3.158 \(\int \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=129 \[ \frac{8 (-1)^{3/4} a^3 \sqrt{d} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}+\frac{8 i a^3 \sqrt{d \tan (e+f x)}}{f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{3/2}}{5 d f} \]

[Out]

(8*(-1)^(3/4)*a^3*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f + ((8*I)*a^3*Sqrt[d*Tan[e + f*x
]])/f - (8*a^3*(d*Tan[e + f*x])^(3/2))/(5*d*f) - (2*(d*Tan[e + f*x])^(3/2)*(a^3 + I*a^3*Tan[e + f*x]))/(5*d*f)

________________________________________________________________________________________

Rubi [A]  time = 0.218687, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3556, 3592, 3528, 3533, 205} \[ \frac{8 (-1)^{3/4} a^3 \sqrt{d} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}+\frac{8 i a^3 \sqrt{d \tan (e+f x)}}{f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{3/2}}{5 d f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Tan[e + f*x]]*(a + I*a*Tan[e + f*x])^3,x]

[Out]

(8*(-1)^(3/4)*a^3*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f + ((8*I)*a^3*Sqrt[d*Tan[e + f*x
]])/f - (8*a^3*(d*Tan[e + f*x])^(3/2))/(5*d*f) - (2*(d*Tan[e + f*x])^(3/2)*(a^3 + I*a^3*Tan[e + f*x]))/(5*d*f)

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3592

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

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))^3 \, dx &=-\frac{2 (d \tan (e+f x))^{3/2} \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f}+\frac{(2 a) \int \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x)) (4 a d+6 i a d \tan (e+f x)) \, dx}{5 d}\\ &=-\frac{8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}-\frac{2 (d \tan (e+f x))^{3/2} \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f}+\frac{(2 a) \int \sqrt{d \tan (e+f x)} \left (10 a^2 d+10 i a^2 d \tan (e+f x)\right ) \, dx}{5 d}\\ &=\frac{8 i a^3 \sqrt{d \tan (e+f x)}}{f}-\frac{8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}-\frac{2 (d \tan (e+f x))^{3/2} \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f}+\frac{(2 a) \int \frac{-10 i a^2 d^2+10 a^2 d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{5 d}\\ &=\frac{8 i a^3 \sqrt{d \tan (e+f x)}}{f}-\frac{8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}-\frac{2 (d \tan (e+f x))^{3/2} \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f}-\frac{\left (80 a^5 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{-10 i a^2 d^3-10 a^2 d^2 x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=\frac{8 (-1)^{3/4} a^3 \sqrt{d} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}+\frac{8 i a^3 \sqrt{d \tan (e+f x)}}{f}-\frac{8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}-\frac{2 (d \tan (e+f x))^{3/2} \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f}\\ \end{align*}

Mathematica [A]  time = 3.98137, size = 122, normalized size = 0.95 \[ \frac{i a^3 \sqrt{d \tan (e+f x)} \left (\sqrt{i \tan (e+f x)} \sec ^2(e+f x) (5 i \sin (2 (e+f x))+21 \cos (2 (e+f x))+19)-40 \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right )}{5 f \sqrt{i \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*Tan[e + f*x]]*(a + I*a*Tan[e + f*x])^3,x]

[Out]

((I/5)*a^3*(-40*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]] + Sec[e + f*x]^2*(19 + 21*
Cos[2*(e + f*x)] + (5*I)*Sin[2*(e + f*x)])*Sqrt[I*Tan[e + f*x]])*Sqrt[d*Tan[e + f*x]])/(f*Sqrt[I*Tan[e + f*x]]
)

________________________________________________________________________________________

Maple [B]  time = 0.021, size = 397, normalized size = 3.1 \begin{align*}{\frac{-{\frac{2\,i}{5}}{a}^{3}}{f{d}^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-2\,{\frac{{a}^{3} \left ( d\tan \left ( fx+e \right ) \right ) ^{3/2}}{df}}+{\frac{8\,i{a}^{3}}{f}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{i{a}^{3}\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{2\,i{a}^{3}\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{2\,i{a}^{3}\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}^{3}d\sqrt{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}}}}}+2\,{\frac{{a}^{3}d\sqrt{2}}{f\sqrt [4]{{d}^{2}}}\arctan \left ({\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) }-2\,{\frac{{a}^{3}d\sqrt{2}}{f\sqrt [4]{{d}^{2}}}\arctan \left ( -{\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) } \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))^3,x)

[Out]

-2/5*I/f*a^3/d^2*(d*tan(f*x+e))^(5/2)-2*a^3*(d*tan(f*x+e))^(3/2)/d/f+8*I*a^3*(d*tan(f*x+e))^(1/2)/f-I/f*a^3*(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)))-2*I/f*a^3*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*ta
n(f*x+e))^(1/2)+1)+2*I/f*a^3*(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^3*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)))+2/f*a^3*d/(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d
*tan(f*x+e))^(1/2)+1)-2/f*a^3*d/(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)

________________________________________________________________________________________

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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.15094, size = 994, normalized size = 7.71 \begin{align*} \frac{5 \, \sqrt{\frac{64 i \, a^{6} d}{f^{2}}}{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{\frac{64 i \, a^{6} 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 )}}{4 \, a^{3}}\right ) - 5 \, \sqrt{\frac{64 i \, a^{6} d}{f^{2}}}{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{\frac{64 i \, a^{6} 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 )}}{4 \, a^{3}}\right ) +{\left (208 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 304 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 128 i \, a^{3}\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}}}{20 \,{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, 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))^3,x, algorithm="fricas")

[Out]

1/20*(5*sqrt(64*I*a^6*d/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(1/4*(-8*I*a^3*d*e^(2*I*
f*x + 2*I*e) + sqrt(64*I*a^6*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^3) - 5*sqrt(64*I*a^6*d/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I
*f*x + 2*I*e) + f)*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + sqrt(64*I*a^6*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^3) + (208*I*a^3*
e^(4*I*f*x + 4*I*e) + 304*I*a^3*e^(2*I*f*x + 2*I*e) + 128*I*a^3)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I
*f*x + 2*I*e) + 1)))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int \sqrt{d \tan{\left (e + f x \right )}}\, dx + \int - 3 \sqrt{d \tan{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int 3 i \sqrt{d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )}\, dx + \int - i \sqrt{d \tan{\left (e + f x \right )}} \tan ^{3}{\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))**3,x)

[Out]

a**3*(Integral(sqrt(d*tan(e + f*x)), x) + Integral(-3*sqrt(d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(3*I*
sqrt(d*tan(e + f*x))*tan(e + f*x), x) + Integral(-I*sqrt(d*tan(e + f*x))*tan(e + f*x)**3, x))

________________________________________________________________________________________

Giac [A]  time = 1.26784, size = 217, normalized size = 1.68 \begin{align*} \frac{8 \, \sqrt{2} a^{3} \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 i \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{10} f^{4} \tan \left (f x + e\right )^{2} + 10 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{10} f^{4} \tan \left (f x + e\right ) - 40 i \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{10} f^{4}}{5 \, d^{10} f^{5}} \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))^3,x, algorithm="giac")

[Out]

8*sqrt(2)*a^3*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)) - 1/5*(2*I*sqrt(d*tan(f*x + e))*a^3*d^10*f^4*tan(f*x + e)^2 + 10*sqrt(d*tan(f*x
+ e))*a^3*d^10*f^4*tan(f*x + e) - 40*I*sqrt(d*tan(f*x + e))*a^3*d^10*f^4)/(d^10*f^5)