\(\int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [344]

Optimal result

Integrand size = 24, antiderivative size = 52 \[ \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

[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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3570, 212} \[ \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

[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 212

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

Rule 3570

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*} \text {integral}& = \frac {(2 i) \text {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} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.35 \[ \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i e^{i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )}{d \sqrt {1+e^{2 i (c+d x)}} \sqrt {a+i a \tan (c+d x)}} \]

[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]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (41 ) = 82\).

Time = 7.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.35

method	result	size
default	\(\frac {\left (i \cos \left (d x +c \right )+i-\sin \left (d x +c \right )\right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(122\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (39) = 78\).

Time = 0.27 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.87 \[ \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {1}{2} i \, \sqrt {2} \sqrt {\frac {1}{a d^{2}}} \log \left (-\frac {4 \, {\left ({\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{d}\right ) + \frac {1}{2} i \, \sqrt {2} \sqrt {\frac {1}{a d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{d}\right ) \]

[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(-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(
1/(a*d^2)) - I)*e^(-I*d*x - I*c)/d) + 1/2*I*sqrt(2)*sqrt(1/(a*d^2))*log(-4*((-I*d*e^(2*I*d*x + 2*I*c) - I*d)*s
qrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) - I)*e^(-I*d*x - I*c)/d)

Sympy [F]

\[ \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]

[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)