3.153 \(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=393 \[ \frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (-2 B+5 i A)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {((77+75 i) A-(30-28 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((1+76 i) A-(29+i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \]

[Out]

(-1/32+1/32*I)*((76+I)*A+(1+29*I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a^3/d*2^(1/2)+(-1/32+1/32*I)*((76+I)*
A+(1+29*I)*B)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/64*((77+75*I)*A+(-30+28*I)*B)*ln(1-2^(1/2)*ta
n(d*x+c)^(1/2)+tan(d*x+c))/a^3/d*2^(1/2)+(-1/64+1/64*I)*((1+76*I)*A-(29+I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+ta
n(d*x+c))/a^3/d*2^(1/2)+15/8*(5*I*A-2*B)/a^3/d/tan(d*x+c)^(1/2)-7/24*(11*A+4*I*B)/a^3/d/tan(d*x+c)^(3/2)+1/6*(
A+I*B)/d/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3+1/4*(2*A+I*B)/a/d/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^2+3/8*(5*
A+2*I*B)/d/tan(d*x+c)^(3/2)/(a^3+I*a^3*tan(d*x+c))

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Rubi [A]  time = 0.86, antiderivative size = 393, normalized size of antiderivative = 1.00, number of steps used = 15, 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} \[ \frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (-2 B+5 i A)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {((77+75 i) A-(30-28 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((1+76 i) A-(29+i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^3),x]

[Out]

((1/16 - I/16)*((76 + I)*A + (1 + 29*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a^3*d) - ((1/16 -
I/16)*((76 + I)*A + (1 + 29*I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a^3*d) + (((77 + 75*I)*A -
(30 - 28*I)*B)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(32*Sqrt[2]*a^3*d) - ((1/32 - I/32)*((1 + 7
6*I)*A - (29 + I)*B)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(Sqrt[2]*a^3*d) - (7*(11*A + (4*I)*B)
)/(24*a^3*d*Tan[c + d*x]^(3/2)) + (15*((5*I)*A - 2*B))/(8*a^3*d*Sqrt[Tan[c + d*x]]) + (A + I*B)/(6*d*Tan[c + d
*x]^(3/2)*(a + I*a*Tan[c + d*x])^3) + (2*A + I*B)/(4*a*d*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^2) + (3*(5*
A + (2*I)*B))/(8*d*Tan[c + d*x]^(3/2)*(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 {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx &=\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {\int \frac {\frac {3}{2} a (5 A+i B)-\frac {9}{2} a (i A-B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {3 a^2 (16 A+5 i B)-21 a^2 (2 i A-B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx}{24 a^4}\\ &=\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \frac {21 a^3 (11 A+4 i B)-45 a^3 (5 i A-2 B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{48 a^6}\\ &=-\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \frac {-45 a^3 (5 i A-2 B)-21 a^3 (11 A+4 i B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{48 a^6}\\ &=-\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (5 i A-2 B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \frac {-21 a^3 (11 A+4 i B)+45 a^3 (5 i A-2 B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{48 a^6}\\ &=-\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (5 i A-2 B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {-21 a^3 (11 A+4 i B)+45 a^3 (5 i A-2 B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{24 a^6 d}\\ &=-\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (5 i A-2 B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {((77+75 i) A-(30-28 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 {\left (\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B)\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^3 d}\\ &=-\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (5 i A-2 B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {((77+75 i) A-(30-28 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 {((77+75 i) A-(30-28 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 {\left (\left (\frac {1}{32}-\frac {i}{32}\right ) ((76+i) A+(1+29 i) B)\right ) \operatorname {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 ) ((76+i) A+(1+29 i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^3 d}\\ &=\frac {((77+75 i) A-(30-28 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {((77+75 i) A-(30-28 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 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (5 i A-2 B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}+-\frac {\left (\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B)\right ) \operatorname {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 ) ((76+i) A+(1+29 i) B)\right ) \operatorname {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 ) ((76+i) A+(1+29 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {((77+75 i) A-(30-28 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {((77+75 i) A-(30-28 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 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (5 i A-2 B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 4.37, size = 306, normalized size = 0.78 \[ \frac {\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^3 (A+B \tan (c+d x)) \left (\frac {1}{3} \csc (c+d x) (\cos (3 d x)-i \sin (3 d x)) (-2 (241 A+90 i B) \cos (2 (c+d x))+(349 A+147 i B) \cos (4 (c+d x))-502 i A \sin (2 (c+d x))+347 i A \sin (4 (c+d x))+69 A+194 B \sin (2 (c+d x))-145 B \sin (4 (c+d x))+33 i B)+(1-i) (\cos (3 c)+i \sin (3 c)) \sqrt {\sin (2 (c+d x))} \sec (c+d x) \left (((76+i) A+(1+29 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+((29+i) B-(1+76 i) A) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^3),x]

