\(\int \frac {1}{\sqrt [3]{a+b \tan (c+d x)}} \, dx\) [692]

Optimal result

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.60 \[ \int \frac {1}{\sqrt [3]{a+b \tan (c+d x)}} \, dx=\frac {i \left (\frac {2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{\sqrt [3]{a-i b}}-\frac {2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{\sqrt [3]{a+i b}}+\frac {\log (i-\tan (c+d x))}{\sqrt [3]{a+i b}}-\frac {\log (i+\tan (c+d x))}{\sqrt [3]{a-i b}}+\frac {3 \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{\sqrt [3]{a-i b}}-\frac {3 \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{\sqrt [3]{a+i b}}\right )}{4 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (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.

\(\displaystyle \int \frac {1}{\sqrt [3]{a+b \tan (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt [3]{a+b \tan (c+d x)}}dx\)

\(\Big \downarrow \) 3966

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

\(\Big \downarrow \) 485

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(b*(-1/2*(Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a - Sqrt[-b^ 
2])^(1/3))/Sqrt[3]])/(Sqrt[-b^2]*(a - Sqrt[-b^2])^(1/3)) + (Sqrt[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*S 
qrt[-b^2]*(a + Sqrt[-b^2])^(1/3)) + Log[Sqrt[-b^2] + b*Tan[c + d*x]]/(4*Sq 
rt[-b^2]*(a - Sqrt[-b^2])^(1/3)) - (3*Log[(a - Sqrt[-b^2])^(1/3) - (a + b* 
Tan[c + d*x])^(1/3)])/(4*Sqrt[-b^2]*(a - Sqrt[-b^2])^(1/3)) + (3*Log[(a + 
Sqrt[-b^2])^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/(4*Sqrt[-b^2]*(a + Sqrt[- 
b^2])^(1/3))))/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 = 0.16 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.14

method	result	size
derivativedivides	\(\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\textit {\_R} \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2 d}\)	\(58\)
default	\(\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\textit {\_R} \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2 d}\)	\(58\)

Input:

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

Output:

1/2/d*b*sum(_R/(_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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1225 vs. \(2 (325) = 650\).

Time = 0.13 (sec) , antiderivative size = 1225, normalized size of antiderivative = 2.95 \[ \int \frac {1}{\sqrt [3]{a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(sqrt(-3) - 1)*(((a^2 + b^2)*d^3*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^ 
6)) + b)/((a^2 + b^2)*d^3))^(1/3)*log(-1/2*(sqrt(-3)*a^2*d^2 + a^2*d^2 + ( 
sqrt(-3)*(a^2*b + b^3)*d^5 + (a^2*b + b^3)*d^5)*sqrt(-a^2/((a^4 + 2*a^2*b^ 
2 + b^4)*d^6)))*(((a^2 + b^2)*d^3*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^6)) 
 + b)/((a^2 + b^2)*d^3))^(2/3) + (b*tan(d*x + c) + a)^(1/3)*a) - 1/4*(sqrt 
(-3) + 1)*(((a^2 + b^2)*d^3*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^6)) + b)/ 
((a^2 + b^2)*d^3))^(1/3)*log(1/2*(sqrt(-3)*a^2*d^2 - a^2*d^2 + (sqrt(-3)*( 
a^2*b + b^3)*d^5 - (a^2*b + b^3)*d^5)*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d 
^6)))*(((a^2 + b^2)*d^3*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^6)) + b)/((a^ 
2 + b^2)*d^3))^(2/3) + (b*tan(d*x + c) + a)^(1/3)*a) + 1/4*(sqrt(-3) - 1)* 
(-((a^2 + b^2)*d^3*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^6)) - b)/((a^2 + b 
^2)*d^3))^(1/3)*log(-1/2*(sqrt(-3)*a^2*d^2 + a^2*d^2 - (sqrt(-3)*(a^2*b + 
b^3)*d^5 + (a^2*b + b^3)*d^5)*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^6)))*(- 
((a^2 + b^2)*d^3*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^6)) - b)/((a^2 + b^2 
)*d^3))^(2/3) + (b*tan(d*x + c) + a)^(1/3)*a) - 1/4*(sqrt(-3) + 1)*(-((a^2 
 + b^2)*d^3*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^6)) - b)/((a^2 + b^2)*d^3 
))^(1/3)*log(1/2*(sqrt(-3)*a^2*d^2 - a^2*d^2 - (sqrt(-3)*(a^2*b + b^3)*d^5 
 - (a^2*b + b^3)*d^5)*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^6)))*(-((a^2 + 
b^2)*d^3*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^6)) - b)/((a^2 + b^2)*d^3))^ 
(2/3) + (b*tan(d*x + c) + a)^(1/3)*a) + 1/2*(((a^2 + b^2)*d^3*sqrt(-a^2...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 4.55 (sec) , antiderivative size = 817, normalized size of antiderivative = 1.97 \[ \int \frac {1}{\sqrt [3]{a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

