\(\int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx\) [684]

Optimal result

Integrand size = 21, antiderivative size = 318 \[ \int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\frac {1}{4} i \sqrt [3]{c-i d} x-\frac {1}{4} i \sqrt [3]{c+i d} x-\frac {\sqrt {3} \sqrt [3]{c-i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )}{2 f}-\frac {\sqrt {3} \sqrt [3]{c+i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt {3}}\right )}{2 f}+\frac {\sqrt [3]{c-i d} \log (\cos (e+f x))}{4 f}+\frac {\sqrt [3]{c+i d} \log (\cos (e+f x))}{4 f}+\frac {3 \sqrt [3]{c-i d} \log \left (\sqrt [3]{c-i d}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 f}+\frac {3 \sqrt [3]{c+i d} \log \left (\sqrt [3]{c+i d}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 f}+\frac {3 \sqrt [3]{c+d \tan (e+f x)}}{f} \]

[Out]

1/4*I*(c-I*d)^(1/3)*x-1/4*I*(c+I*d)^(1/3)*x+1/4*(c-I*d)^(1/3)*ln(cos(f*x+e))/f+1/4*(c+I*d)^(1/3)*ln(cos(f*x+e)
)/f+3/4*(c-I*d)^(1/3)*ln((c-I*d)^(1/3)-(c+d*tan(f*x+e))^(1/3))/f+3/4*(c+I*d)^(1/3)*ln((c+I*d)^(1/3)-(c+d*tan(f
*x+e))^(1/3))/f-1/2*(c-I*d)^(1/3)*arctan(1/3*(1+2*(c+d*tan(f*x+e))^(1/3)/(c-I*d)^(1/3))*3^(1/2))*3^(1/2)/f-1/2
*(c+I*d)^(1/3)*arctan(1/3*(1+2*(c+d*tan(f*x+e))^(1/3)/(c+I*d)^(1/3))*3^(1/2))*3^(1/2)/f+3*(c+d*tan(f*x+e))^(1/
3)/f

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3609, 3620, 3618, 59, 631, 210, 31} \[ \int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=-\frac {\sqrt {3} \sqrt [3]{c-i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )}{2 f}-\frac {\sqrt {3} \sqrt [3]{c+i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt {3}}\right )}{2 f}+\frac {3 \sqrt [3]{c+d \tan (e+f x)}}{f}+\frac {3 \sqrt [3]{c-i d} \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c-i d}\right )}{4 f}+\frac {3 \sqrt [3]{c+i d} \log \left (-\sqrt [3]{c+d \tan (e+f x)}+\sqrt [3]{c+i d}\right )}{4 f}+\frac {\sqrt [3]{c-i d} \log (\cos (e+f x))}{4 f}+\frac {\sqrt [3]{c+i d} \log (\cos (e+f x))}{4 f}+\frac {1}{4} i x \sqrt [3]{c-i d}-\frac {1}{4} i x \sqrt [3]{c+i d} \]

[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 31

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

Rule 59

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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

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 3609

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 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^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 3620

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]

Rubi steps \begin{align*} \text {integral}& = \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) \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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} \arctan \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} \arctan \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] (verified)

Time = 0.26 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.09 \[ \int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=-\frac {2 \sqrt {3} \sqrt [3]{c-i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )+2 \sqrt {3} \sqrt [3]{c+i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt {3}}\right )-2 \sqrt [3]{c-i d} \log \left (\sqrt [3]{c-i d}-\sqrt [3]{c+d \tan (e+f x)}\right )-2 \sqrt [3]{c+i d} \log \left (\sqrt [3]{c+i d}-\sqrt [3]{c+d \tan (e+f x)}\right )+\sqrt [3]{c-i d} \log \left ((c-i d)^{2/3}+\sqrt [3]{c-i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )+\sqrt [3]{c+i d} \log \left ((c+i d)^{2/3}+\sqrt [3]{c+i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )-12 \sqrt [3]{c+d \tan (e+f x)}}{4 f} \]

[In]

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

[Out]

-1/4*(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)*Lo
g[(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))/f

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.51 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.28

method	result	size
derivativedivides	\(\frac {3 \left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\left (\textit {\_R}^{3} c -c^{2}-d^{2}\right ) \ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} c}\right )}{2}}{f}\)	\(88\)
default	\(\frac {3 \left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\left (\textit {\_R}^{3} c -c^{2}-d^{2}\right ) \ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} c}\right )}{2}}{f}\)	\(88\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 476 vs. \(2 (234) = 468\).

