3.477 \(\int \frac {i-\tan (e+f x)}{\sqrt [3]{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=148 \[ -\frac {\sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )}{f \sqrt [3]{c-i d}}-\frac {3 \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c-i d}\right )}{2 f \sqrt [3]{c-i d}}-\frac {\log (\cos (e+f x))}{2 f \sqrt [3]{c-i d}}-\frac {i x}{2 \sqrt [3]{c-i d}} \]

[Out]

-1/2*I*x/(c-I*d)^(1/3)-1/2*ln(cos(f*x+e))/(c-I*d)^(1/3)/f-3/2*ln((c-I*d)^(1/3)-(c+d*tan(f*x+e))^(1/3))/(c-I*d)
^(1/3)/f-arctan(1/3*(1+2*(c+d*tan(f*x+e))^(1/3)/(c-I*d)^(1/3))*3^(1/2))*3^(1/2)/(c-I*d)^(1/3)/f

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Rubi [A]  time = 0.13, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3537, 55, 617, 204, 31} \[ -\frac {\sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )}{f \sqrt [3]{c-i d}}-\frac {3 \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c-i d}\right )}{2 f \sqrt [3]{c-i d}}-\frac {\log (\cos (e+f x))}{2 f \sqrt [3]{c-i d}}-\frac {i x}{2 \sqrt [3]{c-i d}} \]

Antiderivative was successfully verified.

[In]

Int[(I - Tan[e + f*x])/(c + d*Tan[e + f*x])^(1/3),x]

[Out]

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

Rule 31

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

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), 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 3537

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

Rubi steps

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

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Mathematica [C]  time = 1.83, size = 109, normalized size = 0.74 \[ \frac {3 \left (c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}\right )^{2/3} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {i c+\frac {d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}{i c+d}\right )}{2 f (c-i d)} \]

Antiderivative was successfully verified.

[In]

Integrate[(I - Tan[e + f*x])/(c + d*Tan[e + f*x])^(1/3),x]

[Out]

(3*(c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x))))^(2/3)*Hypergeometric2F1[2/3, 1, 5/3, (I*c
+ (d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x))))/(I*c + d)])/(2*(c - I*d)*f)

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fricas [B]  time = 1.03, size = 278, normalized size = 1.88 \[ \frac {1}{2} \, {\left (i \, \sqrt {3} - 1\right )} \left (-\frac {i}{{\left (i \, c + d\right )} f^{3}}\right )^{\frac {1}{3}} \log \left (\frac {1}{2} \, {\left (\sqrt {3} {\left (i \, c + d\right )} f^{2} + {\left (c - i \, d\right )} f^{2}\right )} \left (-\frac {i}{{\left (i \, c + d\right )} f^{3}}\right )^{\frac {2}{3}} + \left (\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {1}{3}}\right ) + \frac {1}{2} \, {\left (-i \, \sqrt {3} - 1\right )} \left (-\frac {i}{{\left (i \, c + d\right )} f^{3}}\right )^{\frac {1}{3}} \log \left (\frac {1}{2} \, {\left (\sqrt {3} {\left (-i \, c - d\right )} f^{2} + {\left (c - i \, d\right )} f^{2}\right )} \left (-\frac {i}{{\left (i \, c + d\right )} f^{3}}\right )^{\frac {2}{3}} + \left (\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {1}{3}}\right ) + \left (-\frac {i}{{\left (i \, c + d\right )} f^{3}}\right )^{\frac {1}{3}} \log \left (-{\left (c - i \, d\right )} f^{2} \left (-\frac {i}{{\left (i \, c + d\right )} f^{3}}\right )^{\frac {2}{3}} + \left (\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {1}{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-tan(f*x+e)+I)/(c+d*tan(f*x+e))^(1/3),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.89, size = 910, normalized size = 6.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-tan(f*x+e)+I)/(c+d*tan(f*x+e))^(1/3),x, algorithm="giac")

[Out]

