3.5.46 \(\int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3} \, dx\) [446]

Optimal. Leaf size=37 \[ \frac {3 i a (d \sec (e+f x))^{2/3}}{f \sqrt [3]{a+i a \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {3574} \begin {gather*} \frac {3 i a (d \sec (e+f x))^{2/3}}{f \sqrt [3]{a+i a \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3574

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

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Mathematica [A]
time = 0.39, size = 47, normalized size = 1.27 \begin {gather*} \frac {3 d^2 (i+\tan (e+f x)) (a+i a \tan (e+f x))^{2/3}}{f (d \sec (e+f x))^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.37, size = 0, normalized size = 0.00 \[\int \left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{\frac {2}{3}}\, dx\]

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))^(2/3),x)

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (31) = 62\).
time = 0.54, size = 114, normalized size = 3.08 \begin {gather*} -\frac {3 \, {\left (-i \cdot 2^{\frac {1}{3}} \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 ) - 2^{\frac {1}{3}} \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}}}{{\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 {1}{6}} f} \end {gather*}

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))^(2/3),x, algorithm="maxima")

[Out]

-3*(-I*2^(1/3)*cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 2^(1/3)*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)^(1/6)*f)

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Fricas [A]
time = 0.36, size = 58, normalized size = 1.57 \begin {gather*} -\frac {3 \cdot 2^{\frac {1}{3}} \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}} {\left (-i \, e^{\left (2 i \, f x + 2 i \, e\right )} - i\right )}}{f} \end {gather*}

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))^(2/3),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d \sec {\left (e + f x \right )}\right )^{\frac {2}{3}} \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {2}{3}}\, dx \end {gather*}

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))**(2/3),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^(2/3),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.81, size = 81, normalized size = 2.19 \begin {gather*} \frac {3\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{2/3}\,\left (\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}+\sin \left (2\,e+2\,f\,x\right )+1{}\mathrm {i}\right )\,{\left (\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}\right )}^{2/3}}{2\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(2/3)*(a + a*tan(e + f*x)*1i)^(2/3),x)

[Out]

(3*(d/cos(e + f*x))^(2/3)*(cos(2*e + 2*f*x)*1i + sin(2*e + 2*f*x) + 1i)*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*
x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(2/3))/(2*f)

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