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

Optimal. Leaf size=340 \[ \frac {i \sqrt {3} \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a-i a \tan (e+f x)}}{\sqrt {3} \sqrt [3]{a}}\right ) (d \sec (e+f x))^{2/3}}{2^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {\sqrt [3]{a} x (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {3 i \sqrt [3]{a} (d \sec (e+f x))^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a-i a \tan (e+f x)}\right )}{2\ 2^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \sqrt [3]{a} (d \sec (e+f x))^{2/3} \log (\cos (e+f x))}{2\ 2^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}} \]

[Out]

-1/4*a^(1/3)*x*(d*sec(f*x+e))^(2/3)*2^(1/3)/(a-I*a*tan(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3)-1/4*I*a^(1/3)*ln
(cos(f*x+e))*(d*sec(f*x+e))^(2/3)*2^(1/3)/f/(a-I*a*tan(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3)-3/4*I*a^(1/3)*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)/f/(a-I*a*tan(f*x+e))^(1/3)/(a+I*a*tan(
f*x+e))^(1/3)+1/2*I*a^(1/3)*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)*3^(1/2)*2^(1/3)/f/(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 = 0.17, antiderivative size = 340, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3492, 3481, 57, 617, 204, 31} \[ \frac {i \sqrt {3} \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a-i a \tan (e+f x)}}{\sqrt {3} \sqrt [3]{a}}\right ) (d \sec (e+f x))^{2/3}}{2^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {\sqrt [3]{a} x (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {3 i \sqrt [3]{a} (d \sec (e+f x))^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a-i a \tan (e+f x)}\right )}{2\ 2^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \sqrt [3]{a} (d \sec (e+f x))^{2/3} \log (\cos (e+f x))}{2\ 2^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^(1/3)*x*(d*Sec[e + f*x])^(2/3))/(2*2^(2/3)*(a - I*a*Tan[e + f*x])^(1/3)*(a + I*a*Tan[e + f*x])^(1/3)) + (I
*Sqrt[3]*a^(1/3)*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)*f*(a - I*a*Tan[e + f*x])^(1/3)*(a + I*a*Tan[e + f*x])^(1/3)) - ((I/2)*a^(1/3)*Log[Cos[e + f*x]]
*(d*Sec[e + f*x])^(2/3))/(2^(2/3)*f*(a - I*a*Tan[e + f*x])^(1/3)*(a + I*a*Tan[e + f*x])^(1/3)) - (((3*I)/2)*a^
(1/3)*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)*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 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 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 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 3481

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Dist[b/d, Subst[Int[(a + x)^(n - 1)/(a - x), x]
, x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

Rule 3492

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

Rubi steps

\begin {align*} \int \frac {(d \sec (e+f x))^{2/3}}{\sqrt [3]{a+i a \tan (e+f x)}} \, dx &=\frac {(d \sec (e+f x))^{2/3} \int \sqrt [3]{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 {\left (i a (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 )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\\ &=-\frac {\sqrt [3]{a} x (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \sqrt [3]{a} \log (\cos (e+f x)) (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac {\left (3 i \sqrt [3]{a} (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} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac {\left (3 i a^{2/3} (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} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\\ &=-\frac {\sqrt [3]{a} x (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \sqrt [3]{a} \log (\cos (e+f x)) (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {3 i \sqrt [3]{a} \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} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {\left (3 i \sqrt [3]{a} (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} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\\ &=-\frac {\sqrt [3]{a} x (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac {i \sqrt {3} \sqrt [3]{a} \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} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \sqrt [3]{a} \log (\cos (e+f x)) (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {3 i \sqrt [3]{a} \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} 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 = 0.67, size = 161, normalized size = 0.47 \[ -\frac {\sqrt [3]{1+e^{2 i (e+f x)}} \left (\frac {d e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{2/3} \left (3 i \log \left (1-\sqrt [3]{1+e^{2 i (e+f x)}}\right )+2 i \sqrt {3} \tan ^{-1}\left (\frac {1+2 \sqrt [3]{1+e^{2 i (e+f x)}}}{\sqrt {3}}\right )+2 f x\right )}{2\ 2^{2/3} f \sqrt [3]{\frac {a e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(((d*E^(I*(e + f*x)))/(1 + E^((2*I)*(e + f*x))))^(2/3)*(1 + E^((2*I)*(e + f*x)))^(1/3)*(2*f*x + (2*I)*Sqr
t[3]*ArcTan[(1 + 2*(1 + E^((2*I)*(e + f*x)))^(1/3))/Sqrt[3]] + (3*I)*Log[1 - (1 + E^((2*I)*(e + f*x)))^(1/3)])
)/(2^(2/3)*((a*E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x))))^(1/3)*f)

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fricas [A]  time = 1.68, size = 369, normalized size = 1.09 \[ \frac {1}{2} \, {\left (i \, \sqrt {3} - 1\right )} \left (\frac {i \, d^{2}}{4 \, a f^{3}}\right )^{\frac {1}{3}} \log \left ({\left (2 \cdot 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 \, \sqrt {3} a f + 2 i \, a f\right )} \left (\frac {i \, d^{2}}{4 \, a f^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) + \frac {1}{2} \, {\left (-i \, \sqrt {3} - 1\right )} \left (\frac {i \, d^{2}}{4 \, a f^{3}}\right )^{\frac {1}{3}} \log \left ({\left (2 \cdot 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (2 \, \sqrt {3} a f - 2 i \, a f\right )} \left (\frac {i \, d^{2}}{4 \, a f^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) + \left (\frac {i \, d^{2}}{4 \, a f^{3}}\right )^{\frac {1}{3}} \log \left ({\left (2 \cdot 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, a f \left (\frac {i \, d^{2}}{4 \, a f^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) \]

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))^(1/3),x, algorithm="fricas")

[Out]

1/2*(I*sqrt(3) - 1)*(1/4*I*d^2/(a*f^3))^(1/3)*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) + (2*sqrt(3)*a*f + 2*I*a*f)*(1/4*I*d^2/(a
*f^3))^(1/3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) + 1/2*(-I*sqrt(3) - 1)*(1/4*I*d^2/(a*f^3))^(1/3)*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) - (2*sqrt(3)*a*f - 2*I*a*f)*(1/4*I*d^2/(a*f^3))^(1/3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I
*e)) + (1/4*I*d^2/(a*f^3))^(1/3)*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) - 4*I*a*f*(1/4*I*d^2/(a*f^3))^(1/3)*e^(2*I*f*x + 2*I*e
))*e^(-2*I*f*x - 2*I*e))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {2}{3}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {1}{3}}}\,{d x} \]

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))^(1/3),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.74, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}}}{\left (a +i a \tan \left (f x +e \right )\right )^{\frac {1}{3}}}\, dx \]

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))^(1/3),x)

[Out]

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

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maxima [B]  time = 1.10, size = 1753, normalized size = 5.16 \[ \text {result too large to display} \]

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))^(1/3),x, algorithm="maxima")

[Out]

1/8*(-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))) - 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), co
s(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)*(sqr
t(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 + s
in(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)^(1/3)*(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), 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^(1/3)*f)

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

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)^(1/3),x)

[Out]

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

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

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))**(1/3),x)

[Out]

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

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