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

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

[Out]

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

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Rubi [A]  time = 0.100693, 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} \[ \frac{2 i a \sqrt{c-i c \tan (e+f x)}}{f} \]

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*Sqrt[c - I*c*Tan[e + f*x]])/f

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

Mathematica [A]  time = 0.796619, size = 25, normalized size = 1. \[ \frac{2 i a \sqrt{c-i c \tan (e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 22, normalized size = 0.9 \begin{align*}{\frac{2\,ia}{f}\sqrt{c-ic\tan \left ( fx+e \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.50945, size = 26, normalized size = 1.04 \begin{align*} \frac{2 i \, \sqrt{-i \, c \tan \left (f x + e\right ) + c} a}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*sqrt(-I*c*tan(f*x + e) + c)*a/f

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Fricas [A]  time = 1.27363, size = 69, normalized size = 2.76 \begin{align*} \frac{2 i \, \sqrt{2} a \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.88737, size = 42, normalized size = 1.68 \begin{align*} \begin{cases} \frac{2 i a \sqrt{- i c \tan{\left (e + f x \right )} + c}}{f} & \text{for}\: f \neq 0 \\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((c-I*c*tan(f*x+e))**(1/2)*(a+I*a*tan(f*x+e)),x)

[Out]

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

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Giac [A]  time = 1.62683, size = 27, normalized size = 1.08 \begin{align*} \frac{2 i \, \sqrt{-i \, c \tan \left (f x + e\right ) + c} a}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*sqrt(-I*c*tan(f*x + e) + c)*a/f