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

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

[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))

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Rubi [A]  time = 0.80, antiderivative size = 364, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3596, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {((30+28 i) A-(7-5 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((1+29 i) A-(6+i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}-\frac {5 (6 A+i B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {7 (4 A+i B)}{24 d \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((29+i) A+(1+6 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((29+i) A+(1+6 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}+\frac {5 A+2 i B}{12 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[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 204

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

Rule 617

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 628

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 1162

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 1165

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 1168

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 3529

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 3534

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 3596

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*} \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 {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 {\operatorname {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) \operatorname {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) \operatorname {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) \operatorname {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) \operatorname {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) \operatorname {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) \operatorname {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) \operatorname {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) \operatorname {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) \tan ^{-1}\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) \tan ^{-1}\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*}

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Mathematica [A]  time = 3.51, size = 278, normalized size = 0.76 \[ \frac {\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^3 (A+B \tan (c+d x)) \left (\frac {2}{3} (\cos (3 d x)-i \sin (3 d x)) ((49 A+19 i B) \cos (c+d x)-(145 A+19 i B) \cos (3 (c+d x))+6 \sin (c+d x) (7 (B-7 i A) \cos (2 (c+d x))-19 i A+2 B))+(-\sin (3 c)+i \cos (3 c)) \sqrt {\sin (2 (c+d x))} \sec (c+d x) \left (((28-30 i) A+(5+7 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))-(1+i) ((29+i) A+(1+6 i) B) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[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]^2*(Cos[d*x] + I*Sin[d*x])^3*((2*(Cos[3*d*x] - I*Sin[3*d*x])*((49*A + (19*I)*B)*Cos[c + d*x] - (1
45*A + (19*I)*B)*Cos[3*(c + d*x)] + 6*((-19*I)*A + 2*B + 7*((-7*I)*A + B)*Cos[2*(c + d*x)])*Sin[c + d*x]))/3 +
(((28 - 30*I)*A + (5 + 7*I)*B)*ArcSin[Cos[c + d*x] - Sin[c + d*x]] - (1 + I)*((29 + I)*A + (1 + 6*I)*B)*Log[C
os[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]])*Sec[c + d*x]*(I*Cos[3*c] - Sin[3*c])*Sqrt[Sin[2*(c + d*x
)]])*(A + B*Tan[c + d*x]))/(32*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^3
)

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fricas [B]  time = 0.48, size = 782, normalized size = 2.15 \[ \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 ({\left (-146 i \, A + 20 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (-105 i \, A + 6 \, 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 )} + {\left (9 i \, A - 6 \, 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 )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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*((-146*I*A + 20*B)*e^(8*I*d*x + 8*I*c) + (-105*I*A + 6*B)*e^(6*I*d*x
+ 6*I*c) + (49*I*A - 19*B)*e^(4*I*d*x + 4*I*c) + (9*I*A - 6*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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \tan \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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maple [A]  time = 0.43, size = 368, normalized size = 1.01 \[ -\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) A}{4 d \,a^{3} \left (\sqrt {2}+i \sqrt {2}\right )}+\frac {i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) B}{4 d \,a^{3} \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {2 A}{d \,a^{3} \sqrt {\tan \left (d x +c \right )}}-\frac {7 A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{4 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {5 i B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {49 i \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) A}{12 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {19 B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{12 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {9 i B \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {5 A \left (\sqrt {\tan }\left (d x +c \right )\right )}{2 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {3 i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) B}{2 d \,a^{3} \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {29 \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) A}{4 d \,a^{3} \left (\sqrt {2}-i \sqrt {2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d/a^3/(2^(1/2)+I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2)))*A+1/4*I/d/a^3/(2^(1/2)+I*2^(1/2)
)*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2)))*B-2/d/a^3*A/tan(d*x+c)^(1/2)-7/4/d/a^3/(tan(d*x+c)-I)^3*A*tan
(d*x+c)^(5/2)-5/8*I/d/a^3/(tan(d*x+c)-I)^3*B*tan(d*x+c)^(5/2)+49/12*I/d/a^3/(tan(d*x+c)-I)^3*tan(d*x+c)^(3/2)*
A-19/12/d/a^3/(tan(d*x+c)-I)^3*B*tan(d*x+c)^(3/2)+9/8*I/d/a^3/(tan(d*x+c)-I)^3*B*tan(d*x+c)^(1/2)+5/2/d/a^3/(t
an(d*x+c)-I)^3*A*tan(d*x+c)^(1/2)-3/2*I/d/a^3/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)
))*B-29/4/d/a^3/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))*A

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 6.83, size = 389, normalized size = 1.07 \[ 2\,\mathrm {atanh}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}}{29\,A}\right )\,\sqrt {-\frac {A^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}-\mathrm {atan}\left (\frac {8\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}}{3\,B}\right )\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\frac {\frac {2\,A}{a^3\,d}+\frac {A\,\mathrm {tan}\left (c+d\,x\right )\,17{}\mathrm {i}}{2\,a^3\,d}-\frac {121\,A\,{\mathrm {tan}\left (c+d\,x\right )}^2}{12\,a^3\,d}-\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^3\,15{}\mathrm {i}}{4\,a^3\,d}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}+{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,3{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}-{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,1{}\mathrm {i}}+\frac {\frac {9\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {5\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,19{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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