\(\int \sqrt [3]{a+b \tan (c+d x)} \, dx\) [691]

Optimal result

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.71 \[ \int \sqrt [3]{a+b \tan (c+d x)} \, dx=\frac {-i \sqrt [3]{a-i b} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )+\log \left ((a-i b)^{2/3}+\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}\right )\right )+i \sqrt [3]{a+i b} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )+\log \left ((a+i b)^{2/3}+\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}\right )\right )}{4 d} \] Input:

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

Output:

((-I)*(a - I*b)^(1/3)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3) 
)/(a - I*b)^(1/3))/Sqrt[3]] - 2*Log[(a - I*b)^(1/3) - (a + b*Tan[c + d*x]) 
^(1/3)] + Log[(a - I*b)^(2/3) + (a - I*b)^(1/3)*(a + b*Tan[c + d*x])^(1/3) 
 + (a + b*Tan[c + d*x])^(2/3)]) + I*(a + I*b)^(1/3)*(2*Sqrt[3]*ArcTan[(1 + 
 (2*(a + b*Tan[c + d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]] - 2*Log[(a + I*b 
)^(1/3) - (a + b*Tan[c + d*x])^(1/3)] + Log[(a + I*b)^(2/3) + (a + I*b)^(1 
/3)*(a + b*Tan[c + d*x])^(1/3) + (a + b*Tan[c + d*x])^(2/3)]))/(4*d)
 

Rubi [A] (verified)

Time = 0.51 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3966

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

\(\Big \downarrow \) 485

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 = 0.42 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.14

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

Input:

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

Output:

1/2/d*b*sum(_R^3/(_R^5-_R^2*a)*ln((a+b*tan(d*x+c))^(1/3)-_R),_R=RootOf(_Z^ 
6-2*_Z^3*a+a^2+b^2))
 

Fricas [A] (verification not implemented)

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

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

Output:

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

Sympy [F]

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

integrate((a+b*tan(d*x+c))**(1/3),x)
 

Output:

Integral((a + b*tan(c + d*x))**(1/3), x)
 

Maxima [F]

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

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

Output:

integrate((b*tan(d*x + c) + a)^(1/3), x)
 

Giac [F]

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

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

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 4.32 (sec) , antiderivative size = 863, normalized size of antiderivative = 2.08 \[ \int \sqrt [3]{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

int((a + b*tan(c + d*x))^(1/3),x)
 

Output:

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

Reduce [F]

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

int((a+b*tan(d*x+c))^(1/3),x)
 

Output:

int((tan(c + d*x)*b + a)**(1/3),x)