\(\int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx\) [546]

Optimal result

Integrand size = 38, antiderivative size = 244 \[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=-\frac {(-1)^{3/4} a^{3/2} (12 A-11 i B) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{4 d}-\frac {(2+2 i) a^{3/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}} \]

[Out]

-1/4*(-1)^(3/4)*a^(3/2)*(12*A-11*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-(2+2*I)*a^(3/2)*(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+1/2*I*a*B*(a+I*a*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(3/2)+1/4*a*(4
*I*A+5*B)*(a+I*a*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=-\frac {(-1)^{3/4} a^{3/2} (12 A-11 i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 d}-\frac {(2+2 i) a^{3/2} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a (5 B+4 i A) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)} \]

[In]

Int[((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]))/Sqrt[Cot[c + d*x]],x]

[Out]

-1/4*((-1)^(3/4)*a^(3/2)*(12*A - (11*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 - ((2 + 2*I)*a^(3/2)*(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 + ((I/2)*a*B*Sqrt[a + I*a*
Tan[c + d*x]])/(d*Cot[c + d*x]^(3/2)) + (a*((4*I)*A + 5*B)*Sqrt[a + I*a*Tan[c + d*x]])/(4*d*Sqrt[Cot[c + d*x]]
)

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))^{3/2} (A+B \tan (c+d x)) \, dx \\ & = \frac {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {1}{2} \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 {1}{2} a (4 A-3 i B)+\frac {1}{2} a (4 i A+5 B) \tan (c+d x)\right ) \, dx \\ & = \frac {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (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 {1}{4} a^2 (4 i A+5 B)+\frac {1}{4} a^2 (12 A-11 i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{2 a} \\ & = \frac {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}-\left (2 a (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}{8} \left ((12 i A+11 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 {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {\left (4 i a^3 (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^2 (12 i A+11 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 )}{8 d} \\ & = -\frac {(2-2 i) a^{3/2} (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 {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {\left (a^2 (12 i A+11 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 )}{4 d} \\ & = -\frac {(2-2 i) a^{3/2} (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 {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {\left (a^2 (12 i A+11 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 )}{4 d} \\ & = -\frac {\sqrt [4]{-1} a^{3/2} (12 i A+11 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)}}{4 d}-\frac {(2-2 i) a^{3/2} (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 {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.72 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.54 \[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac {-\frac {i \left (\frac {1}{2} a^2 (4 A-3 i B)-\frac {1}{2} i a^2 B\right ) \left (-\frac {2 i \sqrt {2} a \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {2 i a^{3/2} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)} \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\right )}{a}+\frac {i a^2 (4 A-3 i B) \left (-\frac {(-1)^{3/4} \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{\sqrt {1+i \tan (c+d x)}}+\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}\right )}{2 d}}{2 a}\right ) \]

[In]

Integrate[((a + I*a*Tan[c + d*x])^(3/2)*(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])^(3/2))/(2*d) + (((-I)*((a^
2*(4*A - (3*I)*B))/2 - (I/2)*a^2*B)*(((-2*I)*Sqrt[2]*a*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*T
an[c + d*x]]]*Sqrt[I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) + ((2*I)*a^(3/2)*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/S
qrt[a]]*Sqrt[1 + I*Tan[c + d*x]]*Sqrt[I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])))/a
+ ((I/2)*a^2*(4*A - (3*I)*B)*(-(((-1)^(3/4)*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sqrt[a + I*a*Tan[c + d*x]]
)/Sqrt[1 + I*Tan[c + d*x]]) + Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]))/d)/(2*a))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 454 vs. \(2 (194 ) = 388\).

Time = 0.66 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.86

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-4 i B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+4 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}-8 i A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+5 B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a -10 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+16 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}-8 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \right )}{8 d \sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \left (1+i \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) \sqrt {i a}\, \sqrt {-i a}}\)	\(455\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-4 i B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+4 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}-8 i A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+5 B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a -10 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+16 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}-8 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \right )}{8 d \sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \left (1+i \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) \sqrt {i a}\, \sqrt {-i a}}\)	\(455\)

[In]

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

[Out]

-1/8/d*(a*(1+I*tan(d*x+c)))^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-4*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c))
)^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)+4*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))*a*(-I*a)^(1/2)-8*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)
^(1/2)+5*B*(-I*a)^(1/2)*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-10*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+16*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)-8*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. 913 vs. \(2 (182) = 364\).

Time = 0.27 (sec) , antiderivative size = 913, normalized size of antiderivative = 3.74 \[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/16*(16*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d
)*log(4*((A - I*B)*a^2*e^(I*d*x + I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^3/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)) - 16*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*
d*x + 2*I*c) + d)*log(4*((A - I*B)*a^2*e^(I*d*x + I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^3/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)) - 4*sqrt(2)*((4*A - 7*I*B)*a*e^(5*I*d*x + 5*I*c) + 4*I*B*a*e^(3*I*d*x +
3*I*c) - (4*A - 3*I*B)*a*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)) - sqrt((144*I*A^2 + 264*A*B - 121*I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*
I*d*x + 2*I*c) + d)*log(-16*(3*(12*I*A + 11*B)*a^2*e^(2*I*d*x + 2*I*c) + (-12*I*A - 11*B)*a^2 + 2*sqrt(2)*sqrt
((144*I*A^2 + 264*A*B - 121*I*B^2)*a^3/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)/(12*I*A + 11*B))
+ sqrt((144*I*A^2 + 264*A*B - 121*I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(-
16*(3*(12*I*A + 11*B)*a^2*e^(2*I*d*x + 2*I*c) + (-12*I*A - 11*B)*a^2 - 2*sqrt(2)*sqrt((144*I*A^2 + 264*A*B - 1
21*I*B^2)*a^3/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)/(12*I*A + 11*B)))/(d*e^(4*I*d*x + 4*I*c)
+ 2*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (c + d x \right )}\right )}{\sqrt {\cot {\left (c + d x \right )}}}\, dx \]

[In]

integrate((a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c))/cot(d*x+c)**(1/2),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(3/2)*(A + B*tan(c + d*x))/sqrt(cot(c + d*x)), x)

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sqrt {\cot \left (d x + c\right )}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)*(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)^(3/2)/sqrt(cot(d*x + c)), x)

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sqrt {\cot \left (d x + c\right )}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)*(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)^(3/2)/sqrt(cot(d*x + c)), x)

Mupad [F(-1)]

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

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2))/cot(c + d*x)^(1/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2))/cot(c + d*x)^(1/2), x)