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

Optimal. Leaf size=90 \[ \frac{4 (-1)^{3/4} a^2 \sqrt{d} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{2 a^2 (d \tan (e+f x))^{3/2}}{3 d f}+\frac{4 i a^2 \sqrt{d \tan (e+f x)}}{f} \]

[Out]

(4*(-1)^(3/4)*a^2*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f + ((4*I)*a^2*Sqrt[d*Tan[e + f*x
]])/f - (2*a^2*(d*Tan[e + f*x])^(3/2))/(3*d*f)

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Rubi [A]  time = 0.132368, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3543, 3528, 3533, 205} \[ \frac{4 (-1)^{3/4} a^2 \sqrt{d} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{2 a^2 (d \tan (e+f x))^{3/2}}{3 d f}+\frac{4 i a^2 \sqrt{d \tan (e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(-1)^(3/4)*a^2*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f + ((4*I)*a^2*Sqrt[d*Tan[e + f*x
]])/f - (2*a^2*(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 3533

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

Rule 205

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

Rubi steps

\begin{align*} \int \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))^2 \, dx &=-\frac{2 a^2 (d \tan (e+f x))^{3/2}}{3 d f}+\int \sqrt{d \tan (e+f x)} \left (2 a^2+2 i a^2 \tan (e+f x)\right ) \, dx\\ &=\frac{4 i a^2 \sqrt{d \tan (e+f x)}}{f}-\frac{2 a^2 (d \tan (e+f x))^{3/2}}{3 d f}+\int \frac{-2 i a^2 d+2 a^2 d \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\\ &=\frac{4 i a^2 \sqrt{d \tan (e+f x)}}{f}-\frac{2 a^2 (d \tan (e+f x))^{3/2}}{3 d f}-\frac{\left (8 a^4 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{-2 i a^2 d^2-2 a^2 d x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=\frac{4 (-1)^{3/4} a^2 \sqrt{d} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}+\frac{4 i a^2 \sqrt{d \tan (e+f x)}}{f}-\frac{2 a^2 (d \tan (e+f x))^{3/2}}{3 d f}\\ \end{align*}

Mathematica [A]  time = 2.0069, size = 102, normalized size = 1.13 \[ \frac{2 i a^2 \sqrt{d \tan (e+f x)} \left ((6+i \tan (e+f x)) \sqrt{i \tan (e+f x)}-6 \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right )}{3 f \sqrt{i \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)/3)*a^2*(-6*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]] + (6 + I*Tan[e + f*x])*
Sqrt[I*Tan[e + f*x]])*Sqrt[d*Tan[e + f*x]])/(f*Sqrt[I*Tan[e + f*x]])

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Maple [B]  time = 0.018, size = 375, normalized size = 4.2 \begin{align*} -{\frac{2\,{a}^{2}}{3\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{4\,i{a}^{2}}{f}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{{\frac{i}{2}}{a}^{2}\sqrt{2}}{f}\sqrt [4]{{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ) }-{\frac{i{a}^{2}\sqrt{2}}{f}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }+{\frac{i{a}^{2}\sqrt{2}}{f}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }+{\frac{{a}^{2}d\sqrt{2}}{2\,f}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{{a}^{2}d\sqrt{2}}{f}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{{a}^{2}d\sqrt{2}}{f}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.08806, size = 842, normalized size = 9.36 \begin{align*} \frac{3 \, \sqrt{\frac{16 i \, a^{4} d}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-4 i \, a^{2} d e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{\frac{16 i \, a^{4} d}{f^{2}}}{\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - 3 \, \sqrt{\frac{16 i \, a^{4} d}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-4 i \, a^{2} d e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{\frac{16 i \, a^{4} d}{f^{2}}}{\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) +{\left (56 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 40 i \, a^{2}\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \,{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16555, size = 174, normalized size = 1.93 \begin{align*} \frac{4 \, \sqrt{2} a^{2} \sqrt{d} \arctan \left (\frac{16 \, \sqrt{d^{2}} \sqrt{d \tan \left (f x + e\right )}}{8 i \, \sqrt{2} d^{\frac{3}{2}} + 8 \, \sqrt{2} \sqrt{d^{2}} \sqrt{d}}\right )}{f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{2 \,{\left (\sqrt{d \tan \left (f x + e\right )} a^{2} d^{3} f^{2} \tan \left (f x + e\right ) - 6 i \, \sqrt{d \tan \left (f x + e\right )} 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((d*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

4*sqrt(2)*a^2*sqrt(d)*arctan(16*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt
(d)))/(f*(I*d/sqrt(d^2) + 1)) - 2/3*(sqrt(d*tan(f*x + e))*a^2*d^3*f^2*tan(f*x + e) - 6*I*sqrt(d*tan(f*x + e))*
a^2*d^3*f^2)/(d^3*f^3)