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

Optimal. Leaf size=222 \[ -\frac{a^2 \sqrt{a+i a \tan (c+d x)}}{2 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{23 (-1)^{3/4} a^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}+\frac{9 i a^2 \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}-\frac{(4+4 i) a^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

[Out]

(-23*(-1)^(3/4)*a^(5/2)*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)*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*Sqrt[a + I*a*Tan[c + d*x]])/(2*d*Cot[c + d*x
]^(3/2)) + (((9*I)/4)*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Cot[c + d*x]])

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Rubi [A]  time = 0.599758, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4241, 3556, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{a^2 \sqrt{a+i a \tan (c+d x)}}{2 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{23 (-1)^{3/4} a^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}+\frac{9 i a^2 \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}-\frac{(4+4 i) a^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-23*(-1)^(3/4)*a^(5/2)*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)*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*Sqrt[a + I*a*Tan[c + d*x]])/(2*d*Cot[c + d*x
]^(3/2)) + (((9*I)/4)*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(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 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3597

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*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(f*(m + n)), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] &
& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 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+i a \tan (c+d x))^{5/2}}{\sqrt{\cot (c+d x)}} \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx\\ &=-\frac{a^2 \sqrt{a+i a \tan (c+d x)}}{2 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{1}{2} \left (a \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (\frac{7 a}{2}+\frac{9}{2} i a \tan (c+d x)\right ) \, dx\\ &=-\frac{a^2 \sqrt{a+i a \tan (c+d x)}}{2 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{9 i a^2 \sqrt{a+i a \tan (c+d x)}}{4 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{9 i a^2}{4}+\frac{23}{4} a^2 \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{a^2 \sqrt{a+i a \tan (c+d x)}}{2 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{9 i a^2 \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}+\frac{1}{8} \left (23 i a \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-\left (4 i a^2 \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{a^2 \sqrt{a+i a \tan (c+d x)}}{2 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{9 i a^2 \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}+\frac{\left (23 i a^3 \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{\left (8 a^4 \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{(4+4 i) a^{5/2} \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 \sqrt{a+i a \tan (c+d x)}}{2 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{9 i a^2 \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}+\frac{\left (23 i a^3 \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} \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 \sqrt{a+i a \tan (c+d x)}}{2 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{9 i a^2 \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}+\frac{\left (23 i a^3 \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{23 (-1)^{3/4} a^{5/2} \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} \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 \sqrt{a+i a \tan (c+d x)}}{2 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{9 i a^2 \sqrt{a+i a \tan (c+d x)}}{4 d \sqrt{\cot (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 7.27034, size = 318, normalized size = 1.43 \[ -\frac{(a+i a \tan (c+d x))^{5/2} \left (\sqrt{2} \sqrt{e^{i d x}} e^{-i (3 c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{\frac{i \left (1+e^{2 i (c+d x)}\right )}{-1+e^{2 i (c+d x)}}} \left (32 \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-23 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )-(\cos (2 c)-i \sin (2 c)) \sqrt{\cot (c+d x)} \sec ^{\frac{5}{2}}(c+d x) \sqrt{\cos (d x)+i \sin (d x)} (9 i \sin (2 (c+d x))+2 \cos (2 (c+d x))-2)\right )}{8 d \sec ^{\frac{5}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((Sqrt[2]*Sqrt[E^(I*d*x)]*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqr
t[(I*(1 + E^((2*I)*(c + d*x))))/(-1 + E^((2*I)*(c + d*x)))]*(32*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c
+ d*x))]] - 23*Sqrt[2]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/E^(I*(3*c + d*x)) -
Sqrt[Cot[c + d*x]]*Sec[c + d*x]^(5/2)*(Cos[2*c] - I*Sin[2*c])*Sqrt[Cos[d*x] + I*Sin[d*x]]*(-2 + 2*Cos[2*(c +
d*x)] + (9*I)*Sin[2*(c + d*x)]))*(a + I*a*Tan[c + d*x])^(5/2))/(8*d*Sec[c + d*x]^(5/2)*(Cos[d*x] + I*Sin[d*x])
^(5/2))

