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

Optimal. Leaf size=415 \[ \frac{\sqrt{3} d \sqrt [3]{c-\sqrt{-d^2}} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-\sqrt{-d^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-d^2} f}-\frac{\sqrt{3} d \sqrt [3]{c+\sqrt{-d^2}} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+\sqrt{-d^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-d^2} f}-\frac{3 d \sqrt [3]{c-\sqrt{-d^2}} \log \left (\sqrt [3]{c-\sqrt{-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt{-d^2} f}+\frac{3 d \sqrt [3]{c+\sqrt{-d^2}} \log \left (\sqrt [3]{c+\sqrt{-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt{-d^2} f}-\frac{d \sqrt [3]{c-\sqrt{-d^2}} \log (\cos (e+f x))}{4 \sqrt{-d^2} f}+\frac{d \sqrt [3]{c+\sqrt{-d^2}} \log (\cos (e+f x))}{4 \sqrt{-d^2} f}-\frac{1}{4} x \sqrt [3]{c-\sqrt{-d^2}}-\frac{1}{4} x \sqrt [3]{c+\sqrt{-d^2}} \]

[Out]

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

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Rubi [A]  time = 0.246101, antiderivative size = 415, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3485, 712, 50, 57, 617, 204, 31} \[ \frac{\sqrt{3} d \sqrt [3]{c-\sqrt{-d^2}} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-\sqrt{-d^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-d^2} f}-\frac{\sqrt{3} d \sqrt [3]{c+\sqrt{-d^2}} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+\sqrt{-d^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-d^2} f}-\frac{3 d \sqrt [3]{c-\sqrt{-d^2}} \log \left (\sqrt [3]{c-\sqrt{-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt{-d^2} f}+\frac{3 d \sqrt [3]{c+\sqrt{-d^2}} \log \left (\sqrt [3]{c+\sqrt{-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt{-d^2} f}-\frac{d \sqrt [3]{c-\sqrt{-d^2}} \log (\cos (e+f x))}{4 \sqrt{-d^2} f}+\frac{d \sqrt [3]{c+\sqrt{-d^2}} \log (\cos (e+f x))}{4 \sqrt{-d^2} f}-\frac{1}{4} x \sqrt [3]{c-\sqrt{-d^2}}-\frac{1}{4} x \sqrt [3]{c+\sqrt{-d^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])^(1/3),x]

[Out]

