\(\int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3} \, dx\) [151]

Optimal result

Integrand size = 36, antiderivative size = 315 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3} \, dx=-\frac {((7-5 i) A-2 i B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {((7-5 i) A-2 i B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {((7+5 i) A-2 i B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((7+5 i) A-2 i B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

1/32*((7-5*I)*A-2*I*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/32*((7-5*I)*A-2*I*B)*arctan(1+2^(1/
2)*tan(d*x+c)^(1/2))/a^3/d*2^(1/2)-1/64*((7+5*I)*A-2*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*((7+5*I)*A-2*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*(A+I*B)*tan(d*x+c)^(1/2
)/d/(a+I*a*tan(d*x+c))^3+1/12*(4*A+I*B)*tan(d*x+c)^(1/2)/a/d/(a+I*a*tan(d*x+c))^2+5/8*A*tan(d*x+c)^(1/2)/d/(a^
3+I*a^3*tan(d*x+c))

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3677, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3} \, dx=-\frac {((7-5 i) A-2 i B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {((7-5 i) A-2 i B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}-\frac {((7+5 i) A-2 i B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {((7+5 i) A-2 i B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3} \]

[In]

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

[Out]

-1/16*(((7 - 5*I)*A - (2*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a^3*d) + (((7 - 5*I)*A - (2*I)
*B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^3*d) - (((7 + 5*I)*A - (2*I)*B)*Log[1 - Sqrt[2]*Sqrt
[Tan[c + d*x]] + Tan[c + d*x]])/(32*Sqrt[2]*a^3*d) + (((7 + 5*I)*A - (2*I)*B)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x
]] + Tan[c + d*x]])/(32*Sqrt[2]*a^3*d) + ((A + I*B)*Sqrt[Tan[c + d*x]])/(6*d*(a + I*a*Tan[c + d*x])^3) + ((4*A
+ I*B)*Sqrt[Tan[c + d*x]])/(12*a*d*(a + I*a*Tan[c + d*x])^2) + (5*A*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 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 {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\frac {1}{2} a (11 A-i B)-\frac {5}{2} a (i A-B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {3 a^2 (6 A-i B)-3 a^2 (4 i A-B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \, dx}{24 a^4} \\ & = \frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \frac {3 a^3 (7 A-2 i B)-15 i a^3 A \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{48 a^6} \\ & = \frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 A \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 (7 A-2 i B)-15 i a^3 A x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{24 a^6 d} \\ & = \frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {((7-5 i) A-2 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 {((7+5 i) A-2 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 {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {((7-5 i) A-2 i B) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^3 d}+\frac {((7-5 i) A-2 i B) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^3 d}-\frac {((7+5 i) A-2 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 {((7+5 i) A-2 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 {((7+5 i) A-2 i B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((7+5 i) A-2 i B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {((7-5 i) A-2 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^3 d}-\frac {((7-5 i) A-2 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^3 d} \\ & = -\frac {((7-5 i) A-2 i B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {((7-5 i) A-2 i B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {((7+5 i) A-2 i B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((7+5 i) A-2 i B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.49 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.53 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3} \, dx=\frac {3 \sqrt [4]{-1} (-6 A+i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\frac {-3 \sqrt [4]{-1} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^3(c+d x) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+\sqrt {\tan (c+d x)} \left (27 i A-6 B-2 (19 A+i B) \tan (c+d x)-15 i A \tan ^2(c+d x)\right )}{(-i+\tan (c+d x))^3}}{24 a^3 d} \]

[In]

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

[Out]

(3*(-1)^(1/4)*(-6*A + I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + (-3*(-1)^(1/4)*(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)]) + Sqrt[Tan[c + d*x]]*((27*I)*A - 6
*B - 2*(19*A + I*B)*Tan[c + d*x] - (15*I)*A*Tan[c + d*x]^2))/(-I + Tan[c + d*x])^3)/(24*a^3*d)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.50

method	result	size
derivativedivides	\(\frac {-\frac {5 i A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (\frac {38 A}{3}+\frac {2 i B}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (-9 i A +2 B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {\left (6 i A +B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}+\frac {4 \left (\frac {i A}{16}+\frac {B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\)	\(157\)
default	\(\frac {-\frac {5 i A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (\frac {38 A}{3}+\frac {2 i B}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (-9 i A +2 B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {\left (6 i A +B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}+\frac {4 \left (\frac {i A}{16}+\frac {B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\)	\(157\)

[In]

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

[Out]

1/d/a^3*(-1/8*(5*I*A*tan(d*x+c)^(5/2)+(38/3*A+2/3*I*B)*tan(d*x+c)^(3/2)+(-9*I*A+2*B)*tan(d*x+c)^(1/2))/(tan(d*
x+c)-I)^3-1/4*(6*I*A+B)/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))+4*(1/16*I*A+1/16*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. 682 vs. \(2 (244) = 488\).

