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

Optimal. Leaf size=184 \[ \frac {11 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {2 \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d} \]

[Out]

11/4*a^(3/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/d-2*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-1/2*I*a^2*cot(d*x+c)/d/(a+I*a*tan(d*x+c))^(1/2)-1/2*a^2*cot(d*x+c)^2/d/(a+I*a*tan(d*x+c))
^(1/2)-5/4*I*a*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.61, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3553, 3596, 3598, 3600, 3480, 206, 3599, 63, 208} \[ \frac {11 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {2 \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3598

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

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 ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {1}{2} \int \frac {\cot ^2(c+d x) \left (-\frac {7 i a^2}{2}+\frac {9}{2} a^2 \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {5 i a^3}{2}+\frac {3}{2} a^3 \tan (c+d x)\right ) \, dx}{2 a^2}\\ &=-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {11 a^4}{4}+\frac {5}{4} i a^4 \tan (c+d x)\right ) \, dx}{2 a^3}\\ &=-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {11}{8} \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx-(2 i a) \int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {\left (11 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}-\frac {\left (4 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 \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 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {(11 i a) \operatorname {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 d}\\ &=\frac {11 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {2 \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 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\\ \end {align*}

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Mathematica [A]  time = 1.94, size = 190, normalized size = 1.03 \[ -\frac {a e^{-\frac {1}{2} i (2 c+3 d x)} \sqrt {1+e^{2 i (c+d x)}} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (\cos \left (\frac {d x}{2}\right )+i \sin \left (\frac {d x}{2}\right )\right ) \left (16 \sinh ^{-1}\left (e^{i (c+d x)}\right )-11 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {1+e^{2 i (c+d x)}}}\right )+\sqrt {1+e^{2 i (c+d x)}} (2 \cot (c+d x)+5 i) \csc (c+d x)\right )}{4 \sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(a*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(16*ArcSinh[E^(I
*(c + d*x))] - 11*Sqrt[2]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[1 + E^((2*I)*(c + d*x))]] + Sqrt[1 + E^((2*I)
*(c + d*x))]*(5*I + 2*Cot[c + d*x])*Csc[c + d*x])*(Cos[(d*x)/2] + I*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 = 546, normalized size = 2.97 \[ -\frac {16 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, 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 (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 ) - 16 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, 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 (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 ) - 11 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, \sqrt {2} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + 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}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 11 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, \sqrt {2} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + 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}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - 4 \, \sqrt {2} {\left (7 \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 4 \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 3 \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{16 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, 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)^3*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/16*(16*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) - 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) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a) - 16*sqr
t(2)*(d*e^(4*I*d*x + 4*I*c) - 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) - (d*e^(2*
I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a) - 11*(d*e^(4*I*d*x +
4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) + 2*sqrt(2)*(d*e^(3*I*d*
x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) + a^2)*e^(-2*I*d*x - 2*I*c)) +
11*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) - 2*
sqrt(2)*(d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) + a^2)*e^(
-2*I*d*x - 2*I*c)) - 4*sqrt(2)*(7*a*e^(5*I*d*x + 5*I*c) + 4*a*e^(3*I*d*x + 3*I*c) - 3*a*e^(I*d*x + I*c))*sqrt(
a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(4*I*d*x + 4*I*c) - 2*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 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(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)^3, x)

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maple [B]  time = 1.46, size = 1142, normalized size = 6.21 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(16*I*cos(d*x+c)^2*2^(1/2)*(-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))+16*I*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+4*I*cos(d*x+c)^2-16*cos(d*x+c)^3*2^
(1/2)*(-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))+11*I*cos(d*x+c)^2*(-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))-11*I*(-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))-11*cos(d*x+c)^3*
(-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))-16*cos(d*x+c)^2*2^(1/2)*(-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))-16*I*cos(d*x+c)*2^(1/2)*(-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))-11*I*cos(d*x+c)*(-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))+11*I*cos(d*x+c)^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))-11*cos(d*x+c)^2*(-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))+16*cos(d*x+c)*2^(1/2)*(-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))-10*I*cos(d*x+c)-16*I*2^(1/2)*(
-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))-14*cos(d*x+c)^2*s
in(d*x+c)+11*cos(d*x+c)*(-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))+16*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+14*I*cos(d*x+c)^3-10*cos(d*x+c)*sin(d*x+c)+11*(-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)))/(I*sin(
d*x+c)+cos(d*x+c)-1)/sin(d*x+c)/(1+cos(d*x+c))*a

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maxima [A]  time = 0.56, size = 178, normalized size = 0.97 \[ \frac {a^{2} {\left (\frac {8 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} - \frac {11 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{\sqrt {a}} + \frac {2 \, {\left (5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + a^{2}}\right )}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a^2*(8*sqrt(2)*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)))/sqrt(a) - 11*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/sqrt(
a) + 2*(5*(I*a*tan(d*x + c) + a)^(3/2) - 3*sqrt(I*a*tan(d*x + c) + a)*a)/((I*a*tan(d*x + c) + a)^2 - 2*(I*a*ta
n(d*x + c) + a)*a + a^2))/d

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mupad [B]  time = 4.07, size = 136, normalized size = 0.74 \[ -\frac {\mathrm {atan}\left (\frac {\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{a^2}\right )\,\sqrt {a^3}\,11{}\mathrm {i}}{4\,d}-\frac {5\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{4\,d\,{\mathrm {tan}\left (c+d\,x\right )}^2}+\frac {3\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{4\,d\,{\mathrm {tan}\left (c+d\,x\right )}^2}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,a^2}\right )\,\sqrt {a^3}\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a*(a + a*tan(c + d*x)*1i)^(1/2))/(4*d*tan(c + d*x)^2) - (5*(a + a*tan(c + d*x)*1i)^(3/2))/(4*d*tan(c + d*x)
^2) - (atan(((a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/a^2)*(a^3)^(1/2)*11i)/(4*d) + (2^(1/2)*atan((2^(1/2
)*(a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^2))*(a^3)^(1/2)*2i)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(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)**3, x)

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