\(\int \frac {b+2 a x}{\sqrt [4]{c+b x+a x^2} (5 c+4 b x+4 a x^2)} \, dx\) [1797]

Optimal result

Integrand size = 37, antiderivative size = 122 \[ \int \frac {b+2 a x}{\sqrt [4]{c+b x+a x^2} \left (5 c+4 b x+4 a x^2\right )} \, dx=-\frac {\arctan \left (1-\frac {2 \sqrt [4]{c+b x+a x^2}}{\sqrt [4]{c}}\right )}{2 \sqrt [4]{c}}+\frac {\arctan \left (1+\frac {2 \sqrt [4]{c+b x+a x^2}}{\sqrt [4]{c}}\right )}{2 \sqrt [4]{c}}-\frac {\text {arctanh}\left (\frac {\frac {\sqrt [4]{c}}{2}+\frac {\sqrt {c+b x+a x^2}}{\sqrt [4]{c}}}{\sqrt [4]{c+b x+a x^2}}\right )}{2 \sqrt [4]{c}} \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.84 \[ \int \frac {b+2 a x}{\sqrt [4]{c+b x+a x^2} \left (5 c+4 b x+4 a x^2\right )} \, dx=-\frac {\arctan \left (1-\frac {2 \sqrt [4]{c+x (b+a x)}}{\sqrt [4]{c}}\right )-\arctan \left (1+\frac {2 \sqrt [4]{c+x (b+a x)}}{\sqrt [4]{c}}\right )+\text {arctanh}\left (\frac {\sqrt {c}+2 \sqrt {c+x (b+a x)}}{2 \sqrt [4]{c} \sqrt [4]{c+x (b+a x)}}\right )}{2 \sqrt [4]{c}} \] Input:

Integrate[(b + 2*a*x)/((c + b*x + a*x^2)^(1/4)*(5*c + 4*b*x + 4*a*x^2)),x]
 

Output:

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

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.

\(\displaystyle \int \frac {2 a x+b}{\sqrt [4]{a x^2+b x+c} \left (4 a x^2+4 b x+5 c\right )} \, dx\)

\(\Big \downarrow \) 1375

\(\displaystyle \int \frac {2 a x+b}{\sqrt [4]{a x^2+b x+c} \left (4 a x^2+4 b x+5 c\right )}dx\)

Input:

Int[(b + 2*a*x)/((c + b*x + a*x^2)^(1/4)*(5*c + 4*b*x + 4*a*x^2)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02

method	result	size
pseudoelliptic	\(\frac {\ln \left (\frac {-2 c^{\frac {1}{4}} \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}}+\sqrt {c}+2 \sqrt {a \,x^{2}+b x +c}}{2 c^{\frac {1}{4}} \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}}+\sqrt {c}+2 \sqrt {a \,x^{2}+b x +c}}\right )+2 \arctan \left (1+\frac {2 \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}}}{c^{\frac {1}{4}}}\right )-2 \arctan \left (1-\frac {2 \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}}}{c^{\frac {1}{4}}}\right )}{4 c^{\frac {1}{4}}}\)	\(125\)

Input:

int((2*a*x+b)/(a*x^2+b*x+c)^(1/4)/(4*a*x^2+4*b*x+5*c),x,method=_RETURNVERB 
OSE)
 

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.29 \[ \int \frac {b+2 a x}{\sqrt [4]{c+b x+a x^2} \left (5 c+4 b x+4 a x^2\right )} \, dx=\frac {1}{2} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (-\frac {1}{c}\right )^{\frac {1}{4}} \log \left (2 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} c \left (-\frac {1}{c}\right )^{\frac {3}{4}} + {\left (a x^{2} + b x + c\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} i \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (-\frac {1}{c}\right )^{\frac {1}{4}} \log \left (2 i \, \left (\frac {1}{4}\right )^{\frac {3}{4}} c \left (-\frac {1}{c}\right )^{\frac {3}{4}} + {\left (a x^{2} + b x + c\right )}^{\frac {1}{4}}\right ) + \frac {1}{2} i \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (-\frac {1}{c}\right )^{\frac {1}{4}} \log \left (-2 i \, \left (\frac {1}{4}\right )^{\frac {3}{4}} c \left (-\frac {1}{c}\right )^{\frac {3}{4}} + {\left (a x^{2} + b x + c\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (-\frac {1}{c}\right )^{\frac {1}{4}} \log \left (-2 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} c \left (-\frac {1}{c}\right )^{\frac {3}{4}} + {\left (a x^{2} + b x + c\right )}^{\frac {1}{4}}\right ) \] Input:

integrate((2*a*x+b)/(a*x^2+b*x+c)^(1/4)/(4*a*x^2+4*b*x+5*c),x, algorithm=" 
fricas")
 

