3.447 \(\int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx\)

Optimal. Leaf size=81 \[ \frac{9 i a^2 (d \sec (e+f x))^{2/3}}{2 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac{3 i a (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{4 f} \]

[Out]

(((9*I)/2)*a^2*(d*Sec[e + f*x])^(2/3))/(f*(a + I*a*Tan[e + f*x])^(1/3)) + (((3*I)/4)*a*(d*Sec[e + f*x])^(2/3)*
(a + I*a*Tan[e + f*x])^(2/3))/f

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Rubi [A]  time = 0.154354, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3494, 3493} \[ \frac{9 i a^2 (d \sec (e+f x))^{2/3}}{2 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac{3 i a (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(5/3),x]

[Out]

(((9*I)/2)*a^2*(d*Sec[e + f*x])^(2/3))/(f*(a + I*a*Tan[e + f*x])^(1/3)) + (((3*I)/4)*a*(d*Sec[e + f*x])^(2/3)*
(a + I*a*Tan[e + f*x])^(2/3))/f

Rule 3494

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
&& IGtQ[Simplify[m/2 + n - 1], 0] &&  !IntegerQ[n]

Rule 3493

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

Rubi steps

\begin{align*} \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx &=\frac{3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{4 f}+\frac{1}{2} (3 a) \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3} \, dx\\ &=\frac{9 i a^2 (d \sec (e+f x))^{2/3}}{2 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac{3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.514008, size = 70, normalized size = 0.86 \[ -\frac{3 a d (\cos (e)-i \sin (e)) (\tan (e+f x)-7 i) (\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x))^{2/3}}{4 f \sqrt [3]{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(5/3),x]

[Out]

(-3*a*d*(Cos[e] - I*Sin[e])*(Cos[f*x] - I*Sin[f*x])*(-7*I + Tan[e + f*x])*(a + I*a*Tan[e + f*x])^(2/3))/(4*f*(
d*Sec[e + f*x])^(1/3))

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Maple [F]  time = 0.14, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{{\frac{2}{3}}} \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3),x)

[Out]

int((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3),x)

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Maxima [B]  time = 2.00544, size = 429, normalized size = 5.3 \begin{align*} \frac{3 \,{\left (-i \cdot 2^{\frac{1}{3}} a \cos \left (\frac{4}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) - 2^{\frac{1}{3}} a \sin \left (\frac{4}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right )\right )} \sqrt{\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1} a^{\frac{2}{3}} d^{\frac{2}{3}} -{\left ({\left (-12 i \cdot 2^{\frac{1}{3}} a \cos \left (2 \, f x + 2 \, e\right )^{2} - 12 i \cdot 2^{\frac{1}{3}} a \sin \left (2 \, f x + 2 \, e\right )^{2} - 24 i \cdot 2^{\frac{1}{3}} a \cos \left (2 \, f x + 2 \, e\right ) - 12 i \cdot 2^{\frac{1}{3}} a\right )} \cos \left (\frac{1}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) - 12 \,{\left (2^{\frac{1}{3}} a \cos \left (2 \, f x + 2 \, e\right )^{2} + 2^{\frac{1}{3}} a \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \cdot 2^{\frac{1}{3}} a \cos \left (2 \, f x + 2 \, e\right ) + 2^{\frac{1}{3}} a\right )} \sin \left (\frac{1}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right )\right )} a^{\frac{2}{3}} d^{\frac{2}{3}}}{2 \,{\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}^{\frac{7}{6}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3),x, algorithm="maxima")

[Out]

1/2*(3*(-I*2^(1/3)*a*cos(4/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 2^(1/3)*a*sin(4/3*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))*sqrt(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*a
^(2/3)*d^(2/3) - ((-12*I*2^(1/3)*a*cos(2*f*x + 2*e)^2 - 12*I*2^(1/3)*a*sin(2*f*x + 2*e)^2 - 24*I*2^(1/3)*a*cos
(2*f*x + 2*e) - 12*I*2^(1/3)*a)*cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 12*(2^(1/3)*a*cos(2
*f*x + 2*e)^2 + 2^(1/3)*a*sin(2*f*x + 2*e)^2 + 2*2^(1/3)*a*cos(2*f*x + 2*e) + 2^(1/3)*a)*sin(1/3*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e) + 1)))*a^(2/3)*d^(2/3))/((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x
+ 2*e) + 1)^(7/6)*f)

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Fricas [A]  time = 2.0753, size = 167, normalized size = 2.06 \begin{align*} \frac{2^{\frac{1}{3}}{\left (12 i \, a e^{\left (2 i \, f x + 2 i \, e\right )} + 9 i \, a\right )} \left (\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{2}{3}} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{2}{3}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3),x, algorithm="fricas")

[Out]

1/2*2^(1/3)*(12*I*a*e^(2*I*f*x + 2*I*e) + 9*I*a)*(a/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(d/(e^(2*I*f*x + 2*I*e) +
1))^(2/3)/f

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(2/3)*(a+I*a*tan(f*x+e))**(5/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{2}{3}}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(2/3)*(I*a*tan(f*x + e) + a)^(5/3), x)