3.777 \(\int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=147 \[ -\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {1}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {i}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4241, 3547, 3546, 3544, 205} \[ -\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {1}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {i}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

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

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 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 3546

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*b*f*m), x] - Dist[(a*c - b*d)/(2*b^2), Int[(a + b*Tan[e
+ f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), 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] && EqQ[m + n, 0] && LeQ[m, -2^(-1)]

Rule 3547

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

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

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Mathematica [A]  time = 1.43, size = 156, normalized size = 1.06 \[ \frac {e^{-4 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {\cot (c+d x)} \left (-e^{2 i (c+d x)}+2 e^{4 i (c+d x)}-3 e^{3 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )-1\right )}{6 \sqrt {2} a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

(Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(-1 - E^((2*I)*(c + d*x)) + 2*E^((4*I)*(c + d*x)) - 3
*E^((3*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*S
qrt[Cot[c + d*x]])/(6*Sqrt[2]*a^2*d*E^((4*I)*(c + d*x)))

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fricas [B]  time = 1.23, size = 355, normalized size = 2.41 \[ -\frac {{\left (3 \, a^{2} d \sqrt {\frac {i}{2 \, a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{2 \, a^{3} d^{2}}} + i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3 \, a^{2} d \sqrt {\frac {i}{2 \, a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{2 \, a^{3} d^{2}}} - i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (2 \, e^{\left (4 i \, d x + 4 i \, c\right )} - e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*a^2*d*sqrt(1/2*I/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(4*(sqrt(2)*(a^2*d*e^(2*I*d*x + 2*I*c) - a^2*d)*sq
rt(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))*sqrt(1/2*I/(a^3*d^
2)) + I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 3*a^2*d*sqrt(1/2*I/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(-4*(sqrt(
2)*(a^2*d*e^(2*I*d*x + 2*I*c) - a^2*d)*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))*sqrt(1/2*I/(a^3*d^2)) - I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - sqrt(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))*(2*e^(4*I*d*x + 4*I*c) - e^(2*
I*d*x + 2*I*c) - 1))*e^(-3*I*d*x - 3*I*c)/(a^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 3.87, size = 482, normalized size = 3.28 \[ \frac {\left (\frac {1}{12}+\frac {i}{12}\right ) \left (6 i \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )+3 i \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 )}}-6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {2}-i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-3 i \cos \left (d x +c \right ) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {2}+3 \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 )}}+3 \sin \left (d x +c \right ) \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )-3 i \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )+\left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+i \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 )}}\right ) \cos \left (d x +c \right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{d \left (2 i \cos \left (d x +c \right ) \sin \left (d x +c \right )+2 \left (\cos ^{2}\left (d x +c \right )\right )-1\right ) \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/12+1/12*I)/d*(6*I*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)^2+3*I*s
in(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)-6*cos(d*x+c)*sin(d*x+c)*arctan((1/2+1/2*I)*((-1+cos(d*
x+c))/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)-I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2-3*I*2^(1/2)*arctan(
(1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)+3*cos(d*x+c)*sin(d*x+c)*((-1+cos(d*x+c))/si
n(d*x+c))^(1/2)+3*sin(d*x+c)*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))-3*I*2^(1/2
)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))+cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)+I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-((-1+cos(d*x+c))/sin(d*x+c))^(1/2))*cos(d*x+c)*(a*(I*sin(d*x+c)+cos(d*
x+c))/cos(d*x+c))^(1/2)/(2*I*cos(d*x+c)*sin(d*x+c)+2*cos(d*x+c)^2-1)/sin(d*x+c)/((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)/(cos(d*x+c)/sin(d*x+c))^(1/2)/a^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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