3.683 \(\int \tan ^2(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=439 \[ -\frac{\sqrt{3} d \sqrt [3]{c-\sqrt{-d^2}} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-\sqrt{-d^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-d^2} f}+\frac{\sqrt{3} d \sqrt [3]{c+\sqrt{-d^2}} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+\sqrt{-d^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-d^2} f}+\frac{3 d \sqrt [3]{c-\sqrt{-d^2}} \log \left (\sqrt [3]{c-\sqrt{-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt{-d^2} f}-\frac{3 d \sqrt [3]{c+\sqrt{-d^2}} \log \left (\sqrt [3]{c+\sqrt{-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt{-d^2} f}+\frac{d \sqrt [3]{c-\sqrt{-d^2}} \log (\cos (e+f x))}{4 \sqrt{-d^2} f}-\frac{d \sqrt [3]{c+\sqrt{-d^2}} \log (\cos (e+f x))}{4 \sqrt{-d^2} f}+\frac{1}{4} x \sqrt [3]{c-\sqrt{-d^2}}+\frac{1}{4} x \sqrt [3]{c+\sqrt{-d^2}}+\frac{3 (c+d \tan (e+f x))^{4/3}}{4 d f} \]

[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) +
(3*(c + d*Tan[e + f*x])^(4/3))/(4*d*f)

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Rubi [A]  time = 0.313167, antiderivative size = 439, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {3543, 3485, 712, 50, 57, 617, 204, 31} \[ -\frac{\sqrt{3} d \sqrt [3]{c-\sqrt{-d^2}} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-\sqrt{-d^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-d^2} f}+\frac{\sqrt{3} d \sqrt [3]{c+\sqrt{-d^2}} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+\sqrt{-d^2}}}+1}{\sqrt{3}}\right )}{2 \sqrt{-d^2} f}+\frac{3 d \sqrt [3]{c-\sqrt{-d^2}} \log \left (\sqrt [3]{c-\sqrt{-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt{-d^2} f}-\frac{3 d \sqrt [3]{c+\sqrt{-d^2}} \log \left (\sqrt [3]{c+\sqrt{-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt{-d^2} f}+\frac{d \sqrt [3]{c-\sqrt{-d^2}} \log (\cos (e+f x))}{4 \sqrt{-d^2} f}-\frac{d \sqrt [3]{c+\sqrt{-d^2}} \log (\cos (e+f x))}{4 \sqrt{-d^2} f}+\frac{1}{4} x \sqrt [3]{c-\sqrt{-d^2}}+\frac{1}{4} x \sqrt [3]{c+\sqrt{-d^2}}+\frac{3 (c+d \tan (e+f x))^{4/3}}{4 d f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^2*(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) +
(3*(c + d*Tan[e + f*x])^(4/3))/(4*d*f)

Rule 3543

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

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 \tan ^2(e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx &=\frac{3 (c+d \tan (e+f x))^{4/3}}{4 d f}-\int \sqrt [3]{c+d \tan (e+f x)} \, dx\\ &=\frac{3 (c+d \tan (e+f x))^{4/3}}{4 d f}-\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{3 (c+d \tan (e+f x))^{4/3}}{4 d 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{3 (c+d \tan (e+f x))^{4/3}}{4 d 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{3 (c+d \tan (e+f x))^{4/3}}{4 d 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{3 (c+d \tan (e+f x))^{4/3}}{4 d 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{3 (c+d \tan (e+f x))^{4/3}}{4 d 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}+\frac{3 (c+d \tan (e+f x))^{4/3}}{4 d f}\\ \end{align*}

Mathematica [C]  time = 0.777114, size = 313, normalized size = 0.71 \[ \frac{\frac{3 (c+d \tan (e+f x))^{4/3}}{d}+i \sqrt [3]{c-i d} \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt{3}}\right )-2 \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c-i d}\right )+\log \left (\sqrt [3]{c-i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}+(c-i d)^{2/3}\right )\right )-i \sqrt [3]{c+i d} \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt{3}}\right )-2 \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c+i d}\right )+\log \left (\sqrt [3]{c+i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}+(c+i d)^{2/3}\right )\right )}{4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^2*(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)]) + (3*(c + d*Tan[e + f*x])^(4/3))/d)/(4*f)

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Maple [C]  time = 0.017, size = 81, normalized size = 0.2 \begin{align*}{\frac{3}{4\,df} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{{\frac{4}{3}}}}-{\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(tan(f*x+e)^2*(c+d*tan(f*x+e))^(1/3),x)

[Out]

3/4*(c+d*tan(f*x+e))^(4/3)/d/f-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}} \tan \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2*(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)^2, x)

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Fricas [B]  time = 2.66595, size = 8288, normalized size = 18.88 \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)^2*(c+d*tan(f*x+e))^(1/3),x, algorithm="fricas")

[Out]

