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

Optimal. Leaf size=279 \[ \frac {((1+3 i) B-(1-3 i) A) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((1+3 i) B-(1-3 i) A) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{16 \sqrt {2} a^2 d}+\frac {(3 B+i A) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {((1+3 i) A+(1-3 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(1-3 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}+\frac {(-B+i A) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

-1/32*((-1+3*I)*A+(1+3*I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a^2/d*2^(1/2)-1/32*((-1+3*I)*A+(1+3*I)*B)*arc
tan(1+2^(1/2)*tan(d*x+c)^(1/2))/a^2/d*2^(1/2)+1/64*((1+3*I)*A+(1-3*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+3*I)*A+(1-3*I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a^2/d*2^(1/2)+1/8*(I*A
+3*B)*tan(d*x+c)^(1/2)/a^2/d/(1+I*tan(d*x+c))+1/4*(I*A-B)*tan(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))^2

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Rubi [A]  time = 0.47, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 12, 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} \[ \frac {((1+3 i) B-(1-3 i) A) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((1+3 i) B-(1-3 i) A) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{16 \sqrt {2} a^2 d}+\frac {(3 B+i A) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {((1+3 i) A+(1-3 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(1-3 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}+\frac {(-B+i A) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-1 + 3*I)*A + (1 + 3*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^2*d) - (((-1 + 3*I)*A + (1
+ 3*I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^2*d) + (((1 + 3*I)*A + (1 - 3*I)*B)*Log[1 - S
qrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(32*Sqrt[2]*a^2*d) - (((1 + 3*I)*A + (1 - 3*I)*B)*Log[1 + Sqrt[2]*S
qrt[Tan[c + d*x]] + Tan[c + d*x]])/(32*Sqrt[2]*a^2*d) + ((I*A + 3*B)*Sqrt[Tan[c + d*x]])/(8*a^2*d*(1 + I*Tan[c
+ d*x])) + ((I*A - B)*Sqrt[Tan[c + d*x]])/(4*d*(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 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))^2} \, dx &=\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\frac {1}{2} a (i A-B)-\frac {1}{2} a (3 A-5 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \, dx}{4 a^2}\\ &=\frac {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\frac {1}{2} a^2 (3 i A+B)-\frac {1}{2} a^2 (A-3 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{8 a^4}\\ &=\frac {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} a^2 (3 i A+B)-\frac {1}{2} a^2 (A-3 i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a^4 d}\\ &=\frac {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {((1+3 i) A+(1-3 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 {((-1+3 i) A+(1+3 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 {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}+\frac {((1+3 i) A+(1-3 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 {((1+3 i) A+(1-3 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 {((-1+3 i) A+(1+3 i) B) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^2 d}-\frac {((-1+3 i) A+(1+3 i) B) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^2 d}\\ &=\frac {((1+3 i) A+(1-3 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(1-3 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {((-1+3 i) A+(1+3 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^2 d}+\frac {((-1+3 i) A+(1+3 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^2 d}\\ &=\frac {((-1+3 i) A+(1+3 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((-1+3 i) A+(1+3 i) B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {((1+3 i) A+(1-3 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(1-3 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}\\ \end {align*}

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Mathematica [A]  time = 2.21, size = 241, normalized size = 0.86 \[ \frac {\sec (c+d x) (\cos (d x)+i \sin (d x))^2 (A+B \tan (c+d x)) \left ((1-i) (-\sin (2 c)+i \cos (2 c)) \sqrt {\sin (2 (c+d x))} \sec (c+d x) \left (((1+2 i) A+(2+i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+((1+2 i) B-(2+i) A) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )-4 \sin (c+d x) (\cos (2 d x)-i \sin (2 d x)) ((A-3 i B) \sin (c+d x)+(-B-3 i A) \cos (c+d x))\right )}{32 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x])^2*(-4*(Cos[2*d*x] - I*Sin[2*d*x])*Sin[c + d*x]*(((-3*I)*A - B)*Cos[c + d
*x] + (A - (3*I)*B)*Sin[c + d*x]) + (1 - I)*(((1 + 2*I)*A + (2 + I)*B)*ArcSin[Cos[c + d*x] - Sin[c + d*x]] + (
(-2 - I)*A + (1 + 2*I)*B)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]])*Sec[c + d*x]*(I*Cos[2*c]
- Sin[2*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])^2)

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fricas [B]  time = 0.54, size = 658, normalized size = 2.36 \[ -\frac {{\left (2 \, a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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^{4} 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 ) - 2 \, a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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^{4} 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 ) - a^{2} d \sqrt {\frac {-i \, A^{2} + 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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^{4} d^{2}}} + A + i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + a^{2} d \sqrt {\frac {-i \, A^{2} + 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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^{4} d^{2}}} - A - i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) - 2 \, {\left ({\left (2 i \, A + 2 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (3 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 (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

-1/32*(2*a^2*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(2*((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((I*A^2 + 2*A*B - I*B^2)/(a^4*d^2)) +
(A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 2*a^2*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^4*d^
2))*e^(4*I*d*x + 4*I*c)*log(-2*((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((I*A^2 + 2*A*B - I*B^2)/(a^4*d^2)) - (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*
c)/(I*A + B)) - a^2*d*sqrt((-I*A^2 + 2*A*B + I*B^2)/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*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((-I*A^2 + 2*A*B + I*B^2)/(
a^4*d^2)) + A + I*B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) + a^2*d*sqrt((-I*A^2 + 2*A*B + I*B^2)/(a^4*d^2))*e^(4*I*d*x
+ 4*I*c)*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((-I*A^2 + 2*A*B + I*B^2)/(a^4*d^2)) - A - I*B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) - 2*((2*I*A + 2*B)*e^
(4*I*d*x + 4*I*c) + (3*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^(-4*I*d*x - 4*I*c)/(a^2*d)

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

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))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.49, size = 294, normalized size = 1.05 \[ \frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) A}{2 d \,a^{2} \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) B}{2 d \,a^{2} \left (\sqrt {2}+i \sqrt {2}\right )}+\frac {\left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) A}{8 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {3 i \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) B}{8 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {3 i \left (\sqrt {\tan }\left (d x +c \right )\right ) A}{8 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) B}{8 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) A}{4 d \,a^{2} \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) B}{4 d \,a^{2} \left (\sqrt {2}-i \sqrt {2}\right )} \]

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

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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))^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 = 9.75, size = 318, normalized size = 1.14 \[ \frac {\frac {3\,A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^2\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,1{}\mathrm {i}}{8\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}-\frac {-\frac {3\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{8\,a^2\,d}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{8\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}-2\,\mathrm {atanh}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}-\mathrm {atan}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}\,2{}\mathrm {i} \]

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)^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {A \sqrt {\tan {\left (c + d x \right )}}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx + \int \frac {B \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]

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))**2,x)

[Out]

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

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