\(\int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx\) [135]

Optimal result

Integrand size = 36, antiderivative size = 275 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {((1-3 i) A+(3+5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((1+2 i) A-(4+i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) ((2+i) A+(1+4 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) ((2+i) A+(1+4 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}-\frac {(A+5 i B) \sqrt {\tan (c+d x)}}{2 a d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))} \]

[Out]

1/8*((1-3*I)*A+(3+5*I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a/d*2^(1/2)-(1/8+1/8*I)*((1+2*I)*A-(4+I)*B)*arct
an(1+2^(1/2)*tan(d*x+c)^(1/2))/a/d*2^(1/2)-(1/16+1/16*I)*((2+I)*A+(1+4*I)*B)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan
(d*x+c))/a/d*2^(1/2)+(1/16+1/16*I)*((2+I)*A+(1+4*I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a/d*2^(1/2)-1
/2*(A+5*I*B)*tan(d*x+c)^(1/2)/a/d+1/2*(I*A-B)*tan(d*x+c)^(3/2)/d/(a+I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3676, 3609, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {((1-3 i) A+(3+5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((1+2 i) A-(4+i) B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(A+5 i B) \sqrt {\tan (c+d x)}}{2 a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) ((2+i) A+(1+4 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) ((2+i) A+(1+4 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d} \]

[In]

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

[Out]

-1/4*(((1 - 3*I)*A + (3 + 5*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a*d) - ((1/4 + I/4)*((1 + 2
*I)*A - (4 + I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a*d) - ((1/8 + I/8)*((2 + I)*A + (1 + 4*I)
*B)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(Sqrt[2]*a*d) + ((1/8 + I/8)*((2 + I)*A + (1 + 4*I)*B)
*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(Sqrt[2]*a*d) - ((A + (5*I)*B)*Sqrt[Tan[c + d*x]])/(2*a*d
) + ((I*A - B)*Tan[c + d*x]^(3/2))/(2*d*(a + I*a*Tan[c + d*x]))

Rule 210

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

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \sqrt {\tan (c+d x)} \left (\frac {3}{2} a (i A-B)+\frac {1}{2} a (A+5 i B) \tan (c+d x)\right ) \, dx}{2 a^2} \\ & = -\frac {(A+5 i B) \sqrt {\tan (c+d x)}}{2 a d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \frac {-\frac {1}{2} a (A+5 i B)+\frac {3}{2} a (i A-B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(A+5 i B) \sqrt {\tan (c+d x)}}{2 a d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\text {Subst}\left (\int \frac {-\frac {1}{2} a (A+5 i B)+\frac {3}{2} a (i A-B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d} \\ & = -\frac {(A+5 i B) \sqrt {\tan (c+d x)}}{2 a d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {((1+3 i) A-(3-5 i) B) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a d}+\frac {((1-3 i) A+(3+5 i) B) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a d} \\ & = -\frac {(A+5 i B) \sqrt {\tan (c+d x)}}{2 a d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {((1+3 i) A-(3-5 i) B) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 \sqrt {2} a d}-\frac {((1+3 i) A-(3-5 i) B) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 \sqrt {2} a d}+\frac {((1-3 i) A+(3+5 i) B) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 a d}+\frac {((1-3 i) A+(3+5 i) B) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 a d} \\ & = -\frac {((1+3 i) A-(3-5 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}+\frac {((1+3 i) A-(3-5 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {(A+5 i B) \sqrt {\tan (c+d x)}}{2 a d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {((1-3 i) A+(3+5 i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}-\frac {((1-3 i) A+(3+5 i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d} \\ & = -\frac {((1-3 i) A+(3+5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((1-3 i) A+(3+5 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}-\frac {((1+3 i) A-(3-5 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}+\frac {((1+3 i) A-(3-5 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {(A+5 i B) \sqrt {\tan (c+d x)}}{2 a d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.56 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.41 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {\sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-2 \sqrt [4]{-1} (A+2 i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\frac {i \sqrt {\tan (c+d x)} (A+5 i B-4 B \tan (c+d x))}{-i+\tan (c+d x)}}{2 a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.52

method	result	size
derivativedivides	\(\frac {-2 i B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {i \left (-\frac {i \left (i A -B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\tan \left (d x +c \right )-i}-\frac {4 \left (2 i B +A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{2}+\frac {4 \left (-\frac {i A}{4}-\frac {B}{4}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\)	\(143\)
default	\(\frac {-2 i B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {i \left (-\frac {i \left (i A -B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\tan \left (d x +c \right )-i}-\frac {4 \left (2 i B +A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{2}+\frac {4 \left (-\frac {i A}{4}-\frac {B}{4}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\)	\(143\)

[In]

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

[Out]

1/d/a*(-2*I*B*tan(d*x+c)^(1/2)+1/2*I*(-I*(I*A-B)*tan(d*x+c)^(1/2)/(tan(d*x+c)-I)-4*(A+2*I*B)/(2^(1/2)-I*2^(1/2
))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2))))+4*(-1/4*I*A-1/4*B)/(2^(1/2)+I*2^(1/2))*arctan(2*tan(d*x+c)^
(1/2)/(2^(1/2)+I*2^(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. 621 vs. \(2 (198) = 396\).