Time = 0.26 (sec) , antiderivative size = 682, normalized size of antiderivative = 2.17 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\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 (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, 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 (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, 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 {36 i \, A^{2} + 12 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\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 {36 i \, A^{2} + 12 \, A B - i \, B^{2}}{a^{6} d^{2}}} + 6 i \, A + B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 3 \, a^{3} d \sqrt {\frac {36 i \, A^{2} + 12 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {{\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 {36 i \, A^{2} + 12 \, A B - i \, B^{2}}{a^{6} d^{2}}} - 6 i \, A - B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 2 \, {\left (2 \, {\left (10 \, A + i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (26 \, A + 5 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (7 \, A + 4 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 (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(1/2)/(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*((I*a^3*d*e^(2*I*d*x + 2*I*
c) + I*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*((-I*a^3*d*e^(2*I*d*x + 2*I*c) - I*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((36*I*A^2 + 12*A*B - I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(1/8*
((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((36*I*A
^2 + 12*A*B - I*B^2)/(a^6*d^2)) + 6*I*A + B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) + 3*a^3*d*sqrt((36*I*A^2 + 12*A*B -
I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-1/8*((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((36*I*A^2 + 12*A*B - I*B^2)/(a^6*d^2)) - 6*I*A - B)*e^(-2*I*d*x - 2*I
*c)/(a^3*d)) - 2*(2*(10*A + I*B)*e^(6*I*d*x + 6*I*c) + (26*A + 5*I*B)*e^(4*I*d*x + 4*I*c) + (7*A + 4*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^(-6*I*d*x - 6*I*c)/
(a^3*d)

Sympy [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >> Invalid comparison of non-real -I

Maxima [F(-2)]

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

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(1/2)/(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 = 1.25 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.43 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (6 i \, A + B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} + \frac {\left (i - 1\right ) \, \sqrt {2} {\left (-i \, A - B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} - \frac {15 i \, A \tan \left (d x + c\right )^{\frac {5}{2}} + 38 \, A \tan \left (d x + c\right )^{\frac {3}{2}} + 2 i \, B \tan \left (d x + c\right )^{\frac {3}{2}} - 27 i \, A \sqrt {\tan \left (d x + c\right )} + 6 \, B \sqrt {\tan \left (d x + c\right )}}{24 \, a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} \]

[In]

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

[Out]

-(1/16*I + 1/16)*sqrt(2)*(6*I*A + B)*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^3*d) + (1/16*I - 1/16
)*sqrt(2)*(-I*A - B)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^3*d) - 1/24*(15*I*A*tan(d*x + c)^(5/
2) + 38*A*tan(d*x + c)^(3/2) + 2*I*B*tan(d*x + c)^(3/2) - 27*I*A*sqrt(tan(d*x + c)) + 6*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 = 308, normalized size of antiderivative = 0.98 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3} \, dx=\frac {\frac {9\,A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {5\,A\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,19{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}+\frac {-\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{12\,a^3\,d}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{4\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}-\frac {{\left (-1\right )}^{1/4}\,B\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{8\,a^3\,d}-\frac {{\left (-1\right )}^{1/4}\,B\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{8\,a^3\,d}-\mathrm {atan}\left (\frac {8\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}}{3\,A}\right )\,\sqrt {\frac {A^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i} \]

[In]

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

[Out]

atan((16*a^3*d*tan(c + d*x)^(1/2)*(-(A^2*1i)/(256*a^6*d^2))^(1/2))/A)*(-(A^2*1i)/(256*a^6*d^2))^(1/2)*2i - ata
n((8*a^3*d*tan(c + d*x)^(1/2)*((A^2*9i)/(64*a^6*d^2))^(1/2))/(3*A))*((A^2*9i)/(64*a^6*d^2))^(1/2)*2i + ((9*A*t
an(c + d*x)^(1/2))/(8*a^3*d) + (A*tan(c + d*x)^(3/2)*19i)/(12*a^3*d) - (5*A*tan(c + d*x)^(5/2))/(8*a^3*d))/(ta
n(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1) + ((B*tan(c + d*x)^(1/2)*1i)/(4*a^3*d) - (B*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)*B*atan((-1)
^(1/4)*tan(c + d*x)^(1/2)))/(8*a^3*d) - ((-1)^(1/4)*B*atanh((-1)^(1/4)*tan(c + d*x)^(1/2)))/(8*a^3*d)