3.21.1 \(\int \frac {b+a x^4}{\sqrt [4]{-b+a x^4} (b+3 a x^4)} \, dx\) [2001]

3.21.1.1 Optimal result
3.21.1.2 Mathematica [A] (verified)
3.21.1.3 Rubi [A] (verified)
3.21.1.4 Maple [A] (verified)
3.21.1.5 Fricas [C] (verification not implemented)
3.21.1.6 Sympy [F]
3.21.1.7 Maxima [F]
3.21.1.8 Giac [F]
3.21.1.9 Mupad [F(-1)]

3.21.1.1 Optimal result

Integrand size = 31, antiderivative size = 141 \[ \int \frac {b+a x^4}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{6 \sqrt [4]{a}}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{3 \sqrt {2} \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{6 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{3 \sqrt {2} \sqrt [4]{a}} \]

output
1/6*arctan(a^(1/4)*x/(a*x^4-b)^(1/4))/a^(1/4)+1/6*arctan(2^(1/2)*a^(1/4)*x 
/(a*x^4-b)^(1/4))*2^(1/2)/a^(1/4)+1/6*arctanh(a^(1/4)*x/(a*x^4-b)^(1/4))/a 
^(1/4)+1/6*arctanh(2^(1/2)*a^(1/4)*x/(a*x^4-b)^(1/4))*2^(1/2)/a^(1/4)
 
3.21.1.2 Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.82 \[ \int \frac {b+a x^4}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{6 \sqrt [4]{a}} \]

input
Integrate[(b + a*x^4)/((-b + a*x^4)^(1/4)*(b + 3*a*x^4)),x]
 
output
(ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] + Sqrt[2]*ArcTan[(Sqrt[2]*a^(1/4)* 
x)/(-b + a*x^4)^(1/4)] + ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] + Sqrt[2] 
*ArcTanh[(Sqrt[2]*a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(6*a^(1/4))
 
3.21.1.3 Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {1026, 770, 756, 216, 219, 902, 756, 216, 219}

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.

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

\(\Big \downarrow \) 1026

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {2}{3} b \int \frac {1}{\sqrt [4]{a x^4-b} \left (3 a x^4+b\right )}dx+\frac {1}{3} \left (\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}\right )\)

\(\Big \downarrow \) 902

\(\displaystyle \frac {2}{3} b \int \frac {1}{b-\frac {4 a b x^4}{a x^4-b}}d\frac {x}{\sqrt [4]{a x^4-b}}+\frac {1}{3} \left (\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}\right )\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {2}{3} b \left (\frac {\int \frac {1}{1-\frac {2 \sqrt {a} x^2}{\sqrt {a x^4-b}}}d\frac {x}{\sqrt [4]{a x^4-b}}}{2 b}+\frac {\int \frac {1}{\frac {2 \sqrt {a} x^2}{\sqrt {a x^4-b}}+1}d\frac {x}{\sqrt [4]{a x^4-b}}}{2 b}\right )+\frac {1}{3} \left (\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {2}{3} b \left (\frac {\int \frac {1}{1-\frac {2 \sqrt {a} x^2}{\sqrt {a x^4-b}}}d\frac {x}{\sqrt [4]{a x^4-b}}}{2 b}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt {2} \sqrt [4]{a} b}\right )+\frac {1}{3} \left (\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}\right )+\frac {2}{3} b \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt {2} \sqrt [4]{a} b}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt {2} \sqrt [4]{a} b}\right )\)

input
Int[(b + a*x^4)/((-b + a*x^4)^(1/4)*(b + 3*a*x^4)),x]
 
output
(ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]/(2*a^(1/4)) + ArcTanh[(a^(1/4)*x)/ 
(-b + a*x^4)^(1/4)]/(2*a^(1/4)))/3 + (2*b*(ArcTan[(Sqrt[2]*a^(1/4)*x)/(-b 
+ a*x^4)^(1/4)]/(2*Sqrt[2]*a^(1/4)*b) + ArcTanh[(Sqrt[2]*a^(1/4)*x)/(-b + 
a*x^4)^(1/4)]/(2*Sqrt[2]*a^(1/4)*b)))/3
 

3.21.1.3.1 Defintions of rubi rules used

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 756
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 770
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

rule 902
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Su 
bst[Int[1/(c - (b*c - a*d)*x^n), x], x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b 
, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
 

rule 1026
Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)* 
(x_)^(n_)), x_Symbol] :> Simp[f/d   Int[(a + b*x^n)^p, x], x] + Simp[(d*e - 
 c*f)/d   Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, 
 p, n}, x]
 