Time = 0.29 (sec) , antiderivative size = 476, normalized size of antiderivative = 1.50 \[ \int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=-\frac {{\left (\sqrt {-3} f + f\right )} \left (\frac {f^{3} \sqrt {-\frac {d^{2}}{f^{6}}} + c}{f^{3}}\right )^{\frac {1}{3}} \log \left (\frac {1}{2} \, {\left (\sqrt {-3} f + f\right )} \left (\frac {f^{3} \sqrt {-\frac {d^{2}}{f^{6}}} + c}{f^{3}}\right )^{\frac {1}{3}} + {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}}\right ) - {\left (\sqrt {-3} f - f\right )} \left (\frac {f^{3} \sqrt {-\frac {d^{2}}{f^{6}}} + c}{f^{3}}\right )^{\frac {1}{3}} \log \left (-\frac {1}{2} \, {\left (\sqrt {-3} f - f\right )} \left (\frac {f^{3} \sqrt {-\frac {d^{2}}{f^{6}}} + c}{f^{3}}\right )^{\frac {1}{3}} + {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}}\right ) - 2 \, f \left (\frac {f^{3} \sqrt {-\frac {d^{2}}{f^{6}}} + c}{f^{3}}\right )^{\frac {1}{3}} \log \left (-f \left (\frac {f^{3} \sqrt {-\frac {d^{2}}{f^{6}}} + c}{f^{3}}\right )^{\frac {1}{3}} + {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}}\right ) + {\left (\sqrt {-3} f + f\right )} \left (-\frac {f^{3} \sqrt {-\frac {d^{2}}{f^{6}}} - c}{f^{3}}\right )^{\frac {1}{3}} \log \left (\frac {1}{2} \, {\left (\sqrt {-3} f + f\right )} \left (-\frac {f^{3} \sqrt {-\frac {d^{2}}{f^{6}}} - c}{f^{3}}\right )^{\frac {1}{3}} + {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}}\right ) - {\left (\sqrt {-3} f - f\right )} \left (-\frac {f^{3} \sqrt {-\frac {d^{2}}{f^{6}}} - c}{f^{3}}\right )^{\frac {1}{3}} \log \left (-\frac {1}{2} \, {\left (\sqrt {-3} f - f\right )} \left (-\frac {f^{3} \sqrt {-\frac {d^{2}}{f^{6}}} - c}{f^{3}}\right )^{\frac {1}{3}} + {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}}\right ) - 2 \, f \left (-\frac {f^{3} \sqrt {-\frac {d^{2}}{f^{6}}} - c}{f^{3}}\right )^{\frac {1}{3}} \log \left (-f \left (-\frac {f^{3} \sqrt {-\frac {d^{2}}{f^{6}}} - c}{f^{3}}\right )^{\frac {1}{3}} + {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}}\right ) - 12 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}}}{4 \, f} \]

[In]

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

[Out]

-1/4*((sqrt(-3)*f + f)*((f^3*sqrt(-d^2/f^6) + c)/f^3)^(1/3)*log(1/2*(sqrt(-3)*f + f)*((f^3*sqrt(-d^2/f^6) + c)
/f^3)^(1/3) + (d*tan(f*x + e) + c)^(1/3)) - (sqrt(-3)*f - f)*((f^3*sqrt(-d^2/f^6) + c)/f^3)^(1/3)*log(-1/2*(sq
rt(-3)*f - f)*((f^3*sqrt(-d^2/f^6) + c)/f^3)^(1/3) + (d*tan(f*x + e) + c)^(1/3)) - 2*f*((f^3*sqrt(-d^2/f^6) +
c)/f^3)^(1/3)*log(-f*((f^3*sqrt(-d^2/f^6) + c)/f^3)^(1/3) + (d*tan(f*x + e) + c)^(1/3)) + (sqrt(-3)*f + f)*(-(
f^3*sqrt(-d^2/f^6) - c)/f^3)^(1/3)*log(1/2*(sqrt(-3)*f + f)*(-(f^3*sqrt(-d^2/f^6) - c)/f^3)^(1/3) + (d*tan(f*x
+ e) + c)^(1/3)) - (sqrt(-3)*f - f)*(-(f^3*sqrt(-d^2/f^6) - c)/f^3)^(1/3)*log(-1/2*(sqrt(-3)*f - f)*(-(f^3*sq
rt(-d^2/f^6) - c)/f^3)^(1/3) + (d*tan(f*x + e) + c)^(1/3)) - 2*f*(-(f^3*sqrt(-d^2/f^6) - c)/f^3)^(1/3)*log(-f*
(-(f^3*sqrt(-d^2/f^6) - c)/f^3)^(1/3) + (d*tan(f*x + e) + c)^(1/3)) - 12*(d*tan(f*x + e) + c)^(1/3))/f

