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

Optimal. Leaf size=267 \[ \frac {((5+3 i) A-(3-i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((3-i) B-(5+3 i) A) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{4 \sqrt {2} a d}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}-\frac {5 A+i B}{2 a d \sqrt {\tan (c+d x)}}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) ((4+i) A+(1+2 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {((5-3 i) A+(3+i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{8 \sqrt {2} a d} \]

[Out]

-1/8*((5+3*I)*A+(-3+I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a/d*2^(1/2)+1/8*((-5-3*I)*A+(3-I)*B)*arctan(1+2^
(1/2)*tan(d*x+c)^(1/2))/a/d*2^(1/2)+(-1/16+1/16*I)*((4+I)*A+(1+2*I)*B)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c
))/a/d*2^(1/2)+1/16*((5-3*I)*A+(3+I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a/d*2^(1/2)+1/2*(-5*A-I*B)/a
/d/tan(d*x+c)^(1/2)+1/2*(A+I*B)/d/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))

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Rubi [A]  time = 0.37, antiderivative size = 267, 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 = {3596, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {((5+3 i) A-(3-i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((3-i) B-(5+3 i) A) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{4 \sqrt {2} a d}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}-\frac {5 A+i B}{2 a d \sqrt {\tan (c+d x)}}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) ((4+i) A+(1+2 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {((5-3 i) A+(3+i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{8 \sqrt {2} a d} \]

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])),x]

[Out]

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

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Mathematica [A]  time = 2.44, size = 217, normalized size = 0.81 \[ \frac {(\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \left ((-4 \cos (d x)+4 i \sin (d x)) (4 A \cos (c+d x)+(-B+5 i A) \sin (c+d x))+(-\sin (c)+i \cos (c)) \sqrt {\sin (2 (c+d x))} \sec (c+d x) \left (((3-5 i) A+(1+3 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))-(1+i) ((4+i) A+(1+2 i) B) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{8 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) (A \cos (c+d x)+B \sin (c+d x))} \]

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])),x]

[Out]

((Cos[d*x] + I*Sin[d*x])*((-4*Cos[d*x] + (4*I)*Sin[d*x])*(4*A*Cos[c + d*x] + ((5*I)*A - B)*Sin[c + d*x]) + (((
3 - 5*I)*A + (1 + 3*I)*B)*ArcSin[Cos[c + d*x] - Sin[c + d*x]] - (1 + I)*((4 + 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[c] - Sin[c])*Sqrt[Sin[2*(c + d*x)]])*(A + B
*Tan[c + d*x]))/(8*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x]))

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fricas [B]  time = 0.75, size = 703, normalized size = 2.63 \[ \frac {{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} \log \left (\frac {2 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a 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^{2} 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 ) - {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} \log \left (-\frac {2 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a 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^{2} 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 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a^{2} d^{2}}} \log \left (\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a 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 {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a^{2} d^{2}}} + 2 \, A + i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) - 2 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a^{2} d^{2}}} \log \left (-\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a 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 {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a^{2} d^{2}}} - 2 \, A - i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 2 \, {\left ({\left (-9 i \, A + B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, A 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}}}{8 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 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)),x, algorithm="fricas")

[Out]

1/8*((a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2))*log(2*((a*d*e
^(2*I*d*x + 2*I*c) + a*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^2*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*d*e^(4*I*d*x + 4*I*c) -
a*d*e^(2*I*d*x + 2*I*c))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2))*log(-2*((a*d*e^(2*I*d*x + 2*I*c) + a*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^2*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*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sq
rt((-4*I*A^2 + 4*A*B + I*B^2)/(a^2*d^2))*log(((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I
)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a^2*d^2)) + 2*A + I*B)*e^(-2*I*d*x - 2*I*c)/(a*d
)) - 2*(a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a^2*d^2))*log(-((a
*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-4*I*A^2 + 4*
A*B + I*B^2)/(a^2*d^2)) - 2*A - I*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) + 2*((-9*I*A + B)*e^(4*I*d*x + 4*I*c) - 8*I*A
*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*d*e^(4*I*d*x
+ 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))

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

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)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.36, size = 254, normalized size = 0.95 \[ -\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) A}{d a \left (\sqrt {2}+i \sqrt {2}\right )}+\frac {i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) B}{d a \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {2 A}{a d \sqrt {\tan \left (d x +c \right )}}-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) A}{2 d a \left (\tan \left (d x +c \right )-i\right )}-\frac {i \left (\sqrt {\tan }\left (d x +c \right )\right ) B}{2 d a \left (\tan \left (d x +c \right )-i\right )}-\frac {2 i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) B}{d a \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {4 \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) A}{d a \left (\sqrt {2}-i \sqrt {2}\right )} \]

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)),x)

[Out]

-1/d/a/(2^(1/2)+I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2)))*A+I/d/a/(2^(1/2)+I*2^(1/2))*arctan(2
*tan(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2)))*B-2*A/a/d/tan(d*x+c)^(1/2)-1/2/d/a*tan(d*x+c)^(1/2)/(tan(d*x+c)-I)*A-1/
2*I/d/a*tan(d*x+c)^(1/2)/(tan(d*x+c)-I)*B-2*I/d/a*B/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2
^(1/2)))-4/d/a/(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)),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 = 7.37, size = 266, normalized size = 1.00 \[ 2\,\mathrm {atanh}\left (\frac {a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{a^2\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{a^2\,d^2}}+2\,\mathrm {atanh}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}-\mathrm {atan}\left (\frac {2\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}-\frac {\frac {2\,A}{a\,d}+\frac {A\,\mathrm {tan}\left (c+d\,x\right )\,5{}\mathrm {i}}{2\,a\,d}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}+{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,1{}\mathrm {i}}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]

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)),x)

[Out]

2*atanh((a*d*tan(c + d*x)^(1/2)*(-(A^2*1i)/(a^2*d^2))^(1/2))/A)*(-(A^2*1i)/(a^2*d^2))^(1/2) + 2*atanh((4*a*d*t
an(c + d*x)^(1/2)*((A^2*1i)/(16*a^2*d^2))^(1/2))/A)*((A^2*1i)/(16*a^2*d^2))^(1/2) - atan((2*a*d*tan(c + d*x)^(
1/2)*((B^2*1i)/(4*a^2*d^2))^(1/2))/B)*((B^2*1i)/(4*a^2*d^2))^(1/2)*2i + atan((4*a*d*tan(c + d*x)^(1/2)*(-(B^2*
1i)/(16*a^2*d^2))^(1/2))/B)*(-(B^2*1i)/(16*a^2*d^2))^(1/2)*2i - ((2*A)/(a*d) + (A*tan(c + d*x)*5i)/(2*a*d))/(t
an(c + d*x)^(1/2) + tan(c + d*x)^(3/2)*1i) + (B*tan(c + d*x)^(1/2))/(2*a*d*(tan(c + d*x)*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)),x)

[Out]

Exception raised: TypeError

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