3.684 \(\int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=318 \[ -\frac{\sqrt{3} \sqrt [3]{c-i d} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt{3}}\right )}{2 f}-\frac{\sqrt{3} \sqrt [3]{c+i d} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt{3}}\right )}{2 f}+\frac{3 \sqrt [3]{c+d \tan (e+f x)}}{f}+\frac{3 \sqrt [3]{c-i d} \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c-i d}\right )}{4 f}+\frac{3 \sqrt [3]{c+i d} \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c+i d}\right )}{4 f}+\frac{\sqrt [3]{c-i d} \log (\cos (e+f x))}{4 f}+\frac{\sqrt [3]{c+i d} \log (\cos (e+f x))}{4 f}+\frac{1}{4} i x \sqrt [3]{c-i d}-\frac{1}{4} i x \sqrt [3]{c+i d} \]

[Out]

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

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Rubi [A]  time = 0.29972, antiderivative size = 318, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3528, 3539, 3537, 57, 617, 204, 31} \[ -\frac{\sqrt{3} \sqrt [3]{c-i d} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt{3}}\right )}{2 f}-\frac{\sqrt{3} \sqrt [3]{c+i d} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt{3}}\right )}{2 f}+\frac{3 \sqrt [3]{c+d \tan (e+f x)}}{f}+\frac{3 \sqrt [3]{c-i d} \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c-i d}\right )}{4 f}+\frac{3 \sqrt [3]{c+i d} \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c+i d}\right )}{4 f}+\frac{\sqrt [3]{c-i d} \log (\cos (e+f x))}{4 f}+\frac{\sqrt [3]{c+i d} \log (\cos (e+f x))}{4 f}+\frac{1}{4} i x \sqrt [3]{c-i d}-\frac{1}{4} i x \sqrt [3]{c+i d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

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]

