3.737 \(\int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx\)
Optimal. Leaf size=68 \[ \frac {\sqrt [4]{-1} \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{2 a d}+\frac {i \sqrt {\cot (c+d x)}}{2 d (a \cot (c+d x)+i a)} \]
[Out]
1/2*(-1)^(1/4)*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/a/d+1/2*I*cot(d*x+c)^(1/2)/d/(I*a+a*cot(d*x+c))
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Rubi [A] time = 0.13, antiderivative size = 68, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used =
{3673, 3549, 3533, 208} \[ \frac {\sqrt [4]{-1} \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{2 a d}+\frac {i \sqrt {\cot (c+d x)}}{2 d (a \cot (c+d x)+i a)} \]
Antiderivative was successfully verified.
[In]
Int[1/(Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])),x]
[Out]
((-1)^(1/4)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/(2*a*d) + ((I/2)*Sqrt[Cot[c + d*x]])/(d*(I*a + a*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 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 3549
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
a*c + b*d)*(c + d*Tan[e + f*x])^n)/(2*(b*c - a*d)*f*(a + b*Tan[e + f*x])), x] + Dist[1/(2*a*(b*c - a*d)), Int[
(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c*d*(n - 1) + b*c^2 + b*d^2*n - d*(b*c - a*d)*(n - 1)*Tan[e + f*x], x], 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] && LtQ[0,
n, 1]
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 {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx &=\int \frac {\sqrt {\cot (c+d x)}}{i a+a \cot (c+d x)} \, dx\\ &=\frac {i \sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac {\int \frac {-\frac {a}{2}+\frac {1}{2} i a \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{2 a^2}\\ &=\frac {i \sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {a}{2}+\frac {1}{2} i a x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{4 d}\\ &=\frac {\sqrt [4]{-1} \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{2 a d}+\frac {i \sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.89, size = 126, normalized size = 1.85 \[ \frac {(\sin (c+d x)+i \cos (c+d x)) \left (\cos (c+d x) \sqrt {i \tan (c+d x)}-\tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) (\cos (c+d x)+i \sin (c+d x))\right )}{2 a d \sqrt {i \tan (c+d x)} \sqrt {\cot (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])),x]
[Out]
((I*Cos[c + d*x] + Sin[c + d*x])*(-(ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*(Cos[c
+ d*x] + I*Sin[c + d*x])) + Cos[c + d*x]*Sqrt[I*Tan[c + d*x]]))/(2*a*d*Sqrt[Cot[c + d*x]]*Sqrt[I*Tan[c + d*x]
])
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fricas [B] time = 1.16, size = 269, normalized size = 3.96 \[ -\frac {{\left (a d \sqrt {\frac {i}{4 \, a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (2 \, {\left (2 \, {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{4 \, a^{2} d^{2}}} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - a d \sqrt {\frac {i}{4 \, a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-2 \, {\left (2 \, {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{4 \, a^{2} d^{2}}} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \]
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)),x, algorithm="fricas")
[Out]
-1/4*(a*d*sqrt(1/4*I/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log(2*(2*(a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt((I*e^(2*I*d*
x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(1/4*I/(a^2*d^2)) + I*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)
) - a*d*sqrt(1/4*I/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log(-2*(2*(a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt((I*e^(2*I*d*x
+ 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(1/4*I/(a^2*d^2)) - I*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c))
- 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) - 1))*e^(-2*I*d*x - 2*I*c)
/(a*d)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )} \sqrt {\cot \left (d x + c\right )}}\,{d x} \]
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)),x, algorithm="giac")
[Out]
integrate(1/((I*a*tan(d*x + c) + a)*sqrt(cot(d*x + c))), x)
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maple [C] time = 1.34, size = 1134, normalized size = 16.68 \[ \text {result too large to display} \]
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)),x)
[Out]
-1/4/a/d*(-1+cos(d*x+c))*(I*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))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(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)-I*cos(d*x+c)*sin(d*x+c)*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))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(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)+I*cos(d*x+c)*sin(d*x+c)*EllipticF(((1-cos(d*x+c)+sin(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)*((-1+cos(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*((1-cos(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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))+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)
)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(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)-sin(d*x+c)*cos(d*x+c)*((1-cos(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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))-I*((1-cos(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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))-I*cos(d*x+c)^2*2^(1/2)+((1-cos(d*x+c)+sin(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))^(1/2)*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))-EllipticF(((1-cos(d*x+c)+sin(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)*((-1+cos(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)+I*cos(
d*x+c)*2^(1/2))*(1+cos(d*x+c))^2/(I*sin(d*x+c)+cos(d*x+c))/(cos(d*x+c)/sin(d*x+c))^(1/2)/sin(d*x+c)^4*2^(1/2)
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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)),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \]
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)),x)
[Out]
int(1/(cot(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \int \frac {1}{\tan {\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}} - i \sqrt {\cot {\left (c + d x \right )}}}\, dx}{a} \]
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)),x)
[Out]
-I*Integral(1/(tan(c + d*x)*sqrt(cot(c + d*x)) - I*sqrt(cot(c + d*x))), x)/a
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