3.444 \(\int \frac{(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{4/3}} \, dx\)
Optimal. Leaf size=378 \[ \frac{i \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a-i a \tan (e+f x)}}{\sqrt{3} \sqrt [3]{a}}\right ) (d \sec (e+f x))^{2/3}}{2^{2/3} \sqrt{3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac{x (d \sec (e+f x))^{2/3}}{6\ 2^{2/3} a^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac{i (d \sec (e+f x))^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a-i a \tan (e+f x)}\right )}{2\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac{i (d \sec (e+f x))^{2/3} \log (\cos (e+f x))}{6\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac{i (d \sec (e+f x))^{2/3}}{2 f (a+i a \tan (e+f x))^{4/3}} \]
[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))
________________________________________________________________________________________
Rubi [A] time = 0.342637, antiderivative size = 378, normalized size of antiderivative = 1.,
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 =
{3505, 3522, 3487, 51, 57, 617, 204, 31} \[ \frac{i \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a-i a \tan (e+f x)}}{\sqrt{3} \sqrt [3]{a}}\right ) (d \sec (e+f x))^{2/3}}{2^{2/3} \sqrt{3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac{x (d \sec (e+f x))^{2/3}}{6\ 2^{2/3} a^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac{i (d \sec (e+f x))^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a-i a \tan (e+f x)}\right )}{2\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac{i (d \sec (e+f x))^{2/3} \log (\cos (e+f x))}{6\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac{i (d \sec (e+f x))^{2/3}}{2 f (a+i a \tan (e+f x))^{4/3}} \]
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 3505
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 3522
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]))
Rule 3487
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 51
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] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]
Rule 57
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 617
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 204
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}
Mathematica [A] time = 1.23268, size = 220, normalized size = 0.58 \[ \frac{e^{-i (e+f x)} (d \sec (e+f x))^{5/3} \left (-2 f x e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}}+3 i e^{2 i (e+f x)}-3 i e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \log \left (1-\sqrt [3]{1+e^{2 i (e+f x)}}\right )-2 i \sqrt{3} e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \tan ^{-1}\left (\frac{1+2 \sqrt [3]{1+e^{2 i (e+f x)}}}{\sqrt{3}}\right )+3 i\right )}{12 d f (a+i a \tan (e+f x))^{4/3}} \]
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))
________________________________________________________________________________________
Maple [F] time = 0.157, 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{4}{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))^(4/3),x)
[Out]
int((d*sec(f*x+e))^(2/3)/(a+I*a*tan(f*x+e))^(4/3),x)
________________________________________________________________________________________
Maxima [B] time = 2.32492, size = 2572, normalized size = 6.8 \begin{align*} \text{result too large to display} \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))^(4/3),x, algorithm="maxima")
[Out]
1/24*((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/3)*((6*I*2^(1/3)*cos(2*f*x + 2*e)
+ 6*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)) - 6*(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
*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 + s
in(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)) + s
qrt(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), c
os(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
*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)*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
*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) - 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*ar
ctan2(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),
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)^(1/3)*((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), c
os(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 = 2.07599, size = 1496, normalized size = 3.96 \begin{align*} \text{result too large to display} \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))^(4/3),x, algorithm="fricas")
[Out]
1/8*(8*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*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)
+ (4*I*sqrt(3)*a^2*f - 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*(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)) + (-4*I*sqrt
(3)*a^2*f - 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*(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)
________________________________________________________________________________________
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))**(4/3),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \sec \left (f x + e\right )\right )^{\frac{2}{3}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{4}{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))^(4/3),x, algorithm="giac")
[Out]
integrate((d*sec(f*x + e))^(2/3)/(I*a*tan(f*x + e) + a)^(4/3), x)