\(\int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\) [150]

Optimal result

Integrand size = 36, antiderivative size = 317 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {(2 i A+(1-i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {(2 i A+(1-i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

(-1/32-1/32*I)*((1+I)*A+B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a^3/d*2^(1/2)-(1/32+1/32*I)*((1+I)*A+B)*arctan(
1+2^(1/2)*tan(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/64*(2*I*A+(1-I)*B)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a^3/d
*2^(1/2)-1/64*(2*I*A+(1-I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a^3/d*2^(1/2)+1/6*(I*A-B)*tan(d*x+c)^(
1/2)/d/(a+I*a*tan(d*x+c))^3+1/12*(I*A+2*B)*tan(d*x+c)^(1/2)/a/d/(a+I*a*tan(d*x+c))^2+1/8*B*tan(d*x+c)^(1/2)/d/
(a^3+I*a^3*tan(d*x+c))

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3676, 3677, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (B+(1+i) A) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (B+(1+i) A) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {(2 i A+(1-i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {(2 i A+(1-i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(-B+i A) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 B+i A) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2} \]

[In]

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

[Out]

((1/16 + I/16)*((1 + I)*A + B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a^3*d) - ((1/16 + I/16)*((1 +
I)*A + B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a^3*d) + (((2*I)*A + (1 - I)*B)*Log[1 - Sqrt[2]*Sqr
t[Tan[c + d*x]] + Tan[c + d*x]])/(32*Sqrt[2]*a^3*d) - (((2*I)*A + (1 - I)*B)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]
] + Tan[c + d*x]])/(32*Sqrt[2]*a^3*d) + ((I*A - B)*Sqrt[Tan[c + d*x]])/(6*d*(a + I*a*Tan[c + d*x])^3) + ((I*A
+ 2*B)*Sqrt[Tan[c + d*x]])/(12*a*d*(a + I*a*Tan[c + d*x])^2) + (B*Sqrt[Tan[c + d*x]])/(8*d*(a^3 + I*a^3*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 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]

Rule 3677

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*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*
f*m*(b*c - a*d))), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*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] && LtQ[m, 0] &&  !GtQ[n,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\frac {1}{2} a (i A-B)-\frac {1}{2} a (5 A-7 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}-\frac {\int \frac {3 i a^2 A-3 a^2 (A-2 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \, dx}{24 a^4} \\ & = \frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \frac {3 a^3 (2 i A+B)+3 i a^3 B \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{48 a^6} \\ & = \frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {3 a^3 (2 i A+B)+3 i a^3 B x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{24 a^6 d} \\ & = \frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+B)\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^3 d}-\frac {(2 i A+(1-i) B) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^3 d} \\ & = \frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\left (\left (\frac {1}{32}+\frac {i}{32}\right ) ((1+i) A+B)\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^3 d}-\frac {\left (\left (\frac {1}{32}+\frac {i}{32}\right ) ((1+i) A+B)\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^3 d}+\frac {(2 i A+(1-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^3 d}+\frac {(2 i A+(1-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^3 d} \\ & = \frac {(2 i A+(1-i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {(2 i A+(1-i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}--\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d} \\ & = \frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {(2 i A+(1-i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {(2 i A+(1-i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.01 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.56 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {3 (-1)^{3/4} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^3(c+d x) (\cos (3 (c+d x))+i \sin (3 (c+d x)))-3 \sqrt [4]{-1} A \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^3(c+d x) (\cos (3 (c+d x))+i \sin (3 (c+d x)))-i \sqrt {\tan (c+d x)} (-3 i+\tan (c+d x)) (2 A-i B+3 B \tan (c+d x))}{24 a^3 d (-i+\tan (c+d x))^3} \]

[In]

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

[Out]

(3*(-1)^(3/4)*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*Sec[c + d*x]^3*(Cos[3*(c + d*x)] + I*Sin[3*(c +
d*x)]) - 3*(-1)^(1/4)*A*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*Sec[c + d*x]^3*(Cos[3*(c + d*x)] + I*Sin[3*(c +
d*x)]) - I*Sqrt[Tan[c + d*x]]*(-3*I + Tan[c + d*x])*(2*A - I*B + 3*B*Tan[c + d*x]))/(24*a^3*d*(-I + Tan[c + d
*x])^3)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.48

method	result	size
derivativedivides	\(\frac {\frac {-i B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (-\frac {10 B}{3}-\frac {2 i A}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (i B -2 A \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {A \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}+\frac {4 \left (\frac {A}{16}-\frac {i B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\)	\(152\)
default	\(\frac {\frac {-i B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (-\frac {10 B}{3}-\frac {2 i A}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (i B -2 A \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {A \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}+\frac {4 \left (\frac {A}{16}-\frac {i B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\)	\(152\)

[In]

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

[Out]

1/d/a^3*(1/8*(-I*B*tan(d*x+c)^(5/2)+(-10/3*B-2/3*I*A)*tan(d*x+c)^(3/2)+(-2*A+I*B)*tan(d*x+c)^(1/2))/(tan(d*x+c
)-I)^3-1/4*A/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))+4*(1/16*A-1/16*I*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. 631 vs. \(2 (244) = 488\).

