3.5.44 \(\int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{4/3}} \, dx\) [444]

Optimal. Leaf size=378 \[ \frac {i (d \sec (e+f x))^{2/3}}{2 f (a+i a \tan (e+f x))^{4/3}}-\frac {x (d \sec (e+f x))^{2/3}}{6\ 2^{2/3} a^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac {i \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a-i a \tan (e+f x)}}{\sqrt {3} \sqrt [3]{a}}\right ) (d \sec (e+f x))^{2/3}}{2^{2/3} \sqrt {3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \log (\cos (e+f x)) (d \sec (e+f x))^{2/3}}{6\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a-i a \tan (e+f x)}\right ) (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}} \]

[Out]

1/2*I*(d*sec(f*x+e))^(2/3)/f/(a+I*a*tan(f*x+e))^(4/3)-1/12*x*(d*sec(f*x+e))^(2/3)*2^(1/3)/a^(2/3)/(a-I*a*tan(f
*x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3)-1/12*I*ln(cos(f*x+e))*(d*sec(f*x+e))^(2/3)*2^(1/3)/a^(2/3)/f/(a-I*a*tan(
f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3)-1/4*I*ln(2^(1/3)*a^(1/3)-(a-I*a*tan(f*x+e))^(1/3))*(d*sec(f*x+e))^(2/3)
*2^(1/3)/a^(2/3)/f/(a-I*a*tan(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3)+1/6*I*arctan(1/3*(a^(1/3)+2^(2/3)*(a-I*a*
tan(f*x+e))^(1/3))/a^(1/3)*3^(1/2))*(d*sec(f*x+e))^(2/3)*2^(1/3)/a^(2/3)/f*3^(1/2)/(a-I*a*tan(f*x+e))^(1/3)/(a
+I*a*tan(f*x+e))^(1/3)

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Rubi [A]
time = 0.25, antiderivative size = 378, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3586, 3603, 3568, 44, 59, 631, 210, 31} \begin {gather*} \frac {i (d \sec (e+f x))^{2/3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a-i a \tan (e+f x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2^{2/3} \sqrt {3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {x (d \sec (e+f x))^{2/3}}{6\ 2^{2/3} a^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i (d \sec (e+f x))^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a-i a \tan (e+f x)}\right )}{2\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i (d \sec (e+f x))^{2/3} \log (\cos (e+f x))}{6\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac {i (d \sec (e+f x))^{2/3}}{2 f (a+i a \tan (e+f x))^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*(d*Sec[e + f*x])^(2/3))/(f*(a + I*a*Tan[e + f*x])^(4/3)) - (x*(d*Sec[e + f*x])^(2/3))/(6*2^(2/3)*a^(2/3
)*(a - I*a*Tan[e + f*x])^(1/3)*(a + I*a*Tan[e + f*x])^(1/3)) + (I*ArcTan[(a^(1/3) + 2^(2/3)*(a - I*a*Tan[e + f
*x])^(1/3))/(Sqrt[3]*a^(1/3))]*(d*Sec[e + f*x])^(2/3))/(2^(2/3)*Sqrt[3]*a^(2/3)*f*(a - I*a*Tan[e + f*x])^(1/3)
*(a + I*a*Tan[e + f*x])^(1/3)) - ((I/6)*Log[Cos[e + f*x]]*(d*Sec[e + f*x])^(2/3))/(2^(2/3)*a^(2/3)*f*(a - I*a*
Tan[e + f*x])^(1/3)*(a + I*a*Tan[e + f*x])^(1/3)) - ((I/2)*Log[2^(1/3)*a^(1/3) - (a - I*a*Tan[e + f*x])^(1/3)]
*(d*Sec[e + f*x])^(2/3))/(2^(2/3)*a^(2/3)*f*(a - I*a*Tan[e + f*x])^(1/3)*(a + I*a*Tan[e + f*x])^(1/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 59

