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)*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]])

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Rubi [A]  time = 0.8251, antiderivative size = 243, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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 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 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 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])

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 = 7.5004, size = 388, normalized size = 1.6 \[ \frac{e^{-2 i (c+d x)} \sqrt{\cot (c+d x)} \sec (c+d x) \left (3 (B+i A) e^{3 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \log \left (\sqrt{-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )+5 i A e^{2 i (c+d x)}-4 i A e^{4 i (c+d x)}-i A-11 B e^{2 i (c+d x)}+10 B e^{4 i (c+d x)}-3 \sqrt{2} B e^{3 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \log \left (-2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )+3 \sqrt{2} B e^{3 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \log \left (2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )+B\right ) (A+B \tan (c+d x))}{12 d (a+i a \tan (c+d x))^{3/2} (A \cos (c+d x)+B \sin (c+d x))} \]

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))

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Maple [B]  time = 1.064, size = 1516, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

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

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Fricas [B]  time = 2.76993, size = 2129, normalized size = 8.76 \begin{align*} \text{result too large to display} \end{align*}

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^(4*I*d*x + 4*I*c)*log((2*sqrt(1/2)*a^2*d*sq
rt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + 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^(I*d*x + I*c)
)*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - 3*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(4*I*d*x + 4*
I*c)*log(-(2*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(2*I*d*x + 2*I*c) - sqrt(2)*((I*A + 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^(I*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - 3*a^2*d*sqrt(4*I*B^2/(a^3*d^2))*e^(4*I*d*x
+ 4*I*c)*log(52/605*(4*sqrt(2)*(B*e^(2*I*d*x + 2*I*c) - 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^(I*d*x + I*c) + (3*a^2*d*e^(2*I*d*x + 2*I*c) - a^2*d)*sqrt(4*I*B^
2/(a^3*d^2)))/(B*e^(2*I*d*x + 2*I*c) + B)) + 3*a^2*d*sqrt(4*I*B^2/(a^3*d^2))*e^(4*I*d*x + 4*I*c)*log(52/605*(4
*sqrt(2)*(B*e^(2*I*d*x + 2*I*c) - 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^(I*d*x + I*c) - (3*a^2*d*e^(2*I*d*x + 2*I*c) - a^2*d)*sqrt(4*I*B^2/(a^3*d^2)))/(B*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^(I*d*x +
 I*c))*e^(-4*I*d*x - 4*I*c)/(a^2*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)