\(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx\) [152]

Optimal result

Integrand size = 36, antiderivative size = 364 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {((30+28 i) A-(7-5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((1+29 i) A-(6+i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((29+i) A+(1+6 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a^3 d}+\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((29+i) A+(1+6 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a^3 d}-\frac {5 (6 A+i B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}+\frac {5 A+2 i B}{12 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {7 (4 A+i B)}{24 d \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

-1/32*((30+28*I)*A+(-7+5*I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a^3/d*2^(1/2)+(-1/32+1/32*I)*((1+29*I)*A-(6
+I)*B)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/a^3/d*2^(1/2)+(-1/64+1/64*I)*((29+I)*A+(1+6*I)*B)*ln(1-2^(1/2)*tan(d
*x+c)^(1/2)+tan(d*x+c))/a^3/d*2^(1/2)+(1/64-1/64*I)*((29+I)*A+(1+6*I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x
+c))/a^3/d*2^(1/2)-5/8*(6*A+I*B)/a^3/d/tan(d*x+c)^(1/2)+1/6*(A+I*B)/d/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^3+1/
12*(5*A+2*I*B)/a/d/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^2+7/24*(4*A+I*B)/d/tan(d*x+c)^(1/2)/(a^3+I*a^3*tan(d*x+
c))

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3677, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {((30+28 i) A-(7-5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((1+29 i) A-(6+i) B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}-\frac {5 (6 A+i B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {7 (4 A+i B)}{24 d \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((29+i) A+(1+6 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((29+i) A+(1+6 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}+\frac {5 A+2 i B}{12 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2} \]

[In]

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

[Out]

(((30 + 28*I)*A - (7 - 5*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^3*d) - ((1/16 - I/16)*((1
+ 29*I)*A - (6 + I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a^3*d) - ((1/32 - I/32)*((29 + I)*A +
(1 + 6*I)*B)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(Sqrt[2]*a^3*d) + ((1/32 - I/32)*((29 + I)*A
+ (1 + 6*I)*B)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(Sqrt[2]*a^3*d) - (5*(6*A + I*B))/(8*a^3*d
*Sqrt[Tan[c + d*x]]) + (A + I*B)/(6*d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^3) + (5*A + (2*I)*B)/(12*a*d*S
qrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^2) + (7*(4*A + I*B))/(24*d*Sqrt[Tan[c + d*x]]*(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 3610

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

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}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}+\frac {\int \frac {\frac {1}{2} a (13 A+i B)-\frac {7}{2} a (i A-B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}+\frac {5 A+2 i B}{12 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {\int \frac {a^2 (31 A+4 i B)-5 a^2 (5 i A-2 B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx}{24 a^4} \\ & = \frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}+\frac {5 A+2 i B}{12 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {7 (4 A+i B)}{24 d \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \frac {15 a^3 (6 A+i B)-21 a^3 (4 i A-B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{48 a^6} \\ & = -\frac {5 (6 A+i B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}+\frac {5 A+2 i B}{12 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {7 (4 A+i B)}{24 d \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \frac {-21 a^3 (4 i A-B)-15 a^3 (6 A+i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{48 a^6} \\ & = -\frac {5 (6 A+i B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}+\frac {5 A+2 i B}{12 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {7 (4 A+i B)}{24 d \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {-21 a^3 (4 i A-B)-15 a^3 (6 A+i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{24 a^6 d} \\ & = -\frac {5 (6 A+i B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}+\frac {5 A+2 i B}{12 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {7 (4 A+i B)}{24 d \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {((30+28 i) A-(7-5 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 {((30-28 i) A+(7+5 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 {5 (6 A+i B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}+\frac {5 A+2 i B}{12 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {7 (4 A+i B)}{24 d \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {((30+28 i) A-(7-5 i) B) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^3 d}-\frac {((30+28 i) A-(7-5 i) B) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^3 d}-\frac {((30-28 i) A+(7+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^3 d}-\frac {((30-28 i) A+(7+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^3 d} \\ & = -\frac {((30-28 i) A+(7+5 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((30-28 i) A+(7+5 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {5 (6 A+i B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}+\frac {5 A+2 i B}{12 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {7 (4 A+i B)}{24 d \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {((30+28 i) A-(7-5 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^3 d}+\frac {((30+28 i) A-(7-5 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^3 d} \\ & = \frac {((30+28 i) A-(7-5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {((30+28 i) A-(7-5 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {((30-28 i) A+(7+5 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((30-28 i) A+(7+5 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {5 (6 A+i B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}+\frac {5 A+2 i B}{12 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {7 (4 A+i B)}{24 d \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 3.03 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.53 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {\sec ^3(c+d x) \left (2 i \cos (c+d x) (7 A+4 i B+(35 A+11 i B) \cos (2 (c+d x))+(33 i A-9 B) \sin (2 (c+d x)))+6 (-29 i A+6 B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-i \tan (c+d x)\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+6 (A-i B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},i \tan (c+d x)\right ) (-i \cos (3 (c+d x))+\sin (3 (c+d x)))\right )}{48 a^3 d \sqrt {\tan (c+d x)} (-i+\tan (c+d x))^3} \]

