3.90 \(\int \frac{1+i \cot (c+d x)}{\sqrt{a+b \cot (c+d x)}} \, dx\)
Optimal. Leaf size=45 \[ \frac{2 i \tanh ^{-1}\left (\frac{\sqrt{a+b \cot (c+d x)}}{\sqrt{a-i b}}\right )}{d \sqrt{a-i b}} \]
[Out]
((2*I)*ArcTanh[Sqrt[a + b*Cot[c + d*x]]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d)
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Rubi [A] time = 0.0652656, antiderivative size = 45, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used =
{3537, 63, 208} \[ \frac{2 i \tanh ^{-1}\left (\frac{\sqrt{a+b \cot (c+d x)}}{\sqrt{a-i b}}\right )}{d \sqrt{a-i b}} \]
Antiderivative was successfully verified.
[In]
Int[(1 + I*Cot[c + d*x])/Sqrt[a + b*Cot[c + d*x]],x]
[Out]
((2*I)*ArcTanh[Sqrt[a + b*Cot[c + d*x]]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d)
Rule 3537
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
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 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]
Rubi steps
\begin{align*} \int \frac{1+i \cot (c+d x)}{\sqrt{a+b \cot (c+d x)}} \, dx &=-\frac{i \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a-i b x}} \, dx,x,i \cot (c+d x)\right )}{d}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i a}{b}+\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \cot (c+d x)}\right )}{b d}\\ &=\frac{2 i \tanh ^{-1}\left (\frac{\sqrt{a+b \cot (c+d x)}}{\sqrt{a-i b}}\right )}{\sqrt{a-i b} d}\\ \end{align*}
Mathematica [A] time = 0.0737778, size = 45, normalized size = 1. \[ \frac{2 i \tanh ^{-1}\left (\frac{\sqrt{a+b \cot (c+d x)}}{\sqrt{a-i b}}\right )}{d \sqrt{a-i b}} \]
Antiderivative was successfully verified.
[In]
Integrate[(1 + I*Cot[c + d*x])/Sqrt[a + b*Cot[c + d*x]],x]
[Out]
((2*I)*ArcTanh[Sqrt[a + b*Cot[c + d*x]]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d)
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Maple [B] time = 0.079, size = 1622, normalized size = 36. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1+I*cot(d*x+c))/(a+b*cot(d*x+c))^(1/2),x)
[Out]
-I/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/(a^2+b^2)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)*ln((a+b*cot(d*x+c))^(1/2)*(2*(a
^2+b^2)^(1/2)+2*a)^(1/2)-b*cot(d*x+c)-a-(a^2+b^2)^(1/2))*a^3+1/2*I/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/(a^2+b^2)^(
1/2)*ln(b*cot(d*x+c)+a+(a+b*cot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*a-1/2/d/(2*(a^2+b
^2)^(1/2)+2*a)^(1/2)/(a^2+b^2)^(1/2)*ln(b*cot(d*x+c)+a+(a+b*cot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a
^2+b^2)^(1/2))*b+I/d/((a^2+b^2)^(1/2)*a+a^2+b^2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)
^(1/2)-2*(a+b*cot(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*b^2-I/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((
2*(a+b*cot(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))-1/d/(a^2+b^2)^(1/2)/(2*
(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*cot(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-
2*a)^(1/2))*b-I/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)*ln((a+b*cot(d*x+c))^(1/2)*(2*(a^2+
b^2)^(1/2)+2*a)^(1/2)-b*cot(d*x+c)-a-(a^2+b^2)^(1/2))*a^2-1/2*I/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/((a^2+b^2)^(1/
2)*a+a^2+b^2)*ln((a+b*cot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*cot(d*x+c)-a-(a^2+b^2)^(1/2))*b^2-I/d/
(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/(a^2+b^2)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)*ln((a+b*cot(d*x+c))^(1/2)*(2*(a^2+b^
2)^(1/2)+2*a)^(1/2)-b*cot(d*x+c)-a-(a^2+b^2)^(1/2))*a*b^2+I/d/(a^2+b^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*ar
ctan((2*(a+b*cot(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+1/2/d/(2*(a^2+b
^2)^(1/2)+2*a)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)*ln((a+b*cot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*cot
(d*x+c)-a-(a^2+b^2)^(1/2))*a*b+1/2/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/(a^2+b^2)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)
*ln((a+b*cot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*cot(d*x+c)-a-(a^2+b^2)^(1/2))*a^2*b+1/2/d/(2*(a^2+b
^2)^(1/2)+2*a)^(1/2)/(a^2+b^2)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)*ln((a+b*cot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+
2*a)^(1/2)-b*cot(d*x+c)-a-(a^2+b^2)^(1/2))*b^3+1/2*I/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln(b*cot(d*x+c)+a+(a+b*co
t(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))+1/d/((a^2+b^2)^(1/2)*a+a^2+b^2)/(2*(a^2+b^2)^(1
/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*cot(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*
a*b+1/d/(a^2+b^2)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2
*a)^(1/2)-2*(a+b*cot(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^2*b+1/d/(a^2+b^2)^(1/2)/((a^2+b^2)^(1/2)*
a+a^2+b^2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*cot(d*x+c))^(1/2))/(2*(a
^2+b^2)^(1/2)-2*a)^(1/2))*b^3
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{i \, \cot \left (d x + c\right ) + 1}{\sqrt{b \cot \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+I*cot(d*x+c))/(a+b*cot(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate((I*cot(d*x + c) + 1)/sqrt(b*cot(d*x + c) + a), x)
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Fricas [B] time = 1.65204, size = 425, normalized size = 9.44 \begin{align*} -\frac{1}{2} \, \sqrt{-\frac{4 i}{{\left (i \, a + b\right )} d^{2}}} \log \left (\frac{1}{2} \,{\left (i \, a + b\right )} d \sqrt{-\frac{4 i}{{\left (i \, a + b\right )} d^{2}}} + \sqrt{\frac{{\left (a + i \, b\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - a + i \, b}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right ) + \frac{1}{2} \, \sqrt{-\frac{4 i}{{\left (i \, a + b\right )} d^{2}}} \log \left (\frac{1}{2} \,{\left (-i \, a - b\right )} d \sqrt{-\frac{4 i}{{\left (i \, a + b\right )} d^{2}}} + \sqrt{\frac{{\left (a + i \, b\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - a + i \, b}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+I*cot(d*x+c))/(a+b*cot(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
-1/2*sqrt(-4*I/((I*a + b)*d^2))*log(1/2*(I*a + b)*d*sqrt(-4*I/((I*a + b)*d^2)) + sqrt(((a + I*b)*e^(2*I*d*x +
2*I*c) - a + I*b)/(e^(2*I*d*x + 2*I*c) - 1))) + 1/2*sqrt(-4*I/((I*a + b)*d^2))*log(1/2*(-I*a - b)*d*sqrt(-4*I/
((I*a + b)*d^2)) + sqrt(((a + I*b)*e^(2*I*d*x + 2*I*c) - a + I*b)/(e^(2*I*d*x + 2*I*c) - 1)))
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{i \cot{\left (c + d x \right )} + 1}{\sqrt{a + b \cot{\left (c + d x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+I*cot(d*x+c))/(a+b*cot(d*x+c))**(1/2),x)
[Out]
Integral((I*cot(c + d*x) + 1)/sqrt(a + b*cot(c + d*x)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{i \, \cot \left (d x + c\right ) + 1}{\sqrt{b \cot \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+I*cot(d*x+c))/(a+b*cot(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((I*cot(d*x + c) + 1)/sqrt(b*cot(d*x + c) + a), x)