Time = 0.28 (sec) , antiderivative size = 621, normalized size of antiderivative = 2.26 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {{\left (a d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} - {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - a d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} - {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) + 2 \, a d \sqrt {\frac {i \, A^{2} - 4 \, A B - 4 i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, A^{2} - 4 \, A B - 4 i \, B^{2}}{a^{2} d^{2}}} + i \, A - 2 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) - 2 \, a d \sqrt {\frac {i \, A^{2} - 4 \, A B - 4 i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, A^{2} - 4 \, A B - 4 i \, B^{2}}{a^{2} d^{2}}} - i \, A + 2 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) - 2 \, {\left ({\left (A + 9 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a d} \]

[In]

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

[Out]

1/8*(a*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log(-2*((I*a*d*e^(2*I*d*x + 2*I*c) + I*a
*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^2*d^2)) - (A
- I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - a*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^2*d^2))*e^
(2*I*d*x + 2*I*c)*log(-2*((-I*a*d*e^(2*I*d*x + 2*I*c) - I*a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x +
2*I*c) + 1))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^2*d^2)) - (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(
I*A + B)) + 2*a*d*sqrt((I*A^2 - 4*A*B - 4*I*B^2)/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log(((a*d*e^(2*I*d*x + 2*I*c)
+ a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*A^2 - 4*A*B - 4*I*B^2)/(a^2*d^2))
+ I*A - 2*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) - 2*a*d*sqrt((I*A^2 - 4*A*B - 4*I*B^2)/(a^2*d^2))*e^(2*I*d*x + 2*I*c)
*log(-((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*A^
2 - 4*A*B - 4*I*B^2)/(a^2*d^2)) - I*A + 2*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) - 2*((A + 9*I*B)*e^(2*I*d*x + 2*I*c)
+ A + I*B)*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)/(a*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.44 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (i \, A - 2 \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{2 \, a d} - \frac {\left (i - 1\right ) \, \sqrt {2} {\left (-i \, A - B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{4 \, a d} - \frac {2 i \, B \sqrt {\tan \left (d x + c\right )}}{a d} - \frac {-i \, A \sqrt {\tan \left (d x + c\right )} + B \sqrt {\tan \left (d x + c\right )}}{2 \, a d {\left (\tan \left (d x + c\right ) - i\right )}} \]

[In]

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

[Out]

-(1/2*I + 1/2)*sqrt(2)*(I*A - 2*B)*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a*d) - (1/4*I - 1/4)*sqrt
(2)*(-I*A - B)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a*d) - 2*I*B*sqrt(tan(d*x + c))/(a*d) - 1/2*
(-I*A*sqrt(tan(d*x + c)) + B*sqrt(tan(d*x + c)))/(a*d*(tan(d*x + c) - I))

Mupad [B] (verification not implemented)

Time = 12.59 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.98 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\mathrm {atan}\left (\frac {2\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{a^2\,d^2}}\,1{}\mathrm {i}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{a^2\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,4{}\mathrm {i}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}-\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{a\,d}-\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}-\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]

[In]

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

[Out]

atan((a*d*tan(c + d*x)^(1/2)*(-(B^2*1i)/(a^2*d^2))^(1/2)*1i)/B)*(-(B^2*1i)/(a^2*d^2))^(1/2)*2i - atan((4*a*d*t
an(c + d*x)^(1/2)*(-(A^2*1i)/(16*a^2*d^2))^(1/2))/A)*(-(A^2*1i)/(16*a^2*d^2))^(1/2)*2i - atan((2*a*d*tan(c + d
*x)^(1/2)*((A^2*1i)/(4*a^2*d^2))^(1/2))/A)*((A^2*1i)/(4*a^2*d^2))^(1/2)*2i - atan((a*d*tan(c + d*x)^(1/2)*((B^
2*1i)/(16*a^2*d^2))^(1/2)*4i)/B)*((B^2*1i)/(16*a^2*d^2))^(1/2)*2i - (B*tan(c + d*x)^(1/2)*2i)/(a*d) - (A*tan(c
+ d*x)^(1/2))/(2*a*d*(tan(c + d*x)*1i + 1)) - (B*tan(c + d*x)^(1/2)*1i)/(2*a*d*(tan(c + d*x)*1i + 1))