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

Optimal. Leaf size=347 \[ \frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}-\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (-5 B+9 i A)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {((49+45 i) A-(25-21 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((2+47 i) A-(23+2 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \]

[Out]

(-1/32+1/32*I)*((47+2*I)*A+(2+23*I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a^2/d*2^(1/2)+(-1/32+1/32*I)*((47+2
*I)*A+(2+23*I)*B)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/a^2/d*2^(1/2)+1/64*((49+45*I)*A+(-25+21*I)*B)*ln(1-2^(1/2
)*tan(d*x+c)^(1/2)+tan(d*x+c))/a^2/d*2^(1/2)+(-1/64+1/64*I)*((2+47*I)*A-(23+2*I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1
/2)+tan(d*x+c))/a^2/d*2^(1/2)+5/8*(9*I*A-5*B)/a^2/d/tan(d*x+c)^(1/2)-7/24*(7*A+3*I*B)/a^2/d/tan(d*x+c)^(3/2)+1
/8*(9*A+5*I*B)/a^2/d/(1+I*tan(d*x+c))/tan(d*x+c)^(3/2)+1/4*(A+I*B)/d/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^2

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Rubi [A]  time = 0.63, antiderivative size = 347, 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} \[ \frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}-\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (-5 B+9 i A)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {((49+45 i) A-(25-21 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((2+47 i) A-(23+2 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/16 - I/16)*((47 + 2*I)*A + (2 + 23*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a^2*d) - ((1/16
- I/16)*((47 + 2*I)*A + (2 + 23*I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a^2*d) + (((49 + 45*I)*
A - (25 - 21*I)*B)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(32*Sqrt[2]*a^2*d) - ((1/32 - I/32)*((2
+ 47*I)*A - (23 + 2*I)*B)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(Sqrt[2]*a^2*d) - (7*(7*A + (3*
I)*B))/(24*a^2*d*Tan[c + d*x]^(3/2)) + (9*A + (5*I)*B)/(8*a^2*d*(1 + I*Tan[c + d*x])*Tan[c + d*x]^(3/2)) + (5*
((9*I)*A - 5*B))/(8*a^2*d*Sqrt[Tan[c + d*x]]) + (A + I*B)/(4*d*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^2)

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))^2} \, dx &=\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a (11 A+3 i B)-\frac {7}{2} a (i A-B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx}{4 a^2}\\ &=\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {7}{2} a^2 (7 A+3 i B)-\frac {5}{2} a^2 (9 i A-5 B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{8 a^4}\\ &=-\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {-\frac {5}{2} a^2 (9 i A-5 B)-\frac {7}{2} a^2 (7 A+3 i B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{8 a^4}\\ &=-\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 i A-5 B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {-\frac {7}{2} a^2 (7 A+3 i B)+\frac {5}{2} a^2 (9 i A-5 B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 i A-5 B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {7}{2} a^2 (7 A+3 i B)+\frac {5}{2} a^2 (9 i A-5 B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a^4 d}\\ &=-\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 i A-5 B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}-\frac {((49+45 i) A-(25-21 i) B) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^2 d}+-\frac {\left (\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B)\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d}\\ &=-\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 i A-5 B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {((49+45 i) A-(25-21 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^2 d}+\frac {((49+45 i) A-(25-21 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^2 d}+-\frac {\left (\left (\frac {1}{32}-\frac {i}{32}\right ) ((47+2 i) A+(2+23 i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d}+-\frac {\left (\left (\frac {1}{32}-\frac {i}{32}\right ) ((47+2 i) A+(2+23 i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d}\\ &=\frac {((49+45 i) A-(25-21 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {((49+45 i) A-(25-21 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 i A-5 B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+-\frac {\left (\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 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^2 d}+\frac {\left (\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 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^2 d}\\ &=\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {((49+45 i) A-(25-21 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {((49+45 i) A-(25-21 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 i A-5 B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}\\ \end {align*}