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Maple [B]  time = 0.388, size = 2352, normalized size = 10.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/d*2^(1/2)*a^2*(-4*I*2^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+22*2^(1/2)*cos(d*x+c)^3*((cos(d*x+c)-1)/sin
(d*x+c))^(1/2)-22*2^(1/2)*cos(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+64*cos(d*x+c)^2*sin(d*x+c)*arctan(((cos
(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)+32*cos(d*x+c)^2*sin(d*x+c)*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))+23*2^(1/2)*cos(d*x+c)^3*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+1)+46*2^(1/2)*cos(d*x+c)^3*arctan(((cos
(d*x+c)-1)/sin(d*x+c))^(1/2))-23*2^(1/2)*cos(d*x+c)^3*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)+64*I*cos(d*x+c)^
3*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)+64*I*cos(d*x+c)^3*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/
2)*2^(1/2)-1)+32*I*cos(d*x+c)^3*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*cos(d*x+c)^2*arctan(
((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-64*I*cos(d*x+c)^2*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2
)-1)+64*cos(d*x+c)^2*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-32*I*cos(d*x+c)^2*ln(-(((c
os(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*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-46*cos(d*x
+c)^2*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2))*2^(1/2)-23*cos(d*x+c)^2*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+1
)*2^(1/2)+23*cos(d*x+c)^2*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)*2^(1/2)-4*cos(d*x+c)^2*((cos(d*x+c)-1)/sin(d
*x+c))^(1/2)*2^(1/2)+4*I*2^(1/2)*cos(d*x+c)^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-64*I*cos(d*x+c)^2*sin(d*x+c)*a
rctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-64*I*cos(d*x+c)^2*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x
+c))^(1/2)*2^(1/2)-1)-32*I*cos(d*x+c)^2*sin(d*x+c)*ln(-(((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)/(((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))-22*I*2
^(1/2)*cos(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-4*I*2^(1/2)*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-2
3*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+1)-46*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a
rctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2))+23*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/
2)-1)-22*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+32*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*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-23*I*2^(1/2)*cos(d*x+c)^2*sin(d*
x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+1)+46*I*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d
*x+c))^(1/2))+23*I*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)+22*I*2^(1/2)*cos(d*
x+c)^2*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-18*I*2^(1/2)*cos(d*x+c)*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x
+c))^(1/2)-18*cos(d*x+c)*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+23*I*2^(1/2)*cos(d*x+c)^3*ln(((c
os(d*x+c)-1)/sin(d*x+c))^(1/2)+1)-46*I*2^(1/2)*cos(d*x+c)^3*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2))-23*I*2^(
1/2)*cos(d*x+c)^3*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)+22*I*2^(1/2)*cos(d*x+c)^3*((cos(d*x+c)-1)/sin(d*x+c)
)^(1/2)-23*I*cos(d*x+c)^2*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+1)*2^(1/2)+46*I*2^(1/2)*cos(d*x+c)^2*arctan(((c
os(d*x+c)-1)/sin(d*x+c))^(1/2))+23*I*2^(1/2)*cos(d*x+c)^2*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)-64*cos(d*x+c
)^3*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-64*cos(d*x+c)^3*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/
2)*2^(1/2)-1)-32*cos(d*x+c)^3*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*cos(d*x+c)^2*arctan(((co
s(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)+64*cos(d*x+c)^2*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1))*
(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(I*cos(d*x+c)+I*sin(d*x+c)+cos(d*x+c)-sin(d*x+c)-1+I)/cos(d*x+c
)/((cos(d*x+c)-1)/sin(d*x+c))^(1/2)/(cos(d*x+c)/sin(d*x+c))^(1/2)/sin(d*x+c)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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((a+I*a*tan(d*x+c))^(5/2)/cot(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.50617, size = 2025, normalized size = 9.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*(11*a^2*e^(4*I*d*x + 4*I*c) - 4*a^2*e^(2*I*d*x + 2*I*c) - 7*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(529/16*I*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(1/23*(23*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) - 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) + 8*sqrt
(529/16*I*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) - 2*sqrt(529/16*I*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(1/23*(23*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) - 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) - 8*sqrt(529/16*I*
a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) - 2*sqrt(32*I*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(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) - 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^5/d^2)*d*e^(2*I*d*x
+ 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) + 2*sqrt(32*I*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(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) - 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^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x
- 2*I*c)/a^2))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + 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+I*a*tan(d*x+c))**(5/2)/cot(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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((a+I*a*tan(d*x+c))^(5/2)/cot(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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