-(c - I*d)^(2/3)*log((d*tan(f*x + e) + c)^(1/3) - (c - I*d)^(1/3))/(c*f - I*d*f) - (sqrt(3)*(c^2 + d^2)^(1/3)*
c*cos(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))^2 - sqrt(3)*(c^2 + d^2)^(1/3)*c*sin(1/6*pi*sgn(c
)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))^2 + 2*(c^2 + d^2)^(1/3)*c*cos(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d)
- 1/3*arctan(d/c))*sin(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c)))*arctan(1/3*sqrt(3)*(2*(d*tan(
f*x + e) + c)^(1/3) + (c - I*d)^(1/3))/(c - I*d)^(1/3))/((c^2 + d^2)*f) - I*(sqrt(3)*(c^2 + d^2)^(1/3)*d*cos(1
/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))^2 - sqrt(3)*(c^2 + d^2)^(1/3)*d*sin(1/6*pi*sgn(c)*sgn(d
) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))^2 + 2*(c^2 + d^2)^(1/3)*d*cos(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*
arctan(d/c))*sin(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c)))*arctan(1/3*sqrt(3)*(2*(d*tan(f*x + e
) + c)^(1/3) + (c - I*d)^(1/3))/(c - I*d)^(1/3))/((c^2 + d^2)*f) - 1/2*(2*sqrt(3)*(c^2 + d^2)^(1/3)*c*cos(1/6*
pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))*sin(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c)
) - (c^2 + d^2)^(1/3)*c*cos(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))^2 + (c^2 + d^2)^(1/3)*c*si
n(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))^2)*log((c^2 + d^2)^(1/3)*cos(1/6*pi*sgn(c)*sgn(d) -
1/6*pi*sgn(d) - 1/3*arctan(d/c))^2 + (c^2 + d^2)^(1/3)*sin(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d
/c))^2 + (d*tan(f*x + e) + c)^(2/3) + (d*tan(f*x + e) + c)^(1/3)*(c - I*d)^(1/3))/((c^2 + d^2)*f) - 1/2*I*(2*s
qrt(3)*(c^2 + d^2)^(1/3)*d*cos(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))*sin(1/6*pi*sgn(c)*sgn(d
) - 1/6*pi*sgn(d) - 1/3*arctan(d/c)) - (c^2 + d^2)^(1/3)*d*cos(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arct
an(d/c))^2 + (c^2 + d^2)^(1/3)*d*sin(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))^2)*log((c^2 + d^2
)^(1/3)*cos(1/6*pi*sgn(c)*sgn(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))^2 + (c^2 + d^2)^(1/3)*sin(1/6*pi*sgn(c)*sg
n(d) - 1/6*pi*sgn(d) - 1/3*arctan(d/c))^2 + (d*tan(f*x + e) + c)^(2/3) + (d*tan(f*x + e) + c)^(1/3)*(c - I*d)^
(1/3))/((c^2 + d^2)*f)

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maple [C]  time = 0.33, size = 42, normalized size = 0.28 \[ -\frac {\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3}+i d -c \right )}{\sum }\frac {\ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-tan(f*x+e)+I)/(c+d*tan(f*x+e))^(1/3),x)

[Out]

-1/f*sum(1/_R*ln((c+d*tan(f*x+e))^(1/3)-_R),_R=RootOf(_Z^3+I*d-c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-tan(f*x+e)+I)/(c+d*tan(f*x+e))^(1/3),x, algorithm="maxima")

[Out]

-integrate((tan(f*x + e) - I)/(d*tan(f*x + e) + c)^(1/3), x)

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mupad [B]  time = 20.67, size = 2982, normalized size = 20.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(tan(e + f*x) - 1i)/(c + d*tan(e + f*x))^(1/3),x)

[Out]