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Mathematica [A]  time = 3.52, size = 282, normalized size = 0.81 \[ -\frac {i \sec (c+d x) (\cos (d x)+i \sin (d x))^2 (A+B \tan (c+d x)) \left (\frac {1}{3} \csc (c+d x) (\cos (2 d x)-i \sin (2 d x)) ((129 B-269 i A) \cos (c+d x)+(-129 B+205 i A) \cos (3 (c+d x))-2 \sin (c+d x) ((199 A+123 i B) \cos (2 (c+d x))-71 A-27 i B))+(1+i) (\cos (2 c)+i \sin (2 c)) \sqrt {\sin (2 (c+d x))} \sec (c+d x) \left (((47+2 i) A+(2+23 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+((23+2 i) B-(2+47 i) A) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

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])^2),x]

[Out]

((-1/32*I)*Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x])^2*((Csc[c + d*x]*(Cos[2*d*x] - I*Sin[2*d*x])*(((-269*I)*A + 12
9*B)*Cos[c + d*x] + ((205*I)*A - 129*B)*Cos[3*(c + d*x)] - 2*(-71*A - (27*I)*B + (199*A + (123*I)*B)*Cos[2*(c
+ d*x)])*Sin[c + d*x]))/3 + (1 + I)*(((47 + 2*I)*A + (2 + 23*I)*B)*ArcSin[Cos[c + d*x] - Sin[c + d*x]] + ((-2
- 47*I)*A + (23 + 2*I)*B)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]])*Sec[c + d*x]*(Cos[2*c] +
I*Sin[2*c])*Sqrt[Sin[2*(c + d*x)]])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*Sqrt[Tan[c + d*
x]]*(a + I*a*Tan[c + d*x])^2)

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fricas [B]  time = 1.18, size = 861, normalized size = 2.48 \[ \text {result too large to display} \]

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))^2,x, algorithm="fricas")

[Out]

-1/96*(6*(a^2*d*e^(8*I*d*x + 8*I*c) - 2*a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))*sqrt((-I*A^2 -
2*A*B + I*B^2)/(a^4*d^2))*log(1/4*((8*I*a^2*d*e^(2*I*d*x + 2*I*c) + 8*I*a^2*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^4*d^2)) + 8*(A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-
2*I*d*x - 2*I*c)/(I*A + B)) - 6*(a^2*d*e^(8*I*d*x + 8*I*c) - 2*a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x +
4*I*c))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^4*d^2))*log(1/4*((-8*I*a^2*d*e^(2*I*d*x + 2*I*c) - 8*I*a^2*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^4*d^2)) + 8*(A - I*B)*e
^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) + 3*(a^2*d*e^(8*I*d*x + 8*I*c) - 2*a^2*d*e^(6*I*d*x + 6*I*
c) + a^2*d*e^(4*I*d*x + 4*I*c))*sqrt((2209*I*A^2 - 2162*A*B - 529*I*B^2)/(a^4*d^2))*log(-1/8*((a^2*d*e^(2*I*d*
x + 2*I*c) + a^2*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((2209*I*A^2 - 2162*A*B -
529*I*B^2)/(a^4*d^2)) + 47*I*A - 23*B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) - 3*(a^2*d*e^(8*I*d*x + 8*I*c) - 2*a^2*d
*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))*sqrt((2209*I*A^2 - 2162*A*B - 529*I*B^2)/(a^4*d^2))*log(1/8*
((a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((2209*I
*A^2 - 2162*A*B - 529*I*B^2)/(a^4*d^2)) - 47*I*A + 23*B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) + 2*(2*(101*A + 63*I*B)
*e^(8*I*d*x + 8*I*c) - (103*A + 27*I*B)*e^(6*I*d*x + 6*I*c) - (269*A + 129*I*B)*e^(4*I*d*x + 4*I*c) + 3*(13*A
+ 9*I*B)*e^(2*I*d*x + 2*I*c) + 3*A + 3*I*B)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(a^2
*d*e^(8*I*d*x + 8*I*c) - 2*a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \tan \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

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))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.36, size = 346, normalized size = 1.00 \[ -\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) B}{2 d \,a^{2} \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) A}{2 d \,a^{2} \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {2 A}{3 d \,a^{2} \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 i A}{d \,a^{2} \sqrt {\tan \left (d x +c \right )}}-\frac {2 B}{d \,a^{2} \sqrt {\tan \left (d x +c \right )}}+\frac {13 i \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) A}{8 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {9 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) B}{8 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {15 \left (\sqrt {\tan }\left (d x +c \right )\right ) A}{8 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {11 i \left (\sqrt {\tan }\left (d x +c \right )\right ) B}{8 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {23 \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) B}{4 d \,a^{2} \left (\sqrt {2}-i \sqrt {2}\right )}+\frac {47 i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) A}{4 d \,a^{2} \left (\sqrt {2}-i \sqrt {2}\right )} \]