Sympy [F]

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

[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)

Maxima [F]

\[ \int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\int { {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}} \tan \left (f x + e\right ) \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 2.60 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.05 \[ \int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\frac {3 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}}}{f} \]

[In]

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

[Out]

3*(d*tan(f*x + e) + c)^(1/3)/f

Mupad [B] (verification not implemented)

Time = 9.02 (sec) , antiderivative size = 830, normalized size of antiderivative = 2.61 \[ \int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

log((c + d*tan(e + f*x))^(1/3) - f*((c - d*1i)/f^3)^(1/3))*((c - d*1i)/(8*f^3))^(1/3) + log((c + d*tan(e + f*x
))^(1/3) - f*((c + d*1i)/f^3)^(1/3))*((c + d*1i)/(8*f^3))^(1/3) + (3*(c + d*tan(e + f*x))^(1/3))/f + log((486*
(d^8 - c^4*d^4)*(c + d*tan(e + f*x))^(1/3))/f^4 - (((3^(1/2)*1i)/2 - 1/2)*((972*(d^8 - c^4*d^4))/f^3 + (((3^(1
/2)*1i)/2 + 1/2)*((3888*c*d^4*(c^2 + d^2)*(c + d*tan(e + f*x))^(1/3))/f - 3888*c*d^4*((3^(1/2)*1i)/2 - 1/2)*((
c - d*1i)/f^3)^(1/3)*(c^2 + d^2))*((c - d*1i)/f^3)^(2/3))/4)*((c - d*1i)/f^3)^(1/3))/2)*((3^(1/2)*1i)/2 - 1/2)
*((c - d*1i)/(8*f^3))^(1/3) + log((486*(d^8 - c^4*d^4)*(c + d*tan(e + f*x))^(1/3))/f^4 - (((3^(1/2)*1i)/2 - 1/
2)*((972*(d^8 - c^4*d^4))/f^3 + (((3^(1/2)*1i)/2 + 1/2)*((3888*c*d^4*(c^2 + d^2)*(c + d*tan(e + f*x))^(1/3))/f
- 3888*c*d^4*((3^(1/2)*1i)/2 - 1/2)*((c + d*1i)/f^3)^(1/3)*(c^2 + d^2))*((c + d*1i)/f^3)^(2/3))/4)*((c + d*1i
)/f^3)^(1/3))/2)*((3^(1/2)*1i)/2 - 1/2)*((c + d*1i)/(8*f^3))^(1/3) - log((((3^(1/2)*1i)/2 + 1/2)*((972*(d^8 -
c^4*d^4))/f^3 - (((3^(1/2)*1i)/2 - 1/2)*((3888*c*d^4*(c^2 + d^2)*(c + d*tan(e + f*x))^(1/3))/f + 3888*c*d^4*((
3^(1/2)*1i)/2 + 1/2)*((c - d*1i)/f^3)^(1/3)*(c^2 + d^2))*((c - d*1i)/f^3)^(2/3))/4)*((c - d*1i)/f^3)^(1/3))/2
+ (486*(d^8 - c^4*d^4)*(c + d*tan(e + f*x))^(1/3))/f^4)*((3^(1/2)*1i)/2 + 1/2)*((c - d*1i)/(8*f^3))^(1/3) - lo
g((((3^(1/2)*1i)/2 + 1/2)*((972*(d^8 - c^4*d^4))/f^3 - (((3^(1/2)*1i)/2 - 1/2)*((3888*c*d^4*(c^2 + d^2)*(c + d
*tan(e + f*x))^(1/3))/f + 3888*c*d^4*((3^(1/2)*1i)/2 + 1/2)*((c + d*1i)/f^3)^(1/3)*(c^2 + d^2))*((c + d*1i)/f^
3)^(2/3))/4)*((c + d*1i)/f^3)^(1/3))/2 + (486*(d^8 - c^4*d^4)*(c + d*tan(e + f*x))^(1/3))/f^4)*((3^(1/2)*1i)/2
+ 1/2)*((c + d*1i)/(8*f^3))^(1/3)