[In]

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

[Out]

(Sec[c + d*x]^3*((2*I)*Cos[c + d*x]*(7*A + (4*I)*B + (35*A + (11*I)*B)*Cos[2*(c + d*x)] + ((33*I)*A - 9*B)*Sin
[2*(c + d*x)]) + 6*((-29*I)*A + 6*B)*Hypergeometric2F1[-1/2, 1, 1/2, (-I)*Tan[c + d*x]]*(Cos[3*(c + d*x)] + I*
Sin[3*(c + d*x)]) + 6*(A - I*B)*Hypergeometric2F1[-1/2, 1, 1/2, I*Tan[c + d*x]]*((-I)*Cos[3*(c + d*x)] + Sin[3
*(c + d*x)])))/(48*a^3*d*Sqrt[Tan[c + d*x]]*(-I + Tan[c + d*x])^3)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.49

method	result	size
derivativedivides	\(\frac {\frac {i \left (14 i A -5 B \right ) \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (\frac {98 i A}{3}-\frac {38 B}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (9 i B +20 A \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {\left (6 i B +29 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {2 A}{\sqrt {\tan \left (d x +c \right )}}+\frac {4 \left (-\frac {A}{16}+\frac {i B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\)	\(177\)
default	\(\frac {\frac {i \left (14 i A -5 B \right ) \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (\frac {98 i A}{3}-\frac {38 B}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (9 i B +20 A \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {\left (6 i B +29 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {2 A}{\sqrt {\tan \left (d x +c \right )}}+\frac {4 \left (-\frac {A}{16}+\frac {i B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\)	\(177\)

[In]

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

[Out]

1/d/a^3*(1/8*(I*(14*I*A-5*B)*tan(d*x+c)^(5/2)+(98/3*I*A-38/3*B)*tan(d*x+c)^(3/2)+(20*A+9*I*B)*tan(d*x+c)^(1/2)
)/(tan(d*x+c)-I)^3-1/4*(6*I*B+29*A)/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))-2*A/tan
(d*x+c)^(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. 784 vs. \(2 (273) = 546\).

Time = 0.28 (sec) , antiderivative size = 784, normalized size of antiderivative = 2.15 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} \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 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} \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 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {-841 i \, A^{2} + 348 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} \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 {-841 i \, A^{2} + 348 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} + 29 \, A + 6 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {-841 i \, A^{2} + 348 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} \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 {-841 i \, A^{2} + 348 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} - 29 \, A - 6 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 2 \, {\left (2 \, {\left (73 i \, A - 10 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 3 \, {\left (35 i \, A - 2 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (49 i \, A - 19 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (-3 i \, A + 2 \, 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}}}{96 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \]

[In]

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

[Out]

1/96*(3*(a^3*d*e^(8*I*d*x + 8*I*c) - a^3*d*e^(6*I*d*x + 6*I*c))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2))*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*e^(8*I
*d*x + 8*I*c) - a^3*d*e^(6*I*d*x + 6*I*c))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2))*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*e^(8*I*d*x + 8*I*c) - a^3*
d*e^(6*I*d*x + 6*I*c))*sqrt((-841*I*A^2 + 348*A*B + 36*I*B^2)/(a^6*d^2))*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((-841*I*A^2 + 348*A*B + 36*I*B^2)/(a
^6*d^2)) + 29*A + 6*I*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - 3*(a^3*d*e^(8*I*d*x + 8*I*c) - a^3*d*e^(6*I*d*x + 6*I
*c))*sqrt((-841*I*A^2 + 348*A*B + 36*I*B^2)/(a^6*d^2))*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((-841*I*A^2 + 348*A*B + 36*I*B^2)/(a^6*d^2)) - 29*A -
6*I*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - 2*(2*(73*I*A - 10*B)*e^(8*I*d*x + 8*I*c) + 3*(35*I*A - 2*B)*e^(6*I*d*x
+ 6*I*c) - (49*I*A - 19*B)*e^(4*I*d*x + 4*I*c) + 3*(-3*I*A + 2*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)))/(a^3*d*e^(8*I*d*x + 8*I*c) - a^3*d*e^(6*I*d*x + 6*I*c))

