3.976 \(\int \frac{a+i a \tan (e+f x)}{\sqrt{c-i c \tan (e+f x)}} \, dx\)

Optimal. Leaf size=25 \[ -\frac{2 i a}{f \sqrt{c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.103065, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {3522, 3487, 32} \[ -\frac{2 i a}{f \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])/Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

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

Rule 3522

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rule 3487

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \frac{a+i a \tan (e+f x)}{\sqrt{c-i c \tan (e+f x)}} \, dx &=(a c) \int \frac{\sec ^2(e+f x)}{(c-i c \tan (e+f x))^{3/2}} \, dx\\ &=\frac{(i a) \operatorname{Subst}\left (\int \frac{1}{(c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{f}\\ &=-\frac{2 i a}{f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 0.903638, size = 64, normalized size = 2.56 \[ \frac{2 a \cos (e+f x) (\cos (f x)-i \sin (f x)) \sqrt{c-i c \tan (e+f x)} (\sin (e+2 f x)-i \cos (e+2 f x))}{c f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])/Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

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

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Maple [A]  time = 0.04, size = 22, normalized size = 0.9 \begin{align*}{\frac{-2\,ia}{f}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*a/f/(c-I*c*tan(f*x+e))^(1/2)

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Maxima [A]  time = 1.32103, size = 26, normalized size = 1.04 \begin{align*} -\frac{2 i \, a}{\sqrt{-i \, c \tan \left (f x + e\right ) + c} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*a/(sqrt(-I*c*tan(f*x + e) + c)*f)

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Fricas [B]  time = 1.38257, size = 111, normalized size = 4.44 \begin{align*} \frac{\sqrt{2}{\left (-i \, a e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.72834, size = 44, normalized size = 1.76 \begin{align*} \begin{cases} - \frac{2 i a}{f \sqrt{- i c \tan{\left (e + f x \right )} + c}} & \text{for}\: f \neq 0 \\\frac{x \left (i a \tan{\left (e \right )} + a\right )}{\sqrt{- i c \tan{\left (e \right )} + c}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))**(1/2),x)

[Out]

Piecewise((-2*I*a/(f*sqrt(-I*c*tan(e + f*x) + c)), Ne(f, 0)), (x*(I*a*tan(e) + a)/sqrt(-I*c*tan(e) + c), True)
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((I*a*tan(f*x + e) + a)/sqrt(-I*c*tan(f*x + e) + c), x)