\(\int \frac {d-c \tan (e+f x)}{(c+d \tan (e+f x))^{2/3}} \, dx\) [478]

Optimal result

Integrand size = 26, antiderivative size = 299 \[ \int \frac {d-c \tan (e+f x)}{(c+d \tan (e+f x))^{2/3}} \, dx=-\frac {1}{4} i \sqrt [3]{c-i d} x+\frac {1}{4} i \sqrt [3]{c+i d} x+\frac {\sqrt {3} \sqrt [3]{c-i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )}{2 f}+\frac {\sqrt {3} \sqrt [3]{c+i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt {3}}\right )}{2 f}-\frac {\sqrt [3]{c-i d} \log (\cos (e+f x))}{4 f}-\frac {\sqrt [3]{c+i d} \log (\cos (e+f x))}{4 f}-\frac {3 \sqrt [3]{c-i d} \log \left (\sqrt [3]{c-i d}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 f}-\frac {3 \sqrt [3]{c+i d} \log \left (\sqrt [3]{c+i d}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 f} \] 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)*arc 
tan(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
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.10 \[ \int \frac {d-c \tan (e+f x)}{(c+d \tan (e+f x))^{2/3}} \, dx=\frac {2 \sqrt {3} \sqrt [3]{c-i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )+2 \sqrt {3} \sqrt [3]{c+i d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt {3}}\right )-2 \sqrt [3]{c-i d} \log \left (\sqrt [3]{c-i d}-\sqrt [3]{c+d \tan (e+f x)}\right )-2 \sqrt [3]{c+i d} \log \left (\sqrt [3]{c+i d}-\sqrt [3]{c+d \tan (e+f x)}\right )+\sqrt [3]{c-i d} \log \left ((c-i d)^{2/3}+\sqrt [3]{c-i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )+\sqrt [3]{c+i d} \log \left ((c+i d)^{2/3}+\sqrt [3]{c+i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )}{4 f} \] Input:

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

Output:

(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*Ta 
n[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)])/(4*f)
 

Rubi [A] (warning: unable to verify)

Time = 0.60 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.79, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

((I/2)*(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*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
 

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 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 = 0.85 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.24

method	result	size
derivativedivides	\(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\left (\textit {\_R}^{3} c -c^{2}-d^{2}\right ) \ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} c}}{2 f}\)	\(72\)
default	\(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\left (\textit {\_R}^{3} c -c^{2}-d^{2}\right ) \ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} c}}{2 f}\)	\(72\)
parts	\(\frac {d^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{2} \left (\textit {\_R}^{3}-c \right )}\right )}{2 f}-\frac {c \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{2 f}\)	\(107\)

Input:

int((d-c*tan(f*x+e))/(c+d*tan(f*x+e))^(2/3),x,method=_RETURNVERBOSE)
 

Output:

-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))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 443 vs. \(2 (217) = 434\).

Time = 0.09 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.48 \[ \int \frac {d-c \tan (e+f x)}{(c+d \tan (e+f x))^{2/3}} \, dx=-\frac {1}{4} \, {\left (\sqrt {-3} + 1\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 ) + \frac {1}{4} \, {\left (\sqrt {-3} - 1\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 ) - \frac {1}{4} \, {\left (\sqrt {-3} + 1\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 ) + \frac {1}{4} \, {\left (\sqrt {-3} - 1\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 ) + \frac {1}{2} \, \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 ) + \frac {1}{2} \, \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 ) \] Input:

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

Output:

-1/4*(sqrt(-3) + 1)*(-(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)) + 1/4*(sqrt(-3) - 1)*(-(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)) - 1/4*(sqrt(-3) + 1)*((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)) + 1/4*(sqrt(-3) - 1)*((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)) + 1/2*(-(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)) + 1/2*((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))
 

Sympy [F]

\[ \int \frac {d-c \tan (e+f x)}{(c+d \tan (e+f x))^{2/3}} \, dx=- \int \left (- \frac {d}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {2}{3}}}\right )\, dx - \int \frac {c \tan {\left (e + f x \right )}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {2}{3}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {d-c \tan (e+f x)}{(c+d \tan (e+f x))^{2/3}} \, dx=\int { -\frac {c \tan \left (f x + e\right ) - d}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {2}{3}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {d-c \tan (e+f x)}{(c+d \tan (e+f x))^{2/3}} \, dx=\int { -\frac {c \tan \left (f x + e\right ) - d}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {2}{3}}} \,d x } \] Input:

