3.113 \(\int \frac{1}{\sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=71 \[ \frac{i}{d \sqrt{a+i a \tan (c+d x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d} \]

[Out]

((-I)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d) + I/(d*Sqrt[a + I*a*Tan[c + d
*x]])

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Rubi [A]  time = 0.0380318, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {3479, 3480, 206} \[ \frac{i}{d \sqrt{a+i a \tan (c+d x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((-I)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d) + I/(d*Sqrt[a + I*a*Tan[c + d
*x]])

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.184683, size = 84, normalized size = 1.18 \[ \frac{i \left (\sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{d \sqrt{1+e^{2 i (c+d x)}} \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*(Sqrt[1 + E^((2*I)*(c + d*x))] - E^(I*(c + d*x))*ArcSinh[E^(I*(c + d*x))]))/(d*Sqrt[1 + E^((2*I)*(c + d*x))
]*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.041, size = 59, normalized size = 0.8 \begin{align*}{\frac{2\,ia}{d} \left ({\frac{1}{2\,a}{\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}-{\frac{\sqrt{2}}{4}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{3}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.06899, size = 737, normalized size = 10.38 \begin{align*} \frac{{\left (-i \, \sqrt{2} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left ({\left (\sqrt{2} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + i \, \sqrt{2} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-{\left (\sqrt{2} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} 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}}{\left (2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{i a \tan{\left (c + d x \right )} + a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(I*a*tan(d*x + c) + a), x)