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

Optimal result

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

1/4*I*(c-I*d)^(1/3)*x-1/4*I*(c+I*d)^(1/3)*x-1/2*3^(1/2)*(c-I*d)^(1/3)*arct 
an(1/3*(1+2*(c+d*tan(f*x+e))^(1/3)/(c-I*d)^(1/3))*3^(1/2))/f-1/2*3^(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) 
)/f+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+3*(c+d*tan(f*x+e))^(1/3)/f
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.09 \[ \int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=-\frac {2 \sqrt {3} \sqrt [3]{c-i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )+2 \sqrt {3} \sqrt [3]{c+i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt {3}}\right )-2 \sqrt [3]{c-i d} \log \left (\sqrt [3]{c-i d}-\sqrt [3]{c+d \tan (e+f x)}\right )-2 \sqrt [3]{c+i d} \log \left (\sqrt [3]{c+i d}-\sqrt [3]{c+d \tan (e+f x)}\right )+\sqrt [3]{c-i d} \log \left ((c-i d)^{2/3}+\sqrt [3]{c-i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )+\sqrt [3]{c+i d} \log \left ((c+i d)^{2/3}+\sqrt [3]{c+i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )-12 \sqrt [3]{c+d \tan (e+f x)}}{4 f} \] Input:

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

Output:

-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)*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))/f
 

Rubi [A] (warning: unable to verify)

Time = 0.64 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.80, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 4011, 3042, 4022, 3042, 4020, 25, 69, 16, 1082, 217}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

\(\displaystyle \int \frac {c \tan (e+f x)-d}{(c+d \tan (e+f x))^{2/3}}dx+\frac {3 \sqrt [3]{c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {c \tan (e+f x)-d}{(c+d \tan (e+f x))^{2/3}}dx+\frac {3 \sqrt [3]{c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {1}{2} (-d+i c) \int \frac {1-i \tan (e+f x)}{(c+d \tan (e+f x))^{2/3}}dx-\frac {1}{2} (d+i c) \int \frac {i \tan (e+f x)+1}{(c+d \tan (e+f x))^{2/3}}dx+\frac {3 \sqrt [3]{c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} (-d+i c) \int \frac {1-i \tan (e+f x)}{(c+d \tan (e+f x))^{2/3}}dx-\frac {1}{2} (d+i c) \int \frac {i \tan (e+f x)+1}{(c+d \tan (e+f x))^{2/3}}dx+\frac {3 \sqrt [3]{c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 4020

\(\displaystyle -\frac {i (d+i c) \int -\frac {1}{(1-i \tan (e+f x)) (c+d \tan (e+f x))^{2/3}}d(i \tan (e+f x))}{2 f}-\frac {i (-d+i c) \int -\frac {1}{(i \tan (e+f x)+1) (c+d \tan (e+f x))^{2/3}}d(-i \tan (e+f x))}{2 f}+\frac {3 \sqrt [3]{c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {i (d+i c) \int \frac {1}{(1-i \tan (e+f x)) (c+d \tan (e+f x))^{2/3}}d(i \tan (e+f x))}{2 f}+\frac {i (-d+i c) \int \frac {1}{(i \tan (e+f x)+1) (c+d \tan (e+f x))^{2/3}}d(-i \tan (e+f x))}{2 f}+\frac {3 \sqrt [3]{c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 69

\(\displaystyle -\frac {i (d+i c) \left (-\frac {3 \int \frac {1}{-\tan ^2(e+f x)+(c-i d)^{2/3}+\sqrt [3]{c-i d} \sqrt [3]{c+d \tan (e+f x)}}d\sqrt [3]{c+d \tan (e+f x)}}{2 \sqrt [3]{c-i d}}-\frac {3 \int \frac {1}{\sqrt [3]{c-i d}-i \tan (e+f x)}d\sqrt [3]{c+d \tan (e+f x)}}{2 (c-i d)^{2/3}}-\frac {\log (1-i \tan (e+f x))}{2 (c-i d)^{2/3}}\right )}{2 f}-\frac {i (-d+i c) \left (-\frac {3 \int \frac {1}{-\tan ^2(e+f x)+(c+i d)^{2/3}+\sqrt [3]{c+i d} \sqrt [3]{c+d \tan (e+f x)}}d\sqrt [3]{c+d \tan (e+f x)}}{2 \sqrt [3]{c+i d}}-\frac {3 \int \frac {1}{i \tan (e+f x)+\sqrt [3]{c+i d}}d\sqrt [3]{c+d \tan (e+f x)}}{2 (c+i d)^{2/3}}-\frac {\log (1+i \tan (e+f x))}{2 (c+i d)^{2/3}}\right )}{2 f}+\frac {3 \sqrt [3]{c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 16

