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

Optimal. Leaf size=52 \[ \frac {i \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3489, 206} \[ \frac {i \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(I*Sqrt[2]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(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 3489

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*a)/(b*f), Subst[
Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^
2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.38, size = 70, normalized size = 1.35 \[ \frac {2 i e^{i (c+d x)} \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )}{d \sqrt {1+e^{2 i (c+d x)}} \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*E^(I*(c + d*x))*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])/(d*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt[a + I*a*T
an[c + d*x]])

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fricas [B]  time = 0.68, size = 149, normalized size = 2.87 \[ \frac {1}{2} i \, \sqrt {2} \sqrt {\frac {1}{a d^{2}}} \log \left (\frac {{\left (2 \, {\left (2 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + 4 i\right )} e^{\left (-i \, d x - i \, c\right )}}{d}\right ) - \frac {1}{2} i \, \sqrt {2} \sqrt {\frac {1}{a d^{2}}} \log \left (\frac {{\left (2 \, {\left (-2 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + 4 i\right )} e^{\left (-i \, d x - i \, c\right )}}{d}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.98, size = 137, normalized size = 2.63 \[ -\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sin \left (d x +c \right ) \sqrt {2}}{d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\cos \left (c+d\,x\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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