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

Mathematica [A]  time = 0.242699, size = 346, normalized size = 1.09 \[ -\frac{2 \sqrt{3} \sqrt [3]{c-i d} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt{3}}\right )+2 \sqrt{3} \sqrt [3]{c+i d} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt{3}}\right )-12 \sqrt [3]{c+d \tan (e+f x)}-2 \sqrt [3]{c-i d} \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c-i d}\right )-2 \sqrt [3]{c+i d} \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c+i d}\right )+\sqrt [3]{c-i d} \log \left (\sqrt [3]{c-i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}+(c-i d)^{2/3}\right )+\sqrt [3]{c+i d} \log \left (\sqrt [3]{c+i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}+(c+i d)^{2/3}\right )}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.016, size = 90, normalized size = 0.3 \begin{align*} 3\,{\frac{\sqrt [3]{c+d\tan \left ( fx+e \right ) }}{f}}+{\frac{1}{2\,f}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{6}-2\,{{\it \_Z}}^{3}c+{c}^{2}+{d}^{2} \right ) }{\frac{{{\it \_R}}^{3}c-{c}^{2}-{d}^{2}}{{{\it \_R}}^{5}-{{\it \_R}}^{2}c}\ln \left ( \sqrt [3]{c+d\tan \left ( fx+e \right ) }-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*(c+d*tan(f*x+e))^(1/3)/f+1/2/f*sum((_R^3*c-c^2-d^2)/(_R^5-_R^2*c)*ln((c+d*tan(f*x+e))^(1/3)-_R),_R=RootOf(_Z
^6-2*_Z^3*c+c^2+d^2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} \tan \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.52244, size = 6801, normalized size = 21.39 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*f*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2))*log(-2
*f*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt((c^
2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)) + f^2*((c^2 + d^2)/f^6)^(1/3) + ((c*cos(f*x + e) + d*sin(f*x + e))/
cos(f*x + e))^(2/3)) + 8*f*((c^2 + d^2)/f^6)^(1/6)*arctan((sqrt(-2*f*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*
x + e))^(1/3)*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)) +
f^2*((c^2 + d^2)/f^6)^(1/3) + ((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(2/3))*f^5*((c^2 + d^2)/f^6)^(
5/6) - f^5*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*((c^2 + d^2)/f^6)^(5/6) + (c^2 + d^2)*cos(2/
3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)))/((c^2 + d^2)*sin(2/3*arctan((f^6*sqrt((c^2 +
d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2))))*sin(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2
)) + 4*(sqrt(3)*f*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2
)) - f*((c^2 + d^2)/f^6)^(1/6)*sin(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)))*arctan(
-(2*sqrt(3)*f^5*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*((c^2 + d^2)/f^6)^(5/6)*cos(2/3*arctan(
(f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)) + 2*(f^5*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x +
e))^(1/3)*((c^2 + d^2)/f^6)^(5/6) + 2*(c^2 + d^2)*cos(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/
f^6)/d^2)))*sin(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)) - 2*(sqrt(3)*f^5*((c^2 + d^
2)/f^6)^(5/6)*cos(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)) + f^5*((c^2 + d^2)/f^6)^(
5/6)*sin(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)))*sqrt(sqrt(3)*f*((c*cos(f*x + e) +
d*sin(f*x + e))/cos(f*x + e))^(1/3)*((c^2 + d^2)/f^6)^(1/6)*sin(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3
)*sqrt(d^2/f^6)/d^2)) + f*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*((c^2 + d^2)/f^6)^(1/6)*cos(2
/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)) + f^2*((c^2 + d^2)/f^6)^(1/3) + ((c*cos(f*x
+ e) + d*sin(f*x + e))/cos(f*x + e))^(2/3)) + sqrt(3)*(c^2 + d^2))/(4*(c^2 + d^2)*cos(2/3*arctan((f^6*sqrt((c^
2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2))^2 - c^2 - d^2)) - 4*(sqrt(3)*f*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arct
an((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)) + f*((c^2 + d^2)/f^6)^(1/6)*sin(2/3*arctan((f^6*sqr
t((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)))*arctan((2*sqrt(3)*f^5*((c*cos(f*x + e) + d*sin(f*x + e))/cos(
f*x + e))^(1/3)*((c^2 + d^2)/f^6)^(5/6)*cos(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2))
- 2*(f^5*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*((c^2 + d^2)/f^6)^(5/6) + 2*(c^2 + d^2)*cos(2
/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)))*sin(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) -
c*f^3)*sqrt(d^2/f^6)/d^2)) - 2*(sqrt(3)*f^5*((c^2 + d^2)/f^6)^(5/6)*cos(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6)
- c*f^3)*sqrt(d^2/f^6)/d^2)) - f^5*((c^2 + d^2)/f^6)^(5/6)*sin(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)
*sqrt(d^2/f^6)/d^2)))*sqrt(-sqrt(3)*f*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*((c^2 + d^2)/f^6)
^(1/6)*sin(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)) + f*((c*cos(f*x + e) + d*sin(f*x
+ e))/cos(f*x + e))^(1/3)*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2
/f^6)/d^2)) + f^2*((c^2 + d^2)/f^6)^(1/3) + ((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(2/3)) + sqrt(3)*
(c^2 + d^2))/(4*(c^2 + d^2)*cos(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2))^2 - c^2 - d
^2)) - (sqrt(3)*f*((c^2 + d^2)/f^6)^(1/6)*sin(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2
)) + f*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)))*log(sqr
t(3)*f*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*((c^2 + d^2)/f^6)^(1/6)*sin(2/3*arctan((f^6*sqrt
((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)) + f*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*((c^
2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)) + f^2*((c^2 + d^2)/
f^6)^(1/3) + ((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(2/3)) + (sqrt(3)*f*((c^2 + d^2)/f^6)^(1/6)*sin(
2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)) - f*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan(
(f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)))*log(-sqrt(3)*f*((c*cos(f*x + e) + d*sin(f*x + e))/cos
(f*x + e))^(1/3)*((c^2 + d^2)/f^6)^(1/6)*sin(2/3*arctan((f^6*sqrt((c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)
) + f*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt(
(c^2 + d^2)/f^6) - c*f^3)*sqrt(d^2/f^6)/d^2)) + f^2*((c^2 + d^2)/f^6)^(1/3) + ((c*cos(f*x + e) + d*sin(f*x + e
))/cos(f*x + e))^(2/3)) + 12*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3))/f

