\(\int \frac {1}{\sqrt {-b+a^2 x^2} \sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}} \, dx\) [2412]

Optimal result

Integrand size = 45, antiderivative size = 193 \[ \int \frac {1}{\sqrt {-b+a^2 x^2} \sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}} \, dx=\frac {4 \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}}{\sqrt {3} \sqrt [3]{c}}\right )}{a \sqrt [3]{c}}+\frac {4 \log \left (-\sqrt [3]{c}+\sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}\right )}{a \sqrt [3]{c}}-\frac {2 \log \left (c^{2/3}+\sqrt [3]{c} \sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}+\left (c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )^{2/3}\right )}{a \sqrt [3]{c}} \] Output:

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

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\sqrt {-b+a^2 x^2} \sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}} \, dx=\frac {4 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}}{\sqrt [3]{c}}}{\sqrt {3}}\right )+4 \log \left (-\sqrt [3]{c}+\sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}\right )-2 \log \left (c^{2/3}+\sqrt [3]{c} \sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}+\left (c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )^{2/3}\right )}{a \sqrt [3]{c}} \] Input:

Integrate[1/(Sqrt[-b + a^2*x^2]*(c + (a*x + Sqrt[-b + a^2*x^2])^(1/4))^(1/ 
3)),x]
 

Output:

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

Rubi [F]

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.

\(\displaystyle \int \frac {1}{\sqrt {a^2 x^2-b} \sqrt [3]{\sqrt [4]{\sqrt {a^2 x^2-b}+a x}+c}} \, dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \frac {1}{\sqrt {a^2 x^2-b} \sqrt [3]{\sqrt [4]{\sqrt {a^2 x^2-b}+a x}+c}}dx\)

Input:

Int[1/(Sqrt[-b + a^2*x^2]*(c + (a*x + Sqrt[-b + a^2*x^2])^(1/4))^(1/3)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {\frac {4 \ln \left (-c^{\frac {1}{3}}+{\left (c +\left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{4}}\right )}^{\frac {1}{3}}\right )}{c^{\frac {1}{3}}}-\frac {2 \ln \left (c^{\frac {2}{3}}+c^{\frac {1}{3}} {\left (c +\left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{4}}\right )}^{\frac {1}{3}}+{\left (c +\left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{4}}\right )}^{\frac {2}{3}}\right )}{c^{\frac {1}{3}}}+\frac {4 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 {\left (c +\left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{4}}\right )}^{\frac {1}{3}}}{c^{\frac {1}{3}}}+1\right )}{3}\right )}{c^{\frac {1}{3}}}}{a}\)	\(144\)
default	\(\frac {\frac {4 \ln \left (-c^{\frac {1}{3}}+{\left (c +\left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{4}}\right )}^{\frac {1}{3}}\right )}{c^{\frac {1}{3}}}-\frac {2 \ln \left (c^{\frac {2}{3}}+c^{\frac {1}{3}} {\left (c +\left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{4}}\right )}^{\frac {1}{3}}+{\left (c +\left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{4}}\right )}^{\frac {2}{3}}\right )}{c^{\frac {1}{3}}}+\frac {4 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 {\left (c +\left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{4}}\right )}^{\frac {1}{3}}}{c^{\frac {1}{3}}}+1\right )}{3}\right )}{c^{\frac {1}{3}}}}{a}\)	\(144\)

Input:

int(1/(a^2*x^2-b)^(1/2)/(c+(a*x+(a^2*x^2-b)^(1/2))^(1/4))^(1/3),x,method=_ 
RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.76 \[ \int \frac {1}{\sqrt {-b+a^2 x^2} \sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}} \, dx =\text {Too large to display} \] Input:

integrate(1/(a^2*x^2-b)^(1/2)/(c+(a*x+(a^2*x^2-b)^(1/2))^(1/4))^(1/3),x, a 
lgorithm="fricas")
 

Output:

[2*(sqrt(3)*c*sqrt(-1/c^(2/3))*log(2*sqrt(3)*(a*c^(2/3)*x - sqrt(a^2*x^2 - 
 b)*c^(2/3))*(a*x + sqrt(a^2*x^2 - b))^(3/4)*(c + (a*x + sqrt(a^2*x^2 - b) 
)^(1/4))^(2/3)*sqrt(-1/c^(2/3)) - (3*a*c^(2/3)*x + sqrt(3)*(a*c*x - sqrt(a 
^2*x^2 - b)*c)*sqrt(-1/c^(2/3)) - 3*sqrt(a^2*x^2 - b)*c^(2/3))*(a*x + sqrt 
(a^2*x^2 - b))^(3/4)*(c + (a*x + sqrt(a^2*x^2 - b))^(1/4))^(1/3) + (3*a*c* 
x - sqrt(3)*(a*c^(4/3)*x - sqrt(a^2*x^2 - b)*c^(4/3))*sqrt(-1/c^(2/3)) - 3 
*sqrt(a^2*x^2 - b)*c)*(a*x + sqrt(a^2*x^2 - b))^(3/4) + 2*b) - c^(2/3)*log 
((c + (a*x + sqrt(a^2*x^2 - b))^(1/4))^(2/3) + (c + (a*x + sqrt(a^2*x^2 - 
b))^(1/4))^(1/3)*c^(1/3) + c^(2/3)) + 2*c^(2/3)*log((c + (a*x + sqrt(a^2*x 
^2 - b))^(1/4))^(1/3) - c^(1/3)))/(a*c), 2*(2*sqrt(3)*c^(2/3)*arctan(1/3*s 
qrt(3) + 2/3*sqrt(3)*(c + (a*x + sqrt(a^2*x^2 - b))^(1/4))^(1/3)/c^(1/3)) 
- c^(2/3)*log((c + (a*x + sqrt(a^2*x^2 - b))^(1/4))^(2/3) + (c + (a*x + sq 
rt(a^2*x^2 - b))^(1/4))^(1/3)*c^(1/3) + c^(2/3)) + 2*c^(2/3)*log((c + (a*x 
 + sqrt(a^2*x^2 - b))^(1/4))^(1/3) - c^(1/3)))/(a*c)]
 

Sympy [F]

\[ \int \frac {1}{\sqrt {-b+a^2 x^2} \sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}} \, dx=\int \frac {1}{\sqrt [3]{c + \sqrt [4]{a x + \sqrt {a^{2} x^{2} - b}}} \sqrt {a^{2} x^{2} - b}}\, dx \] Input:

integrate(1/(a**2*x**2-b)**(1/2)/(c+(a*x+(a**2*x**2-b)**(1/2))**(1/4))**(1 
/3),x)
 

Output:

Integral(1/((c + (a*x + sqrt(a**2*x**2 - b))**(1/4))**(1/3)*sqrt(a**2*x**2 
 - b)), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {-b+a^2 x^2} \sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}} \, dx=\int { \frac {1}{\sqrt {a^{2} x^{2} - b} {\left (c + {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}}\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(1/(a^2*x^2-b)^(1/2)/(c+(a*x+(a^2*x^2-b)^(1/2))^(1/4))^(1/3),x, a 
lgorithm="maxima")
 

Output:

integrate(1/(sqrt(a^2*x^2 - b)*(c + (a*x + sqrt(a^2*x^2 - b))^(1/4))^(1/3) 
), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {-b+a^2 x^2} \sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}} \, dx=\text {Timed out} \] Input:

integrate(1/(a^2*x^2-b)^(1/2)/(c+(a*x+(a^2*x^2-b)^(1/2))^(1/4))^(1/3),x, a 
lgorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {-b+a^2 x^2} \sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}} \, dx=\int \frac {1}{{\left (c+{\left (a\,x+\sqrt {a^2\,x^2-b}\right )}^{1/4}\right )}^{1/3}\,\sqrt {a^2\,x^2-b}} \,d x \] Input:

int(1/((c + (a*x + (a^2*x^2 - b)^(1/2))^(1/4))^(1/3)*(a^2*x^2 - b)^(1/2)), 
x)
                                                                                    
                                                                                    
 

Output:

int(1/((c + (a*x + (a^2*x^2 - b)^(1/2))^(1/4))^(1/3)*(a^2*x^2 - b)^(1/2)), 
 x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt {-b+a^2 x^2} \sqrt [3]{c+\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}} \, dx=\int \frac {1}{\sqrt {a^{2} x^{2}-b}\, {\left (c +\left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{4}}\right )}^{\frac {1}{3}}}d x \] Input:

int(1/(a^2*x^2-b)^(1/2)/(c+(a*x+(a^2*x^2-b)^(1/2))^(1/4))^(1/3),x)
 

Output:

int(1/(a^2*x^2-b)^(1/2)/(c+(a*x+(a^2*x^2-b)^(1/2))^(1/4))^(1/3),x)