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))^2,x)

[Out]

-1/2/d/a^2/(2^(1/2)+I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2)))*B-1/2*I/d/a^2/(2^(1/2)+I*2^(1/2)
)*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2)))*A-2/3/d/a^2*A/tan(d*x+c)^(3/2)+4*I/d/a^2/tan(d*x+c)^(1/2)*A-2
/d/a^2/tan(d*x+c)^(1/2)*B+13/8*I/d/a^2/(tan(d*x+c)-I)^2*tan(d*x+c)^(3/2)*A-9/8/d/a^2/(tan(d*x+c)-I)^2*tan(d*x+
c)^(3/2)*B+15/8/d/a^2/(tan(d*x+c)-I)^2*tan(d*x+c)^(1/2)*A+11/8*I/d/a^2/(tan(d*x+c)-I)^2*tan(d*x+c)^(1/2)*B-23/
4/d/a^2/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))*B+47/4*I/d/a^2/(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))^2,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 = 10.48, size = 373, normalized size = 1.07 \[ -\mathrm {atan}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,2209{}\mathrm {i}}{256\,a^4\,d^2}}}{47\,A}\right )\,\sqrt {\frac {A^2\,2209{}\mathrm {i}}{256\,a^4\,d^2}}\,2{}\mathrm {i}+2\,\mathrm {atanh}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,529{}\mathrm {i}}{256\,a^4\,d^2}}}{23\,B}\right )\,\sqrt {-\frac {B^2\,529{}\mathrm {i}}{256\,a^4\,d^2}}+\frac {\frac {8\,A\,\mathrm {tan}\left (c+d\,x\right )}{3\,a^2\,d}-\frac {45\,A\,{\mathrm {tan}\left (c+d\,x\right )}^3}{8\,a^2\,d}+\frac {A\,2{}\mathrm {i}}{3\,a^2\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^2\,221{}\mathrm {i}}{24\,a^2\,d}}{2\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}-{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,1{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,1{}\mathrm {i}}-\frac {\frac {43\,B\,\mathrm {tan}\left (c+d\,x\right )}{8\,a^2\,d}-\frac {B\,2{}\mathrm {i}}{a^2\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^2\,25{}\mathrm {i}}{8\,a^2\,d}}{2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,1{}\mathrm {i}} \]

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)^2),x)

[Out]

atan((16*a^2*d*tan(c + d*x)^(1/2)*((A^2*2209i)/(256*a^4*d^2))^(1/2))/(47*A))*((A^2*2209i)/(256*a^4*d^2))^(1/2)
*2i - atan((8*a^2*d*tan(c + d*x)^(1/2)*(-(A^2*1i)/(64*a^4*d^2))^(1/2))/A)*(-(A^2*1i)/(64*a^4*d^2))^(1/2)*2i +
2*atanh((8*a^2*d*tan(c + d*x)^(1/2)*((B^2*1i)/(64*a^4*d^2))^(1/2))/B)*((B^2*1i)/(64*a^4*d^2))^(1/2) + 2*atanh(
(16*a^2*d*tan(c + d*x)^(1/2)*(-(B^2*529i)/(256*a^4*d^2))^(1/2))/(23*B))*(-(B^2*529i)/(256*a^4*d^2))^(1/2) + ((
A*2i)/(3*a^2*d) + (8*A*tan(c + d*x))/(3*a^2*d) + (A*tan(c + d*x)^2*221i)/(24*a^2*d) - (45*A*tan(c + d*x)^3)/(8
*a^2*d))/(2*tan(c + d*x)^(5/2) - tan(c + d*x)^(3/2)*1i + tan(c + d*x)^(7/2)*1i) - ((43*B*tan(c + d*x))/(8*a^2*
d) - (B*2i)/(a^2*d) + (B*tan(c + d*x)^2*25i)/(8*a^2*d))/(2*tan(c + d*x)^(3/2) - tan(c + d*x)^(1/2)*1i + tan(c
+ d*x)^(5/2)*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))**2,x)

[Out]

Timed out

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