3.1120 \(\int \frac{(a+i a \tan (e+f x))^2}{\sqrt{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=74 \[ -\frac{2 a^2 \sqrt{c+d \tan (e+f x)}}{d f}-\frac{4 i a^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}} \]

[Out]

((-4*I)*a^2*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) - (2*a^2*Sqrt[c + d*Tan[e + f*x
]])/(d*f)

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Rubi [A]  time = 0.151448, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3543, 3537, 63, 208} \[ -\frac{2 a^2 \sqrt{c+d \tan (e+f x)}}{d f}-\frac{4 i a^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*I)*a^2*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) - (2*a^2*Sqrt[c + d*Tan[e + f*x
]])/(d*f)

Rule 3543

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

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))^2}{\sqrt{c+d \tan (e+f x)}} \, dx &=-\frac{2 a^2 \sqrt{c+d \tan (e+f x)}}{d f}+\int \frac{2 a^2+2 i a^2 \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=-\frac{2 a^2 \sqrt{c+d \tan (e+f x)}}{d f}+\frac{\left (4 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-4 a^4+2 a^2 x\right ) \sqrt{c-\frac{i d x}{2 a^2}}} \, dx,x,2 i a^2 \tan (e+f x)\right )}{f}\\ &=-\frac{2 a^2 \sqrt{c+d \tan (e+f x)}}{d f}-\frac{\left (16 a^6\right ) \operatorname{Subst}\left (\int \frac{1}{-4 a^4-\frac{4 i a^4 c}{d}+\frac{4 i a^4 x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{d f}\\ &=-\frac{4 i a^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d} f}-\frac{2 a^2 \sqrt{c+d \tan (e+f x)}}{d f}\\ \end{align*}

Mathematica [A]  time = 2.75475, size = 95, normalized size = 1.28 \[ \frac{2 a^2 \left (-\frac{\sqrt{c+d \tan (e+f x)}}{d}-\frac{2 i \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.049, size = 1698, normalized size = 23. \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))^2/(c+d*tan(f*x+e))^(1/2),x)

[Out]

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

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

[Out]

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

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Fricas [B]  time = 1.85841, size = 890, normalized size = 12.03 \begin{align*} \frac{d \sqrt{-\frac{16 i \, a^{4}}{{\left (i \, c + d\right )} f^{2}}} f \log \left (\frac{{\left (4 \, a^{2} 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{16 i \, a^{4}}{{\left (i \, c + d\right )} f^{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}} +{\left (4 \, a^{2} c - 4 i \, a^{2} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - d \sqrt{-\frac{16 i \, a^{4}}{{\left (i \, c + d\right )} f^{2}}} f \log \left (\frac{{\left (4 \, a^{2} 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{16 i \, a^{4}}{{\left (i \, c + d\right )} f^{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}} +{\left (4 \, a^{2} c - 4 i \, a^{2} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - 8 \, a^{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}}}{4 \, d f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(d*sqrt(-16*I*a^4/((I*c + d)*f^2))*f*log(1/2*(4*a^2*c + ((I*c + d)*f*e^(2*I*f*x + 2*I*e) + (I*c + d)*f)*sq
rt(-16*I*a^4/((I*c + d)*f^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)) + (4*a
^2*c - 4*I*a^2*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^2) - d*sqrt(-16*I*a^4/((I*c + d)*f^2))*f*log(1/2
*(4*a^2*c + ((-I*c - d)*f*e^(2*I*f*x + 2*I*e) + (-I*c - d)*f)*sqrt(-16*I*a^4/((I*c + d)*f^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)) + (4*a^2*c - 4*I*a^2*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I
*f*x - 2*I*e)/a^2) - 8*a^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)))/(d*f)

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

[Out]

a**2*(Integral(-tan(e + f*x)**2/sqrt(c + d*tan(e + f*x)), x) + Integral(2*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 = 1.50993, size = 250, normalized size = 3.38 \begin{align*} -\frac{2 \, \sqrt{d \tan \left (f x + e\right ) + c} a^{2}}{d f} + \frac{16 i \, a^{2} \arctan \left (\frac{4 \,{\left (\sqrt{d \tan \left (f x + e\right ) + c} c - \sqrt{c^{2} + d^{2}} \sqrt{d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} - i \, \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d - \sqrt{c^{2} + d^{2}} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}}\right )}{\sqrt{-8 \, c + 8 \, \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))^2/(c+d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-2*sqrt(d*tan(f*x + e) + c)*a^2/(d*f) + 16*I*a^2*arctan(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d
*tan(f*x + e) + c))/(c*sqrt(-8*c + 8*sqrt(c^2 + d^2)) - I*sqrt(-8*c + 8*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*s
qrt(-8*c + 8*sqrt(c^2 + d^2))))/(sqrt(-8*c + 8*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1))