\(\int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx\) [693]

Optimal result

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 313, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx=\frac {i \left (\frac {2 \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{(a-i b)^{2/3}}-\frac {2 \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{(a+i b)^{2/3}}-\frac {2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )+\log \left ((a-i b)^{2/3}+\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}\right )}{(a-i b)^{2/3}}+\frac {2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )+\log \left ((a+i b)^{2/3}+\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}\right )}{(a+i b)^{2/3}}\right )}{4 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (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.

\(\displaystyle \int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+b \tan (c+d x))^{2/3}}dx\)

\(\Big \downarrow \) 3966

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

\(\Big \downarrow \) 485

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(b*((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])^(2/3)) - (Sqrt[3]*ArcTan[(1 
 + (2*(a + b*Tan[c + d*x])^(1/3))/(a + Sqrt[-b^2])^(1/3))/Sqrt[3]])/(2*Sqr 
t[-b^2]*(a + Sqrt[-b^2])^(2/3)) - Log[Sqrt[-b^2] - b*Tan[c + d*x]]/(4*Sqrt 
[-b^2]*(a + Sqrt[-b^2])^(2/3)) + Log[Sqrt[-b^2] + b*Tan[c + d*x]]/(4*Sqrt[ 
-b^2]*(a - Sqrt[-b^2])^(2/3)) - (3*Log[(a - Sqrt[-b^2])^(1/3) - (a + b*Tan 
[c + d*x])^(1/3)])/(4*Sqrt[-b^2]*(a - Sqrt[-b^2])^(2/3)) + (3*Log[(a + Sqr 
t[-b^2])^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/(4*Sqrt[-b^2]*(a + Sqrt[-b^2 
])^(2/3))))/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 = 0.14 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.14

method	result	size
derivativedivides	\(\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2 d}\)	\(57\)
default	\(\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2 d}\)	\(57\)

Input:

int(1/(a+b*tan(d*x+c))^(2/3),x,method=_RETURNVERBOSE)
 

Output:

1/2/d*b*sum(1/(_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. 2139 vs. \(2 (325) = 650\).

Time = 0.14 (sec) , antiderivative size = 2139, normalized size of antiderivative = 5.15 \[ \int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*tan(d*x+c))^(2/3),x, algorithm="fricas")
 

Output:

-1/4*(sqrt(-3) + 1)*(((a^4 + 2*a^2*b^2 + b^4)*d^3*sqrt(-(a^4 - 2*a^2*b^2 + 
 b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^6)) + 2*a*b)/((a^ 
4 + 2*a^2*b^2 + b^4)*d^3))^(1/3)*log(-(a^2 - b^2)*(b*tan(d*x + c) + a)^(1/ 
3) + 1/2*(sqrt(-3)*(a^2*b - b^3)*d + (a^2*b - b^3)*d - (sqrt(-3)*(a^5 + 2* 
a^3*b^2 + a*b^4)*d^4 + (a^5 + 2*a^3*b^2 + a*b^4)*d^4)*sqrt(-(a^4 - 2*a^2*b 
^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^6)))*(((a^4 + 
 2*a^2*b^2 + b^4)*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6* 
a^4*b^4 + 4*a^2*b^6 + b^8)*d^6)) + 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^3))^( 
1/3)) + 1/4*(sqrt(-3) - 1)*(((a^4 + 2*a^2*b^2 + b^4)*d^3*sqrt(-(a^4 - 2*a^ 
2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^6)) + 2*a* 
b)/((a^4 + 2*a^2*b^2 + b^4)*d^3))^(1/3)*log(-(a^2 - b^2)*(b*tan(d*x + c) + 
 a)^(1/3) - 1/2*(sqrt(-3)*(a^2*b - b^3)*d - (a^2*b - b^3)*d - (sqrt(-3)*(a 
^5 + 2*a^3*b^2 + a*b^4)*d^4 - (a^5 + 2*a^3*b^2 + a*b^4)*d^4)*sqrt(-(a^4 - 
2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^6)))*( 
((a^4 + 2*a^2*b^2 + b^4)*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b 
^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^6)) + 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)* 
d^3))^(1/3)) - 1/4*(sqrt(-3) + 1)*(-((a^4 + 2*a^2*b^2 + b^4)*d^3*sqrt(-(a^ 
4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^6) 
) - 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^3))^(1/3)*log(-(a^2 - b^2)*(b*tan(d* 
x + c) + a)^(1/3) + 1/2*(sqrt(-3)*(a^2*b - b^3)*d + (a^2*b - b^3)*d + (...
 

