I*a*d)^(1/2)*a*d^4*tan(f*x+e)+18*
(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*d+2*(a*(1+I*tan(f*x+e
))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*d^3-16*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))*(I*a*
d)^(1/2)*a*c^3*d-16*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))*(I*a*d)^(1/2)*a*c*d^3)*2^(1/2)/(a*(1+I*tan(f*x+e))*(c+d*tan(
f*x+e)))^(1/2)/(I*a*d)^(1/2)/(-a*(I*d-c))^(1/2)/(I*c-d)/(-tan(f*x+e)+I)

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. 1117 vs. \(2 (193) = 386\).

Time = 0.29 (sec) , antiderivative size = 1117, normalized size of antiderivative = 4.35 \[ \int \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx=\text {Too large to display} \]

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/8*(4*sqrt(2)*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(a*c^5 - 5*I*a*c^4*d - 10*a*c^3*d^2 + 10*I*a*c^2*d^3 + 5*a*c
*d^4 - I*a*d^5)/f^2)*log(-(I*sqrt(2)*f*sqrt(-(a*c^5 - 5*I*a*c^4*d - 10*a*c^3*d^2 + 10*I*a*c^2*d^3 + 5*a*c*d^4
- I*a*d^5)/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))*sq
rt(((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^(
-I*f*x - I*e)/(c^2 - 2*I*c*d - d^2)) - 4*sqrt(2)*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(a*c^5 - 5*I*a*c^4*d - 10*a
*c^3*d^2 + 10*I*a*c^2*d^3 + 5*a*c*d^4 - I*a*d^5)/f^2)*log(-(-I*sqrt(2)*f*sqrt(-(a*c^5 - 5*I*a*c^4*d - 10*a*c^3
*d^2 + 10*I*a*c^2*d^3 + 5*a*c*d^4 - I*a*d^5)/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)))*e^(-I*f*x - I*e)/(c^2 - 2*I*c*d - d^2)) - 2*sqrt(2)*(3*(3*c*d - I*d^2)*e^(3*I*f
*x + 3*I*e) + (9*c*d + I*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)) + (f*e^(2*I*f*x + 2*I*e) + f)*sqrt((225*I*a*c^4*d + 300*a*c^3*d^2
- 310*I*a*c^2*d^3 - 140*a*c*d^4 + 49*I*a*d^5)/f^2)*log((sqrt(2)*(15*c^2 - 10*I*c*d - 7*d^2 + (15*c^2 - 10*I*c
*d - 7*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))*sqr
t(a/(e^(2*I*f*x + 2*I*e) + 1)) + 2*I*f*sqrt((225*I*a*c^4*d + 300*a*c^3*d^2 - 310*I*a*c^2*d^3 - 140*a*c*d^4 + 4
9*I*a*d^5)/f^2)*e^(I*f*x + I*e))*e^(-I*f*x - I*e)/(15*c^2 - 10*I*c*d - 7*d^2)) - (f*e^(2*I*f*x + 2*I*e) + f)*s
qrt((225*I*a*c^4*d + 300*a*c^3*d^2 - 310*I*a*c^2*d^3 - 140*a*c*d^4 + 49*I*a*d^5)/f^2)*log((sqrt(2)*(15*c^2 - 1
0*I*c*d - 7*d^2 + (15*c^2 - 10*I*c*d - 7*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)) - 2*I*f*sqrt((225*I*a*c^4*d + 300*a*c^3*d^2 -
310*I*a*c^2*d^3 - 140*a*c*d^4 + 49*I*a*d^5)/f^2)*e^(I*f*x + I*e))*e^(-I*f*x - I*e)/(15*c^2 - 10*I*c*d - 7*d^2
)))/(f*e^(2*I*f*x + 2*I*e) + f)

Sympy [F]

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

[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)

Maxima [F]

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

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

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

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(c+d*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:Unable to divide, perhaps due to rounding error%%%{18,[0,10,0,0,0]%%%}+%%%{-120,[0,9,0,0,1]%%%}+%%%{%%%{%%{
[360,0]:[1,

Mupad [F(-1)]

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

[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)