log(d^5*f*(c + d*tan(e + f*x))^(1/3)*243i + ((1944*d^4*f^4*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/3) - 3*2^(1/3)*
c*d^4*f^6*(c^2 + d^2)*(-(11664*f^3*((c^2*d^12*(c^2 + d^2)^2)/f^6)^(1/2) + d^9*11664i + c^2*d^7*11664i)/(d^6*f^
3*(c^2 + d^2)^2))^(2/3))*(11664*f^3*((c^2*d^12*(c^2 + d^2)^2)/f^6)^(1/2) + d^9*11664i + c^2*d^7*11664i))/(9331
2*d^6*f^3*(c^2 + d^2)^2))*(-(f^3*((4*(729*d^8 + 729*c^2*d^6)*(46656*d^10 + 93312*c^2*d^8 + 46656*c^4*d^6))/f^6
- (11664*d^9 + 11664*c^2*d^7)^2/f^6)^(1/2) + d^9*11664i + c^2*d^7*11664i)/(93312*f^3*(d^10 + 2*c^2*d^8 + c^4*
d^6)))^(1/3) + log(d^5*f*(c + d*tan(e + f*x))^(1/3)*243i + ((1944*d^4*f^4*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/
3) - 3*2^(1/3)*c*d^4*f^6*(c^2 + d^2)*(-(d^9*11664i - 11664*f^3*((c^2*d^12*(c^2 + d^2)^2)/f^6)^(1/2) + c^2*d^7*
11664i)/(d^6*f^3*(c^2 + d^2)^2))^(2/3))*(d^9*11664i - 11664*f^3*((c^2*d^12*(c^2 + d^2)^2)/f^6)^(1/2) + c^2*d^7
*11664i))/(93312*d^6*f^3*(c^2 + d^2)^2))*(-(d^9*11664i - f^3*((4*(729*d^8 + 729*c^2*d^6)*(46656*d^10 + 93312*c
^2*d^8 + 46656*c^4*d^6))/f^6 - (11664*d^9 + 11664*c^2*d^7)^2/f^6)^(1/2) + c^2*d^7*11664i)/(93312*f^3*(d^10 + 2
*c^2*d^8 + c^4*d^6)))^(1/3) + (log(- ((-1/(f^3*(c - d*1i)))^(2/3)*(((-1/(f^3*(c - d*1i)))^(1/3)*((1944*d^4*(c^
2 - d^2)*(c + d*tan(e + f*x))^(1/3))/f^2 - 1944*c*d^4*(-1/(f^3*(c - d*1i)))^(2/3)*(c^2 + d^2)))/2 - (972*d^4*(
c^2 + d^2))/f^3))/4 - (243*c*d^4*(c + d*tan(e + f*x))^(1/3))/f^5)*(-1/(c*f^3 - d*f^3*1i))^(1/3))/2 + log(((777
6*c*d^4*(c^2 + d^2)*(-1i/(8*f^3*(c*1i - d)))^(2/3) - (1944*d^4*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/3))/f^2)*(-
1i/(8*f^3*(c*1i - d)))^(1/3) + (972*d^4*(c^2 + d^2))/f^3)*(-1i/(8*f^3*(c*1i - d)))^(2/3) - (243*c*d^4*(c + d*t
an(e + f*x))^(1/3))/f^5)*(-1i/(8*(c*f^3*1i - d*f^3)))^(1/3) + (log(- ((-1/(f^3*(c - d*1i)))^(2/3)*(3^(1/2)*1i
- 1)^2*(((-1/(f^3*(c - d*1i)))^(1/3)*(3^(1/2)*1i - 1)*((1944*d^4*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/3))/f^2 -
486*c*d^4*(-1/(f^3*(c - d*1i)))^(2/3)*(3^(1/2)*1i - 1)^2*(c^2 + d^2)))/4 - (972*d^4*(c^2 + d^2))/f^3))/16 - (
243*c*d^4*(c + d*tan(e + f*x))^(1/3))/f^5)*(3^(1/2)*1i - 1)*(-1/(8*(c*f^3 - d*f^3*1i)))^(1/3))/2 - (log(((-1/(
f^3*(c - d*1i)))^(2/3)*(3^(1/2)*1i + 1)^2*(((-1/(f^3*(c - d*1i)))^(1/3)*(3^(1/2)*1i + 1)*((1944*d^4*(c^2 - d^2
)*(c + d*tan(e + f*x))^(1/3))/f^2 - 486*c*d^4*(-1/(f^3*(c - d*1i)))^(2/3)*(3^(1/2)*1i + 1)^2*(c^2 + d^2)))/4 +
(972*d^4*(c^2 + d^2))/f^3))/16 - (243*c*d^4*(c + d*tan(e + f*x))^(1/3))/f^5)*(3^(1/2)*1i + 1)*(-1/(8*(c*f^3 -
d*f^3*1i)))^(1/3))/2 - log(d^5*f*(c + d*tan(e + f*x))^(1/3)*243i + ((1944*d^4*f^4*(c^2 - d^2)*(c + d*tan(e +
f*x))^(1/3) - (3*2^(1/3)*c*d^4*f^6*(3^(1/2)*1i - 1)*(c^2 + d^2)*(-(11664*f^3*((c^2*d^12*(c^2 + d^2)^2)/f^6)^(1
/2) + d^9*11664i + c^2*d^7*11664i)/(d^6*f^3*(c^2 + d^2)^2))^(2/3))/2)*(11664*f^3*((c^2*d^12*(c^2 + d^2)^2)/f^6
)^(1/2) + d^9*11664i + c^2*d^7*11664i))/(93312*d^6*f^3*(c^2 + d^2)^2))*((3^(1/2)*1i)/2 + 1/2)*(-(f^3*((4*(729*
