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

Optimal. Leaf size=216 \[ -\frac {3 (-1)^{3/4} a^{3/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(2+2 i) a^{3/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {a^2}{d \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {i a^2}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-3*(-1)^(3/4)*a^(3/2)*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)*ta
n(d*x+c)^(1/2)/d-(2+2*I)*a^(3/2)*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)/d-a^2/d/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(1/2)+I*a^2/d/cot(d*x+c)^(1/2)/(a+I*a*tan(d*
x+c))^(1/2)

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Rubi [A]  time = 0.61, antiderivative size = 216, normalized size of antiderivative = 1.00, 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, 3595, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac {a^2}{d \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {3 (-1)^{3/4} a^{3/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {i a^2}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(2+2 i) a^{3/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(-1)^(3/4)*a^(3/2)*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]])/d - ((2 + 2*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Ta
n[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d - a^2/(d*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])
+ (I*a^2)/(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 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 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+i a \tan (c+d x))^{3/2}}{\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))^{3/2} \, dx\\ &=-\frac {a^2}{d \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\left (a \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {\tan (c+d x)} \left (\frac {5 a}{2}+\frac {3}{2} i a \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=-\frac {a^2}{d \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {i a^2}{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 {i a^2}{2}-\frac {3}{2} a^2 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{a}\\ &=-\frac {a^2}{d \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {i a^2}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {1}{2} \left (3 i \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 (2 i a \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}{d \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {i a^2}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (3 i a^2 \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 )}{2 d}-\frac {\left (4 a^3 \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 {(2+2 i) a^{3/2} \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}{d \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {i a^2}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (3 i a^2 \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 )}{d}\\ &=-\frac {(2+2 i) a^{3/2} \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}{d \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {i a^2}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (3 i a^2 \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 )}{d}\\ &=-\frac {3 (-1)^{3/4} a^{3/2} \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)}}{d}-\frac {(2+2 i) a^{3/2} \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}{d \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {i a^2}{d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 2.50, size = 228, normalized size = 1.06 \[ \frac {a e^{-i (c+d x)} \sqrt {\cot (c+d x)} \left (\sqrt {2} e^{i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )-2 \sqrt {2} \left (1+e^{2 i (c+d x)}\right ) \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+3 \left (1+e^{2 i (c+d x)}\right ) \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{\sqrt {2} d \left (1+e^{2 i (c+d x)}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(Sqrt[2]*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x))) - 2*Sqrt[2]*Sqrt[-1 + E^((2*I)*(c + d*x))]*(1 + E^((2*I
)*(c + d*x)))*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]] + 3*Sqrt[-1 + E^((2*I)*(c + d*x))]*(1 +
E^((2*I)*(c + d*x)))*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Cot[c + d*x]]*Sqr
t[a + I*a*Tan[c + d*x]])/(Sqrt[2]*d*E^(I*(c + d*x))*(1 + E^((2*I)*(c + d*x))))

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fricas [B]  time = 4.13, size = 635, normalized size = 2.94 \[ \frac {4 \, \sqrt {2} {\left (a e^{\left (3 i \, d x + 3 i \, c\right )} - a 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}} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {9 i \, a^{3}}{d^{2}}} \log \left (\frac {1}{3} \, {\left (\sqrt {2} {\left (32 i \, d e^{\left (3 i \, d x + 3 i \, c\right )} - 32 i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {9 i \, a^{3}}{d^{2}}} \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}} - 144 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 48 \, a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {9 i \, a^{3}}{d^{2}}} \log \left (\frac {1}{3} \, {\left (\sqrt {2} {\left (-32 i \, d e^{\left (3 i \, d x + 3 i \, c\right )} + 32 i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {9 i \, a^{3}}{d^{2}}} \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}} - 144 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 48 \, a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {32 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {32 i \, a^{3}}{d^{2}}} \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}} + 8 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {32 i \, a^{3}}{d^{2}}} \log \left (-\frac {{\left (\sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {32 i \, a^{3}}{d^{2}}} \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}} - 8 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right )}{4 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/cot(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