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 3568

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

Rule 3586

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

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps

\begin {align*} \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{4/3}} \, dx &=\frac {(d \sec (e+f x))^{2/3} \int \frac {\sqrt [3]{a-i a \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx}{\sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\\ &=\frac {(d \sec (e+f x))^{2/3} \int \cos ^2(e+f x) (a-i a \tan (e+f x))^{4/3} \, dx}{a^2 \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\\ &=\frac {\left (i a (d \sec (e+f x))^{2/3}\right ) \text {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{2/3}} \, dx,x,-i a \tan (e+f x)\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\\ &=\frac {i (d \sec (e+f x))^{2/3}}{2 f (a+i a \tan (e+f x))^{4/3}}+\frac {\left (i (d \sec (e+f x))^{2/3}\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{2/3}} \, dx,x,-i a \tan (e+f x)\right )}{3 f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\\ &=\frac {i (d \sec (e+f x))^{2/3}}{2 f (a+i a \tan (e+f x))^{4/3}}-\frac {x (d \sec (e+f x))^{2/3}}{6\ 2^{2/3} a^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \log (\cos (e+f x)) (d \sec (e+f x))^{2/3}}{6\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac {\left (i (d \sec (e+f x))^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a-i a \tan (e+f x)}\right )}{2\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac {\left (i (d \sec (e+f x))^{2/3}\right ) \text {Subst}\left (\int \frac {1}{2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a-i a \tan (e+f x)}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\\ &=\frac {i (d \sec (e+f x))^{2/3}}{2 f (a+i a \tan (e+f x))^{4/3}}-\frac {x (d \sec (e+f x))^{2/3}}{6\ 2^{2/3} a^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \log (\cos (e+f x)) (d \sec (e+f x))^{2/3}}{6\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a-i a \tan (e+f x)}\right ) (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {\left (i (d \sec (e+f x))^{2/3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2^{2/3} \sqrt [3]{a-i a \tan (e+f x)}}{\sqrt [3]{a}}\right )}{2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\\ &=\frac {i (d \sec (e+f x))^{2/3}}{2 f (a+i a \tan (e+f x))^{4/3}}-\frac {x (d \sec (e+f x))^{2/3}}{6\ 2^{2/3} a^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac {i \tan ^{-1}\left (\frac {1+\frac {2^{2/3} \sqrt [3]{a-i a \tan (e+f x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right ) (d \sec (e+f x))^{2/3}}{2^{2/3} \sqrt {3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \log (\cos (e+f x)) (d \sec (e+f x))^{2/3}}{6\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a-i a \tan (e+f x)}\right ) (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 1.39, size = 220, normalized size = 0.58 \begin {gather*} \frac {e^{-i (e+f x)} \left (3 i+3 i e^{2 i (e+f x)}-2 e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} f x-2 i \sqrt {3} e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \text {ArcTan}\left (\frac {1+2 \sqrt [3]{1+e^{2 i (e+f x)}}}{\sqrt {3}}\right )-3 i e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \log \left (1-\sqrt [3]{1+e^{2 i (e+f x)}}\right )\right ) (d \sec (e+f x))^{5/3}}{12 d f (a+i a \tan (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])^(4/3),x]

[Out]

((3*I + (3*I)*E^((2*I)*(e + f*x)) - 2*E^((2*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^(1/3)*f*x - (2*I)*Sqrt[3]*
E^((2*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^(1/3)*ArcTan[(1 + 2*(1 + E^((2*I)*(e + f*x)))^(1/3))/Sqrt[3]] -
(3*I)*E^((2*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^(1/3)*Log[1 - (1 + E^((2*I)*(e + f*x)))^(1/3)])*(d*Sec[e +
f*x])^(5/3))/(12*d*E^(I*(e + f*x))*f*(a + I*a*Tan[e + f*x])^(4/3))

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Maple [F]
time = 0.33, size = 0, normalized size = 0.00 \[\int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}}}{\left (a +i a \tan \left (f x +e \right )\right )^{\frac {4}{3}}}\, dx\]

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))^(4/3),x)

[Out]

int((d*sec(f*x+e))^(2/3)/(a+I*a*tan(f*x+e))^(4/3),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. 2045 vs. \(2 (296) = 592\).
time = 0.60, size = 2045, normalized size = 5.41 \begin {gather*} \text {Too large to display} \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))^(4/3),x, algorithm="maxima")

[Out]

