3.260 \(\int \frac {1}{\sqrt {a \cos (c+d x)+i a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=31 \[ \frac {2 i}{d \sqrt {a \cos (c+d x)+i a \sin (c+d x)}} \]

[Out]

2*I/d/(a*cos(d*x+c)+I*a*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {3071} \[ \frac {2 i}{d \sqrt {a \cos (c+d x)+i a \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a*Cos[c + d*x] + I*a*Sin[c + d*x]],x]

[Out]

(2*I)/(d*Sqrt[a*Cos[c + d*x] + I*a*Sin[c + d*x]])

Rule 3071

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a*Cos[c + d*x]
+ b*Sin[c + d*x])^n)/(b*d*n), x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a \cos (c+d x)+i a \sin (c+d x)}} \, dx &=\frac {2 i}{d \sqrt {a \cos (c+d x)+i a \sin (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 30, normalized size = 0.97 \[ \frac {2 i}{d \sqrt {a (\cos (c+d x)+i \sin (c+d x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a*Cos[c + d*x] + I*a*Sin[c + d*x]],x]

[Out]

(2*I)/(d*Sqrt[a*(Cos[c + d*x] + I*Sin[c + d*x])])

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fricas [A]  time = 1.04, size = 17, normalized size = 0.55 \[ \frac {2 i \, e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{\sqrt {a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2*I*e^(-1/2*I*d*x - 1/2*I*c)/(sqrt(a)*d)

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giac [A]  time = 0.33, size = 37, normalized size = 1.19 \[ \frac {2 i}{d \sqrt {-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

2*I/(d*sqrt(-(a*tan(1/2*d*x + 1/2*c) - I*a)/(tan(1/2*d*x + 1/2*c) + I)))

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maple [A]  time = 0.22, size = 28, normalized size = 0.90 \[ \frac {2 i}{d \sqrt {a \cos \left (d x +c \right )+i a \sin \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^(1/2),x)

[Out]

2*I/d/(a*cos(d*x+c)+I*a*sin(d*x+c))^(1/2)

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maxima [B]  time = 0.43, size = 51, normalized size = 1.65 \[ \frac {2 i \, \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + i}}{\sqrt {a} d \sqrt {-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

2*I*sqrt(sin(d*x + c)/(cos(d*x + c) + 1) + I)/(sqrt(a)*d*sqrt(-sin(d*x + c)/(cos(d*x + c) + 1) + I))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {1}{\sqrt {a\,\cos \left (c+d\,x\right )+a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cos(c + d*x) + a*sin(c + d*x)*1i)^(1/2),x)

[Out]

int(1/(a*cos(c + d*x) + a*sin(c + d*x)*1i)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {i a \sin {\left (c + d x \right )} + a \cos {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))**(1/2),x)

[Out]

Integral(1/sqrt(I*a*sin(c + d*x) + a*cos(c + d*x)), x)

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