[Out]

(Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])^3*((1 - I)*(((76 + I)*A + (1 + 29*I)*B)*ArcSin[Cos[c + d*x] - Sin[c +
d*x]] + ((-1 - 76*I)*A + (29 + I)*B)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]])*Sec[c + d*x]*(
Cos[3*c] + I*Sin[3*c])*Sqrt[Sin[2*(c + d*x)]] + (Csc[c + d*x]*(Cos[3*d*x] - I*Sin[3*d*x])*(69*A + (33*I)*B - 2
*(241*A + (90*I)*B)*Cos[2*(c + d*x)] + (349*A + (147*I)*B)*Cos[4*(c + d*x)] - (502*I)*A*Sin[2*(c + d*x)] + 194
*B*Sin[2*(c + d*x)] + (347*I)*A*Sin[4*(c + d*x)] - 145*B*Sin[4*(c + d*x)]))/3)*(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 = 0.68, size = 875, normalized size = 2.23 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/96*(3*(a^3*d*e^(10*I*d*x + 10*I*c) - 2*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(1/8*((16*I*a^3*d*e^(2*I*d*x + 2*I*c) + 16*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)) + 16*(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^(10*I*d*x + 10*I*c) - 2*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(1/8*((-16*I*a^3*d*e^(2*I*d*x + 2*I*c) - 16*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)) + 16*(
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^(10*I*d*x + 10*I*c) - 2*a^3*d*e^(8*
I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))*sqrt((5776*I*A^2 - 4408*A*B - 841*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((5776*I*A^2
- 4408*A*B - 841*I*B^2)/(a^6*d^2)) + 76*I*A - 29*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - 3*(a^3*d*e^(10*I*d*x + 10*
I*c) - 2*a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))*sqrt((5776*I*A^2 - 4408*A*B - 841*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((5776*I*A^2 - 4408*A*B - 841*I*B^2)/(a^6*d^2)) - 76*I*A + 29*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) + 2*(2*(1
74*A + 73*I*B)*e^(10*I*d*x + 10*I*c) - (144*A + 41*I*B)*e^(8*I*d*x + 8*I*c) - (423*A + 154*I*B)*e^(6*I*d*x + 6
*I*c) + (79*A + 40*I*B)*e^(4*I*d*x + 4*I*c) + (11*A + 8*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)))/(a^3*d*e^(10*I*d*x + 10*I*c) - 2*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 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \tan \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(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)^(5/2)), x)

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maple [A]  time = 0.47, size = 403, normalized size = 1.03 \[ -\frac {i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) A}{4 d \,a^{3} \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) B}{4 d \,a^{3} \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {2 A}{3 d \,a^{3} \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {6 i A}{d \,a^{3} \sqrt {\tan \left (d x +c \right )}}-\frac {2 B}{d \,a^{3} \sqrt {\tan \left (d x +c \right )}}+\frac {27 i A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {7 B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{4 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {91 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) A}{12 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {49 i \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) B}{12 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {5 B \left (\sqrt {\tan }\left (d x +c \right )\right )}{2 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {35 i A \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {29 \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) B}{4 d \,a^{3} \left (\sqrt {2}-i \sqrt {2}\right )}+\frac {19 i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) A}{d \,a^{3} \left (\sqrt {2}-i \sqrt {2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^3,x)

