\(\int \sqrt [3]{c+d \tan (e+f x)} \, dx\) [685]

Optimal result

Integrand size = 14, antiderivative size = 415 \[ \int \sqrt [3]{c+d \tan (e+f x)} \, dx=-\frac {1}{4} \sqrt [3]{c-\sqrt {-d^2}} x-\frac {1}{4} \sqrt [3]{c+\sqrt {-d^2}} x+\frac {\sqrt {3} d \sqrt [3]{c-\sqrt {-d^2}} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-\sqrt {-d^2}}}}{\sqrt {3}}\right )}{2 \sqrt {-d^2} f}-\frac {\sqrt {3} d \sqrt [3]{c+\sqrt {-d^2}} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+\sqrt {-d^2}}}}{\sqrt {3}}\right )}{2 \sqrt {-d^2} f}-\frac {d \sqrt [3]{c-\sqrt {-d^2}} \log (\cos (e+f x))}{4 \sqrt {-d^2} f}+\frac {d \sqrt [3]{c+\sqrt {-d^2}} \log (\cos (e+f x))}{4 \sqrt {-d^2} f}-\frac {3 d \sqrt [3]{c-\sqrt {-d^2}} \log \left (\sqrt [3]{c-\sqrt {-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt {-d^2} f}+\frac {3 d \sqrt [3]{c+\sqrt {-d^2}} \log \left (\sqrt [3]{c+\sqrt {-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt {-d^2} f} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.71 \[ \int \sqrt [3]{c+d \tan (e+f x)} \, dx=\frac {-i \sqrt [3]{c-i d} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c-i d}-\sqrt [3]{c+d \tan (e+f x)}\right )+\log \left ((c-i d)^{2/3}+\sqrt [3]{c-i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )\right )+i \sqrt [3]{c+i d} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c+i d}-\sqrt [3]{c+d \tan (e+f x)}\right )+\log \left ((c+i d)^{2/3}+\sqrt [3]{c+i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )\right )}{4 f} \] Input:

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

Output:

((-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)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3966, 485, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3966

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

\(\Big \downarrow \) 485

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d \left (\frac {\sqrt {3} \sqrt [3]{c-\sqrt {-d^2}} \arctan \left (\frac {\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-\sqrt {-d^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-d^2}}-\frac {\sqrt {3} \sqrt [3]{c+\sqrt {-d^2}} \arctan \left (\frac {\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+\sqrt {-d^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-d^2}}-\frac {\sqrt [3]{c+\sqrt {-d^2}} \log \left (\sqrt {-d^2}-d \tan (e+f x)\right )}{4 \sqrt {-d^2}}+\frac {\sqrt [3]{c-\sqrt {-d^2}} \log \left (\sqrt {-d^2}+d \tan (e+f x)\right )}{4 \sqrt {-d^2}}-\frac {3 \sqrt [3]{c-\sqrt {-d^2}} \log \left (\sqrt [3]{c-\sqrt {-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt {-d^2}}+\frac {3 \sqrt [3]{c+\sqrt {-d^2}} \log \left (\sqrt [3]{c+\sqrt {-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt {-d^2}}\right )}{f}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 485

Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[Expand 
Integrand[(c + d*x)^n, 1/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d, n}, x] & 
&  !IntegerQ[2*n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3966

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Su 
bst[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]
 

Maple [C] (verified)

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

Time = 0.56 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.14

method	result	size
derivativedivides	\(\frac {d \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\textit {\_R}^{3} \ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} c}\right )}{2 f}\)	\(60\)
default	\(\frac {d \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\textit {\_R}^{3} \ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} c}\right )}{2 f}\)	\(60\)

Input:

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

Output:

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))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 536, normalized size of antiderivative = 1.29 \[ \int \sqrt [3]{c+d \tan (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 4.19 (sec) , antiderivative size = 863, normalized size of antiderivative = 2.08 \[ \int \sqrt [3]{c+d \tan (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

log((c + d*tan(e + f*x))^(1/3) + f*(-(c*1i + d)/f^3)^(1/3)*1i)*(-(c*1i + d 
)/(8*f^3))^(1/3) + log(d*(c + d*tan(e + f*x))^(1/3)*1i - c*(c + d*tan(e + 
f*x))^(1/3) + f^4*((c*1i - d)/f^3)^(4/3) + 2*d*f*((c*1i - d)/f^3)^(1/3))*( 
(c*1i - d)/(8*f^3))^(1/3) - log((((c*1i - d)/f^3)^(1/3)*((3^(1/2)*1i)/2 + 
1/2)*((((c*1i - d)/f^3)^(2/3)*((3^(1/2)*1i)/2 - 1/2)*((3888*d^5*(c^2 + d^2 
)*(c + d*tan(e + f*x))^(1/3))/f - 3888*c*d^4*((c*1i - d)/f^3)^(1/3)*((3^(1 
/2)*1i)/2 + 1/2)*(c^2 + d^2)))/4 + (1944*c*d^5*(c^2 + d^2))/f^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*1i - d)/(8*f^3))^(1/3) + log((486*(d^8 - c^4*d^4)*(c + d*tan(e + f*x))^ 
(1/3))/f^4 - (((c*1i - d)/f^3)^(1/3)*((3^(1/2)*1i)/2 - 1/2)*((((c*1i - d)/ 
f^3)^(2/3)*((3^(1/2)*1i)/2 + 1/2)*((3888*d^5*(c^2 + d^2)*(c + d*tan(e + f* 
x))^(1/3))/f + 3888*c*d^4*((c*1i - d)/f^3)^(1/3)*((3^(1/2)*1i)/2 - 1/2)*(c 
^2 + d^2)))/4 - (1944*c*d^5*(c^2 + d^2))/f^3))/2)*((3^(1/2)*1i)/2 - 1/2)*( 
(c*1i - d)/(8*f^3))^(1/3) - log((((3^(1/2)*1i)/2 + 1/2)*((((3888*d^5*(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*1i + d)/f^3)^(1/3)*(c^2 + d^2))*((3^(1/2)*1i)/2 - 1/2)*(-(c*1i + d)/f^ 
3)^(2/3))/4 + (1944*c*d^5*(c^2 + d^2))/f^3)*(-(c*1i + d)/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*1i + d)/(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)*((((3888*d^5*(c^2 + d^2)*(c + ...
 

Reduce [F]

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

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

Output:

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