3.21.1.4 Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.05

method result size
pseudoelliptic \(\frac {\ln \left (\frac {-x \,a^{\frac {1}{4}} \sqrt {2}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{x \,a^{\frac {1}{4}} \sqrt {2}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right ) \sqrt {2}-2 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{4}-b \right )^{\frac {1}{4}}}{2 a^{\frac {1}{4}} x}\right ) \sqrt {2}-2 \arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\ln \left (\frac {a^{\frac {1}{4}} x +\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right )}{12 a^{\frac {1}{4}}}\) \(148\)

input
int((a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x,method=_RETURNVERBOSE)
 
output
1/12*(ln((-x*a^(1/4)*2^(1/2)-(a*x^4-b)^(1/4))/(x*a^(1/4)*2^(1/2)-(a*x^4-b) 
^(1/4)))*2^(1/2)-2*arctan(1/2*2^(1/2)/a^(1/4)/x*(a*x^4-b)^(1/4))*2^(1/2)-2 
*arctan(1/a^(1/4)/x*(a*x^4-b)^(1/4))+ln((a^(1/4)*x+(a*x^4-b)^(1/4))/(-a^(1 
/4)*x+(a*x^4-b)^(1/4))))/a^(1/4)
 
3.21.1.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.79 \[ \int \frac {b+a x^4}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx=\frac {\left (\frac {1}{4}\right )^{\frac {1}{4}} \log \left (\frac {4 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} a^{\frac {1}{4}} x + {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}\right )}{6 \, a^{\frac {1}{4}}} - \frac {\left (\frac {1}{4}\right )^{\frac {1}{4}} \log \left (-\frac {4 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} a^{\frac {1}{4}} x - {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}\right )}{6 \, a^{\frac {1}{4}}} - \frac {i \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \log \left (\frac {4 i \, \left (\frac {1}{4}\right )^{\frac {3}{4}} a^{\frac {1}{4}} x + {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}\right )}{6 \, a^{\frac {1}{4}}} + \frac {i \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \log \left (\frac {-4 i \, \left (\frac {1}{4}\right )^{\frac {3}{4}} a^{\frac {1}{4}} x + {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}\right )}{6 \, a^{\frac {1}{4}}} + \frac {\log \left (\frac {a^{\frac {1}{4}} x + {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}\right )}{12 \, a^{\frac {1}{4}}} - \frac {\log \left (-\frac {a^{\frac {1}{4}} x - {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}\right )}{12 \, a^{\frac {1}{4}}} - \frac {i \, \log \left (\frac {i \, a^{\frac {1}{4}} x + {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}\right )}{12 \, a^{\frac {1}{4}}} + \frac {i \, \log \left (\frac {-i \, a^{\frac {1}{4}} x + {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}\right )}{12 \, a^{\frac {1}{4}}} \]

input
integrate((a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x, algorithm="fricas")
 
output
1/6*(1/4)^(1/4)*log((4*(1/4)^(3/4)*a^(1/4)*x + (a*x^4 - b)^(1/4))/x)/a^(1/ 
4) - 1/6*(1/4)^(1/4)*log(-(4*(1/4)^(3/4)*a^(1/4)*x - (a*x^4 - b)^(1/4))/x) 
/a^(1/4) - 1/6*I*(1/4)^(1/4)*log((4*I*(1/4)^(3/4)*a^(1/4)*x + (a*x^4 - b)^ 
(1/4))/x)/a^(1/4) + 1/6*I*(1/4)^(1/4)*log((-4*I*(1/4)^(3/4)*a^(1/4)*x + (a 
*x^4 - b)^(1/4))/x)/a^(1/4) + 1/12*log((a^(1/4)*x + (a*x^4 - b)^(1/4))/x)/ 
a^(1/4) - 1/12*log(-(a^(1/4)*x - (a*x^4 - b)^(1/4))/x)/a^(1/4) - 1/12*I*lo 
g((I*a^(1/4)*x + (a*x^4 - b)^(1/4))/x)/a^(1/4) + 1/12*I*log((-I*a^(1/4)*x 
+ (a*x^4 - b)^(1/4))/x)/a^(1/4)
 
3.21.1.6 Sympy [F]

\[ \int \frac {b+a x^4}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx=\int \frac {a x^{4} + b}{\sqrt [4]{a x^{4} - b} \left (3 a x^{4} + b\right )}\, dx \]

input
integrate((a*x**4+b)/(a*x**4-b)**(1/4)/(3*a*x**4+b),x)
 
output
Integral((a*x**4 + b)/((a*x**4 - b)**(1/4)*(3*a*x**4 + b)), x)
 
3.21.1.7 Maxima [F]

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

input
integrate((a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x, algorithm="maxima")
 
output
integrate((a*x^4 + b)/((3*a*x^4 + b)*(a*x^4 - b)^(1/4)), x)
 
3.21.1.8 Giac [F]

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

input
integrate((a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x, algorithm="giac")
 
output
integrate((a*x^4 + b)/((3*a*x^4 + b)*(a*x^4 - b)^(1/4)), x)
 
3.21.1.9 Mupad [F(-1)]

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

input
int((b + a*x^4)/((a*x^4 - b)^(1/4)*(b + 3*a*x^4)),x)
 
output
int((b + a*x^4)/((a*x^4 - b)^(1/4)*(b + 3*a*x^4)), x)