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

Rule 3485

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

Rule 712

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 \sqrt [3]{c+d \tan (e+f x)} \, dx &=\frac{d \operatorname{Subst}\left (\int \frac{\sqrt [3]{c+x}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}\\ &=\frac{d \operatorname{Subst}\left (\int \left (\frac{\sqrt{-d^2} \sqrt [3]{c+x}}{2 d^2 \left (\sqrt{-d^2}-x\right )}+\frac{\sqrt{-d^2} \sqrt [3]{c+x}}{2 d^2 \left (\sqrt{-d^2}+x\right )}\right ) \, dx,x,d \tan (e+f x)\right )}{f}\\ &=-\frac{d \operatorname{Subst}\left (\int \frac{\sqrt [3]{c+x}}{\sqrt{-d^2}-x} \, dx,x,d \tan (e+f x)\right )}{2 \sqrt{-d^2} f}-\frac{d \operatorname{Subst}\left (\int \frac{\sqrt [3]{c+x}}{\sqrt{-d^2}+x} \, dx,x,d \tan (e+f x)\right )}{2 \sqrt{-d^2} f}\\ &=-\frac{\left (d \left (c+\sqrt{-d^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{-d^2}-x\right ) (c+x)^{2/3}} \, dx,x,d \tan (e+f x)\right )}{2 \sqrt{-d^2} f}+\frac{\left (d^2+c \sqrt{-d^2}\right ) \operatorname{Subst}\left (\int \frac{1}{(c+x)^{2/3} \left (\sqrt{-d^2}+x\right )} \, dx,x,d \tan (e+f x)\right )}{2 d f}\\ &=-\frac{1}{4} \sqrt [3]{c-\sqrt{-d^2}} x-\frac{1}{4} \sqrt [3]{c+\sqrt{-d^2}} x+\frac{\sqrt{-d^2} \sqrt [3]{c-\sqrt{-d^2}} \log (\cos (e+f x))}{4 d f}+\frac{d \sqrt [3]{c+\sqrt{-d^2}} \log (\cos (e+f x))}{4 \sqrt{-d^2} f}-\frac{\left (3 d \sqrt [3]{c+\sqrt{-d^2}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{c+\sqrt{-d^2}}-x} \, dx,x,\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt{-d^2} f}-\frac{\left (3 d \left (c+\sqrt{-d^2}\right )^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (c+\sqrt{-d^2}\right )^{2/3}+\sqrt [3]{c+\sqrt{-d^2}} x+x^2} \, dx,x,\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt{-d^2} f}-\frac{\left (3 \left (d^2+c \sqrt{-d^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{c-\sqrt{-d^2}}-x} \, dx,x,\sqrt [3]{c+d \tan (e+f x)}\right )}{4 d \left (c-\sqrt{-d^2}\right )^{2/3} f}-\frac{\left (3 \left (d^2+c \sqrt{-d^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (c-\sqrt{-d^2}\right )^{2/3}+\sqrt [3]{c-\sqrt{-d^2}} x+x^2} \, dx,x,\sqrt [3]{c+d \tan (e+f x)}\right )}{4 d \sqrt [3]{c-\sqrt{-d^2}} f}\\ &=-\frac{1}{4} \sqrt [3]{c-\sqrt{-d^2}} x-\frac{1}{4} \sqrt [3]{c+\sqrt{-d^2}} x+\frac{\sqrt{-d^2} \sqrt [3]{c-\sqrt{-d^2}} \log (\cos (e+f x))}{4 d f}+\frac{d \sqrt [3]{c+\sqrt{-d^2}} \log (\cos (e+f x))}{4 \sqrt{-d^2} f}+\frac{3 \sqrt{-d^2} \sqrt [3]{c-\sqrt{-d^2}} \log \left (\sqrt [3]{c-\sqrt{-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 d f}+\frac{3 d \sqrt [3]{c+\sqrt{-d^2}} \log \left (\sqrt [3]{c+\sqrt{-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt{-d^2} f}+\frac{\left (3 d \sqrt [3]{c+\sqrt{-d^2}}\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+\sqrt{-d^2}}}\right )}{2 \sqrt{-d^2} f}+\frac{\left (3 \left (d^2+c \sqrt{-d^2}\right )\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-\sqrt{-d^2}}}\right )}{2 d \left (c-\sqrt{-d^2}\right )^{2/3} f}\\ &=-\frac{1}{4} \sqrt [3]{c-\sqrt{-d^2}} x-\frac{1}{4} \sqrt [3]{c+\sqrt{-d^2}} x-\frac{\sqrt{3} \sqrt{-d^2} \sqrt [3]{c-\sqrt{-d^2}} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-\sqrt{-d^2}}}}{\sqrt{3}}\right )}{2 d f}-\frac{\sqrt{3} d \sqrt [3]{c+\sqrt{-d^2}} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+\sqrt{-d^2}}}}{\sqrt{3}}\right )}{2 \sqrt{-d^2} f}+\frac{\sqrt{-d^2} \sqrt [3]{c-\sqrt{-d^2}} \log (\cos (e+f x))}{4 d f}+\frac{d \sqrt [3]{c+\sqrt{-d^2}} \log (\cos (e+f x))}{4 \sqrt{-d^2} f}+\frac{3 \sqrt{-d^2} \sqrt [3]{c-\sqrt{-d^2}} \log \left (\sqrt [3]{c-\sqrt{-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 d f}+\frac{3 d \sqrt [3]{c+\sqrt{-d^2}} \log \left (\sqrt [3]{c+\sqrt{-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt{-d^2} f}\\ \end{align*}

