\(\int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx\) [141]

Optimal result

Integrand size = 36, antiderivative size = 277 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {((1+3 i) A+(9+5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(9+5 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {((1-3 i) A-(9-5 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {((1-3 i) A-(9-5 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {(A+5 i B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

-1/32*((1+3*I)*A+(9+5*I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a^2/d*2^(1/2)-1/32*((1+3*I)*A+(9+5*I)*B)*arcta
n(1+2^(1/2)*tan(d*x+c)^(1/2))/a^2/d*2^(1/2)+1/64*((1-3*I)*A+(-9+5*I)*B)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+
c))/a^2/d*2^(1/2)-1/64*((1-3*I)*A+(-9+5*I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a^2/d*2^(1/2)+1/8*(A+5
*I*B)*tan(d*x+c)^(1/2)/a^2/d/(1+I*tan(d*x+c))+1/4*(I*A-B)*tan(d*x+c)^(3/2)/d/(a+I*a*tan(d*x+c))^2

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3676, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {((1+3 i) A+(9+5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(9+5 i) B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{16 \sqrt {2} a^2 d}+\frac {(A+5 i B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {((1-3 i) A-(9-5 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}-\frac {((1-3 i) A-(9-5 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}+\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[In]

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

[Out]

(((1 + 3*I)*A + (9 + 5*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^2*d) - (((1 + 3*I)*A + (9 +
5*I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^2*d) + (((1 - 3*I)*A - (9 - 5*I)*B)*Log[1 - Sqr
t[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(32*Sqrt[2]*a^2*d) - (((1 - 3*I)*A - (9 - 5*I)*B)*Log[1 + Sqrt[2]*Sqr
t[Tan[c + d*x]] + Tan[c + d*x]])/(32*Sqrt[2]*a^2*d) + ((A + (5*I)*B)*Sqrt[Tan[c + d*x]])/(8*a^2*d*(1 + I*Tan[c
+ d*x])) + ((I*A - B)*Tan[c + d*x]^(3/2))/(4*d*(a + I*a*Tan[c + d*x])^2)

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 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)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\sqrt {\tan (c+d x)} \left (\frac {3}{2} a (i A-B)-\frac {1}{2} a (A-7 i B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = \frac {(A+5 i B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {-\frac {1}{2} a^2 (A+5 i B)-\frac {3}{2} a^2 (i A+3 B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{8 a^4} \\ & = \frac {(A+5 i B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\text {Subst}\left (\int \frac {-\frac {1}{2} a^2 (A+5 i B)-\frac {3}{2} a^2 (i A+3 B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a^4 d} \\ & = \frac {(A+5 i B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+2 i) A+(2-7 i) B)\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d}-\frac {((1+3 i) A+(9+5 i) B) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^2 d} \\ & = \frac {(A+5 i B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {((1-3 i) A-(9-5 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 )}{32 \sqrt {2} a^2 d}+\frac {((1-3 i) A-(9-5 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 )}{32 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(9+5 i) B) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^2 d}-\frac {((1+3 i) A+(9+5 i) B) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^2 d} \\ & = \frac {((1-3 i) A-(9-5 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {((1-3 i) A-(9-5 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {(A+5 i B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {((1+3 i) A+(9+5 i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {((1+3 i) A+(9+5 i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d} \\ & = \frac {((1+3 i) A+(9+5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(9+5 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {((1-3 i) A-(9-5 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {((1-3 i) A-(9-5 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {(A+5 i B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.23 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.64 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {-2 \sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt [4]{-1} (A-7 i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt {\tan (c+d x)} (-A-5 i B+(-3 i A+7 B) \tan (c+d x))}{8 a^2 d (-i+\tan (c+d x))^2} \]

[In]

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

[Out]

(-2*(-1)^(1/4)*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*Sec[c + d*x]^2*(Cos[2*(c + d*x)] + I*Sin[2*(c +
d*x)]) + (-1)^(1/4)*(A - (7*I)*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*Sec[c + d*x]^2*(Cos[2*(c + d*x)] + I
*Sin[2*(c + d*x)]) + Sqrt[Tan[c + d*x]]*(-A - (5*I)*B + ((-3*I)*A + 7*B)*Tan[c + d*x]))/(8*a^2*d*(-I + Tan[c +
d*x])^2)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.53

method	result	size
derivativedivides	\(\frac {\frac {i \left (\frac {\left (-\frac {7 i B}{2}-\frac {3 A}{2}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (-\frac {5 B}{2}+\frac {i A}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\left (-7 i B +A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4}-\frac {i \left (-i B +A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2 \left (\sqrt {2}+i \sqrt {2}\right )}}{d \,a^{2}}\)	\(147\)
default	\(\frac {\frac {i \left (\frac {\left (-\frac {7 i B}{2}-\frac {3 A}{2}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (-\frac {5 B}{2}+\frac {i A}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\left (-7 i B +A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4}-\frac {i \left (-i B +A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2 \left (\sqrt {2}+i \sqrt {2}\right )}}{d \,a^{2}}\)	\(147\)

[In]

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

[Out]

1/d/a^2*(1/4*I*(((-7/2*I*B-3/2*A)*tan(d*x+c)^(3/2)+(-5/2*B+1/2*I*A)*tan(d*x+c)^(1/2))/(tan(d*x+c)-I)^2-(-7*I*B
+A)/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2))))-1/2*I*(A-I*B)/(2^(1/2)+I*2^(1/2))*arct
an(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. 664 vs. \(2 (210) = 420\).