-1/24*(6*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/3)*((-I*2^(1/3)*cos(2*f*x + 2*e
) - 2^(1/3)*sin(2*f*x + 2*e))*cos(2/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + (2^(1/3)*cos(2*f*x +
2*e) - I*2^(1/3)*sin(2*f*x + 2*e))*sin(2/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))*d^(2/3) - (-2*I*s
qrt(3)*2^(1/3)*arctan2(2/3*sqrt(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)*co
s(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 1/3*sqrt(3), 1/3*sqrt(3)*(2*(cos(2*f*x + 2*e)^2 + sin
(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + sqr
t(3))) - 2*I*sqrt(3)*2^(1/3)*arctan2(2/3*sqrt(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)*cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 1/3*sqrt(3), -1/3*sqrt(3)*(2*(cos(2*f*x
+ 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2
*e) + 1)) - sqrt(3))) + sqrt(3)*2^(1/3)*log(4/3*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e)
+ 1)^(1/3)*(cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2 + sin(1/3*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e) + 1))^2) + 4/3*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*(sqrt(3)
*sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
+ 1))) + 4/3) - sqrt(3)*2^(1/3)*log(4/3*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1
/3)*(cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2 + sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e) + 1))^2) - 4/3*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*(sqrt(3)*sin(1/
3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))
+ 4/3) - 2*2^(1/3)*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/3)*sin(2/3*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e
) + 1)^(1/6)*sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^
2 + 2*cos(2*f*x + 2*e) + 1)^(1/3)*cos(2/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + (cos(2*f*x + 2*e)
^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1
)) + 1) + 4*2^(1/3)*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*sin(1/3*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e)
+ 1)^(1/6)*cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 1) - 2*I*2^(1/3)*log((cos(2*f*x + 2*e)^
2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/3)*cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)
)^2 + (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/3)*sin(1/3*arctan2(sin(2*f*x + 2*e
), cos(2*f*x + 2*e) + 1))^2 - 2*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*cos(1
/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 1) + I*2^(1/3)*log((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e
)^2 + 2*cos(2*f*x + 2*e) + 1)^(2/3)*(cos(2/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2 + sin(2/3*arct
an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2) + (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e
) + 1)^(1/3)*(cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2 + sin(1/3*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e) + 1))^2) + 2*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/3)*((cos(2*
f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*(cos(2/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e) + 1))*cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + sin(2/3*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e) + 1))*sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))) + cos(2/3*arctan2(sin(2*f*x + 2*
e), cos(2*f*x + 2*e) + 1))) + 2*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*cos(1
/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 1))*d^(2/3))/(a^(4/3)*f)

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Fricas [A]
time = 0.41, size = 542, normalized size = 1.43 \begin {gather*} \frac {{\left (4 \, a^{2} f \left (\frac {i \, d^{2}}{108 \, a^{4} f^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-2 \, {\left (6 i \, a^{2} f \left (\frac {i \, d^{2}}{108 \, a^{4} f^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, f x + 2 i \, e\right )} - 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 (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) + 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 (4 i \, f x + 4 i \, e\right )} + 2 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 \, {\left (-i \, \sqrt {3} a^{2} f + a^{2} f\right )} \left (\frac {i \, d^{2}}{108 \, a^{4} f^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (2 \, {\left (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 (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 \, {\left (\sqrt {3} a^{2} f + i \, a^{2} f\right )} \left (\frac {i \, d^{2}}{108 \, a^{4} f^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - 2 \, {\left (i \, \sqrt {3} a^{2} f + a^{2} f\right )} \left (\frac {i \, d^{2}}{108 \, a^{4} f^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (2 \, {\left (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 (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 \, {\left (\sqrt {3} a^{2} f - i \, a^{2} f\right )} \left (\frac {i \, d^{2}}{108 \, a^{4} f^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right )\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{4 \, a^{2} 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))^(4/3),x, algorithm="fricas")

[Out]

1/4*(4*a^2*f*(1/108*I*d^2/(a^4*f^3))^(1/3)*e^(4*I*f*x + 4*I*e)*log(-2*(6*I*a^2*f*(1/108*I*d^2/(a^4*f^3))^(1/3)
*e^(2*I*f*x + 2*I*e) - 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)*(e^(2*I
*f*x + 2*I*e) + 1)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) + 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^(4*I*f*x + 4*I*e) + 2*I*e^(2*I*f*x + 2*I*e) + I)*e^(2*I*f*x + 2*I*e) -
2*(-I*sqrt(3)*a^2*f + a^2*f)*(1/108*I*d^2/(a^4*f^3))^(1/3)*e^(4*I*f*x + 4*I*e)*log(2*(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)*(e^(2*I*f*x + 2*I*e) + 1)*e^(2*I*f*x + 2*I*e) + 3*(sqr
t(3)*a^2*f + I*a^2*f)*(1/108*I*d^2/(a^4*f^3))^(1/3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 2*(I*sqrt(3)*
a^2*f + a^2*f)*(1/108*I*d^2/(a^4*f^3))^(1/3)*e^(4*I*f*x + 4*I*e)*log(2*(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)*(e^(2*I*f*x + 2*I*e) + 1)*e^(2*I*f*x + 2*I*e) - 3*(sqrt(3)*a^2*f - I
*a^2*f)*(1/108*I*d^2/(a^4*f^3))^(1/3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)))*e^(-4*I*f*x - 4*I*e)/(a^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\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 {4}{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))**(4/3),x)

[Out]

Integral((d*sec(e + f*x))**(2/3)/(I*a*(tan(e + f*x) - I))**(4/3), 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))^(4/3),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(2/3)/(I*a*tan(f*x + e) + a)^(4/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{2/3}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{4/3}} \,d x \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)^(4/3),x)

[Out]

int((d/cos(e + f*x))^(2/3)/(a + a*tan(e + f*x)*1i)^(4/3), x)

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