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

Optimal. Leaf size=46 \[ -\frac {i \sqrt {2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3480, 206} \[ -\frac {i \sqrt {2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 64, normalized size = 1.39 \[ -\frac {i e^{-i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.42, size = 159, normalized size = 3.46 \[ \frac {1}{2} \, \sqrt {2} \sqrt {-\frac {a}{d^{2}}} \log \left (4 \, {\left ({\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \frac {1}{2} \, \sqrt {2} \sqrt {-\frac {a}{d^{2}}} \log \left (4 \, {\left ({\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.14, size = 36, normalized size = 0.78 \[ -\frac {i \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, \sqrt {a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 60, normalized size = 1.30 \[ \frac {i \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.95, size = 39, normalized size = 0.85 \[ -\frac {\sqrt {2}\,\sqrt {-a}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

-(2^(1/2)*(-a)^(1/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*1i)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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