\(\int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx\) [553]

Optimal result

Integrand size = 38, antiderivative size = 292 \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=-\frac {(-1)^{3/4} a^{5/2} (46 A-45 i B) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{8 d}-\frac {(4+4 i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {a^2 (2 A-3 i B) \sqrt {a+i a \tan (c+d x)}}{4 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a^2 (18 i A+19 B) \sqrt {a+i a \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {i a B (a+i a \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)} \]

[Out]

-1/8*(-1)^(3/4)*a^(5/2)*(46*A-45*I*B)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot
(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d-(4+4*I)*a^(5/2)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x
+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d-1/4*a^2*(2*A-3*I*B)*(a+I*a*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(3/2
)+1/8*a^2*(18*I*A+19*B)*(a+I*a*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(1/2)+1/3*I*a*B*(a+I*a*tan(d*x+c))^(3/2)/d/cot(d
*x+c)^(3/2)

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {4326, 3675, 3678, 3682, 3625, 211, 3680, 65, 223, 209} \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=-\frac {(-1)^{3/4} a^{5/2} (46 A-45 i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d}-\frac {(4+4 i) a^{5/2} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {a^2 (2 A-3 i B) \sqrt {a+i a \tan (c+d x)}}{4 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a^2 (19 B+18 i A) \sqrt {a+i a \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {i a B (a+i a \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)} \]

[In]

Int[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Sqrt[Cot[c + d*x]],x]

[Out]

-1/8*((-1)^(3/4)*a^(5/2)*(46*A - (45*I)*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c +
d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d - ((4 + 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[
Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d - (a^2*(2*A - (3*I)*B)*Sqr
t[a + I*a*Tan[c + d*x]])/(4*d*Cot[c + d*x]^(3/2)) + (a^2*((18*I)*A + 19*B)*Sqrt[a + I*a*Tan[c + d*x]])/(8*d*Sq
rt[Cot[c + d*x]]) + ((I/3)*a*B*(a + I*a*Tan[c + d*x])^(3/2))/(d*Cot[c + d*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 209

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[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 3675

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*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*
(m + n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*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] && GtQ[m, 1] &&  !LtQ[n, -1]

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]

Rule 4326

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx \\ & = \frac {i a B (a+i a \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} \left (\frac {3}{2} a (2 A-i B)+\frac {3}{2} a (2 i A+3 B) \tan (c+d x)\right ) \, dx \\ & = -\frac {a^2 (2 A-3 i B) \sqrt {a+i a \tan (c+d x)}}{4 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {i a B (a+i a \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {1}{6} \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (\frac {3}{4} a^2 (14 A-13 i B)+\frac {3}{4} a^2 (18 i A+19 B) \tan (c+d x)\right ) \, dx \\ & = -\frac {a^2 (2 A-3 i B) \sqrt {a+i a \tan (c+d x)}}{4 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a^2 (18 i A+19 B) \sqrt {a+i a \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {i a B (a+i a \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{8} a^3 (18 i A+19 B)+\frac {3}{8} a^3 (46 A-45 i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{6 a} \\ & = -\frac {a^2 (2 A-3 i B) \sqrt {a+i a \tan (c+d x)}}{4 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a^2 (18 i A+19 B) \sqrt {a+i a \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {i a B (a+i a \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\left (4 a^2 (i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx+\frac {1}{16} \left (a (46 i A+45 B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\frac {a^2 (2 A-3 i B) \sqrt {a+i a \tan (c+d x)}}{4 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a^2 (18 i A+19 B) \sqrt {a+i a \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {i a B (a+i a \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (8 i a^4 (i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {\left (a^3 (46 i A+45 B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{16 d} \\ & = -\frac {(4-4 i) a^{5/2} (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {a^2 (2 A-3 i B) \sqrt {a+i a \tan (c+d x)}}{4 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a^2 (18 i A+19 B) \sqrt {a+i a \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {i a B (a+i a \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (a^3 (46 i A+45 B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 d} \\ & = -\frac {(4-4 i) a^{5/2} (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {a^2 (2 A-3 i B) \sqrt {a+i a \tan (c+d x)}}{4 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a^2 (18 i A+19 B) \sqrt {a+i a \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {i a B (a+i a \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (a^3 (46 i A+45 B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d} \\ & = -\frac {\sqrt [4]{-1} a^{5/2} (46 i A+45 B) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{8 d}-\frac {(4-4 i) a^{5/2} (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {a^2 (2 A-3 i B) \sqrt {a+i a \tan (c+d x)}}{4 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a^2 (18 i A+19 B) \sqrt {a+i a \tan (c+d x)}}{8 d \sqrt {\cot (c+d x)}}+\frac {i a B (a+i a \tan (c+d x))^{3/2}}{3 d \cot ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.27 (sec) , antiderivative size = 524, normalized size of antiderivative = 1.79 \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}{3 d}+\frac {\frac {i a^3 (6 A-5 i B) \sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} (-1)^{3/4} \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right )+\frac {5}{4} \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)}+\frac {1}{2} i \sqrt {1+i \tan (c+d x)} \tan ^{\frac {3}{2}}(c+d x)\right )}{2 d \sqrt {1+i \tan (c+d x)}}+\frac {a \left (\frac {1}{2} a^2 (6 A-5 i B)-\frac {1}{2} i a^2 B\right ) \left (-\frac {4 i \sqrt {2} a \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}}{\sqrt {i a \tan (c+d x)}}+\frac {4 i a^{3/2} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}+\frac {i \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{\sqrt {1+i \tan (c+d x)} \sqrt {i a \tan (c+d x)}}\right )}{d}}{3 a}\right ) \]