Output:

1/2*(1/4)^(1/4)*(-1/c)^(1/4)*log(2*(1/4)^(3/4)*c*(-1/c)^(3/4) + (a*x^2 + b 
*x + c)^(1/4)) - 1/2*I*(1/4)^(1/4)*(-1/c)^(1/4)*log(2*I*(1/4)^(3/4)*c*(-1/ 
c)^(3/4) + (a*x^2 + b*x + c)^(1/4)) + 1/2*I*(1/4)^(1/4)*(-1/c)^(1/4)*log(- 
2*I*(1/4)^(3/4)*c*(-1/c)^(3/4) + (a*x^2 + b*x + c)^(1/4)) - 1/2*(1/4)^(1/4 
)*(-1/c)^(1/4)*log(-2*(1/4)^(3/4)*c*(-1/c)^(3/4) + (a*x^2 + b*x + c)^(1/4) 
)
 

Sympy [F]

\[ \int \frac {b+2 a x}{\sqrt [4]{c+b x+a x^2} \left (5 c+4 b x+4 a x^2\right )} \, dx=\int \frac {2 a x + b}{\sqrt [4]{a x^{2} + b x + c} \left (4 a x^{2} + 4 b x + 5 c\right )}\, dx \] Input:

integrate((2*a*x+b)/(a*x**2+b*x+c)**(1/4)/(4*a*x**2+4*b*x+5*c),x)
 

Output:

Integral((2*a*x + b)/((a*x**2 + b*x + c)**(1/4)*(4*a*x**2 + 4*b*x + 5*c)), 
 x)
 

Maxima [F]

\[ \int \frac {b+2 a x}{\sqrt [4]{c+b x+a x^2} \left (5 c+4 b x+4 a x^2\right )} \, dx=\int { \frac {2 \, a x + b}{{\left (4 \, a x^{2} + 4 \, b x + 5 \, c\right )} {\left (a x^{2} + b x + c\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate((2*a*x+b)/(a*x^2+b*x+c)^(1/4)/(4*a*x^2+4*b*x+5*c),x, algorithm=" 
maxima")
 

Output:

integrate((2*a*x + b)/((4*a*x^2 + 4*b*x + 5*c)*(a*x^2 + b*x + c)^(1/4)), x 
)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (92) = 184\).

Time = 0.20 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.66 \[ \int \frac {b+2 a x}{\sqrt [4]{c+b x+a x^2} \left (5 c+4 b x+4 a x^2\right )} \, dx=\frac {4^{\frac {3}{4}} \sqrt {2} \arctan \left (\frac {2 \, \sqrt {2} \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (\sqrt {2} \left (\frac {1}{4}\right )^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, {\left (a x^{2} + b x + c\right )}^{\frac {1}{4}}\right )}}{c^{\frac {1}{4}}}\right )}{8 \, c^{\frac {1}{4}}} + \frac {4^{\frac {3}{4}} \sqrt {2} \arctan \left (-\frac {2 \, \sqrt {2} \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (\sqrt {2} \left (\frac {1}{4}\right )^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, {\left (a x^{2} + b x + c\right )}^{\frac {1}{4}}\right )}}{c^{\frac {1}{4}}}\right )}{8 \, c^{\frac {1}{4}}} - \frac {4^{\frac {3}{4}} \sqrt {2} \log \left (\sqrt {2} \left (\frac {1}{4}\right )^{\frac {1}{4}} {\left (a x^{2} + b x + c\right )}^{\frac {1}{4}} c^{\frac {1}{4}} + \sqrt {a x^{2} + b x + c} + \frac {1}{2} \, \sqrt {c}\right )}{16 \, c^{\frac {1}{4}}} + \frac {4^{\frac {3}{4}} \sqrt {2} \log \left (-\sqrt {2} \left (\frac {1}{4}\right )^{\frac {1}{4}} {\left (a x^{2} + b x + c\right )}^{\frac {1}{4}} c^{\frac {1}{4}} + \sqrt {a x^{2} + b x + c} + \frac {1}{2} \, \sqrt {c}\right )}{16 \, c^{\frac {1}{4}}} \] Input:

