Integrand size = 23, antiderivative size = 55 \[ \int \frac {b+a x^2}{-a^4+b^4 x^4} \, dx=\frac {\left (a^3-b^3\right ) \arctan \left (\frac {b x}{a}\right )}{2 a^3 b^3}+\frac {\left (-a^3-b^3\right ) \text {arctanh}\left (\frac {b x}{a}\right )}{2 a^3 b^3} \] Output:
1/2*(a^3-b^3)*arctan(b*x/a)/a^3/b^3+1/2*(-a^3-b^3)*arctanh(b*x/a)/a^3/b^3
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98 \[ \int \frac {b+a x^2}{-a^4+b^4 x^4} \, dx=\frac {2 \left (a^3-b^3\right ) \arctan \left (\frac {b x}{a}\right )+\left (a^3+b^3\right ) (\log (-a+b x)-\log (a+b x))}{4 a^3 b^3} \] Input:
Integrate[(b + a*x^2)/(-a^4 + b^4*x^4),x]
Output:
(2*(a^3 - b^3)*ArcTan[(b*x)/a] + (a^3 + b^3)*(Log[-a + b*x] - Log[a + b*x] ))/(4*a^3*b^3)
Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1481, 218, 221}
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.
\(\displaystyle \int \frac {a x^2+b}{b^4 x^4-a^4} \, dx\) |
\(\Big \downarrow \) 1481 |
\(\displaystyle \frac {1}{2} \left (\frac {b^3}{a^2}+a\right ) \int \frac {1}{b^4 x^2-a^2 b^2}dx+\frac {1}{2} \left (a-\frac {b^3}{a^2}\right ) \int \frac {1}{x^2 b^4+a^2 b^2}dx\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {1}{2} \left (\frac {b^3}{a^2}+a\right ) \int \frac {1}{b^4 x^2-a^2 b^2}dx+\frac {\left (a-\frac {b^3}{a^2}\right ) \arctan \left (\frac {b x}{a}\right )}{2 a b^3}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\left (a-\frac {b^3}{a^2}\right ) \arctan \left (\frac {b x}{a}\right )}{2 a b^3}-\frac {\left (\frac {b^3}{a^2}+a\right ) \text {arctanh}\left (\frac {b x}{a}\right )}{2 a b^3}\) |
Input:
Int[(b + a*x^2)/(-a^4 + b^4*x^4),x]
Output:
((a - b^3/a^2)*ArcTan[(b*x)/a])/(2*a*b^3) - ((a + b^3/a^2)*ArcTanh[(b*x)/a ])/(2*a*b^3)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ (-a)*c, 2]}, Simp[(e/2 + c*(d/(2*q))) Int[1/(-q + c*x^2), x], x] + Simp[( e/2 - c*(d/(2*q))) Int[1/(q + c*x^2), x], x]] /; FreeQ[{a, c, d, e}, x] & & NeQ[c*d^2 - a*e^2, 0] && PosQ[(-a)*c]
Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.33
method | result | size |
default | \(\frac {\left (-a^{3}-b^{3}\right ) \ln \left (b x +a \right )}{4 a^{3} b^{3}}+\frac {\left (a^{3}-b^{3}\right ) \arctan \left (\frac {b x}{a}\right )}{2 a^{3} b^{3}}+\frac {\left (a^{3}+b^{3}\right ) \ln \left (-b x +a \right )}{4 a^{3} b^{3}}\) | \(73\) |
parallelrisch | \(-\frac {i \ln \left (-i a +b x \right ) a^{3}-i \ln \left (-i a +b x \right ) b^{3}-i \ln \left (i a +b x \right ) a^{3}+i \ln \left (i a +b x \right ) b^{3}-\ln \left (b x -a \right ) a^{3}-\ln \left (b x -a \right ) b^{3}+\ln \left (b x +a \right ) a^{3}+\ln \left (b x +a \right ) b^{3}}{4 a^{3} b^{3}}\) | \(116\) |
risch | \(-\frac {\ln \left (-b x -a \right )}{4 b^{3}}-\frac {\ln \left (-b x -a \right )}{4 a^{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{6} b^{6} \textit {\_Z}^{2}+a^{6}-2 a^{3} b^{3}+b^{6}\right )}{\sum }\textit {\_R} \ln \left (\left (a^{6} b^{4}+b^{10}\right ) \textit {\_R} x -a^{4} \textit {\_R}^{2} b^{9}-a^{7}+a \,b^{6}\right )\right )}{4}+\frac {\ln \left (b x -a \right )}{4 b^{3}}+\frac {\ln \left (b x -a \right )}{4 a^{3}}\) | \(126\) |
