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

Optimal. Leaf size=107 \[ -\frac{8 \sqrt [4]{-1} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{d} f}-\frac{16 a^3 \sqrt{d \tan (e+f x)}}{3 d f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt{d \tan (e+f x)}}{3 d f} \]

[Out]

(-8*(-1)^(1/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[d]*f) - (16*a^3*Sqrt[d*Tan[e + f*x
]])/(3*d*f) - (2*Sqrt[d*Tan[e + f*x]]*(a^3 + I*a^3*Tan[e + f*x]))/(3*d*f)

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Rubi [A]  time = 0.181989, antiderivative size = 107, 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 = {3556, 3592, 3533, 205} \[ -\frac{8 \sqrt [4]{-1} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{d} f}-\frac{16 a^3 \sqrt{d \tan (e+f x)}}{3 d f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt{d \tan (e+f x)}}{3 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(-1)^(1/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[d]*f) - (16*a^3*Sqrt[d*Tan[e + f*x
]])/(3*d*f) - (2*Sqrt[d*Tan[e + f*x]]*(a^3 + I*a^3*Tan[e + f*x]))/(3*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 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 \frac{(a+i a \tan (e+f x))^3}{\sqrt{d \tan (e+f x)}} \, dx &=-\frac{2 \sqrt{d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f}+\frac{(2 a) \int \frac{(a+i a \tan (e+f x)) (2 a d+4 i a d \tan (e+f x))}{\sqrt{d \tan (e+f x)}} \, dx}{3 d}\\ &=-\frac{16 a^3 \sqrt{d \tan (e+f x)}}{3 d f}-\frac{2 \sqrt{d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f}+\frac{(2 a) \int \frac{6 a^2 d+6 i a^2 d \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{3 d}\\ &=-\frac{16 a^3 \sqrt{d \tan (e+f x)}}{3 d f}-\frac{2 \sqrt{d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f}+\frac{\left (48 a^5 d\right ) \operatorname{Subst}\left (\int \frac{1}{6 a^2 d^2-6 i a^2 d x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=-\frac{8 \sqrt [4]{-1} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{d} f}-\frac{16 a^3 \sqrt{d \tan (e+f x)}}{3 d f}-\frac{2 \sqrt{d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f}\\ \end{align*}

Mathematica [A]  time = 3.03517, size = 154, normalized size = 1.44 \[ -\frac{2 a^3 e^{-3 i (e+f x)} \sqrt{d \tan (e+f x)} (\cos (3 (e+f x))+i \sin (3 (e+f x))) \left ((9+i \tan (e+f x)) \sqrt{i \tan (e+f x)}-12 \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right )}{3 d f \sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*(-12*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e
+ f*x)))]] + (9 + I*Tan[e + f*x])*Sqrt[I*Tan[e + f*x]])*Sqrt[d*Tan[e + f*x]])/(3*d*E^((3*I)*(e + f*x))*Sqrt[(-
1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]*f)

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Maple [B]  time = 0.036, size = 384, normalized size = 3.6 \begin{align*}{\frac{-{\frac{2\,i}{3}}{a}^{3}}{f{d}^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-6\,{\frac{{a}^{3}\sqrt{d\tan \left ( fx+e \right ) }}{df}}+{\frac{{a}^{3}\sqrt{2}}{df}\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 ) }+2\,{\frac{{a}^{3}\sqrt [4]{{d}^{2}}\sqrt{2}}{df}\arctan \left ({\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) }-2\,{\frac{{a}^{3}\sqrt [4]{{d}^{2}}\sqrt{2}}{df}\arctan \left ( -{\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) }+{\frac{i{a}^{3}\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}}}}}+{\frac{2\,i{a}^{3}\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{2\,i{a}^{3}\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((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(1/2),x)

[Out]

-2/3*I/f*a^3/d^2*(d*tan(f*x+e))^(3/2)-6*a^3*(d*tan(f*x+e))^(1/2)/d/f+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)+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*tan(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)

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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))^3/(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.29203, size = 869, normalized size = 8.12 \begin{align*} \frac{3 \, \sqrt{-\frac{64 i \, a^{6}}{d f^{2}}}{\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d 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 (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d 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 ) - 3 \, \sqrt{-\frac{64 i \, a^{6}}{d f^{2}}}{\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d 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 (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d 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 ) - 16 \,{\left (5 \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, 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}}}{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((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/12*(3*sqrt(-64*I*a^6/(d*f^2))*(d*f*e^(2*I*f*x + 2*I*e) + d*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))*(d*f*e^(2*I*f*x + 2*I*e) + d*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) - 3*sqrt(-64*I*a^6/(d*f^2))*(d*f*e^(2*I*f*x + 2*I*e) + d*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))*(d*f*e^(2*I*f*x + 2*I*e) + d*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) - 16*(5*a^3*e^(2*I*f*x + 2*I*e) + 4*a^3)*sqrt(
(-I*d*e^(2*I*f*x + 2*I*e) + 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^{3} \left (\int \frac{1}{\sqrt{d \tan{\left (e + f x \right )}}}\, dx + \int - \frac{3 \tan ^{2}{\left (e + f x \right )}}{\sqrt{d \tan{\left (e + f x \right )}}}\, dx + \int \frac{3 i \tan{\left (e + f x \right )}}{\sqrt{d \tan{\left (e + f x \right )}}}\, dx + \int - \frac{i \tan ^{3}{\left (e + f x \right )}}{\sqrt{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))**3/(d*tan(f*x+e))**(1/2),x)

[Out]

a**3*(Integral(1/sqrt(d*tan(e + f*x)), x) + Integral(-3*tan(e + f*x)**2/sqrt(d*tan(e + f*x)), x) + Integral(3*
I*tan(e + f*x)/sqrt(d*tan(e + f*x)), x) + Integral(-I*tan(e + f*x)**3/sqrt(d*tan(e + f*x)), x))

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Giac [A]  time = 1.34835, size = 176, normalized size = 1.64 \begin{align*} \frac{8 i \, \sqrt{2} a^{3} \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 )}{\sqrt{d} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{2 i \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{5} f^{2} \tan \left (f x + e\right ) + 18 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{5} f^{2}}{3 \, d^{6} f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

8*I*sqrt(2)*a^3*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)))/
(sqrt(d)*f*(I*d/sqrt(d^2) + 1)) - 1/3*(2*I*sqrt(d*tan(f*x + e))*a^3*d^5*f^2*tan(f*x + e) + 18*sqrt(d*tan(f*x +
e))*a^3*d^5*f^2)/(d^6*f^3)