integrate((2*a*x+b)/(a*x^2+b*x+c)^(1/4)/(4*a*x^2+4*b*x+5*c),x, algorithm=" 
giac")
 

Output:

1/8*4^(3/4)*sqrt(2)*arctan(2*sqrt(2)*(1/4)^(3/4)*(sqrt(2)*(1/4)^(1/4)*c^(1 
/4) + 2*(a*x^2 + b*x + c)^(1/4))/c^(1/4))/c^(1/4) + 1/8*4^(3/4)*sqrt(2)*ar 
ctan(-2*sqrt(2)*(1/4)^(3/4)*(sqrt(2)*(1/4)^(1/4)*c^(1/4) - 2*(a*x^2 + b*x 
+ c)^(1/4))/c^(1/4))/c^(1/4) - 1/16*4^(3/4)*sqrt(2)*log(sqrt(2)*(1/4)^(1/4 
)*(a*x^2 + b*x + c)^(1/4)*c^(1/4) + sqrt(a*x^2 + b*x + c) + 1/2*sqrt(c))/c 
^(1/4) + 1/16*4^(3/4)*sqrt(2)*log(-sqrt(2)*(1/4)^(1/4)*(a*x^2 + b*x + c)^( 
1/4)*c^(1/4) + sqrt(a*x^2 + b*x + c) + 1/2*sqrt(c))/c^(1/4)
 

Mupad [B] (verification not implemented)

Time = 9.16 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.47 \[ \int \frac {b+2 a x}{\sqrt [4]{c+b x+a x^2} \left (5 c+4 b x+4 a x^2\right )} \, dx=\frac {\sqrt {2}\,\left (\mathrm {atan}\left (\frac {\sqrt {2}\,{\left (a\,x^2+b\,x+c\right )}^{1/4}}{{\left (-c\right )}^{1/4}}\right )-\mathrm {atanh}\left (\frac {\sqrt {2}\,{\left (a\,x^2+b\,x+c\right )}^{1/4}}{{\left (-c\right )}^{1/4}}\right )\right )}{2\,{\left (-c\right )}^{1/4}} \] Input:

int((b + 2*a*x)/((c + b*x + a*x^2)^(1/4)*(5*c + 4*b*x + 4*a*x^2)),x)
 

Output:

(2^(1/2)*(atan((2^(1/2)*(c + b*x + a*x^2)^(1/4))/(-c)^(1/4)) - atanh((2^(1 
/2)*(c + b*x + a*x^2)^(1/4))/(-c)^(1/4))))/(2*(-c)^(1/4))
 

Reduce [F]

\[ \int \frac {b+2 a x}{\sqrt [4]{c+b x+a x^2} \left (5 c+4 b x+4 a x^2\right )} \, dx=2 \left (\int \frac {x}{4 \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}} a \,x^{2}+4 \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}} b x +5 \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}} c}d x \right ) a +\left (\int \frac {1}{4 \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}} a \,x^{2}+4 \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}} b x +5 \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}} c}d x \right ) b \] Input:

int((2*a*x+b)/(a*x^2+b*x+c)^(1/4)/(4*a*x^2+4*b*x+5*c),x)
 

Output:

2*int(x/(4*(a*x**2 + b*x + c)**(1/4)*a*x**2 + 4*(a*x**2 + b*x + c)**(1/4)* 
b*x + 5*(a*x**2 + b*x + c)**(1/4)*c),x)*a + int(1/(4*(a*x**2 + b*x + c)**( 
1/4)*a*x**2 + 4*(a*x**2 + b*x + c)**(1/4)*b*x + 5*(a*x**2 + b*x + c)**(1/4 
)*c),x)*b