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

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 17.80 (sec) , antiderivative size = 4308, normalized size of antiderivative = 14.41 \[ \int \frac {d-c \tan (e+f x)}{(c+d \tan (e+f x))^{2/3}} \, dx=\text {Too large to display} \] Input:

int((d - c*tan(e + f*x))/(c + d*tan(e + f*x))^(2/3),x)
 

Output:

log(((((1944*c*d^4*(c^2 + d^2)*((8*(-d^6*f^6*(c^2 - d^2)^2)^(1/2) + 16*c*d 
^4*f^3)/(f^6*(c^2 + d^2)^2))^(1/3) + (7776*c*d^6*(c + d*tan(e + f*x))^(1/3 
))/f)*((8*(-d^6*f^6*(c^2 - d^2)^2)^(1/2) + 16*c*d^4*f^3)/(f^6*(c^2 + d^2)^ 
2))^(2/3))/16 - (972*d^8)/f^3)*((8*(-d^6*f^6*(c^2 - d^2)^2)^(1/2) + 16*c*d 
^4*f^3)/(f^6*(c^2 + d^2)^2))^(1/3))/4 - (486*d^8*(c + d*tan(e + f*x))^(1/3 
))/f^4)*(((256*c^2*d^8*f^6 - d^6*(64*c^4*f^6 + 64*d^4*f^6 + 128*c^2*d^2*f^ 
6))^(1/2) + 16*c*d^4*f^3)/(64*(c^4*f^6 + d^4*f^6 + 2*c^2*d^2*f^6)))^(1/3) 
+ log(((((1944*c*d^4*(c^2 + d^2)*(-(8*(-d^6*f^6*(c^2 - d^2)^2)^(1/2) - 16* 
c*d^4*f^3)/(f^6*(c^2 + d^2)^2))^(1/3) + (7776*c*d^6*(c + d*tan(e + f*x))^( 
1/3))/f)*(-(8*(-d^6*f^6*(c^2 - d^2)^2)^(1/2) - 16*c*d^4*f^3)/(f^6*(c^2 + d 
^2)^2))^(2/3))/16 - (972*d^8)/f^3)*(-(8*(-d^6*f^6*(c^2 - d^2)^2)^(1/2) - 1 
6*c*d^4*f^3)/(f^6*(c^2 + d^2)^2))^(1/3))/4 - (486*d^8*(c + d*tan(e + f*x)) 
^(1/3))/f^4)*(-((256*c^2*d^8*f^6 - d^6*(64*c^4*f^6 + 64*d^4*f^6 + 128*c^2* 
d^2*f^6))^(1/2) - 16*c*d^4*f^3)/(64*(c^4*f^6 + d^4*f^6 + 2*c^2*d^2*f^6)))^ 
(1/3) + log(- (486*c^4*d^4*(c + d*tan(e + f*x))^(1/3))/f^4 - (((16*(-c^8*d 
^2*f^6)^(1/2) - 8*c^5*f^3 + 8*c^3*d^2*f^3)/(f^6*(c^2 + d^2)^2))^(1/3)*(194 
4*c*d^4*(-c^8*d^2*f^6)^(1/2) + 1944*c^4*d^6*f^3 + 243*c^5*d^4*f^5*(c + d*t 
an(e + f*x))^(1/3)*((16*(-c^8*d^2*f^6)^(1/2) - 8*c^5*f^3 + 8*c^3*d^2*f^3)/ 
(f^6*(c^2 + d^2)^2))^(2/3) - 243*c*d^8*f^5*(c + d*tan(e + f*x))^(1/3)*((16 
*(-c^8*d^2*f^6)^(1/2) - 8*c^5*f^3 + 8*c^3*d^2*f^3)/(f^6*(c^2 + d^2)^2))...
 

Reduce [F]

\[ \int \frac {d-c \tan (e+f x)}{(c+d \tan (e+f x))^{2/3}} \, dx=-\left (\int \frac {\tan \left (f x +e \right )}{\left (d \tan \left (f x +e \right )+c \right )^{\frac {2}{3}}}d x \right ) c +\left (\int \frac {1}{\left (d \tan \left (f x +e \right )+c \right )^{\frac {2}{3}}}d x \right ) d \] Input:

int((d-c*tan(f*x+e))/(c+d*tan(f*x+e))^(2/3),x)
 

Output:

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