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

Optimal. Leaf size=63 \[ -\frac{2 i \sqrt{a} \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.117107, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {3523, 63, 217, 203} \[ -\frac{2 i \sqrt{a} \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3523

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 1.46309, size = 74, normalized size = 1.17 \[ -\frac{i \sqrt{2} c e^{-i (e+f x)} \tan ^{-1}\left (e^{i (e+f x)}\right ) \sqrt{a+i a \tan (e+f x)}}{f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[2]*c*ArcTan[E^(I*(e + f*x))]*Sqrt[a + I*a*Tan[e + f*x]])/(E^(I*(e + f*x))*Sqrt[c/(1 + E^((2*I)*(e +
f*x)))]*f)

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Maple [A]  time = 0.072, size = 96, normalized size = 1.5 \begin{align*}{\frac{ac}{f}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}}{\frac{1}{\sqrt{ac}}}} \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))^(1/2),x)

[Out]

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

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Maxima [B]  time = 1.90178, size = 140, normalized size = 2.22 \begin{align*} \frac{\sqrt{a} \sqrt{c}{\left (-2 i \, \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) - 2 i \, \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) + \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right )\right )}}{2 \, 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))^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(a)*sqrt(c)*(-2*I*arctan2(cos(f*x + e), sin(f*x + e) + 1) - 2*I*arctan2(cos(f*x + e), -sin(f*x + e) +
1) + log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f
*x + e) + 1))/f

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Fricas [B]  time = 1.56819, size = 593, normalized size = 9.41 \begin{align*} -\frac{1}{2} \, \sqrt{\frac{a c}{f^{2}}} \log \left (\frac{2 \,{\left (4 \, \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (i \, f x + i \, e\right )} +{\left (2 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, f\right )} \sqrt{\frac{a c}{f^{2}}}\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right ) + \frac{1}{2} \, \sqrt{\frac{a c}{f^{2}}} \log \left (\frac{2 \,{\left (4 \, \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (i \, f x + i \, e\right )} +{\left (-2 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, f\right )} \sqrt{\frac{a c}{f^{2}}}\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right ) \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))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (i \tan{\left (e + f x \right )} + 1\right )} \sqrt{- c \left (i \tan{\left (e + f x \right )} - 1\right )}\, dx \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))**(1/2),x)

[Out]

Integral(sqrt(a*(I*tan(e + f*x) + 1))*sqrt(-c*(I*tan(e + f*x) - 1)), x)

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

[Out]

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