3.552 \(\int \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=246 \[ -\frac{(-1)^{3/4} a^{5/2} (23 B+20 i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}-\frac{a^2 (4 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}+\frac{(4-4 i) a^{5/2} (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a B (a+i a \tan (c+d x))^{3/2}}{2 d \sqrt{\cot (c+d x)}} \]

[Out]

-((-1)^(3/4)*a^(5/2)*((20*I)*A + 23*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]])/(4*d) + ((4 - 4*I)*a^(5/2)*(A - I*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]])/d - (a^2*(4*A - (7*I)*B)*Sqr
t[a + I*a*Tan[c + d*x]])/(4*d*Sqrt[Cot[c + d*x]]) + ((I/2)*a*B*(a + I*a*Tan[c + d*x])^(3/2))/(d*Sqrt[Cot[c + d
*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.881914, antiderivative size = 246, 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, 3594, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{(-1)^{3/4} a^{5/2} (23 B+20 i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}-\frac{a^2 (4 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}+\frac{(4-4 i) a^{5/2} (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a B (a+i a \tan (c+d x))^{3/2}}{2 d \sqrt{\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

-((-1)^(3/4)*a^(5/2)*((20*I)*A + 23*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]])/(4*d) + ((4 - 4*I)*a^(5/2)*(A - I*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]])/d - (a^2*(4*A - (7*I)*B)*Sqr
t[a + I*a*Tan[c + d*x]])/(4*d*Sqrt[Cot[c + d*x]]) + ((I/2)*a*B*(a + I*a*Tan[c + d*x])^(3/2))/(d*Sqrt[Cot[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 3594

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f
*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*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] && GtQ[m, 1] &&  !LtQ[n, -1]

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 \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{i a B (a+i a \tan (c+d x))^{3/2}}{2 d \sqrt{\cot (c+d x)}}+\frac{1}{2} \left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+i a \tan (c+d x))^{3/2} \left (\frac{1}{2} a (4 A-i B)+\frac{1}{2} a (4 i A+7 B) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{a^2 (4 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}+\frac{i a B (a+i a \tan (c+d x))^{3/2}}{2 d \sqrt{\cot (c+d x)}}+\frac{1}{2} \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 (4 A-3 i B)+\frac{1}{4} a^2 (20 i A+23 B) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{a^2 (4 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}+\frac{i a B (a+i a \tan (c+d x))^{3/2}}{2 d \sqrt{\cot (c+d x)}}+\left (4 a^2 (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-\frac{1}{8} \left (a (20 A-23 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\\ &=-\frac{a^2 (4 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}+\frac{i a B (a+i a \tan (c+d x))^{3/2}}{2 d \sqrt{\cot (c+d x)}}-\frac{\left (8 i a^4 (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 )}{d}-\frac{\left (a^3 (20 A-23 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 )}{8 d}\\ &=-\frac{(4+4 i) a^{5/2} (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)}}{d}-\frac{a^2 (4 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}+\frac{i a B (a+i a \tan (c+d x))^{3/2}}{2 d \sqrt{\cot (c+d x)}}-\frac{\left (a^3 (20 A-23 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 )}{4 d}\\ &=-\frac{(4+4 i) a^{5/2} (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)}}{d}-\frac{a^2 (4 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}+\frac{i a B (a+i a \tan (c+d x))^{3/2}}{2 d \sqrt{\cot (c+d x)}}-\frac{\left (a^3 (20 A-23 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 )}{4 d}\\ &=\frac{\sqrt [4]{-1} a^{5/2} (20 A-23 i 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)}}{4 d}-\frac{(4+4 i) a^{5/2} (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)}}{d}-\frac{a^2 (4 A-7 i B) \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}+\frac{i a B (a+i a \tan (c+d x))^{3/2}}{2 d \sqrt{\cot (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 8.02878, size = 447, normalized size = 1.82 \[ \frac{\cos ^3(c+d x) \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \left (\sqrt{2} (\cos (3 c+d x)-i \sin (3 c+d x)) \sqrt{i \sin ^2(c+d x) (\cot (c+d x)+i)} \left (\sqrt{2} (-23 B-20 i A) \log \left (-\frac{2 e^{\frac{7 i c}{2}} \left (-2 \sqrt{-1+e^{2 i (c+d x)}}+\sqrt{2} e^{i (c+d x)}+i \sqrt{2}\right )}{(20 A-23 i B) \left (e^{i (c+d x)}-i\right )}\right )+\sqrt{2} (23 B+20 i A) \log \left (-\frac{2 e^{\frac{7 i c}{2}} \left (2 \sqrt{-1+e^{2 i (c+d x)}}+\sqrt{2} e^{i (c+d x)}-i \sqrt{2}\right )}{(20 A-23 i B) \left (e^{i (c+d x)}+i\right )}\right )-64 i (A-i B) \log \left ((\cos (c)-i \sin (c)) \left (i \sin (c+d x)+\cos (c+d x)+\sqrt{i \sin (2 (c+d x))+\cos (2 (c+d x))-1}\right )\right )\right )-4 (\cos (2 c)-i \sin (2 c)) \tan (c+d x) (4 A+2 B \tan (c+d x)-9 i B)\right )}{16 d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