Time = 0.26 (sec) , antiderivative size = 631, normalized size of antiderivative = 1.99 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (3 \, a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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^{6} 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 ) - 3 \, a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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^{6} 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 ) - 24 \, a^{3} d \sqrt {-\frac {i \, A^{2}}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left (8 \, {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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}}{64 \, a^{6} d^{2}}} + A\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 24 \, a^{3} d \sqrt {-\frac {i \, A^{2}}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {{\left (8 \, {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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}}{64 \, a^{6} d^{2}}} - A\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 2 \, {\left (2 \, {\left (-i \, A - 2 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (5 i \, A + 4 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (4 i \, A - B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + 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 (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

[In]

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

[Out]

-1/96*(3*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(2*((a^3*d*e^(2*I*d*x + 2*I*c) +
a^3*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^6*d^2)) +
(A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 3*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^
2))*e^(6*I*d*x + 6*I*c)*log(-2*((a^3*d*e^(2*I*d*x + 2*I*c) + a^3*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^6*d^2)) - (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*
c)/(I*A + B)) - 24*a^3*d*sqrt(-1/64*I*A^2/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(1/8*(8*(a^3*d*e^(2*I*d*x + 2*I*c)
+ a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-1/64*I*A^2/(a^6*d^2)) + A)*e^(-2*
I*d*x - 2*I*c)/(a^3*d)) + 24*a^3*d*sqrt(-1/64*I*A^2/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-1/8*(8*(a^3*d*e^(2*I*d
*x + 2*I*c) + a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-1/64*I*A^2/(a^6*d^2))
- A)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) + 2*(2*(-I*A - 2*B)*e^(6*I*d*x + 6*I*c) - (5*I*A + 4*B)*e^(4*I*d*x + 4*I*c)
- (4*I*A - B)*e^(2*I*d*x + 2*I*c) - I*A + B)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^
(-6*I*d*x - 6*I*c)/(a^3*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.68 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.41 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} A \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} - \frac {\left (i - 1\right ) \, \sqrt {2} {\left (A - i \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} - \frac {3 i \, B \tan \left (d x + c\right )^{\frac {5}{2}} + 2 i \, A \tan \left (d x + c\right )^{\frac {3}{2}} + 10 \, B \tan \left (d x + c\right )^{\frac {3}{2}} + 6 \, A \sqrt {\tan \left (d x + c\right )} - 3 i \, B \sqrt {\tan \left (d x + c\right )}}{24 \, a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} \]

[In]

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

[Out]

-(1/16*I + 1/16)*sqrt(2)*A*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^3*d) - (1/16*I - 1/16)*sqrt(2)*
(A - I*B)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^3*d) - 1/24*(3*I*B*tan(d*x + c)^(5/2) + 2*I*A*t
an(d*x + c)^(3/2) + 10*B*tan(d*x + c)^(3/2) + 6*A*sqrt(tan(d*x + c)) - 3*I*B*sqrt(tan(d*x + c)))/(a^3*d*(tan(d
*x + c) - I)^3)

Mupad [B] (verification not implemented)

Time = 7.87 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,5{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}+\frac {-\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{12\,a^3\,d}+\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{4\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}-\frac {{\left (-1\right )}^{1/4}\,A\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{8\,a^3\,d}-\frac {{\left (-1\right )}^{1/4}\,A\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{8\,a^3\,d}-\frac {\sqrt {-\frac {1}{256}{}\mathrm {i}}\,B\,\mathrm {atan}\left (16\,\sqrt {-\frac {1}{256}{}\mathrm {i}}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )\,2{}\mathrm {i}}{a^3\,d} \]

[In]

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

[Out]

((B*tan(c + d*x)^(1/2))/(8*a^3*d) + (B*tan(c + d*x)^(3/2)*5i)/(12*a^3*d) - (B*tan(c + d*x)^(5/2))/(8*a^3*d))/(
tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1) + ((A*tan(c + d*x)^(1/2)*1i)/(4*a^3*d) - (A*tan(c
+ d*x)^(3/2))/(12*a^3*d))/(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1) - ((-1)^(1/4)*A*atan((-
1)^(1/4)*tan(c + d*x)^(1/2)))/(8*a^3*d) - ((-1)^(1/4)*A*atanh((-1)^(1/4)*tan(c + d*x)^(1/2)))/(8*a^3*d) - ((-1
i/256)^(1/2)*B*atan(16*(-1i/256)^(1/2)*tan(c + d*x)^(1/2))*2i)/(a^3*d)