3.683 \(\int \frac{\sqrt{e \cos (c+d x)}}{\sqrt{a+i a \tan (c+d x)}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.208784, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3515, 3502, 3488} \[ \frac{2 i \sqrt{e \cos (c+d x)}}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{4 i \sqrt{a+i a \tan (c+d x)} \sqrt{e \cos (c+d x)}}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3515

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

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

Mathematica [A]  time = 0.18951, size = 48, normalized size = 0.6 \[ \frac{2 (2 \tan (c+d x)-i) \sqrt{e \cos (c+d x)}}{3 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.364, size = 74, normalized size = 0.9 \begin{align*}{\frac{2\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -4\,i}{3\,ad}\sqrt{e\cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 3.144, size = 108, normalized size = 1.35 \begin{align*} \frac{\sqrt{e}{\left (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 )\right )}}{3 \, \sqrt{a} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(e)*(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)

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Fricas [A]  time = 2.09765, size = 204, normalized size = 2.55 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-3 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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