3.1102 \(\int (a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=100 \[ -\frac{2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac{4 i a^2 \sqrt{c+d \tan (e+f x)}}{f}-\frac{4 i a^2 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

[Out]

((-4*I)*a^2*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + ((4*I)*a^2*Sqrt[c + d*Tan[e + f
*x]])/f - (2*a^2*(c + d*Tan[e + f*x])^(3/2))/(3*d*f)

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Rubi [A]  time = 0.238907, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3543, 3528, 3537, 63, 208} \[ -\frac{2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac{4 i a^2 \sqrt{c+d \tan (e+f x)}}{f}-\frac{4 i a^2 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

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*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + ((4*I)*a^2*Sqrt[c + d*Tan[e + f
*x]])/f - (2*a^2*(c + d*Tan[e + f*x])^(3/2))/(3*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 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 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 (a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)} \, dx &=-\frac{2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\int \left (2 a^2+2 i a^2 \tan (e+f x)\right ) \sqrt{c+d \tan (e+f x)} \, dx\\ &=\frac{4 i a^2 \sqrt{c+d \tan (e+f x)}}{f}-\frac{2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\int \frac{2 a^2 (c-i d)+2 a^2 (i c+d) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=\frac{4 i a^2 \sqrt{c+d \tan (e+f x)}}{f}-\frac{2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac{\left (4 i a^4 (c-i d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (4 a^4 (i c+d)^2+2 a^2 (c-i d) x\right ) \sqrt{c+\frac{d x}{2 a^2 (i c+d)}}} \, dx,x,2 a^2 (i c+d) \tan (e+f x)\right )}{f}\\ &=\frac{4 i a^2 \sqrt{c+d \tan (e+f x)}}{f}-\frac{2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}-\frac{\left (16 a^6 (c-i d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{4 a^4 c (c-i d) (i c+d)}{d}+4 a^4 (i c+d)^2+\frac{4 a^4 (c-i d) (i c+d) x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{d f}\\ &=-\frac{4 i a^2 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{4 i a^2 \sqrt{c+d \tan (e+f x)}}{f}-\frac{2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}\\ \end{align*}

Mathematica [A]  time = 3.3048, size = 155, normalized size = 1.55 \[ -\frac{2 a^2 e^{-2 i e} (\cos (2 (e+f x))+i \sin (2 (e+f x))) \left (\sqrt{c+d \tan (e+f x)} (c+d \tan (e+f x)-6 i d)+6 i d \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )\right )}{3 d f (\cos (f x)+i \sin (f x))^2} \]

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*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*((6*I)*Sqrt[c - I*d]*d*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 + d*Tan[e + f*x]]*(c - (6*I)*d + d*Tan[e + f*x])
))/(3*d*E^((2*I)*e)*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [B]  time = 0.032, size = 1229, normalized size = 12.3 \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))^2,x)

[Out]

-2/3*a^2*(c+d*tan(f*x+e))^(3/2)/d/f+4*I*a^2*(c+d*tan(f*x+e))^(1/2)/f-2*I/f*a^2/(4*(c^2+d^2)^(1/2)+4*c)*ln(d*ta
n(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*(c^2+d^2)^(1/2)+2*c)^(1/2)
*c+2/f*a^2*d/(4*(c^2+d^2)^(1/2)+4*c)*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*(c^2+d^2)^(1/2)+2*c)^(1/2)-8*I/f*a^2*d^2/(4*(c^2+d^2)^(1/2)+4*c)/(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/(4
*(c^2+d^2)^(1/2)+4*c)*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*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)-8/f*a^2*d/(4*(c^2+d^2)^(1/2)+4*c)/(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^2)^(1/2)
-8/f*a^2*d/(4*(c^2+d^2)^(1/2)+4*c)/(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+8*I/f*a^2*d^2/(4*(c^2+d^2)^(1/2)+4*c)/(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/(4*(c^2+d^2)^(1/2)+4*c)*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*(c^2+d^2)^(1/2)+2*c)^(1/2)*c-2/f*a^2*d/(4*(c^2+d^2)^(1/2)+4*c)*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*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*I/f*a^2/(4*(c^2+d^2)^(1/2)+
4*c)*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*(c^2+d^2)^(1/2
)+2*c)^(1/2)*(c^2+d^2)^(1/2)+8/f*a^2*d/(4*(c^2+d^2)^(1/2)+4*c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*ta
n(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^2+d^2)^(1/2)+8/f*a^2*d/(4*(c^
2+d^2)^(1/2)+4*c)/(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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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((c+d*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^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 = 2.2782, size = 1062, normalized size = 10.62 \begin{align*} \frac{3 \,{\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt{-\frac{16 \, a^{4} c - 16 i \, a^{4} d}{f^{2}}} \log \left (\frac{{\left (4 \, a^{2} c +{\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, 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{16 \, a^{4} c - 16 i \, a^{4} d}{f^{2}}} +{\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 ) - 3 \,{\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt{-\frac{16 \, a^{4} c - 16 i \, a^{4} d}{f^{2}}} \log \left (\frac{{\left (4 \, a^{2} c +{\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, 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{16 \, a^{4} c - 16 i \, a^{4} d}{f^{2}}} +{\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 ) -{\left (8 \, a^{2} c - 40 i \, a^{2} d +{\left (8 \, a^{2} c - 56 i \, a^{2} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\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}}}{12 \,{\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )}} \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))^2,x, algorithm="fricas")

[Out]

1/12*(3*(d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(16*a^4*c - 16*I*a^4*d)/f^2)*log(1/2*(4*a^2*c + (I*f*e^(2*I*f*x
+ 2*I*e) + I*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(-(16*a^4*c - 16
*I*a^4*d)/f^2) + (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) - 3*(d*f*e^(2*I*f*x + 2*
I*e) + d*f)*sqrt(-(16*a^4*c - 16*I*a^4*d)/f^2)*log(1/2*(4*a^2*c + (-I*f*e^(2*I*f*x + 2*I*e) - I*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(-(16*a^4*c - 16*I*a^4*d)/f^2) + (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*c - 40*I*a^2*d + (8*a^2*c - 56*I*a^2*d)*e^
(2*I*f*x + 2*I*e))*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*e^(2*I*f*x
+ 2*I*e) + d*f)

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

[Out]

a**2*(Integral(-sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(2*I*sqrt(c + d*tan(e + f*x))*tan(e + f
*x), x) + Integral(sqrt(c + d*tan(e + f*x)), x))

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Giac [B]  time = 1.4047, size = 306, normalized size = 3.06 \begin{align*} \frac{2 \,{\left (8 i \, a^{2} c + 8 \, a^{2} d\right )} \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 )}} - \frac{2 \,{\left ({\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} a^{2} d^{2} f^{2} - 6 i \, \sqrt{d \tan \left (f x + e\right ) + c} a^{2} d^{3} f^{2}\right )}}{3 \, d^{3} f^{3}} \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))^2,x, algorithm="giac")

[Out]

2*(8*I*a^2*c + 8*a^2*d)*arctan(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sq
rt(-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))))/(sqrt(-8*c + 8*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) - 2/3*((d*tan(f*x + e) + c)^(3/2)*a
^2*d^2*f^2 - 6*I*sqrt(d*tan(f*x + e) + c)*a^2*d^3*f^2)/(d^3*f^3)