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

Optimal. Leaf size=80 \[ \frac{2 i}{3 d \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i \sqrt{a+i a \tan (c+d x)}}{3 a d \sqrt{e \sec (c+d x)}} \]

[Out]

((2*I)/3)/(d*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((4*I)/3)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*Sq
rt[e*Sec[c + d*x]])

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Rubi [A]  time = 0.138318, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3502, 3488} \[ \frac{2 i}{3 d \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i \sqrt{a+i a \tan (c+d x)}}{3 a d \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((2*I)/3)/(d*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((4*I)/3)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*Sq
rt[e*Sec[c + d*x]])

Rule 3502

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rule 3488

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(a*f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{2 i}{3 d \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{2 \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}} \, dx}{3 a}\\ &=\frac{2 i}{3 d \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{4 i \sqrt{a+i a \tan (c+d x)}}{3 a d \sqrt{e \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.101304, size = 48, normalized size = 0.6 \[ \frac{4 \tan (c+d x)-2 i}{3 d \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

(-2*I + 4*Tan[c + d*x])/(3*d*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.319, size = 85, normalized size = 1.1 \begin{align*} -{\frac{2\,i\cos \left ( dx+c \right ) -4\,\sin \left ( dx+c \right ) }{3\,ad \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}{\frac{1}{\sqrt{{\frac{e}{\cos \left ( dx+c \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/d/a*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(I*cos(d*x+c)-2*sin(d*x+c))/(I*sin(d*x+c)+cos(d*x+c))/
(e/cos(d*x+c))^(1/2)

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Maxima [A]  time = 1.87588, size = 108, normalized size = 1.35 \begin{align*} \frac{i \, \cos \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) - 3 i \, \cos \left (\frac{1}{3} \, \arctan \left (\sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ), \cos \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right )\right )\right ) + \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 3 \, \sin \left (\frac{1}{3} \, \arctan \left (\sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ), \cos \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right )\right )\right )}{3 \, \sqrt{a} d \sqrt{e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(I*cos(3/2*d*x + 3/2*c) - 3*I*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sin(3/2*d*x +
3/2*c) + 3*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))/(sqrt(a)*d*sqrt(e))

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Fricas [A]  time = 2.02834, size = 220, normalized size = 2.75 \begin{align*} \frac{\sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-3 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-\frac{3}{2} i \, d x - \frac{3}{2} i \, c\right )}}{3 \, a d e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{e \sec \left (d x + c\right )} \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/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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