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

Optimal. Leaf size=317 \[ \frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (B+(1+i) A) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (B+(1+i) A) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {(2 i A+(1-i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {(2 i A+(1-i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(-B+i A) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 B+i A) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2} \]

[Out]

(-1/32-1/32*I)*((1+I)*A+B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a^3/d*2^(1/2)-(1/32+1/32*I)*((1+I)*A+B)*arctan(
1+2^(1/2)*tan(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/64*(2*I*A+(1-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*(2*I*A+(1-I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a^3/d*2^(1/2)+1/6*(I*A-B)*tan(d*x+c)^(
1/2)/d/(a+I*a*tan(d*x+c))^3+1/12*(I*A+2*B)*tan(d*x+c)^(1/2)/a/d/(a+I*a*tan(d*x+c))^2+1/8*B*tan(d*x+c)^(1/2)/d/
(a^3+I*a^3*tan(d*x+c))

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Rubi [A]  time = 0.62, antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3595, 3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (B+(1+i) A) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (B+(1+i) A) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {(2 i A+(1-i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {(2 i A+(1-i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(-B+i A) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 B+i A) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/16 + I/16)*((1 + I)*A + B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a^3*d) - ((1/16 + I/16)*((1 +
I)*A + B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a^3*d) + (((2*I)*A + (1 - I)*B)*Log[1 - Sqrt[2]*Sqr
t[Tan[c + d*x]] + Tan[c + d*x]])/(32*Sqrt[2]*a^3*d) - (((2*I)*A + (1 - I)*B)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]
] + Tan[c + d*x]])/(32*Sqrt[2]*a^3*d) + ((I*A - B)*Sqrt[Tan[c + d*x]])/(6*d*(a + I*a*Tan[c + d*x])^3) + ((I*A
+ 2*B)*Sqrt[Tan[c + d*x]])/(12*a*d*(a + I*a*Tan[c + d*x])^2) + (B*Sqrt[Tan[c + d*x]])/(8*d*(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 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 3595

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*b - a*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*a*f*
m), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d*n
) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A,
B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 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 {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx &=\frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\frac {1}{2} a (i A-B)-\frac {1}{2} a (5 A-7 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}-\frac {\int \frac {3 i a^2 A-3 a^2 (A-2 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \, dx}{24 a^4}\\ &=\frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \frac {3 a^3 (2 i A+B)+3 i a^3 B \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{48 a^6}\\ &=\frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {3 a^3 (2 i A+B)+3 i a^3 B x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{24 a^6 d}\\ &=\frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+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 {(2 i A+(1-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 {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\left (\left (\frac {1}{32}+\frac {i}{32}\right ) ((1+i) A+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 ) ((1+i) A+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 {(2 i A+(1-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 {(2 i A+(1-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 {(2 i A+(1-i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {(2 i A+(1-i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}--\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+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 ) ((1+i) A+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 ) ((1+i) A+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 ) ((1+i) A+B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {(2 i A+(1-i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {(2 i A+(1-i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(i A+2 B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 3.65, size = 272, normalized size = 0.86 \[ \frac {e^{-4 i (c+d x)} \sec (c+d x) (\cos (3 (c+d x))-i \sin (3 (c+d x))) \left (\left (-2 e^{2 i (c+d x)}+e^{4 i (c+d x)}+2 e^{6 i (c+d x)}-1\right ) \left (A e^{2 i (c+d x)}+A-2 i B e^{2 i (c+d x)}+i B\right )-6 (A-i B) e^{6 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )-3 A e^{6 i (c+d x)} \sqrt {-1+e^{4 i (c+d x)}} \tan ^{-1}\left (\sqrt {-1+e^{4 i (c+d x)}}\right )\right )}{96 a^3 d \sqrt {\tan (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((A + I*B + A*E^((2*I)*(c + d*x)) - (2*I)*B*E^((2*I)*(c + d*x)))*(-1 - 2*E^((2*I)*(c + d*x)) + E^((4*I)*(c +
d*x)) + 2*E^((6*I)*(c + d*x))) - 3*A*E^((6*I)*(c + d*x))*Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4
*I)*(c + d*x))]] - 6*(A - I*B)*E^((6*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))
]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])*Sec[c + d*x]*(Cos[3*(c + d*x)] - I*Sin[
3*(c + d*x)]))/(96*a^3*d*E^((4*I)*(c + d*x))*Sqrt[Tan[c + d*x]])

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fricas [B]  time = 0.93, size = 630, normalized size = 1.99 \[ -\frac {{\left (3 \, a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (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 \, a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (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 ) - 24 \, a^{3} d \sqrt {-\frac {i \, A^{2}}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left (8 \, {\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}}{64 \, a^{6} d^{2}}} + A\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 24 \, a^{3} d \sqrt {-\frac {i \, A^{2}}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {{\left (8 \, {\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}}{64 \, a^{6} d^{2}}} - A\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 2 \, {\left ({\left (2 i \, A + 4 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (5 i \, A + 4 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (4 i \, A - 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}}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(3*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(2*((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*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^
2))*e^(6*I*d*x + 6*I*c)*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)) - 24*a^3*d*sqrt(-1/64*I*A^2/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(1/8*(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(-1/64*I*A^2/(a^6*d^2)) + A)*e^(-2*
I*d*x - 2*I*c)/(a^3*d)) + 24*a^3*d*sqrt(-1/64*I*A^2/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-1/8*(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(-1/64*I*A^2/(a^6*d^2))
- A)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - 2*((2*I*A + 4*B)*e^(6*I*d*x + 6*I*c) + (5*I*A + 4*B)*e^(4*I*d*x + 4*I*c)
+ (4*I*A - 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)))*e^(
-6*I*d*x - 6*I*c)/(a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.47, size = 278, normalized size = 0.88 \[ \frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) A}{4 d \,a^{3} \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) B}{4 d \,a^{3} \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {i B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {5 B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{12 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) A}{12 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {i B \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {A \left (\sqrt {\tan }\left (d x +c \right )\right )}{4 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) A}{4 d \,a^{3} \left (\sqrt {2}-i \sqrt {2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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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(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(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 = 6.59, size = 239, normalized size = 0.75 \[ \frac {\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,5{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}+\frac {-\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{12\,a^3\,d}+\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{4\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}-\frac {{\left (-1\right )}^{1/4}\,A\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{8\,a^3\,d}-\frac {{\left (-1\right )}^{1/4}\,A\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{8\,a^3\,d}-\frac {\sqrt {-\frac {1}{256}{}\mathrm {i}}\,B\,\mathrm {atan}\left (16\,\sqrt {-\frac {1}{256}{}\mathrm {i}}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )\,2{}\mathrm {i}}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((B*tan(c + d*x)^(1/2))/(8*a^3*d) + (B*tan(c + d*x)^(3/2)*5i)/(12*a^3*d) - (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) + ((A*tan(c + d*x)^(1/2)*1i)/(4*a^3*d) - (A*tan(c
+ d*x)^(3/2))/(12*a^3*d))/(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1) - ((-1)^(1/4)*A*atan((-
1)^(1/4)*tan(c + d*x)^(1/2)))/(8*a^3*d) - ((-1)^(1/4)*A*atanh((-1)^(1/4)*tan(c + d*x)^(1/2)))/(8*a^3*d) - ((-1
i/256)^(1/2)*B*atan(16*(-1i/256)^(1/2)*tan(c + d*x)^(1/2))*2i)/(a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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