(log((243*b^5*(a + b*tan(c + d*x))^(1/3))/d^5 + ((1944*b^4*(a^2 - b^2)*(a 
+ b*tan(c + d*x))^(1/3))/d^2 + 1944*a*b^4*(1/(d^3*(a*1i + b)))^(2/3)*(a^2 
+ b^2))/(8*d^3*(a*1i + b)))*(1/(a*d^3*1i + b*d^3))^(1/3))/2 + log((243*b^5 
*(a + b*tan(c + d*x))^(1/3))/d^5 + (((1944*b^4*(a^2 - b^2)*(a + b*tan(c + 
d*x))^(1/3))/d^2 + 1944*a*b^4*(a^2 + b^2)*((a*1i + b)/(d^3*(a^2 + b^2)))^( 
2/3))*(a*1i + b))/(8*d^3*(a^2 + b^2)))*((a*1i + b)/(8*a^2*d^3 + 8*b^2*d^3) 
)^(1/3) + (log((243*b^5*(a + b*tan(c + d*x))^(1/3))/d^5 + (((1944*b^4*(a^2 
 - b^2)*(a + b*tan(c + d*x))^(1/3))/d^2 + 486*a*b^4*(1/(d^3*(a*1i + b)))^( 
2/3)*(3^(1/2)*1i - 1)^2*(a^2 + b^2))*(3^(1/2)*1i - 1)^3)/(64*d^3*(a*1i + b 
)))*(3^(1/2)*1i - 1)*(1/(a*d^3*1i + b*d^3))^(1/3))/4 - (log((243*b^5*(a + 
b*tan(c + d*x))^(1/3))/d^5 - (((1944*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^ 
(1/3))/d^2 + 486*a*b^4*(1/(d^3*(a*1i + b)))^(2/3)*(3^(1/2)*1i + 1)^2*(a^2 
+ b^2))*(3^(1/2)*1i + 1)^3)/(64*d^3*(a*1i + b)))*(3^(1/2)*1i + 1)*(1/(a*d^ 
3*1i + b*d^3))^(1/3))/4 + (log((243*b^5*(a + b*tan(c + d*x))^(1/3))/d^5 + 
(((1944*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d^2 + 1944*a*b^4*(3^(1 
/2)*1i - 1)^2*(a^2 + b^2)*(1i/(8*d^3*(a + b*1i)))^(2/3))*(3^(1/2)*1i - 1)^ 
3*1i)/(64*d^3*(a + b*1i)))*(3^(1/2)*1i - 1)*(1i/(8*(a*d^3 + b*d^3*1i)))^(1 
/3))/2 - (log((243*b^5*(a + b*tan(c + d*x))^(1/3))/d^5 - (((1944*b^4*(a^2 
- b^2)*(a + b*tan(c + d*x))^(1/3))/d^2 + 1944*a*b^4*(3^(1/2)*1i + 1)^2*(a^ 
2 + b^2)*(1i/(8*d^3*(a + b*1i)))^(2/3))*(3^(1/2)*1i + 1)^3*1i)/(64*d^3*...
 

Reduce [F]

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

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

Output:

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