[Out]

-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)))*A-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)))*B-2/3/d/a^3*A/tan(d*x+c)^(3/2)+6*I/d/a^3/tan(d*x+c)^(1/2)*A-2
/d/a^3/tan(d*x+c)^(1/2)*B+27/8*I/d/a^3/(tan(d*x+c)-I)^3*A*tan(d*x+c)^(5/2)-7/4/d/a^3/(tan(d*x+c)-I)^3*B*tan(d*
x+c)^(5/2)+91/12/d/a^3/(tan(d*x+c)-I)^3*tan(d*x+c)^(3/2)*A+49/12*I/d/a^3/(tan(d*x+c)-I)^3*tan(d*x+c)^(3/2)*B+5
/2/d/a^3/(tan(d*x+c)-I)^3*B*tan(d*x+c)^(1/2)-35/8*I/d/a^3/(tan(d*x+c)-I)^3*A*tan(d*x+c)^(1/2)-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)))*B+19*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)))*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)^(5/2)/(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 = 8.53, size = 425, normalized size = 1.08 \[ -\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {4\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,361{}\mathrm {i}}{16\,a^6\,d^2}}}{19\,A}\right )\,\sqrt {\frac {A^2\,361{}\mathrm {i}}{16\,a^6\,d^2}}\,2{}\mathrm {i}-\frac {\frac {A\,2{}\mathrm {i}}{3\,a^3\,d}+\frac {4\,A\,\mathrm {tan}\left (c+d\,x\right )}{a^3\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^2\,163{}\mathrm {i}}{8\,a^3\,d}-\frac {299\,A\,{\mathrm {tan}\left (c+d\,x\right )}^3}{12\,a^3\,d}-\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^4\,75{}\mathrm {i}}{8\,a^3\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^{9/2}-3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}+{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,1{}\mathrm {i}-{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,3{}\mathrm {i}}+2\,\mathrm {atanh}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}}{29\,B}\right )\,\sqrt {-\frac {B^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}-\frac {\frac {2\,B}{a^3\,d}+\frac {B\,\mathrm {tan}\left (c+d\,x\right )\,17{}\mathrm {i}}{2\,a^3\,d}-\frac {121\,B\,{\mathrm {tan}\left (c+d\,x\right )}^2}{12\,a^3\,d}-\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^3\,15{}\mathrm {i}}{4\,a^3\,d}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}+{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,3{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}-{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(c + d*x))/(tan(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^3),x)

[Out]

atan((4*a^3*d*tan(c + d*x)^(1/2)*((A^2*361i)/(16*a^6*d^2))^(1/2))/(19*A))*((A^2*361i)/(16*a^6*d^2))^(1/2)*2i -
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 - ((
A*2i)/(3*a^3*d) + (4*A*tan(c + d*x))/(a^3*d) + (A*tan(c + d*x)^2*163i)/(8*a^3*d) - (299*A*tan(c + d*x)^3)/(12*
a^3*d) - (A*tan(c + d*x)^4*75i)/(8*a^3*d))/(tan(c + d*x)^(3/2)*1i - 3*tan(c + d*x)^(5/2) - tan(c + d*x)^(7/2)*
3i + tan(c + d*x)^(9/2)) + 2*atanh((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) + 2*atanh((16*a^3*d*tan(c + d*x)^(1/2)*(-(B^2*841i)/(256*a^6*d^2))^(1/2))/(29*B))*(-(B^2*8
41i)/(256*a^6*d^2))^(1/2) - ((2*B)/(a^3*d) + (B*tan(c + d*x)*17i)/(2*a^3*d) - (121*B*tan(c + d*x)^2)/(12*a^3*d
) - (B*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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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