Input:
int((a*x^2+b)/(b^4*x^4-a^4),x,method=_RETURNVERBOSE)
Output:
1/4*(-a^3-b^3)/a^3/b^3*ln(b*x+a)+1/2*(a^3-b^3)*arctan(b*x/a)/a^3/b^3+1/4*( a^3+b^3)/a^3/b^3*ln(-b*x+a)
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.05 \[ \int \frac {b+a x^2}{-a^4+b^4 x^4} \, dx=\frac {2 \, {\left (a^{3} - b^{3}\right )} \arctan \left (\frac {b x}{a}\right ) - {\left (a^{3} + b^{3}\right )} \log \left (b x + a\right ) + {\left (a^{3} + b^{3}\right )} \log \left (b x - a\right )}{4 \, a^{3} b^{3}} \] Input:
integrate((a*x^2+b)/(b^4*x^4-a^4),x, algorithm="fricas")
Output:
1/4*(2*(a^3 - b^3)*arctan(b*x/a) - (a^3 + b^3)*log(b*x + a) + (a^3 + b^3)* log(b*x - a))/(a^3*b^3)
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 394, normalized size of antiderivative = 7.16 \[ \int \frac {b+a x^2}{-a^4+b^4 x^4} \, dx=- \frac {i \left (a - b\right ) \left (a^{2} + a b + b^{2}\right ) \log {\left (x + \frac {- 3 i a^{7} b^{2} \left (a - b\right ) \left (a^{2} + a b + b^{2}\right ) - \frac {i a^{4} \left (a - b\right )^{3} \left (a^{2} + a b + b^{2}\right )^{3}}{b} - i a b^{8} \left (a - b\right ) \left (a^{2} + a b + b^{2}\right )}{a^{12} - b^{12}} \right )}}{4 a^{3} b^{3}} + \frac {i \left (a - b\right ) \left (a^{2} + a b + b^{2}\right ) \log {\left (x + \frac {3 i a^{7} b^{2} \left (a - b\right ) \left (a^{2} + a b + b^{2}\right ) + \frac {i a^{4} \left (a - b\right )^{3} \left (a^{2} + a b + b^{2}\right )^{3}}{b} + i a b^{8} \left (a - b\right ) \left (a^{2} + a b + b^{2}\right )}{a^{12} - b^{12}} \right )}}{4 a^{3} b^{3}} - \frac {\left (a + b\right ) \left (a^{2} - a b + b^{2}\right ) \log {\left (x + \frac {- 3 a^{7} b^{2} \left (a + b\right ) \left (a^{2} - a b + b^{2}\right ) + \frac {a^{4} \left (a + b\right )^{3} \left (a^{2} - a b + b^{2}\right )^{3}}{b} - a b^{8} \left (a + b\right ) \left (a^{2} - a b + b^{2}\right )}{a^{12} - b^{12}} \right )}}{4 a^{3} b^{3}} + \frac {\left (a + b\right ) \left (a^{2} - a b + b^{2}\right ) \log {\left (x + \frac {3 a^{7} b^{2} \left (a + b\right ) \left (a^{2} - a b + b^{2}\right ) - \frac {a^{4} \left (a + b\right )^{3} \left (a^{2} - a b + b^{2}\right )^{3}}{b} + a b^{8} \left (a + b\right ) \left (a^{2} - a b + b^{2}\right )}{a^{12} - b^{12}} \right )}}{4 a^{3} b^{3}} \] Input:
integrate((a*x**2+b)/(b**4*x**4-a**4),x)
Output:
-I*(a - b)*(a**2 + a*b + b**2)*log(x + (-3*I*a**7*b**2*(a - b)*(a**2 + a*b + b**2) - I*a**4*(a - b)**3*(a**2 + a*b + b**2)**3/b - I*a*b**8*(a - b)*( a**2 + a*b + b**2))/(a**12 - b**12))/(4*a**3*b**3) + I*(a - b)*(a**2 + a*b + b**2)*log(x + (3*I*a**7*b**2*(a - b)*(a**2 + a*b + b**2) + I*a**4*(a - b)**3*(a**2 + a*b + b**2)**3/b + I*a*b**8*(a - b)*(a**2 + a*b + b**2))/(a* *12 - b**12))/(4*a**3*b**3) - (a + b)*(a**2 - a*b + b**2)*log(x + (-3*a**7 *b**2*(a + b)*(a**2 - a*b + b**2) + a**4*(a + b)**3*(a**2 - a*b + b**2)**3 /b - a*b**8*(a + b)*(a**2 - a*b + b**2))/(a**12 - b**12))/(4*a**3*b**3) + (a + b)*(a**2 - a*b + b**2)*log(x + (3*a**7*b**2*(a + b)*(a**2 - a*b + b** 2) - a**4*(a + b)**3*(a**2 - a*b + b**2)**3/b + a*b**8*(a + b)*(a**2 - a*b + b**2))/(a**12 - b**12))/(4*a**3*b**3)
Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.25 \[ \int \frac {b+a x^2}{-a^4+b^4 x^4} \, dx=\frac {{\left (a^{3} - b^{3}\right )} \arctan \left (\frac {b x}{a}\right )}{2 \, a^{3} b^{3}} - \frac {{\left (a^{3} + b^{3}\right )} \log \left (b x + a\right )}{4 \, a^{3} b^{3}} + \frac {{\left (a^{3} + b^{3}\right )} \log \left (b x - a\right )}{4 \, a^{3} b^{3}} \] Input:
integrate((a*x^2+b)/(b^4*x^4-a^4),x, algorithm="maxima")
Output:
1/2*(a^3 - b^3)*arctan(b*x/a)/(a^3*b^3) - 1/4*(a^3 + b^3)*log(b*x + a)/(a^ 3*b^3) + 1/4*(a^3 + b^3)*log(b*x - a)/(a^3*b^3)
Time = 0.13 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.36 \[ \int \frac {b+a x^2}{-a^4+b^4 x^4} \, dx=\frac {{\left (a^{3} - b^{3}\right )} \arctan \left (\frac {b x}{a}\right )}{2 \, a^{3} b^{3}} - \frac {{\left (a^{3} b + b^{4}\right )} \log \left ({\left | b x + a \right |}\right )}{4 \, a^{3} b^{4}} + \frac {{\left (a^{3} b + b^{4}\right )} \log \left ({\left | b x - a \right |}\right )}{4 \, a^{3} b^{4}} \] Input:
integrate((a*x^2+b)/(b^4*x^4-a^4),x, algorithm="giac")
Output:
1/2*(a^3 - b^3)*arctan(b*x/a)/(a^3*b^3) - 1/4*(a^3*b + b^4)*log(abs(b*x + a))/(a^3*b^4) + 1/4*(a^3*b + b^4)*log(abs(b*x - a))/(a^3*b^4)
Time = 0.22 (sec) , antiderivative size = 167, normalized size of antiderivative = 3.04 \[ \int \frac {b+a x^2}{-a^4+b^4 x^4} \, dx=-\mathrm {atanh}\left (\frac {2\,x\,\left (4\,a^6\,b^8+4\,b^{14}\right )\,\left (\frac {1}{4\,a^3}+\frac {1}{4\,b^3}\right )}{2\,a^7\,b^4-2\,a\,b^{10}+32\,a^4\,b^{13}\,{\left (\frac {1}{4\,a^3}+\frac {1}{4\,b^3}\right )}^2}\right )\,\left (\frac {1}{2\,a^3}+\frac {1}{2\,b^3}\right )-\mathrm {atanh}\left (\frac {2\,x\,\left (4\,a^6\,b^8+4\,b^{14}\right )\,\left (\frac {1{}\mathrm {i}}{4\,a^3}-\frac {1{}\mathrm {i}}{4\,b^3}\right )}{2\,a^7\,b^4-2\,a\,b^{10}+32\,a^4\,b^{13}\,{\left (\frac {1{}\mathrm {i}}{4\,a^3}-\frac {1{}\mathrm {i}}{4\,b^3}\right )}^2}\right )\,\left (\frac {1{}\mathrm {i}}{2\,a^3}-\frac {1{}\mathrm {i}}{2\,b^3}\right ) \] Input:
int(-(b + a*x^2)/(a^4 - b^4*x^4),x)
Output:
- atanh((2*x*(4*b^14 + 4*a^6*b^8)*(1/(4*a^3) + 1/(4*b^3)))/(2*a^7*b^4 - 2* a*b^10 + 32*a^4*b^13*(1/(4*a^3) + 1/(4*b^3))^2))*(1/(2*a^3) + 1/(2*b^3)) - atanh((2*x*(4*b^14 + 4*a^6*b^8)*(1i/(4*a^3) - 1i/(4*b^3)))/(2*a^7*b^4 - 2 *a*b^10 + 32*a^4*b^13*(1i/(4*a^3) - 1i/(4*b^3))^2))*(1i/(2*a^3) - 1i/(2*b^ 3))
Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.40 \[ \int \frac {b+a x^2}{-a^4+b^4 x^4} \, dx=\frac {2 \mathit {atan} \left (\frac {b x}{a}\right ) a^{3}-2 \mathit {atan} \left (\frac {b x}{a}\right ) b^{3}+\mathrm {log}\left (-b x +a \right ) a^{3}+\mathrm {log}\left (-b x +a \right ) b^{3}-\mathrm {log}\left (b x +a \right ) a^{3}-\mathrm {log}\left (b x +a \right ) b^{3}}{4 a^{3} b^{3}} \] Input:
int((a*x^2+b)/(b^4*x^4-a^4),x)
Output:
(2*atan((b*x)/a)*a**3 - 2*atan((b*x)/a)*b**3 + log(a - b*x)*a**3 + log(a - b*x)*b**3 - log(a + b*x)*a**3 - log(a + b*x)*b**3)/(4*a**3*b**3)