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

Optimal. Leaf size=91 \[ \frac {4 i a^2}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {2 a^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2}{d \sqrt {\cot (c+d x)}}-\frac {4 (-1)^{3/4} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3673, 3542, 3529, 3533, 208} \[ \frac {4 i a^2}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {2 a^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2}{d \sqrt {\cot (c+d x)}}-\frac {4 (-1)^{3/4} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 3542

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Ta
n[e + f*x])^(m + 1)*Simp[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Tan[e + f*x], x], x], x] /; FreeQ
[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rule 3673

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rubi steps

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

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Mathematica [A]  time = 4.01, size = 152, normalized size = 1.67 \[ \frac {a^2 e^{-2 i c} (\sin (2 (c+d x))-i \cos (2 (c+d x))) \left (2 i \sec ^2(c+d x) (10 i \sin (2 (c+d x))+33 \cos (2 (c+d x))+27)-\frac {120 i \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt {i \tan (c+d x)}}\right )}{30 d \sqrt {\cot (c+d x)} (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*((-I)*Cos[2*(c + d*x)] + Sin[2*(c + d*x)])*((2*I)*Sec[c + d*x]^2*(27 + 33*Cos[2*(c + d*x)] + (10*I)*Sin[2
*(c + d*x)]) - ((120*I)*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])/Sqrt[I*Tan[c + d*
x]]))/(30*d*E^((2*I)*c)*Sqrt[Cot[c + d*x]]*(Cos[d*x] + I*Sin[d*x])^2)

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fricas [B]  time = 2.14, size = 391, normalized size = 4.30 \[ -\frac {15 \, \sqrt {-\frac {16 i \, a^{4}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {16 i \, a^{4}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{2}}\right ) - 15 \, \sqrt {-\frac {16 i \, a^{4}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {16 i \, a^{4}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{2}}\right ) - {\left (-344 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 88 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 248 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 184 i \, a^{2}\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}}{60 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*sqrt(-16*I*a^4/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*
log(1/2*(4*I*a^2*e^(2*I*d*x + 2*I*c) + sqrt(-16*I*a^4/d^2)*(I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt((I*e^(2*I*d*x
+ 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c)/a^2) - 15*sqrt(-16*I*a^4/d^2)*(d*e^(6*I*d*x + 6
*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(1/2*(4*I*a^2*e^(2*I*d*x + 2*I*c) + sqrt(-16
*I*a^4/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-
2*I*d*x - 2*I*c)/a^2) - (-344*I*a^2*e^(6*I*d*x + 6*I*c) - 88*I*a^2*e^(4*I*d*x + 4*I*c) + 248*I*a^2*e^(2*I*d*x
+ 2*I*c) + 184*I*a^2)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(6*I*d*x + 6*I*c) + 3*
d*e^(4*I*d*x + 4*I*c) + 3*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 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}{\cot \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.28, size = 511, normalized size = 5.62 \[ \frac {a^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (-30 i \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+30 \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-30 \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )+10 i \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-10 i \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+33 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )-33 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}-3 \cos \left (d x +c \right ) \sqrt {2}+3 \sqrt {2}\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {2}}{15 d \left (\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*a^2/d*(-1+cos(d*x+c))*(-30*I*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/si
n(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi(((1-cos(d*x+c)+sin(d*x+c)
)/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+30*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+si
n(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi(((1-cos(d*x+c
)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-30*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+c
os(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticF(((
1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+10*I*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)-10*I*cos(d*x+c)*s
in(d*x+c)*2^(1/2)+33*2^(1/2)*cos(d*x+c)^3-33*cos(d*x+c)^2*2^(1/2)-3*cos(d*x+c)*2^(1/2)+3*2^(1/2))*(1+cos(d*x+c
))^2/(cos(d*x+c)/sin(d*x+c))^(3/2)/sin(d*x+c)^5/cos(d*x+c)*2^(1/2)

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maxima [B]  time = 0.96, size = 161, normalized size = 1.77 \[ -\frac {4 \, {\left (3 \, a^{2} - \frac {10 i \, a^{2}}{\tan \left (d x + c\right )} - \frac {30 \, a^{2}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac {5}{2}} - 15 \, {\left (\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \left (i - 1\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \left (i - 1\right ) \, \sqrt {2} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{2}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(4*(3*a^2 - 10*I*a^2/tan(d*x + c) - 30*a^2/tan(d*x + c)^2)*tan(d*x + c)^(5/2) - 15*((2*I + 2)*sqrt(2)*ar
ctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + (2*I + 2)*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(t
an(d*x + c)))) + (I - 1)*sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - (I - 1)*sqrt(2)*log(-s
qrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^2)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \frac {\tan ^{2}{\left (c + d x \right )}}{\cot ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \left (- \frac {2 i \tan {\left (c + d x \right )}}{\cot ^{\frac {3}{2}}{\left (c + d x \right )}}\right )\, dx + \int \left (- \frac {1}{\cot ^{\frac {3}{2}}{\left (c + d x \right )}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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