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

Optimal. Leaf size=243 \[ \frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {2 (-1)^{3/4} B \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {-B+i A}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[Out]

2*(-1)^(3/4)*B*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c
)^(1/2)/a^(3/2)/d+(1/4+1/4*I)*(I*A+B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x
+c)^(1/2)*tan(d*x+c)^(1/2)/a^(3/2)/d+1/2*(A+3*I*B)/a/d/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2)+1/3*(I*A-B)/d
/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 0.83, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {4241, 3595, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {2 (-1)^{3/4} B \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {-B+i A}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-1)^(3/4)*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*
Sqrt[Tan[c + d*x]])/(a^(3/2)*d) + ((1/4 + I/4)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a +
I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/(a^(3/2)*d) + (I*A - B)/(3*d*Cot[c + d*x]^(3/2)*(a
+ I*a*Tan[c + d*x])^(3/2)) + (A + (3*I)*B)/(2*a*d*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^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 3599

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*B)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]

Rule 3601

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), x], x] /; FreeQ[{a, b
, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rubi steps

\begin {align*} \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx\\ &=\frac {i A-B}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {\tan (c+d x)} \left (\frac {3}{2} a (i A-B)+3 i a B \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac {i A-B}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} a^2 (A+3 i B)-3 a^2 B \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{3 a^4}\\ &=\frac {i A-B}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left ((A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{4 a^2}-\frac {\left (i B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{a^3}\\ &=\frac {i A-B}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (i (A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{2 d}-\frac {\left (i B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (i A+B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{3/2} d}+\frac {i A-B}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (2 i B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a d}\\ &=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (i A+B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{3/2} d}+\frac {i A-B}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (2 i B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a d}\\ &=\frac {2 (-1)^{3/4} B \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{3/2} d}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (i A+B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{3/2} d}+\frac {i A-B}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 8.01, size = 388, normalized size = 1.60 \[ \frac {e^{-2 i (c+d x)} \sqrt {\cot (c+d x)} \sec (c+d x) \left (3 (B+i A) e^{3 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \log \left (\sqrt {-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )+5 i A e^{2 i (c+d x)}-4 i A e^{4 i (c+d x)}-i A-11 B e^{2 i (c+d x)}+10 B e^{4 i (c+d x)}-3 \sqrt {2} B e^{3 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \log \left (-2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )+3 \sqrt {2} B e^{3 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \log \left (2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )+B\right ) (A+B \tan (c+d x))}{12 d (a+i a \tan (c+d x))^{3/2} (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cot[c + d*x]]*((-I)*A + B + (5*I)*A*E^((2*I)*(c + d*x)) - 11*B*E^((2*I)*(c + d*x)) - (4*I)*A*E^((4*I)*(c
+ d*x)) + 10*B*E^((4*I)*(c + d*x)) + 3*(I*A + B)*E^((3*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*Log[E^(I*
(c + d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]] - 3*Sqrt[2]*B*E^((3*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*
Log[1 - 3*E^((2*I)*(c + d*x)) - 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]] + 3*Sqrt[2]*B*E^((3*
I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*Log[1 - 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[-1
+ E^((2*I)*(c + d*x))]])*Sec[c + d*x]*(A + B*Tan[c + d*x]))/(12*d*E^((2*I)*(c + d*x))*(A*Cos[c + d*x] + B*Sin
[c + d*x])*(a + I*a*Tan[c + d*x])^(3/2))