Sympy [F(-2)]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(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 = 1.68 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.46 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (29 \, A + 6 i \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} + \frac {\left (i - 1\right ) \, \sqrt {2} {\left (A - i \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} - \frac {2 \, A}{a^{3} d \sqrt {\tan \left (d x + c\right )}} - \frac {42 i \, A \tan \left (d x + c\right )^{\frac {5}{2}} - 15 \, B \tan \left (d x + c\right )^{\frac {5}{2}} + 98 \, A \tan \left (d x + c\right )^{\frac {3}{2}} + 38 i \, B \tan \left (d x + c\right )^{\frac {3}{2}} - 60 i \, A \sqrt {\tan \left (d x + c\right )} + 27 \, B \sqrt {\tan \left (d x + c\right )}}{24 \, a^{3} d {\left (-i \, \tan \left (d x + c\right ) - 1\right )}^{3}} \]

[In]

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

[Out]

-(1/16*I + 1/16)*sqrt(2)*(29*A + 6*I*B)*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) - 2*A/(a^3*d*sqrt(tan(d*x + c
))) - 1/24*(42*I*A*tan(d*x + c)^(5/2) - 15*B*tan(d*x + c)^(5/2) + 98*A*tan(d*x + c)^(3/2) + 38*I*B*tan(d*x + c
)^(3/2) - 60*I*A*sqrt(tan(d*x + c)) + 27*B*sqrt(tan(d*x + c)))/(a^3*d*(-I*tan(d*x + c) - 1)^3)

Mupad [B] (verification not implemented)

Time = 8.40 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.07 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=2\,\mathrm {atanh}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}}{29\,A}\right )\,\sqrt {-\frac {A^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}-\mathrm {atan}\left (\frac {8\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}}{3\,B}\right )\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\frac {\frac {2\,A}{a^3\,d}+\frac {A\,\mathrm {tan}\left (c+d\,x\right )\,17{}\mathrm {i}}{2\,a^3\,d}-\frac {121\,A\,{\mathrm {tan}\left (c+d\,x\right )}^2}{12\,a^3\,d}-\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^3\,15{}\mathrm {i}}{4\,a^3\,d}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}+{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,3{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}-{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,1{}\mathrm {i}}+\frac {\frac {9\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {5\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,19{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1} \]

[In]

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

[Out]

2*atanh((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) + 2*ata
nh((16*a^3*d*tan(c + d*x)^(1/2)*(-(A^2*841i)/(256*a^6*d^2))^(1/2))/(29*A))*(-(A^2*841i)/(256*a^6*d^2))^(1/2) -
atan((8*a^3*d*tan(c + d*x)^(1/2)*((B^2*9i)/(64*a^6*d^2))^(1/2))/(3*B))*((B^2*9i)/(64*a^6*d^2))^(1/2)*2i + ata
n((16*a^3*d*tan(c + d*x)^(1/2)*(-(B^2*1i)/(256*a^6*d^2))^(1/2))/B)*(-(B^2*1i)/(256*a^6*d^2))^(1/2)*2i - ((2*A)
/(a^3*d) + (A*tan(c + d*x)*17i)/(2*a^3*d) - (121*A*tan(c + d*x)^2)/(12*a^3*d) - (A*tan(c + d*x)^3*15i)/(4*a^3*
d))/(tan(c + d*x)^(1/2) + tan(c + d*x)^(3/2)*3i - 3*tan(c + d*x)^(5/2) - tan(c + d*x)^(7/2)*1i) + ((9*B*tan(c
+ d*x)^(1/2))/(8*a^3*d) + (B*tan(c + d*x)^(3/2)*19i)/(12*a^3*d) - (5*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)