\(\displaystyle -\frac {i (d+i c) \left (-\frac {3 \int \frac {1}{-\tan ^2(e+f x)+(c-i d)^{2/3}+\sqrt [3]{c-i d} \sqrt [3]{c+d \tan (e+f x)}}d\sqrt [3]{c+d \tan (e+f x)}}{2 \sqrt [3]{c-i d}}-\frac {\log (1-i \tan (e+f x))}{2 (c-i d)^{2/3}}+\frac {3 \log \left (\sqrt [3]{c-i d}-i \tan (e+f x)\right )}{2 (c-i d)^{2/3}}\right )}{2 f}-\frac {i (-d+i c) \left (-\frac {3 \int \frac {1}{-\tan ^2(e+f x)+(c+i d)^{2/3}+\sqrt [3]{c+i d} \sqrt [3]{c+d \tan (e+f x)}}d\sqrt [3]{c+d \tan (e+f x)}}{2 \sqrt [3]{c+i d}}-\frac {\log (1+i \tan (e+f x))}{2 (c+i d)^{2/3}}+\frac {3 \log \left (\sqrt [3]{c+i d}+i \tan (e+f x)\right )}{2 (c+i d)^{2/3}}\right )}{2 f}+\frac {3 \sqrt [3]{c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 1082

\(\displaystyle -\frac {i (d+i c) \left (\frac {3 \int \frac {1}{\tan ^2(e+f x)-3}d\left (\frac {2 i \tan (e+f x)}{\sqrt [3]{c-i d}}+1\right )}{(c-i d)^{2/3}}-\frac {\log (1-i \tan (e+f x))}{2 (c-i d)^{2/3}}+\frac {3 \log \left (\sqrt [3]{c-i d}-i \tan (e+f x)\right )}{2 (c-i d)^{2/3}}\right )}{2 f}-\frac {i (-d+i c) \left (\frac {3 \int \frac {1}{\tan ^2(e+f x)-3}d\left (1-\frac {2 i \tan (e+f x)}{\sqrt [3]{c+i d}}\right )}{(c+i d)^{2/3}}-\frac {\log (1+i \tan (e+f x))}{2 (c+i d)^{2/3}}+\frac {3 \log \left (\sqrt [3]{c+i d}+i \tan (e+f x)\right )}{2 (c+i d)^{2/3}}\right )}{2 f}+\frac {3 \sqrt [3]{c+d \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {i (d+i c) \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {\tan (e+f x)}{\sqrt {3}}\right )}{(c-i d)^{2/3}}-\frac {\log (1-i \tan (e+f x))}{2 (c-i d)^{2/3}}+\frac {3 \log \left (\sqrt [3]{c-i d}-i \tan (e+f x)\right )}{2 (c-i d)^{2/3}}\right )}{2 f}-\frac {i (-d+i c) \left (\frac {i \sqrt {3} \text {arctanh}\left (\frac {\tan (e+f x)}{\sqrt {3}}\right )}{(c+i d)^{2/3}}-\frac {\log (1+i \tan (e+f x))}{2 (c+i d)^{2/3}}+\frac {3 \log \left (\sqrt [3]{c+i d}+i \tan (e+f x)\right )}{2 (c+i d)^{2/3}}\right )}{2 f}+\frac {3 \sqrt [3]{c+d \tan (e+f x)}}{f}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 16

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

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 69

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), 
 x] + (-Simp[3/(2*b*q)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] - Simp[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 217

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 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4011

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 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[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 4022

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

Maple [C] (verified)

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

Time = 0.16 (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	\(\frac {3 \left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\left (\textit {\_R}^{3} c -c^{2}-d^{2}\right ) \ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} c}\right )}{2}}{f}\)	\(88\)

Input:

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

Output:

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 = 0.11 (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} \] Input:

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

Output:

-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*(sqrt 
(-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*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*t 
an(f*x + e) + c)^(1/3)) - 12*(d*tan(f*x + e) + c)^(1/3))/f
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 1.50 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.05 \[ \int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\frac {3 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}}}{f} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 5.14 (sec) , antiderivative size = 830, normalized size of antiderivative = 2.61 \[ \int \tan (e+f x) \sqrt [3]{c+d \tan (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

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)*((9 
72*(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) - 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)*...
 

Reduce [F]

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

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

Output:

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