3.1158 \(\int \frac{1}{\sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=174 \[ \frac{2 d \sqrt{c+d \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt{a+i a \tan (e+f x)}}-\frac{\sqrt{c+d \tan (e+f x)}}{f (d+i c) \sqrt{a+i a \tan (e+f x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{\sqrt{2} \sqrt{a} f \sqrt{c-i d}} \]

[Out]

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

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Rubi [A]  time = 0.318434, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3548, 3546, 3544, 208} \[ \frac{2 d \sqrt{c+d \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt{a+i a \tan (e+f x)}}-\frac{\sqrt{c+d \tan (e+f x)}}{f (d+i c) \sqrt{a+i a \tan (e+f x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{\sqrt{2} \sqrt{a} f \sqrt{c-i d}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]),x]

[Out]

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

Rule 3548

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

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

Mathematica [A]  time = 2.9812, size = 195, normalized size = 1.12 \[ \frac{\frac{2 (d-i c) e^{i (e+f x)} \log \left (2 \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{\sqrt{1+e^{2 i (e+f x)}}}+2 i \sqrt{c-i d} \sqrt{c+d \tan (e+f x)}}{2 f \sqrt{c-i d} (c+i d) \sqrt{a+i a \tan (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]),x]

[Out]

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

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Maple [B]  time = 0.096, size = 1738, normalized size = 10. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)/a*(-I*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))*2^(1/2)*(-a*(I*d-
c))^(1/2)*tan(f*x+e)^2*d^3+3*I*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2*c^2*d-3*I
*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*d-2*I*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))*2^(1/2)*(-a*
(I*d-c))^(1/2)*tan(f*x+e)*c^3+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))*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2*c^3-3*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2*c*d^2+4*I*tan(f*x+e)*c^3*(a*(c+d*tan(f*x+e))
*(1+I*tan(f*x+e)))^(1/2)+4*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^2*d+6*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))
*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)*c^2*d-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))*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*
x+e)*d^3+I*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*d^3+6*I*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))*2^(1
/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)*c*d^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))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^3+3*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*c*d^2+4*I*tan(f*x+e)*c*d^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+
e)))^(1/2)+4*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^3-4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*t
an(f*x+e)*c^2*d-4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)*d^3+4*c^3*(a*(c+d*tan(f*x+e))*(1+I*ta
n(f*x+e)))^(1/2)+4*c*d^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1
/2)/(c+I*d)^2/(I*d-c)/(I*c-d)/(-tan(f*x+e)+I)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.65055, size = 1146, normalized size = 6.59 \begin{align*} -\frac{{\left ({\left (-i \, a c + a d\right )} f \sqrt{-\frac{2 i}{{\left (i \, a c + a d\right )} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left ({\left ({\left (i \, a c + a d\right )} f \sqrt{-\frac{2 i}{{\left (i \, a c + a d\right )} 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 ) +{\left (i \, a c - a d\right )} f \sqrt{-\frac{2 i}{{\left (i \, a c + a d\right )} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left ({\left ({\left (-i \, a c - a d\right )} f \sqrt{-\frac{2 i}{{\left (i \, a c + a d\right )} 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 ) + 2 \, \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 (-2 i \, f x - 2 i \, e\right )}}{4 \,{\left (i \, a c - a d\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out