Mathematica [C]  time = 0.228753, size = 294, normalized size = 0.71 \[ \frac{i \sqrt [3]{c+i d} \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt{3}}\right )-2 \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c+i d}\right )+\log \left (\sqrt [3]{c+i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}+(c+i d)^{2/3}\right )\right )-i \sqrt [3]{c-i d} \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt{3}}\right )-2 \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c-i d}\right )+\log \left (\sqrt [3]{c-i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}+(c-i d)^{2/3}\right )\right )}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.014, size = 60, normalized size = 0.1 \begin{align*}{\frac{d}{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}}{{{\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((c+d*tan(f*x+e))^(1/3),x)

[Out]

1/2/f*d*sum(_R^3/(_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}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.72146, size = 7907, normalized size = 19.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2
))*log(2*c*f^4*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(1/6)*si
n(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) + c^2*f^2*((c^2 + d^2)/f^6)
^(1/3) + c^2*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(2/3)) + 2*((c^2 + d^2)/f^6)^(1/6)*arctan(-(c*f^
8*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(5/6) - sqrt(2*c*f^4*
((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(1/6)*sin(2/3*arctan((f
^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) + c^2*f^2*((c^2 + d^2)/f^6)^(1/3) + c^2*((
c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(2/3))*f^8*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(5/6) + (c^4 + c^2*d
^2)*sin(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)))/((c^4 + c^2*d^2)*cos
(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2))))*sin(2/3*arctan((f^6*sqrt(c
^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) + (sqrt(3)*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((
f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) - ((c^2 + d^2)/f^6)^(1/6)*sin(2/3*arctan(
(f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)))*arctan(-(2*c*f^8*((c*cos(f*x + e) + d*s
in(f*x + e))/cos(f*x + e))^(1/3)*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(5/6)*cos(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt(
(c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) - 2*(sqrt(3)*c*f^8*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x +
e))^(1/3)*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(5/6) + 2*(c^4 + c^2*d^2)*cos(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^
2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)))*sin(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sq
rt(c^2/f^6))/c^2)) + 2*(sqrt(3)*f^8*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(5/6)*sin(2/3*arctan((f^6*sqrt(c^2/f^6)*sq
rt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) - f^8*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(5/6)*cos(2/3*arctan((f
^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)))*sqrt(sqrt(3)*c*f^4*((c*cos(f*x + e) + d*s
in(f*x + e))/cos(f*x + e))^(1/3)*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt(
(c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) - c*f^4*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*s
qrt(c^2/f^6)*((c^2 + d^2)/f^6)^(1/6)*sin(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/
f^6))/c^2)) + c^2*f^2*((c^2 + d^2)/f^6)^(1/3) + c^2*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(2/3)) +
sqrt(3)*(c^4 + c^2*d^2))/(3*c^4 + 3*c^2*d^2 - 4*(c^4 + c^2*d^2)*cos(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 +
d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2))^2)) + (sqrt(3)*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt(c^2/f^6
)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) + ((c^2 + d^2)/f^6)^(1/6)*sin(2/3*arctan((f^6*sqrt(c^2/f^
6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)))*arctan((2*c*f^8*((c*cos(f*x + e) + d*sin(f*x + e))/cos(
f*x + e))^(1/3)*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(5/6)*cos(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6)
+ d*f^3*sqrt(c^2/f^6))/c^2)) + 2*(sqrt(3)*c*f^8*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*sqrt(c^
2/f^6)*((c^2 + d^2)/f^6)^(5/6) - 2*(c^4 + c^2*d^2)*cos(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d
*f^3*sqrt(c^2/f^6))/c^2)))*sin(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)
) - 2*(sqrt(3)*f^8*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(5/6)*sin(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^
6) + d*f^3*sqrt(c^2/f^6))/c^2)) + f^8*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(5/6)*cos(2/3*arctan((f^6*sqrt(c^2/f^6)*
sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)))*sqrt(-sqrt(3)*c*f^4*((c*cos(f*x + e) + d*sin(f*x + e))/cos
(f*x + e))^(1/3)*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6)
+ d*f^3*sqrt(c^2/f^6))/c^2)) - c*f^4*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*sqrt(c^2/f^6)*((c
^2 + d^2)/f^6)^(1/6)*sin(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) + c^
2*f^2*((c^2 + d^2)/f^6)^(1/3) + c^2*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(2/3)) - sqrt(3)*(c^4 + c
^2*d^2))/(3*c^4 + 3*c^2*d^2 - 4*(c^4 + c^2*d^2)*cos(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^
3*sqrt(c^2/f^6))/c^2))^2)) - 1/4*(sqrt(3)*((c^2 + d^2)/f^6)^(1/6)*sin(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2
+ d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) + ((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2
+ d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)))*log(sqrt(3)*c*f^4*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^
(1/3)*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sq
rt(c^2/f^6))/c^2)) - c*f^4*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*sqrt(c^2/f^6)*((c^2 + d^2)/f
^6)^(1/6)*sin(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) + c^2*f^2*((c^2
+ d^2)/f^6)^(1/3) + c^2*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(2/3)) + 1/4*(sqrt(3)*((c^2 + d^2)/f
^6)^(1/6)*sin(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) - ((c^2 + d^2)/
f^6)^(1/6)*cos(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)))*log(-sqrt(3)*
c*f^4*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3)*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(1/6)*cos(2/3*arc
tan((f^6*sqrt(c^2/f^6)*sqrt((c^2 + d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) - c*f^4*((c*cos(f*x + e) + d*sin(f*x
+ e))/cos(f*x + e))^(1/3)*sqrt(c^2/f^6)*((c^2 + d^2)/f^6)^(1/6)*sin(2/3*arctan((f^6*sqrt(c^2/f^6)*sqrt((c^2 +
d^2)/f^6) + d*f^3*sqrt(c^2/f^6))/c^2)) + c^2*f^2*((c^2 + d^2)/f^6)^(1/3) + c^2*((c*cos(f*x + e) + d*sin(f*x +
e))/cos(f*x + e))^(2/3))