(Cos[c + d*x]^3*Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x])*(Sqrt[2]*(Sqrt[2]*((-20*I
)*A - 23*B)*Log[(-2*E^(((7*I)/2)*c)*(I*Sqrt[2] + Sqrt[2]*E^(I*(c + d*x)) - 2*Sqrt[-1 + E^((2*I)*(c + d*x))]))/
((20*A - (23*I)*B)*(-I + E^(I*(c + d*x))))] + Sqrt[2]*((20*I)*A + 23*B)*Log[(-2*E^(((7*I)/2)*c)*((-I)*Sqrt[2]
+ Sqrt[2]*E^(I*(c + d*x)) + 2*Sqrt[-1 + E^((2*I)*(c + d*x))]))/((20*A - (23*I)*B)*(I + E^(I*(c + d*x))))] - (6
4*I)*(A - I*B)*Log[(Cos[c] - I*Sin[c])*(Cos[c + d*x] + I*Sin[c + d*x] + Sqrt[-1 + Cos[2*(c + d*x)] + I*Sin[2*(
c + d*x)]])])*Sqrt[I*(I + Cot[c + d*x])*Sin[c + d*x]^2]*(Cos[3*c + d*x] - I*Sin[3*c + d*x]) - 4*(Cos[2*c] - I*
Sin[2*c])*Tan[c + d*x]*(4*A - (9*I)*B + 2*B*Tan[c + d*x])))/(16*d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] +
B*Sin[c + d*x]))

________________________________________________________________________________________

Maple [B]  time = 0.589, size = 1530, 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(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)

[Out]

1/16/d*2^(1/2)*a^2*(cos(d*x+c)-1)*(22*I*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-8*A*
2^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)*sin(d*x+c)-32*A*cos(d*x+c)^2*ln(-(((cos(d*x+c)-1)/sin(d*x
+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+c
os(d*x+c)+sin(d*x+c)-1))-64*B*cos(d*x+c)^2*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-64*B*cos(d*x+c)
^2*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)-32*B*cos(d*x+c)^2*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2
)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)
-sin(d*x+c)+1))-4*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-64*A*cos(d*x+c)^2*arctan(((cos(d*x+c)-1)/sin(d*x
+c))^(1/2)*2^(1/2)+1)-64*A*cos(d*x+c)^2*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)-64*I*A*cos(d*x+c)^
2*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-64*I*A*cos(d*x+c)^2*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(
1/2)*2^(1/2)-1)-32*I*A*cos(d*x+c)^2*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d
*x+c)-1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))+64*I*B*cos(d*x+c)^2*a
rctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)+64*I*B*cos(d*x+c)^2*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2
)*2^(1/2)-1)+32*I*B*cos(d*x+c)^2*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+
c)+1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))+8*I*A*cos(d*x+c)^2*2^(1/
2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+40*I*A*cos(d*x+c)^2*2^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2))-20*
I*A*cos(d*x+c)^2*2^(1/2)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)+20*I*A*cos(d*x+c)^2*2^(1/2)*ln(((cos(d*x+c)-1
)/sin(d*x+c))^(1/2)+1)-46*I*B*cos(d*x+c)^2*2^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2))-23*I*B*cos(d*x+c)
^2*2^(1/2)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)+23*I*B*cos(d*x+c)^2*2^(1/2)*ln(((cos(d*x+c)-1)/sin(d*x+c))^
(1/2)+1)+8*I*A*cos(d*x+c)*2^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+4*I*B*sin(d*x+c)*2^(1/2)*((cos(d*x+c)-1)/s
in(d*x+c))^(1/2)+22*B*cos(d*x+c)^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-20*A*cos(d*x+c)^2*2^(1/2)*ln(((co
s(d*x+c)-1)/sin(d*x+c))^(1/2)+1)+40*A*cos(d*x+c)^2*2^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2))+20*A*cos(
d*x+c)^2*2^(1/2)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)+23*B*cos(d*x+c)^2*2^(1/2)*ln(((cos(d*x+c)-1)/sin(d*x+
c))^(1/2)+1)+46*B*cos(d*x+c)^2*2^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2))-23*B*cos(d*x+c)^2*2^(1/2)*ln(
((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)+18*B*2^(1/2)*cos(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2))*(cos(d*x+c)/si
n(d*x+c))^(1/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)^2/((cos(
d*x+c)-1)/sin(d*x+c))^(1/2)

________________________________________________________________________________________

Maxima [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(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [B]  time = 1.59075, size = 2566, normalized size = 10.43 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/8*(2*sqrt(2)*((4*I*A + 11*B)*a^2*e^(4*I*d*x + 4*I*c) - 4*B*a^2*e^(2*I*d*x + 2*I*c) + (-4*I*A - 7*B)*a^2)*sqr
t(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) + s
qrt((-400*I*A^2 - 920*A*B + 529*I*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqr
t(2)*((20*I*A + 23*B)*a^2*e^(2*I*d*x + 2*I*c) + (-20*I*A - 23*B)*a^2)*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) + 2*sqrt((-400*I*A^2 - 920*A*B + 529*I*B
^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((20*I*A + 23*B)*a^2)) - sqrt((-400*I*A^2 - 920*A*B +
529*I*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((20*I*A + 23*B)*a^2*e
^(2*I*d*x + 2*I*c) + (-20*I*A - 23*B)*a^2)*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) - 2*sqrt((-400*I*A^2 - 920*A*B + 529*I*B^2)*a^5/d^2)*d*e^(2*I*d*x +
2*I*c))*e^(-2*I*d*x - 2*I*c)/((20*I*A + 23*B)*a^2)) - 4*sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/d^2)*(d*e^(4
*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (-4*I*A -
4*B)*a^2)*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) + sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A
+ 4*B)*a^2)) + 4*sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*
c) + d)*log((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (-4*I*A - 4*B)*a^2)*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) - sqrt((-32*I*A^2 - 64*A*B +
32*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^2)))/(d*e^(4*I*d*x + 4*I*c) +
2*d*e^(2*I*d*x + 2*I*c) + d)

________________________________________________________________________________________

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(cot(d*x+c)**(1/2)*(a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sqrt{\cot \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)*sqrt(cot(d*x + c)), x)