d^8 + 729*c^2*d^6)*(46656*d^10 + 93312*c^2*d^8 + 46656*c^4*d^6))/f^6 - (11664*d^9 + 11664*c^2*d^7)^2/f^6)^(1/2
) + d^9*11664i + c^2*d^7*11664i)/(93312*f^3*(d^10 + 2*c^2*d^8 + c^4*d^6)))^(1/3) + log(d^5*f*(c + d*tan(e + f*
x))^(1/3)*243i + ((1944*d^4*f^4*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/3) + (3*2^(1/3)*c*d^4*f^6*(3^(1/2)*1i + 1)
*(c^2 + d^2)*(-(11664*f^3*((c^2*d^12*(c^2 + d^2)^2)/f^6)^(1/2) + d^9*11664i + c^2*d^7*11664i)/(d^6*f^3*(c^2 +
d^2)^2))^(2/3))/2)*(11664*f^3*((c^2*d^12*(c^2 + d^2)^2)/f^6)^(1/2) + d^9*11664i + c^2*d^7*11664i))/(93312*d^6*
f^3*(c^2 + d^2)^2))*((3^(1/2)*1i)/2 - 1/2)*(-(f^3*((4*(729*d^8 + 729*c^2*d^6)*(46656*d^10 + 93312*c^2*d^8 + 46
656*c^4*d^6))/f^6 - (11664*d^9 + 11664*c^2*d^7)^2/f^6)^(1/2) + d^9*11664i + c^2*d^7*11664i)/(93312*f^3*(d^10 +
2*c^2*d^8 + c^4*d^6)))^(1/3) - log(d^5*f*(c + d*tan(e + f*x))^(1/3)*243i + ((1944*d^4*f^4*(c^2 - d^2)*(c + d*
tan(e + f*x))^(1/3) - (3*2^(1/3)*c*d^4*f^6*(3^(1/2)*1i - 1)*(c^2 + d^2)*(-(d^9*11664i - 11664*f^3*((c^2*d^12*(
c^2 + d^2)^2)/f^6)^(1/2) + c^2*d^7*11664i)/(d^6*f^3*(c^2 + d^2)^2))^(2/3))/2)*(d^9*11664i - 11664*f^3*((c^2*d^
12*(c^2 + d^2)^2)/f^6)^(1/2) + c^2*d^7*11664i))/(93312*d^6*f^3*(c^2 + d^2)^2))*((3^(1/2)*1i)/2 + 1/2)*(-(d^9*1
1664i - f^3*((4*(729*d^8 + 729*c^2*d^6)*(46656*d^10 + 93312*c^2*d^8 + 46656*c^4*d^6))/f^6 - (11664*d^9 + 11664
*c^2*d^7)^2/f^6)^(1/2) + c^2*d^7*11664i)/(93312*f^3*(d^10 + 2*c^2*d^8 + c^4*d^6)))^(1/3) + log(d^5*f*(c + d*ta
n(e + f*x))^(1/3)*243i + ((1944*d^4*f^4*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/3) + (3*2^(1/3)*c*d^4*f^6*(3^(1/2)
*1i + 1)*(c^2 + d^2)*(-(d^9*11664i - 11664*f^3*((c^2*d^12*(c^2 + d^2)^2)/f^6)^(1/2) + c^2*d^7*11664i)/(d^6*f^3
*(c^2 + d^2)^2))^(2/3))/2)*(d^9*11664i - 11664*f^3*((c^2*d^12*(c^2 + d^2)^2)/f^6)^(1/2) + c^2*d^7*11664i))/(93
312*d^6*f^3*(c^2 + d^2)^2))*((3^(1/2)*1i)/2 - 1/2)*(-(d^9*11664i - f^3*((4*(729*d^8 + 729*c^2*d^6)*(46656*d^10
+ 93312*c^2*d^8 + 46656*c^4*d^6))/f^6 - (11664*d^9 + 11664*c^2*d^7)^2/f^6)^(1/2) + c^2*d^7*11664i)/(93312*f^3
*(d^10 + 2*c^2*d^8 + c^4*d^6)))^(1/3) + (log((((972*d^4*(c^2 + d^2))/f^3 - ((3^(1/2)*1i - 1)*((1944*d^4*(c^2 -
d^2)*(c + d*tan(e + f*x))^(1/3))/f^2 - 1944*c*d^4*(3^(1/2)*1i - 1)^2*(c^2 + d^2)*(-1i/(8*f^3*(c*1i - d)))^(2/
3))*(-1i/(8*f^3*(c*1i - d)))^(1/3))/2)*(3^(1/2)*1i - 1)^2*(-1i/(8*f^3*(c*1i - d)))^(2/3))/4 - (243*c*d^4*(c +
d*tan(e + f*x))^(1/3))/f^5)*(3^(1/2)*1i - 1)*(-1i/(8*(c*f^3*1i - d*f^3)))^(1/3))/2 - (log((((972*d^4*(c^2 + d^
2))/f^3 + ((3^(1/2)*1i + 1)*((1944*d^4*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/3))/f^2 - 1944*c*d^4*(3^(1/2)*1i +
1)^2*(c^2 + d^2)*(-1i/(8*f^3*(c*1i - d)))^(2/3))*(-1i/(8*f^3*(c*1i - d)))^(1/3))/2)*(3^(1/2)*1i + 1)^2*(-1i/(8
*f^3*(c*1i - d)))^(2/3))/4 - (243*c*d^4*(c + d*tan(e + f*x))^(1/3))/f^5)*(3^(1/2)*1i + 1)*(-1i/(8*(c*f^3*1i -
d*f^3)))^(1/3))/2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((I-tan(f*x+e))/(c+d*tan(f*x+e))**(1/3),x)

[Out]

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

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