________________________________________________________________________________________

fricas [B]  time = 0.52, size = 781, normalized size = 3.21 \[ -\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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^{3} d^{2}}} + {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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^{3} d^{2}}} - {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + 3 \, a^{2} d \sqrt {\frac {4 i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {{\left (48 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 16 \, B a^{2} - \sqrt {2} {\left (16 i \, a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} - 16 i \, a^{3} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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 \, B^{2}}{a^{3} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B}\right ) - 3 \, a^{2} d \sqrt {\frac {4 i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {{\left (48 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 16 \, B a^{2} - \sqrt {2} {\left (-16 i \, a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + 16 i \, a^{3} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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 \, B^{2}}{a^{3} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B}\right ) - \sqrt {2} {\left ({\left (-4 i \, A + 10 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (5 i \, A - 11 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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 (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(-4*(sqrt(2)*sqrt(1/2
)*(a^2*d*e^(2*I*d*x + 2*I*c) - a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*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^3*d^2)) + (A - I*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)
/(I*A + B)) - 3*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(4*(sqrt(2)*sq
rt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c) - a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*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^3*d^2)) - (A - I*B)*a*e^(I*d*x + I*c))*e^(-I*d*x
- I*c)/(I*A + B)) + 3*a^2*d*sqrt(4*I*B^2/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(-(48*B*a^2*e^(2*I*d*x + 2*I*c) - 1
6*B*a^2 - sqrt(2)*(16*I*a^3*d*e^(3*I*d*x + 3*I*c) - 16*I*a^3*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(4*I*B^2/(a^3*d^2)))*e^(-2*I*d*x - 2*I*c)/
B) - 3*a^2*d*sqrt(4*I*B^2/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(-(48*B*a^2*e^(2*I*d*x + 2*I*c) - 16*B*a^2 - sqrt(
2)*(-16*I*a^3*d*e^(3*I*d*x + 3*I*c) + 16*I*a^3*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^
(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(4*I*B^2/(a^3*d^2)))*e^(-2*I*d*x - 2*I*c)/B) - sqrt(2)*(
(-4*I*A + 10*B)*e^(4*I*d*x + 4*I*c) + (5*I*A - 11*B)*e^(2*I*d*x + 2*I*c) - I*A + B)*sqrt(a/(e^(2*I*d*x + 2*I*c
) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-3*I*d*x - 3*I*c)/(a^2*d)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 4.08, size = 1516, normalized size = 6.24 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/12-1/12*I)/d*(-6*I*A*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)*sin(d*x+c)*
2^(1/2)+3*A*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))+3*A*cos(d*x+c)*2^(1/2)*arct
an((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))+3*B*sin(d*x+c)*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(
d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))-5*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+11*B*((-1+cos(d*x+c))/sin(d*x+c))^(1
/2)+12*I*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+I)*cos(d*x+c)*sin(d*x+c)-9*I*B*cos(d*x+c)*sin(d*x+c)*((-1+cos
(d*x+c))/sin(d*x+c))^(1/2)-12*I*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)*sin(d*x+c)+12*I*B*ln(((-
1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)*sin(d*x+c)-12*I*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)*cos(d
*x+c)*sin(d*x+c)+3*I*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)*sin(d*x+c)+5*I*A*((-1+cos(d*x+c))/sin(d*x
+c))^(1/2)*cos(d*x+c)^2+11*I*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2-6*I*B*ln(((-1+cos(d*x+c))/sin(d
*x+c))^(1/2)+I)*sin(d*x+c)+6*I*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*sin(d*x+c)-6*I*B*ln(((-1+cos(d*x+c))
/sin(d*x+c))^(1/2)+1)*sin(d*x+c)+6*I*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)*sin(d*x+c)-3*A*cos(d*x+c)*sin(
d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-9*B*cos(d*x+c)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-6*A*arc
tan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)^2*2^(1/2)-3*I*B*2^(1/2)*arctan((1/2+1/2
*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))-6*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+I)+6*B*ln(((-1+cos(d
*x+c))/sin(d*x+c))^(1/2)-1)-6*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)+6*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)-I)+5*A*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-11*B*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)-5*I*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-11*I*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+12*B*ln(((-1+cos(d*x+c))
/sin(d*x+c))^(1/2)+I)*cos(d*x+c)^2-12*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)^2+12*B*ln(((-1+cos
(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)^2-12*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)*cos(d*x+c)^2-6*B*ln((
(-1+cos(d*x+c))/sin(d*x+c))^(1/2)+I)*cos(d*x+c)+6*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)-6*B*ln
(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)+6*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)*cos(d*x+c)+6*I*
B*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)^2*2^(1/2)-6*B*arctan((1/2+1/2*I)*(
(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)*sin(d*x+c)*2^(1/2)-3*I*B*arctan((1/2+1/2*I)*((-1+cos(d*x
+c))/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)*2^(1/2)+3*I*A*sin(d*x+c)*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c)
)/sin(d*x+c))^(1/2)*2^(1/2)))*cos(d*x+c)^2*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(2*I*cos(d*x+c)*sin(
d*x+c)+2*cos(d*x+c)^2-1)/((-1+cos(d*x+c))/sin(d*x+c))^(1/2)/(cos(d*x+c)/sin(d*x+c))^(3/2)/sin(d*x+c)^2/a^2

________________________________________________________________________________________

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))/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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))/cot(d*x+c)**(3/2)/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________