\(\int (a+b \tan (c+d x))^{2/3} \, dx\) [690]

Optimal result

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

-1/4*(a-(-b^2)^(1/2))^(2/3)*x-1/4*(a+(-b^2)^(1/2))^(2/3)*x-1/2*3^(1/2)*b*( 
a-(-b^2)^(1/2))^(2/3)*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))^(2/3)*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))^(2/3)*ln(cos(d*x+c))/(-b^2)^(1/2)/d+1/4* 
b*(a+(-b^2)^(1/2))^(2/3)*ln(cos(d*x+c))/(-b^2)^(1/2)/d-3/4*b*(a-(-b^2)^(1/ 
2))^(2/3)*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))^(2/3)*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 = 0.20 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.54 \[ \int (a+b \tan (c+d x))^{2/3} \, dx=\frac {\frac {(i a+b) \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )-\log (i+\tan (c+d x))+3 \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )\right )}{\sqrt [3]{a-i b}}+\frac {(-i a+b) \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )-\log (i-\tan (c+d x))+3 \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )\right )}{\sqrt [3]{a+i b}}}{4 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.57 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3966

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

\(\Big \downarrow \) 485

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(b*(-1/2*(Sqrt[3]*(a - Sqrt[-b^2])^(2/3)*ArcTan[(1 + (2*(a + b*Tan[c + d*x 
])^(1/3))/(a - Sqrt[-b^2])^(1/3))/Sqrt[3]])/Sqrt[-b^2] + (Sqrt[3]*(a + Sqr 
t[-b^2])^(2/3)*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)*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]) - (3*(a - Sqrt[-b^2])^(2/3)*Log[(a - Sq 
rt[-b^2])^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/(4*Sqrt[-b^2]) + (3*(a + Sq 
rt[-b^2])^(2/3)*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 = 0.32 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.14

method	result	size
derivativedivides	\(\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\textit {\_R}^{4} \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2 d}\)	\(60\)
default	\(\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\textit {\_R}^{4} \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2 d}\)	\(60\)

Input:

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

Output:

1/2/d*b*sum(_R^4/(_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. 1011 vs. \(2 (325) = 650\).

Time = 0.12 (sec) , antiderivative size = 1011, normalized size of antiderivative = 2.44 \[ \int (a+b \tan (c+d x))^{2/3} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 5.44 (sec) , antiderivative size = 1229, normalized size of antiderivative = 2.96 \[ \int (a+b \tan (c+d x))^{2/3} \, dx=\text {Too large to display} \] Input:

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

Output:

log((((((1944*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2 + 1944*a*b 
^4*(a^2 + b^2)*(-((a - b*1i)^2*1i)/d^3)^(2/3))*(-((a - b*1i)^2*1i)/d^3)^(1 
/3))/2 + (972*b^5*(3*a^4 - b^4 + 2*a^2*b^2))/d^3)*(-((a - b*1i)^2*1i)/d^3) 
^(2/3))/4 + (486*a*b^5*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5)*(-(2 
*a*b + a^2*1i - b^2*1i)/(8*d^3))^(1/3) + log(((((1944*a*b^4*(a^2 + b^2)*(- 
((a*1i - b)^2*1i)/d^3)^(2/3) + (1944*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x) 
)^(1/3))/d^2)*(-((a*1i - b)^2*1i)/d^3)^(1/3))/2 + (972*b^5*(3*a^4 - b^4 + 
2*a^2*b^2))/d^3)*(-((a*1i - b)^2*1i)/d^3)^(2/3))/4 + (486*a*b^5*(a^2 + b^2 
)^2*(a + b*tan(c + d*x))^(1/3))/d^5)*(-(2*a*b - a^2*1i + b^2*1i)/(8*d^3))^ 
(1/3) - log((((3^(1/2)*1i)/2 - 1/2)*((972*b^5*(3*a^4 - b^4 + 2*a^2*b^2))/d 
^3 - (((3^(1/2)*1i)/2 + 1/2)*((1944*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x)) 
^(1/3))/d^2 + 1944*a*b^4*((3^(1/2)*1i)/2 - 1/2)*(a^2 + b^2)*(-((a - b*1i)^ 
2*1i)/d^3)^(2/3))*(-((a - b*1i)^2*1i)/d^3)^(1/3))/2)*(-((a - b*1i)^2*1i)/d 
^3)^(2/3))/4 + (486*a*b^5*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5)*( 
(3^(1/2)*1i)/2 + 1/2)*(-(2*a*b + a^2*1i - b^2*1i)/(8*d^3))^(1/3) + log((48 
6*a*b^5*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5 - (((3^(1/2)*1i)/2 + 
 1/2)*((972*b^5*(3*a^4 - b^4 + 2*a^2*b^2))/d^3 + (((3^(1/2)*1i)/2 - 1/2)*( 
(1944*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2 - 1944*a*b^4*((3^( 
1/2)*1i)/2 + 1/2)*(a^2 + b^2)*(-((a - b*1i)^2*1i)/d^3)^(2/3))*(-((a - b*1i 
)^2*1i)/d^3)^(1/3))/2)*(-((a - b*1i)^2*1i)/d^3)^(2/3))/4)*((3^(1/2)*1i)...
 

Reduce [F]

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

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

Output:

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