3.1140 \(\int \frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{a+i a \tan (e+f x)}} \, dx\)

Optimal. Leaf size=121 \[ \frac{i \sqrt{c+d \tan (e+f x)}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{\sqrt{2} \sqrt{a} f} \]

[Out]

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

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Rubi [A]  time = 0.21492, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {3546, 3544, 208} \[ \frac{i \sqrt{c+d \tan (e+f x)}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{\sqrt{2} \sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3546

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

Rule 3544

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

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 3.17367, size = 182, normalized size = 1.5 \[ \frac{i \left (\sqrt{1+e^{2 i (e+f x)}} \sqrt{c+d \tan (e+f x)}-\sqrt{c-i d} e^{i (e+f x)} \log \left (2 \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )\right )}{f \sqrt{1+e^{2 i (e+f x)}} \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.105, size = 877, normalized size = 7.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [B]  time = 1.79314, size = 1084, normalized size = 8.96 \begin{align*} \frac{{\left (\sqrt{2} a f \sqrt{-\frac{c - i \, d}{a f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left ({\left (i \, \sqrt{2} a f \sqrt{-\frac{c - i \, d}{a f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{2} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{a}{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 )}\right )} e^{\left (-i \, f x - i \, e\right )}\right ) - \sqrt{2} a f \sqrt{-\frac{c - i \, d}{a f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left ({\left (-i \, \sqrt{2} a f \sqrt{-\frac{c - i \, d}{a f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{2} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{a}{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 )}\right )} e^{\left (-i \, f x - i \, e\right )}\right ) + \sqrt{2} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (2 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError