Integrand size = 37, antiderivative size = 194 \[ \int \frac {b-a x^4+2 x^8}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx=\frac {x \left (-b+a x^4\right )^{3/4}}{6 a^2}+\frac {\left (-6 a^2-b\right ) \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{36 a^{9/4}}+\frac {\left (6 a^2+b\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{9 \sqrt {2} a^{9/4}}+\frac {\left (-6 a^2-b\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{36 a^{9/4}}+\frac {\left (6 a^2+b\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{9 \sqrt {2} a^{9/4}} \] Output:
1/6*x*(a*x^4-b)^(3/4)/a^2+1/36*(-6*a^2-b)*arctan(a^(1/4)*x/(a*x^4-b)^(1/4) )/a^(9/4)+1/18*(6*a^2+b)*arctan(2^(1/2)*a^(1/4)*x/(a*x^4-b)^(1/4))*2^(1/2) /a^(9/4)+1/36*(-6*a^2-b)*arctanh(a^(1/4)*x/(a*x^4-b)^(1/4))/a^(9/4)+1/18*( 6*a^2+b)*arctanh(2^(1/2)*a^(1/4)*x/(a*x^4-b)^(1/4))*2^(1/2)/a^(9/4)
Time = 1.07 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.88 \[ \int \frac {b-a x^4+2 x^8}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx=\frac {6 \sqrt [4]{a} x \left (-b+a x^4\right )^{3/4}-\left (6 a^2+b\right ) \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+2 \sqrt {2} \left (6 a^2+b\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )-\left (6 a^2+b\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+2 \sqrt {2} \left (6 a^2+b\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{36 a^{9/4}} \] Input:
Integrate[(b - a*x^4 + 2*x^8)/((-b + a*x^4)^(1/4)*(b + 3*a*x^4)),x]
Output:
(6*a^(1/4)*x*(-b + a*x^4)^(3/4) - (6*a^2 + b)*ArcTan[(a^(1/4)*x)/(-b + a*x ^4)^(1/4)] + 2*Sqrt[2]*(6*a^2 + b)*ArcTan[(Sqrt[2]*a^(1/4)*x)/(-b + a*x^4) ^(1/4)] - (6*a^2 + b)*ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] + 2*Sqrt[2]* (6*a^2 + b)*ArcTanh[(Sqrt[2]*a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(36*a^(9/4))
Time = 0.70 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.32, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {7276, 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.
\(\displaystyle \int \frac {-a x^4+b+2 x^8}{\sqrt [4]{a x^4-b} \left (3 a x^4+b\right )} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {2 \left (6 a^2 b+b^2\right )}{9 a^2 \sqrt [4]{a x^4-b} \left (3 a x^4+b\right )}+\frac {-\frac {2 b}{a^2}-3}{9 \sqrt [4]{a x^4-b}}+\frac {2 x^4}{3 a \sqrt [4]{a x^4-b}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{12 a^{9/4}}+\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{12 a^{9/4}}+\frac {x \left (a x^4-b\right )^{3/4}}{6 a^2}-\frac {\left (3 a^2+2 b\right ) \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{18 a^{9/4}}+\frac {\left (6 a^2+b\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{9 \sqrt {2} a^{9/4}}-\frac {\left (3 a^2+2 b\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{18 a^{9/4}}+\frac {\left (6 a^2+b\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{9 \sqrt {2} a^{9/4}}\) |
Input:
Int[(b - a*x^4 + 2*x^8)/((-b + a*x^4)^(1/4)*(b + 3*a*x^4)),x]
Output:
(x*(-b + a*x^4)^(3/4))/(6*a^2) + (b*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] )/(12*a^(9/4)) - ((3*a^2 + 2*b)*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(1 8*a^(9/4)) + ((6*a^2 + b)*ArcTan[(Sqrt[2]*a^(1/4)*x)/(-b + a*x^4)^(1/4)])/ (9*Sqrt[2]*a^(9/4)) + (b*ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(12*a^(9 /4)) - ((3*a^2 + 2*b)*ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(18*a^(9/4) ) + ((6*a^2 + b)*ArcTanh[(Sqrt[2]*a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(9*Sqrt[ 2]*a^(9/4))
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 0.96 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.90
method | result | size |
pseudoelliptic | \(\frac {x \left (a \,x^{4}-b \right )^{\frac {3}{4}} a^{\frac {1}{4}}+\left (a^{2}+\frac {b}{6}\right ) \left (\left (\ln \left (\frac {-\sqrt {2}\, x \,a^{\frac {1}{4}}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{\sqrt {2}\, x \,a^{\frac {1}{4}}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right )-2 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{4}-b \right )^{\frac {1}{4}}}{2 a^{\frac {1}{4}} x}\right )\right ) \sqrt {2}-\frac {\ln \left (\frac {-x \,a^{\frac {1}{4}}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{x \,a^{\frac {1}{4}}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right )}{2}+\arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )\right )}{6 a^{\frac {9}{4}}}\) | \(175\) |
Input:
int((2*x^8-a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x,method=_RETURNVERBOSE)
Output:
1/6*(x*(a*x^4-b)^(3/4)*a^(1/4)+(a^2+1/6*b)*((ln((-2^(1/2)*x*a^(1/4)-(a*x^4 -b)^(1/4))/(2^(1/2)*x*a^(1/4)-(a*x^4-b)^(1/4)))-2*arctan(1/2*2^(1/2)/a^(1/ 4)/x*(a*x^4-b)^(1/4)))*2^(1/2)-1/2*ln((-x*a^(1/4)-(a*x^4-b)^(1/4))/(x*a^(1 /4)-(a*x^4-b)^(1/4)))+arctan(1/a^(1/4)/x*(a*x^4-b)^(1/4))))/a^(9/4)
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 1053, normalized size of antiderivative = 5.43 \[ \int \frac {b-a x^4+2 x^8}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx=\text {Too large to display} \] Input:
integrate((2*x^8-a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x, algorithm="fricas ")
Output:
1/72*(4*(1/4)^(1/4)*a^2*((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(1/4)*log((4*(1/4)^(3/4)*a^7*x*((1296*a^8 + 864*a^6*b + 216*a^ 4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(3/4) + (216*a^6 + 108*a^4*b + 18*a^2*b^2 + b^3)*(a*x^4 - b)^(1/4))/x) - 4*(1/4)^(1/4)*a^2*((1296*a^8 + 864*a^6*b + 2 16*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(1/4)*log(-(4*(1/4)^(3/4)*a^7*x*((1296 *a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(3/4) - (216*a^6 + 108*a^4*b + 18*a^2*b^2 + b^3)*(a*x^4 - b)^(1/4))/x) - 4*I*(1/4)^(1/4)*a^2 *((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(1/4)*log(( 4*I*(1/4)^(3/4)*a^7*x*((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(3/4) + (216*a^6 + 108*a^4*b + 18*a^2*b^2 + b^3)*(a*x^4 - b)^(1/ 4))/x) + 4*I*(1/4)^(1/4)*a^2*((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2 *b^3 + b^4)/a^9)^(1/4)*log((-4*I*(1/4)^(3/4)*a^7*x*((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(3/4) + (216*a^6 + 108*a^4*b + 18*a ^2*b^2 + b^3)*(a*x^4 - b)^(1/4))/x) - a^2*((1296*a^8 + 864*a^6*b + 216*a^4 *b^2 + 24*a^2*b^3 + b^4)/a^9)^(1/4)*log((a^7*x*((1296*a^8 + 864*a^6*b + 21 6*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(3/4) + (216*a^6 + 108*a^4*b + 18*a^2*b ^2 + b^3)*(a*x^4 - b)^(1/4))/x) + a^2*((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(1/4)*log(-(a^7*x*((1296*a^8 + 864*a^6*b + 216*a ^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(3/4) - (216*a^6 + 108*a^4*b + 18*a^2*b^2 + b^3)*(a*x^4 - b)^(1/4))/x) + I*a^2*((1296*a^8 + 864*a^6*b + 216*a^4*b...
\[ \int \frac {b-a x^4+2 x^8}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx=\int \frac {- a x^{4} + b + 2 x^{8}}{\sqrt [4]{a x^{4} - b} \left (3 a x^{4} + b\right )}\, dx \] Input:
integrate((2*x**8-a*x**4+b)/(a*x**4-b)**(1/4)/(3*a*x**4+b),x)
Output:
Integral((-a*x**4 + b + 2*x**8)/((a*x**4 - b)**(1/4)*(3*a*x**4 + b)), x)
\[ \int \frac {b-a x^4+2 x^8}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx=\int { \frac {2 \, x^{8} - a x^{4} + b}{{\left (3 \, a x^{4} + b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate((2*x^8-a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x, algorithm="maxima ")
Output:
integrate((2*x^8 - a*x^4 + b)/((3*a*x^4 + b)*(a*x^4 - b)^(1/4)), x)
\[ \int \frac {b-a x^4+2 x^8}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx=\int { \frac {2 \, x^{8} - a x^{4} + b}{{\left (3 \, a x^{4} + b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate((2*x^8-a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x, algorithm="giac")
Output:
integrate((2*x^8 - a*x^4 + b)/((3*a*x^4 + b)*(a*x^4 - b)^(1/4)), x)
Timed out. \[ \int \frac {b-a x^4+2 x^8}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx=\int \frac {2\,x^8-a\,x^4+b}{{\left (a\,x^4-b\right )}^{1/4}\,\left (3\,a\,x^4+b\right )} \,d x \] Input:
int((b - a*x^4 + 2*x^8)/((a*x^4 - b)^(1/4)*(b + 3*a*x^4)),x)
Output:
int((b - a*x^4 + 2*x^8)/((a*x^4 - b)^(1/4)*(b + 3*a*x^4)), x)
\[ \int \frac {b-a x^4+2 x^8}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx=2 \left (\int \frac {x^{8}}{3 \left (a \,x^{4}-b \right )^{\frac {1}{4}} a \,x^{4}+\left (a \,x^{4}-b \right )^{\frac {1}{4}} b}d x \right )-\left (\int \frac {x^{4}}{3 \left (a \,x^{4}-b \right )^{\frac {1}{4}} a \,x^{4}+\left (a \,x^{4}-b \right )^{\frac {1}{4}} b}d x \right ) a +\left (\int \frac {1}{3 \left (a \,x^{4}-b \right )^{\frac {1}{4}} a \,x^{4}+\left (a \,x^{4}-b \right )^{\frac {1}{4}} b}d x \right ) b \] Input:
int((2*x^8-a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x)
Output:
2*int(x**8/(3*(a*x**4 - b)**(1/4)*a*x**4 + (a*x**4 - b)**(1/4)*b),x) - int (x**4/(3*(a*x**4 - b)**(1/4)*a*x**4 + (a*x**4 - b)**(1/4)*b),x)*a + int(1/ (3*(a*x**4 - b)**(1/4)*a*x**4 + (a*x**4 - b)**(1/4)*b),x)*b