1/4*(2*d*f*((c^2 + d^2)/f^6)^(1/6)*cos(f*x + e)*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)*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)) + 8*d*f*((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))))*co
s(f*x + e)*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)) + 4*(sqrt(3)*d
*f*((c^2 + d^2)/f^6)^(1/6)*cos(f*x + e)*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)) - d*f*((c^2 + d^2)/f^6)^(1/6)*cos(f*x + e)*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*s
qrt(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)) + 4*(sqrt(3)*d*f*((c^2 + d^2)/f^6)^(1/6)*cos(f*x + e)*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)) + d*f*((c^2 + d^2)/f^6)^(1/6)*cos(f*x + e)*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)*s
qrt((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)) - (sqrt(3)*d*f*((c^2 + d^2)/f^6)^(1/6)*cos(f*x + e)*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)) + d*f*((c^2 + d^2)/f^6)^(1/6)*cos(f*x + e)*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)*sqr
t((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)*d*f*((c^2 + d^2)/f^6)^(1/6)*cos(f*x + e)*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)) - d*f*((c^2 + d^2)/f^6)^(1/6)*cos(f*x + e)*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*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)) + 3*(c*cos(f*x + e) + d*s
in(f*x + e))*((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))^(1/3))/(d*f*cos(f*x + e))

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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 ^{2}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**2*(c+d*tan(f*x+e))**(1/3),x)

[Out]

Integral((c + d*tan(e + f*x))**(1/3)*tan(e + f*x)**2, x)

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Giac [C]  time = 2.90627, size = 574, normalized size = 1.31 \begin{align*} -\frac{1}{4} \,{\left (i \, \sqrt{3} + 1\right )} \left (-\frac{i \, c - d}{f^{3}}\right )^{\frac{1}{3}} \log \left ({\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} d^{5} f^{4}{\left (i \, \sqrt{3} + 1\right )} +{\left (c + i \, d\right )}^{\frac{1}{3}} d^{5} f^{4}{\left (i \, \sqrt{3} - 1\right )}\right ) - \frac{1}{4} \,{\left (-i \, \sqrt{3} + 1\right )} \left (-\frac{-i \, c - d}{f^{3}}\right )^{\frac{1}{3}} \log \left ({\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} d^{5} f^{4}{\left (i \, \sqrt{3} - 1\right )} +{\left (-c + i \, d\right )}^{\frac{1}{3}} d^{5} f^{4}{\left (-i \, \sqrt{3} - 1\right )}\right ) - \frac{1}{4} \,{\left (i \, \sqrt{3} + 1\right )} \left (-\frac{-i \, c - d}{f^{3}}\right )^{\frac{1}{3}} \log \left ({\left (-c + i \, d\right )}^{\frac{1}{3}} d^{5} f^{4}{\left (i \, \sqrt{3} - 1\right )} +{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} d^{5} f^{4}{\left (-i \, \sqrt{3} - 1\right )}\right ) - \frac{1}{4} \,{\left (-i \, \sqrt{3} + 1\right )} \left (-\frac{i \, c - d}{f^{3}}\right )^{\frac{1}{3}} \log \left ({\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} d^{5} f^{4}{\left (-i \, \sqrt{3} + 1\right )} +{\left (c + i \, d\right )}^{\frac{1}{3}} d^{5} f^{4}{\left (-i \, \sqrt{3} - 1\right )}\right ) + \frac{1}{2} \, \left (-\frac{-i \, c - d}{f^{3}}\right )^{\frac{1}{3}} \log \left ({\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} d^{5} f^{4} +{\left (-c + i \, d\right )}^{\frac{1}{3}} d^{5} f^{4}\right ) + \frac{1}{2} \, \left (-\frac{i \, c - d}{f^{3}}\right )^{\frac{1}{3}} \log \left (-{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{1}{3}} d^{5} f^{4} +{\left (c + i \, d\right )}^{\frac{1}{3}} d^{5} f^{4}\right ) + \frac{3 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{4}{3}}}{4 \, d f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2*(c+d*tan(f*x+e))^(1/3),x, algorithm="giac")

[Out]

-1/4*(I*sqrt(3) + 1)*(-(I*c - d)/f^3)^(1/3)*log((d*tan(f*x + e) + c)^(1/3)*d^5*f^4*(I*sqrt(3) + 1) + (c + I*d)
^(1/3)*d^5*f^4*(I*sqrt(3) - 1)) - 1/4*(-I*sqrt(3) + 1)*(-(-I*c - d)/f^3)^(1/3)*log((d*tan(f*x + e) + c)^(1/3)*
d^5*f^4*(I*sqrt(3) - 1) + (-c + I*d)^(1/3)*d^5*f^4*(-I*sqrt(3) - 1)) - 1/4*(I*sqrt(3) + 1)*(-(-I*c - d)/f^3)^(
1/3)*log((-c + I*d)^(1/3)*d^5*f^4*(I*sqrt(3) - 1) + (d*tan(f*x + e) + c)^(1/3)*d^5*f^4*(-I*sqrt(3) - 1)) - 1/4
*(-I*sqrt(3) + 1)*(-(I*c - d)/f^3)^(1/3)*log((d*tan(f*x + e) + c)^(1/3)*d^5*f^4*(-I*sqrt(3) + 1) + (c + I*d)^(
1/3)*d^5*f^4*(-I*sqrt(3) - 1)) + 1/2*(-(-I*c - d)/f^3)^(1/3)*log((d*tan(f*x + e) + c)^(1/3)*d^5*f^4 + (-c + I*
d)^(1/3)*d^5*f^4) + 1/2*(-(I*c - d)/f^3)^(1/3)*log(-(d*tan(f*x + e) + c)^(1/3)*d^5*f^4 + (c + I*d)^(1/3)*d^5*f
^4) + 3/4*(d*tan(f*x + e) + c)^(4/3)/(d*f)