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

3.18.97.1 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)
 

3.18.97.2 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)
 

3.18.97.3 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.

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

\(\Big \downarrow \) 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
 

3.18.97.3.1 Defintions of rubi rules used

rule 1375

Int[((g_.) + (h_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + 
(e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Unintegrable[(g + h*x)*(a + b 
*x + c*x^2)^p*(d + e*x + f*x^2)^q, x] /; FreeQ[{a, b, c, d, e, f, g, h, p, 
q}, x]
 

3.18.97.4 Maple [A] (verified)

Time = 1.14 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07

method	result	size
pseudoelliptic	\(\frac {\ln \left (\frac {-2 c^{\frac {1}{4}} \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}}+\sqrt {c}+2 \sqrt {a \,x^{2}+b x +c}}{2 c^{\frac {1}{4}} \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}}+\sqrt {c}+2 \sqrt {a \,x^{2}+b x +c}}\right )+2 \arctan \left (\frac {c^{\frac {1}{4}}+2 \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}}}{c^{\frac {1}{4}}}\right )-2 \arctan \left (\frac {c^{\frac {1}{4}}-2 \left (a \,x^{2}+b x +c \right )^{\frac {1}{4}}}{c^{\frac {1}{4}}}\right )}{4 c^{\frac {1}{4}}}\)	\(131\)

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

3.18.97.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (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) 
)
 

3.18.97.6 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)
 

3.18.97.7 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 
)
 

3.18.97.8 Giac [B] (verification not implemented)

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

Time = 0.29 (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)
 

3.18.97.9 Mupad [B] (verification not implemented)

Time = 6.34 (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))