Time = 0.26 (sec) , antiderivative size = 664, normalized size of antiderivative = 2.40 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (2 \, a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} 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^{4} 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^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} 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^{4} 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^{2} d \sqrt {\frac {i \, A^{2} + 14 \, A B - 49 i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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} + 14 \, A B - 49 i \, B^{2}}{a^{4} d^{2}}} + i \, A + 7 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) - a^{2} d \sqrt {\frac {i \, A^{2} + 14 \, A B - 49 i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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} + 14 \, A B - 49 i \, B^{2}}{a^{4} d^{2}}} - i \, A - 7 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + 2 \, {\left (2 \, {\left (A + 3 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (A + 5 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 (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \]

[In]

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

[Out]

1/32*(2*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-2*((I*a^2*d*e^(2*I*d*x + 2*I*c
) + I*a^2*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^4*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^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/
(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-2*((-I*a^2*d*e^(2*I*d*x + 2*I*c) - I*a^2*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^4*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^2*d*sqrt((I*A^2 + 14*A*B - 49*I*B^2)/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(1/8*((a
^2*d*e^(2*I*d*x + 2*I*c) + a^2*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 + 1
4*A*B - 49*I*B^2)/(a^4*d^2)) + I*A + 7*B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) - a^2*d*sqrt((I*A^2 + 14*A*B - 49*I*B^
2)/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-1/8*((a^2*d*e^(2*I*d*x + 2*I*c) + a^2*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 + 14*A*B - 49*I*B^2)/(a^4*d^2)) - I*A - 7*B)*e^(-2*I*d*x - 2*I*c)/(
a^2*d)) + 2*(2*(A + 3*I*B)*e^(4*I*d*x + 4*I*c) + (A + 5*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^(-4*I*d*x - 4*I*c)/(a^2*d)

Sympy [F]

\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=- \frac {\int \frac {A \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx + \int \frac {B \tan ^{\frac {5}{2}}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]

[In]

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

[Out]

-(Integral(A*tan(c + d*x)**(3/2)/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), x) + Integral(B*tan(c + d*x)**(5/2)
/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), x))/a**2

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.65 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.45 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (i \, A + 7 \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{2} d} + \frac {\left (i - 1\right ) \, \sqrt {2} {\left (i \, A + B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{8 \, a^{2} d} - \frac {3 i \, A \tan \left (d x + c\right )^{\frac {3}{2}} - 7 \, B \tan \left (d x + c\right )^{\frac {3}{2}} + A \sqrt {\tan \left (d x + c\right )} + 5 i \, B \sqrt {\tan \left (d x + c\right )}}{8 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \]

[In]

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

[Out]

-(1/16*I + 1/16)*sqrt(2)*(I*A + 7*B)*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^2*d) + (1/8*I - 1/8)*
sqrt(2)*(I*A + B)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^2*d) - 1/8*(3*I*A*tan(d*x + c)^(3/2) -
7*B*tan(d*x + c)^(3/2) + A*sqrt(tan(d*x + c)) + 5*I*B*sqrt(tan(d*x + c)))/(a^2*d*(tan(d*x + c) - I)^2)

Mupad [B] (verification not implemented)

Time = 11.37 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.15 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {-\frac {3\,A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{8\,a^2\,d}+\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{8\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}+\frac {\frac {5\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^2\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,7{}\mathrm {i}}{8\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}-\mathrm {atan}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}\,2{}\mathrm {i}+2\,\mathrm {atanh}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,49{}\mathrm {i}}{256\,a^4\,d^2}}}{7\,B}\right )\,\sqrt {-\frac {B^2\,49{}\mathrm {i}}{256\,a^4\,d^2}} \]

[In]

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

[Out]

((5*B*tan(c + d*x)^(1/2))/(8*a^2*d) + (B*tan(c + d*x)^(3/2)*7i)/(8*a^2*d))/(2*tan(c + d*x) + tan(c + d*x)^2*1i
- 1i) - ((A*tan(c + d*x)^(1/2)*1i)/(8*a^2*d) - (3*A*tan(c + d*x)^(3/2))/(8*a^2*d))/(2*tan(c + d*x) + tan(c +
d*x)^2*1i - 1i) - atan((8*a^2*d*tan(c + d*x)^(1/2)*(-(A^2*1i)/(64*a^4*d^2))^(1/2))/A)*(-(A^2*1i)/(64*a^4*d^2))
^(1/2)*2i - atan((16*a^2*d*tan(c + d*x)^(1/2)*((A^2*1i)/(256*a^4*d^2))^(1/2))/A)*((A^2*1i)/(256*a^4*d^2))^(1/2
)*2i + 2*atanh((8*a^2*d*tan(c + d*x)^(1/2)*((B^2*1i)/(64*a^4*d^2))^(1/2))/B)*((B^2*1i)/(64*a^4*d^2))^(1/2) + 2
*atanh((16*a^2*d*tan(c + d*x)^(1/2)*(-(B^2*49i)/(256*a^4*d^2))^(1/2))/(7*B))*(-(B^2*49i)/(256*a^4*d^2))^(1/2)