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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 )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 3.96367, size = 273, normalized size = 0.66 \begin{align*} -\frac{1}{24} \,{\left ({\left (i \, \sqrt{3} + 1\right )} \left (\frac{216 i \, c - 216 \, d}{d^{3} f^{3}}\right )^{\frac{1}{3}} \log \left (f\right ) +{\left (-i \, \sqrt{3} + 1\right )} \left (\frac{216 i \, c - 216 \, d}{d^{3} f^{3}}\right )^{\frac{1}{3}} \log \left (f\right ) +{\left (i \, \sqrt{3} + 1\right )} \left (\frac{-216 i \, c - 216 \, d}{d^{3} f^{3}}\right )^{\frac{1}{3}} \log \left (f\right ) +{\left (-i \, \sqrt{3} + 1\right )} \left (\frac{-216 i \, c - 216 \, d}{d^{3} f^{3}}\right )^{\frac{1}{3}} \log \left (f\right ) - 2 \, \left (\frac{216 i \, c - 216 \, d}{d^{3} f^{3}}\right )^{\frac{1}{3}} \log \left (i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} f +{\left (i \, c - d\right )}^{\frac{1}{3}} f\right ) - 2 \, \left (\frac{-216 i \, c - 216 \, d}{d^{3} f^{3}}\right )^{\frac{1}{3}} \log \left (-i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} f +{\left (-i \, c - d\right )}^{\frac{1}{3}} f\right )\right )} d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*((I*sqrt(3) + 1)*((216*I*c - 216*d)/(d^3*f^3))^(1/3)*log(f) + (-I*sqrt(3) + 1)*((216*I*c - 216*d)/(d^3*f
^3))^(1/3)*log(f) + (I*sqrt(3) + 1)*((-216*I*c - 216*d)/(d^3*f^3))^(1/3)*log(f) + (-I*sqrt(3) + 1)*((-216*I*c
- 216*d)/(d^3*f^3))^(1/3)*log(f) - 2*((216*I*c - 216*d)/(d^3*f^3))^(1/3)*log(I*(d*tan(f*x + e) + c)^(1/3)*f +
(I*c - d)^(1/3)*f) - 2*((-216*I*c - 216*d)/(d^3*f^3))^(1/3)*log(-I*(d*tan(f*x + e) + c)^(1/3)*f + (-I*c - d)^(
1/3)*f))*d