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

Optimal. Leaf size=141 \[ -\frac {3 i a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 i \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {i a^2}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.41, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3553, 3596, 3600, 3480, 206, 3599, 63, 208} \[ -\frac {i a^2}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 i a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 i \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*I)*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + ((2*I)*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*T
an[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (I*a^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (a^2*Cot[c + d*x])/(d*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 206

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

Rule 208

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

Rule 3480

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

Rule 3553

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(a^2*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] +
Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(m
- 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; Fr
eeQ[{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] && GtQ[m, 1] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3596

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*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*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] && 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 3600

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

Rubi steps

\begin {align*} \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac {a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\int \frac {\cot (c+d x) \left (-\frac {3 i a^2}{2}+\frac {5}{2} a^2 \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=-\frac {i a^2}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {3 i a^3}{2}+\frac {1}{2} a^3 \tan (c+d x)\right ) \, dx}{a^2}\\ &=-\frac {i a^2}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {3}{2} i \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx-(2 a) \int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=-\frac {i a^2}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (3 i a^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (4 i a^2\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=\frac {2 i \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {i a^2}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {(3 a) \operatorname {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=-\frac {3 i a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 i \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {i a^2}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.33, size = 178, normalized size = 1.26 \[ \frac {a e^{-\frac {1}{2} i (2 c+3 d x)} \sqrt {1+e^{2 i (c+d x)}} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (\sin \left (\frac {d x}{2}\right )-i \cos \left (\frac {d x}{2}\right )\right ) \left (-i \sqrt {1+e^{2 i (c+d x)}} \csc (c+d x)-4 \sinh ^{-1}\left (e^{i (c+d x)}\right )+3 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {1+e^{2 i (c+d x)}}}\right )\right )}{\sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(-4*ArcSinh[E^(I*(c +
d*x))] + 3*Sqrt[2]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[1 + E^((2*I)*(c + d*x))]] - I*Sqrt[1 + E^((2*I)*(c
+ d*x))]*Csc[c + d*x])*((-I)*Cos[(d*x)/2] + Sin[(d*x)/2]))/(Sqrt[2]*d*E^((I/2)*(2*c + 3*d*x)))

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fricas [B]  time = 0.45, size = 501, normalized size = 3.55 \[ -\frac {4 \, \sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \log \left (\frac {4 \, {\left (a^{2} e^{\left (i \, d x + i \, c\right )} + {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - 4 \, \sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \log \left (\frac {4 \, {\left (a^{2} e^{\left (i \, d x + i \, c\right )} + {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - 3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \log \left ({\left (48 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \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 {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 16 \, a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \log \left ({\left (48 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \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 {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 16 \, a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - \sqrt {2} {\left (-4 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 4 i \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{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(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/4*(4*sqrt(2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-a^3/d^2)*log(4*(a^2*e^(I*d*x + I*c) + (I*d*e^(2*I*d*x + 2*I*
c) + I*d)*sqrt(-a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a) - 4*sqrt(2)*(d*e^(2*I*d*x + 2*
I*c) - d)*sqrt(-a^3/d^2)*log(4*(a^2*e^(I*d*x + I*c) + (-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(-a^3/d^2)*sqrt(a/(
e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a) - 3*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-a^3/d^2)*log((48*a^2*e^(2
*I*d*x + 2*I*c) + sqrt(2)*(32*I*d*e^(3*I*d*x + 3*I*c) + 32*I*d*e^(I*d*x + I*c))*sqrt(-a^3/d^2)*sqrt(a/(e^(2*I*
d*x + 2*I*c) + 1)) + 16*a^2)*e^(-2*I*d*x - 2*I*c)) + 3*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-a^3/d^2)*log((48*a^2*
e^(2*I*d*x + 2*I*c) + sqrt(2)*(-32*I*d*e^(3*I*d*x + 3*I*c) - 32*I*d*e^(I*d*x + I*c))*sqrt(-a^3/d^2)*sqrt(a/(e^
(2*I*d*x + 2*I*c) + 1)) + 16*a^2)*e^(-2*I*d*x - 2*I*c)) - sqrt(2)*(-4*I*a*e^(3*I*d*x + 3*I*c) - 4*I*a*e^(I*d*x
+ I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(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 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.35, size = 631, normalized size = 4.48 \[ \frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (4 i \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}+4 i \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \sqrt {2}+3 i \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (-\frac {-\sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+4 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}+3 i \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (-\frac {-\sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )+4 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sin \left (d x +c \right ) \sqrt {2}+3 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sin \left (d x +c \right )-2 i \left (\cos ^{2}\left (d x +c \right )\right )-2 i \cos \left (d x +c \right )+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) a}{2 d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \left (1+\cos \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(4*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2
*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c)*sin(d*x+c)*2^(1/2)+4*I*(-2*cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(
d*x+c)*2^(1/2)+3*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+
cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)+4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)*sin(d*x+c)*2^(1/2)+3*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(
-sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)+4*(-2*cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*2^(1/2)+3*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)*sin(d*x+c)+3*(-2*cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)-2*I*cos(d*x+c)^2-2*I*cos(d*x+c)+2*
cos(d*x+c)*sin(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)/(1+cos(d*x+c))*a

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maxima [A]  time = 0.59, size = 132, normalized size = 0.94 \[ -\frac {i \, {\left (2 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 3 \, \sqrt {a} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right ) - \frac {2 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\tan \left (d x + c\right )}\right )} a}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*(2*sqrt(2)*sqrt(a)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(
d*x + c) + a))) - 3*sqrt(a)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))
- 2*I*sqrt(I*a*tan(d*x + c) + a)/tan(d*x + c))*a/d

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mupad [B]  time = 4.20, size = 112, normalized size = 0.79 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2}\right )\,\sqrt {-a^3}\,3{}\mathrm {i}}{d}-\frac {a\,\mathrm {cot}\left (c+d\,x\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,a^2}\right )\,\sqrt {-a^3}\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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