3.1121 \(\int \frac{a+i a \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=46 \[ -\frac{2 i a \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}} \]

[Out]

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

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Rubi [A]  time = 0.0724957, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3537, 63, 208} \[ -\frac{2 i a \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^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 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{a+i a \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-a^2+a x\right ) \sqrt{c-\frac{i d x}{a}}} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=-\frac{\left (2 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{-a^2-\frac{i a^2 c}{d}+\frac{i a^2 x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{d f}\\ &=-\frac{2 i a \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d} f}\\ \end{align*}

Mathematica [A]  time = 1.22909, size = 71, normalized size = 1.54 \[ -\frac{2 i a \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.047, size = 1641, normalized size = 35.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.70413, size = 725, normalized size = 15.76 \begin{align*} \frac{1}{4} \, \sqrt{-\frac{4 i \, a^{2}}{{\left (i \, c + d\right )} f^{2}}} \log \left (\frac{{\left (2 \, a c +{\left ({\left (i \, c + d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, c + d\right )} f\right )} \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{4 i \, a^{2}}{{\left (i \, c + d\right )} f^{2}}} +{\left (2 \, a c - 2 i \, a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) - \frac{1}{4} \, \sqrt{-\frac{4 i \, a^{2}}{{\left (i \, c + d\right )} f^{2}}} \log \left (\frac{{\left (2 \, a c +{\left ({\left (-i \, c - d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-i \, c - d\right )} f\right )} \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{4 i \, a^{2}}{{\left (i \, c + d\right )} f^{2}}} +{\left (2 \, a c - 2 i \, a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.39228, size = 207, normalized size = 4.5 \begin{align*} \frac{2 \, \sqrt{2} a \arctan \left (\frac{-16 i \, \sqrt{d \tan \left (f x + e\right ) + c} c - 16 i \, \sqrt{c^{2} + d^{2}} \sqrt{d \tan \left (f x + e\right ) + c}}{8 \, \sqrt{2} \sqrt{c + \sqrt{c^{2} + d^{2}}} c - 8 i \, \sqrt{2} \sqrt{c + \sqrt{c^{2} + d^{2}}} d + 8 \, \sqrt{2} \sqrt{c^{2} + d^{2}} \sqrt{c + \sqrt{c^{2} + d^{2}}}}\right )}{\sqrt{c + \sqrt{c^{2} + d^{2}}} f{\left (-\frac{i \, d}{c + \sqrt{c^{2} + d^{2}}} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(2)*a*arctan((-16*I*sqrt(d*tan(f*x + e) + c)*c - 16*I*sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(8*sqrt(
2)*sqrt(c + sqrt(c^2 + d^2))*c - 8*I*sqrt(2)*sqrt(c + sqrt(c^2 + d^2))*d + 8*sqrt(2)*sqrt(c^2 + d^2)*sqrt(c +
sqrt(c^2 + d^2))))/(sqrt(c + sqrt(c^2 + d^2))*f*(-I*d/(c + sqrt(c^2 + d^2)) + 1))