1/4*(4*sqrt(2)*(a*e^(3*I*d*x + 3*I*c) - a*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)) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(9*I*a^3/d^2)*log(1/3*(sqrt(2)*(3
2*I*d*e^(3*I*d*x + 3*I*c) - 32*I*d*e^(I*d*x + I*c))*sqrt(9*I*a^3/d^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)) - 144*a^2*e^(2*I*d*x + 2*I*c) + 48*a^2)*e^(-2*I*d*x - 2*
I*c)) - (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(9*I*a^3/d^2)*log(1/3*(sqrt(2)*(-32*I*d*e^(3*I*d*x + 3*I*c) + 32*I*d*e
^(I*d*x + I*c))*sqrt(9*I*a^3/d^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)) - 144*a^2*e^(2*I*d*x + 2*I*c) + 48*a^2)*e^(-2*I*d*x - 2*I*c)) - (d*e^(2*I*d*x + 2*I*c) + d)*
sqrt(32*I*a^3/d^2)*log(1/2*(sqrt(2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(32*I*a^3/d^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)) + 8*I*a^2*e^(I*d*x + I*c))*e^(-I*d*x - I*c)
/a) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(32*I*a^3/d^2)*log(-1/2*(sqrt(2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(32*I*a
^3/d^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)) - 8*I*a^
2*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a))/(d*e^(2*I*d*x + 2*I*c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/cot(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 2.06, size = 2091, normalized size = 9.68 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d*(-2*2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-2*2^(1/2)*cos(d*x+c)*sin(d*x+c)*((-1+cos(d*x+c
))/sin(d*x+c))^(1/2)+2*I*2^(1/2)*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-2*I*2^(1/2)*sin(d*x+c)*((-1+c
os(d*x+c))/sin(d*x+c))^(1/2)-3*2^(1/2)*cos(d*x+c)*sin(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)-6*2^(1/2
)*cos(d*x+c)*sin(d*x+c)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))+3*2^(1/2)*cos(d*x+c)*sin(d*x+c)*ln(((-1+cos
(d*x+c))/sin(d*x+c))^(1/2)-1)-8*cos(d*x+c)^2*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)-8*cos(d*x+c)
^2*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)+8*cos(d*x+c)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c
))^(1/2)-1)-2*2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-3*I*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+
c)^2*2^(1/2)+3*I*2^(1/2)*cos(d*x+c)^2*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)-6*I*arctan(((-1+cos(d*x+c))/sin
(d*x+c))^(1/2))*cos(d*x+c)^2*2^(1/2)-8*I*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)*sin(d
*x+c)-8*I*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)*sin(d*x+c)-4*I*ln(-(2^(1/2)*((-1+cos
(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin
(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))*cos(d*x+c)*sin(d*x+c)+3*I*2^(1/2)*cos(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))
^(1/2)-1)-3*I*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)*2^(1/2)+6*I*arctan(((-1+cos(d*x+c))/sin(d*x+
c))^(1/2))*cos(d*x+c)*2^(1/2)+3*2^(1/2)*cos(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)+8*I*arctan(2^(1/2)
*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)^2+8*I*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*c
os(d*x+c)^2+4*I*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(2^(1/2)*(
(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))*cos(d*x+c)^2-8*I*arctan(2^(1/2)*((-1+co
s(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)-8*I*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)-
4*I*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(2^(1/2)*((-1+cos(d*x+
c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))*cos(d*x+c)+4*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+
c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c
)-sin(d*x+c)+1))*cos(d*x+c)*sin(d*x+c)+8*cos(d*x+c)*sin(d*x+c)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)-1)+8*cos(d*x+c)*sin(d*x+c)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)+3*2^(1/2)*cos(d*x+c)^2*ln((
(-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)+6*2^(1/2)*cos(d*x+c)^2*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))-3*2^(1/
2)*cos(d*x+c)^2*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)-3*2^(1/2)*cos(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^
(1/2)+1)-6*2^(1/2)*cos(d*x+c)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))-2*I*2^(1/2)*((-1+cos(d*x+c))/sin(d*x+
c))^(1/2)+2*2^(1/2)*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+2*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos
(d*x+c)*sin(d*x+c)*2^(1/2)+3*I*2^(1/2)*cos(d*x+c)*sin(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)-3*I*ln((
(-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)*sin(d*x+c)*2^(1/2)+6*I*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/
2))*cos(d*x+c)*sin(d*x+c)*2^(1/2)-4*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(
d*x+c)-1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))*cos(d*x+c)^2+4*ln(-
(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(2^(1/2)*((-1+cos(d*x+c))/sin(
d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))*cos(d*x+c)+8*cos(d*x+c)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin
(d*x+c))^(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)-1+I+cos(d*x+c)-si
n(d*x+c))/(cos(d*x+c)/sin(d*x+c))^(1/2)/((-1+cos(d*x+c))/sin(d*x+c))^(1/2)/sin(d*x+c)*2^(1/2)*a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/cot(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(3/2)/cot(d*x+c)**(1/2),x)

[Out]

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

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