Sympy [F]

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

integrate(1/(a+b*tan(d*x+c))**(2/3),x)
 

Output:

Integral((a + b*tan(c + d*x))**(-2/3), x)
 

Maxima [F]

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

integrate(1/(a+b*tan(d*x+c))^(2/3),x, algorithm="maxima")
 

Output:

integrate((b*tan(d*x + c) + a)^(-2/3), x)
 

Giac [F]

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

integrate(1/(a+b*tan(d*x+c))^(2/3),x, algorithm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 5.59 (sec) , antiderivative size = 1048, normalized size of antiderivative = 2.53 \[ \int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx=\text {Too large to display} \] Input:

int(1/(a + b*tan(c + d*x))^(2/3),x)
 

Output:

(log(((-1i/(d^3*(a*1i - b)^2))^(4/3)*(a^2*b^5*d*486i - b^7*d*486i + 486*a^ 
3*b^4*d - 486*a*b^6*d + (972*a*b^5*(a + b*tan(c + d*x))^(1/3))/(-1i/(d^3*( 
a*1i - b)^2))^(1/3)))/d - (486*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*(1/(b^ 
2*d^3*1i - a^2*d^3*1i + 2*a*b*d^3))^(1/3))/2 + log((((7776*a*b^5*(a + b*ta 
n(c + d*x))^(1/3))/d + 7776*a*b^4*(a^2 + b^2)*(1i/(8*d^3*(a*1i + b)^2))^(1 
/3))*(1i/(8*d^3*(a*1i + b)^2))^(2/3) - (972*b^5)/d^3)*(1i/(8*d^3*(a*1i + b 
)^2))^(1/3) - (486*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*(1i/(8*(b^2*d^3 - 
a^2*d^3 + a*b*d^3*2i)))^(1/3) + (log((486*b^4*(a + b*tan(c + d*x))^(1/3))/ 
d^4 + ((3^(1/2)*1i - 1)*((972*b^5)/d^3 - ((3^(1/2)*1i - 1)^2*((7776*a*b^5* 
(a + b*tan(c + d*x))^(1/3))/d + 1944*a*b^4*(3^(1/2)*1i - 1)*(a^2 + b^2)*(- 
1i/(d^3*(a*1i - b)^2))^(1/3))*(-1i/(d^3*(a*1i - b)^2))^(2/3))/16)*(-1i/(d^ 
3*(a*1i - b)^2))^(1/3))/4)*(3^(1/2)*1i - 1)*(1/(b^2*d^3*1i - a^2*d^3*1i + 
2*a*b*d^3))^(1/3))/4 - (log((486*b^4*(a + b*tan(c + d*x))^(1/3))/d^4 - ((3 
^(1/2)*1i + 1)*((972*b^5)/d^3 - ((3^(1/2)*1i + 1)^2*((7776*a*b^5*(a + b*ta 
n(c + d*x))^(1/3))/d - 1944*a*b^4*(3^(1/2)*1i + 1)*(a^2 + b^2)*(-1i/(d^3*( 
a*1i - b)^2))^(1/3))*(-1i/(d^3*(a*1i - b)^2))^(2/3))/16)*(-1i/(d^3*(a*1i - 
 b)^2))^(1/3))/4)*(3^(1/2)*1i + 1)*(1/(b^2*d^3*1i - a^2*d^3*1i + 2*a*b*d^3 
))^(1/3))/4 + (log((486*b^4*(a + b*tan(c + d*x))^(1/3))/d^4 + ((3^(1/2)*1i 
 - 1)*((972*b^5)/d^3 - ((3^(1/2)*1i - 1)^2*((7776*a*b^5*(a + b*tan(c + d*x 
))^(1/3))/d + 3888*a*b^4*(3^(1/2)*1i - 1)*(a^2 + b^2)*(1i/(8*d^3*(a*1i ...
 

Reduce [F]

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

int(1/(a+b*tan(d*x+c))^(2/3),x)
 

Output:

int(1/(tan(c + d*x)*b + a)**(2/3),x)