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{c + d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 4.41641, size = 487, normalized size = 1.53 \begin{align*} -\frac{{\left (i \, \sqrt{3} + 1\right )} \left (\frac{c d^{3} - i \, d^{4}}{f^{3}}\right )^{\frac{1}{3}} \log \left (-{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} f{\left (\sqrt{3} + i\right )} + 2 \,{\left (i \, c + d\right )}^{\frac{1}{3}} f\right ) +{\left (i \, \sqrt{3} + 1\right )} \left (\frac{c d^{3} + i \, d^{4}}{f^{3}}\right )^{\frac{1}{3}} \log \left (-{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} f{\left (\sqrt{3} + i\right )} + 2 \,{\left (i \, c - d\right )}^{\frac{1}{3}} f\right ) +{\left (-i \, \sqrt{3} + 1\right )} \left (\frac{c d^{3} - i \, d^{4}}{f^{3}}\right )^{\frac{1}{3}} \log \left ({\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} f{\left (\sqrt{3} - i\right )} + 2 \,{\left (i \, c + d\right )}^{\frac{1}{3}} f\right ) +{\left (-i \, \sqrt{3} + 1\right )} \left (\frac{c d^{3} + i \, d^{4}}{f^{3}}\right )^{\frac{1}{3}} \log \left ({\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} f{\left (\sqrt{3} - i\right )} + 2 \,{\left (i \, c - d\right )}^{\frac{1}{3}} f\right ) - 2 \, \left (\frac{c d^{3} - i \, d^{4}}{f^{3}}\right )^{\frac{1}{3}} \log \left (i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} f^{2} +{\left (i \, c + d\right )}^{\frac{1}{3}} f^{2}\right ) - 2 \, \left (\frac{c d^{3} + i \, d^{4}}{f^{3}}\right )^{\frac{1}{3}} \log \left (i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} f^{2} +{\left (i \, c - d\right )}^{\frac{1}{3}} f^{2}\right ) - \frac{12 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} d}{f}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((I*sqrt(3) + 1)*((c*d^3 - I*d^4)/f^3)^(1/3)*log(-(d*tan(f*x + e) + c)^(1/3)*f*(sqrt(3) + I) + 2*(I*c + d
)^(1/3)*f) + (I*sqrt(3) + 1)*((c*d^3 + I*d^4)/f^3)^(1/3)*log(-(d*tan(f*x + e) + c)^(1/3)*f*(sqrt(3) + I) + 2*(
I*c - d)^(1/3)*f) + (-I*sqrt(3) + 1)*((c*d^3 - I*d^4)/f^3)^(1/3)*log((d*tan(f*x + e) + c)^(1/3)*f*(sqrt(3) - I
) + 2*(I*c + d)^(1/3)*f) + (-I*sqrt(3) + 1)*((c*d^3 + I*d^4)/f^3)^(1/3)*log((d*tan(f*x + e) + c)^(1/3)*f*(sqrt
(3) - I) + 2*(I*c - d)^(1/3)*f) - 2*((c*d^3 - I*d^4)/f^3)^(1/3)*log(I*(d*tan(f*x + e) + c)^(1/3)*f^2 + (I*c +
d)^(1/3)*f^2) - 2*((c*d^3 + I*d^4)/f^3)^(1/3)*log(I*(d*tan(f*x + e) + c)^(1/3)*f^2 + (I*c - d)^(1/3)*f^2) - 12
*(d*tan(f*x + e) + c)^(1/3)*d/f)/d