[In]

Integrate[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Sqrt[Cot[c + d*x]],x]

[Out]

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((B*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2))/(3*d) + (((I/2)*a^3
*(6*A - (5*I)*B)*Sqrt[a + I*a*Tan[c + d*x]]*((-3*(-1)^(3/4)*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]])/4 + (5*Sqr
t[1 + I*Tan[c + d*x]]*Sqrt[Tan[c + d*x]])/4 + (I/2)*Sqrt[1 + I*Tan[c + d*x]]*Tan[c + d*x]^(3/2)))/(d*Sqrt[1 +
I*Tan[c + d*x]]) + (a*((a^2*(6*A - (5*I)*B))/2 - (I/2)*a^2*B)*(((-4*I)*Sqrt[2]*a*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan
[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Tan[c + d*x]])/Sqrt[I*a*Tan[c + d*x]] + ((4*I)*a^(3/2)*ArcSinh[Sq
rt[I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[1 + I*Tan[c + d*x]]*Sqrt[Tan[c + d*x]])/(Sqrt[I*a*Tan[c + d*x]]*Sqrt[a + I*
a*Tan[c + d*x]]) + I*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]] + (I*Sqrt[a]*ArcSinh[Sqrt[I*a*Tan[c + d*x]]
/Sqrt[a]]*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[1 + I*Tan[c + d*x]]*Sqrt[I*a*Tan[c + d*x]])))/d
)/(3*a))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 541 vs. \(2 (234 ) = 468\).

Time = 0.58 (sec) , antiderivative size = 542, normalized size of antiderivative = 1.86

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (57 B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a -114 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-52 i B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+16 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}+54 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}-108 i A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+24 A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+192 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}-96 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \right )}{48 d \sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \left (1+i \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) \sqrt {i a}\, \sqrt {-i a}}\)	\(542\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (57 B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a -114 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-52 i B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+16 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}+54 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}-108 i A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+24 A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+192 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}-96 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \right )}{48 d \sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \left (1+i \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) \sqrt {i a}\, \sqrt {-i a}}\)	\(542\)

[In]

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

[Out]

-1/48/d*(a*(1+I*tan(d*x+c)))^(1/2)*a*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(57*B*(-I*a)^(1/2)*ln(1/2*(2*I*a*ta
n(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a-114*B*(a*tan(d*x+c)*(1+I*tan(d*
x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)-52*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*ta
n(d*x+c)+16*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+54*I*A*ln(1/2*(2*I*a
*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a-108*I*A*(I*a)^(
1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+24*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d
*x+c)))^(1/2)*tan(d*x+c)+192*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)
/(I*a)^(1/2))*a*(-I*a)^(1/2)-96*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a
-3*a*tan(d*x+c))/(tan(d*x+c)+I))*(I*a)^(1/2)*a)/(1/tan(d*x+c))^(1/2)/(1+I*tan(d*x+c))/tan(d*x+c)/(I*a)^(1/2)/(
-I*a)^(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. 1013 vs. \(2 (220) = 440\).

Time = 0.29 (sec) , antiderivative size = 1013, normalized size of antiderivative = 3.47 \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/96*(192*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) +
3*d*e^(2*I*d*x + 2*I*c) + d)*log(4*((A - I*B)*a^3*e^(I*d*x + I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/d^2)*(I
*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x +
2*I*c) - 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 192*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/d^2)*(d*e^(
6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(4*((A - I*B)*a^3*e^(I*d*x + I*c)
- sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*
sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 4*sqrt(2)*((
66*A - 91*I*B)*a^2*e^(7*I*d*x + 7*I*c) + 7*(6*A - I*B)*a^2*e^(5*I*d*x + 5*I*c) - (66*A - 59*I*B)*a^2*e^(3*I*d*
x + 3*I*c) - 3*(14*A - 13*I*B)*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I
*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) - 3*sqrt((2116*I*A^2 + 4140*A*B - 2025*I*B^2)*a^5/d^2)*(d*e^(6*I*d*x + 6*I
*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(-16*(3*(46*I*A + 45*B)*a^3*e^(2*I*d*x + 2*I*c
) + (-46*I*A - 45*B)*a^3 + 2*sqrt(2)*sqrt((2116*I*A^2 + 4140*A*B - 2025*I*B^2)*a^5/d^2)*(d*e^(3*I*d*x + 3*I*c)
- d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c)
- 1)))*e^(-2*I*d*x - 2*I*c)/((46*I*A + 45*B)*a)) + 3*sqrt((2116*I*A^2 + 4140*A*B - 2025*I*B^2)*a^5/d^2)*(d*e^(
6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(-16*(3*(46*I*A + 45*B)*a^3*e^(2*
I*d*x + 2*I*c) + (-46*I*A - 45*B)*a^3 - 2*sqrt(2)*sqrt((2116*I*A^2 + 4140*A*B - 2025*I*B^2)*a^5/d^2)*(d*e^(3*I
*d*x + 3*I*c) - d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*
d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c)/((46*I*A + 45*B)*a)))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c
) + 3*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate((a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c))/cot(d*x+c)**(1/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)/sqrt(cot(d*x + c)), x)

Giac [F]

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

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)/sqrt(cot(d*x + c)), x)

Mupad [F(-1)]

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

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2))/cot(c + d*x)^(1/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2))/cot(c + d*x)^(1/2), x)