Integrand size = 58, antiderivative size = 68 \[ \int \frac {x \left (1+a+a^2+x^2+a x^2+b x^2+2 a b x^2+b x^4+b^2 x^4\right )}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx=-\frac {1}{3 b \left (a+b x^2\right )^{3/2}}-\frac {1}{b \sqrt {a+b x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a-b}}\right )}{\sqrt {a-b}} \]
Time = 0.06 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.91 \[ \int \frac {x \left (1+a+a^2+x^2+a x^2+b x^2+2 a b x^2+b x^4+b^2 x^4\right )}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx=\frac {-1-3 a-3 b x^2}{3 b \left (a+b x^2\right )^{3/2}}+\frac {\arctan \left (\frac {\sqrt {a+b x^2}}{\sqrt {-a+b}}\right )}{\sqrt {-a+b}} \]
Integrate[(x*(1 + a + a^2 + x^2 + a*x^2 + b*x^2 + 2*a*b*x^2 + b*x^4 + b^2* x^4))/((1 + x^2)*(a + b*x^2)^(5/2)),x]
(-1 - 3*a - 3*b*x^2)/(3*b*(a + b*x^2)^(3/2)) + ArcTan[Sqrt[a + b*x^2]/Sqrt [-a + b]]/Sqrt[-a + b]
Time = 0.68 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.81, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {6, 6, 6, 6, 7266, 1192, 1584, 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.
\(\displaystyle \int \frac {x \left (a^2+2 a b x^2+a x^2+a+b^2 x^4+b x^4+b x^2+x^2+1\right )}{\left (x^2+1\right ) \left (a+b x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x \left (a^2+2 a b x^2+(a+1) x^2+a+b^2 x^4+b x^4+b x^2+1\right )}{\left (x^2+1\right ) \left (a+b x^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x \left (a^2+2 a b x^2+x^2 (a+b+1)+a+b^2 x^4+b x^4+1\right )}{\left (x^2+1\right ) \left (a+b x^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x \left (a^2+x^2 (2 a b+a+b+1)+a+b^2 x^4+b x^4+1\right )}{\left (x^2+1\right ) \left (a+b x^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x \left (a^2+x^2 (2 a b+a+b+1)+a+\left (b^2+b\right ) x^4+1\right )}{\left (x^2+1\right ) \left (a+b x^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 7266 |
\(\displaystyle \frac {1}{2} \int \frac {b (b+1) x^4+(2 b a+a+b+1) x^2+a^2+a+1}{\left (x^2+1\right ) \left (b x^2+a\right )^{5/2}}dx^2\) |
\(\Big \downarrow \) 1192 |
\(\displaystyle \frac {\int \frac {-b (b+1) x^8-b (-a+b+1) x^4+(a-b) b}{x^8 \left (-x^4+a-b\right )}d\sqrt {b x^2+a}}{b^2}\) |
\(\Big \downarrow \) 1584 |
\(\displaystyle \frac {\int \left (\frac {b^2}{x^4-a+b}+\frac {b}{x^4}+\frac {b}{x^8}\right )d\sqrt {b x^2+a}}{b^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}-\frac {b}{3 x^6}-\frac {b}{x^2}}{b^2}\) |
Int[(x*(1 + a + a^2 + x^2 + a*x^2 + b*x^2 + 2*a*b*x^2 + b*x^4 + b^2*x^4))/ ((1 + x^2)*(a + b*x^2)^(5/2)),x]
3.7.90.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^( 2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function OfQ[x^(m + 1), u, x]
Time = 1.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00
method | result | size |
pseudoelliptic | \(\frac {\arctan \left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {-a +b}}\right ) b \left (b \,x^{2}+a \right )^{\frac {3}{2}}-\sqrt {-a +b}\, \left (b \,x^{2}+a +\frac {1}{3}\right )}{\sqrt {-a +b}\, \left (b \,x^{2}+a \right )^{\frac {3}{2}} b}\) | \(68\) |
default | \(\left (b^{2}+b \right ) \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )-\frac {2 a b -b^{2}+a +1}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\left (a^{2}-2 a b +b^{2}\right ) \left (\frac {1}{\left (a -b \right )^{2} \sqrt {b \,x^{2}+a}}+\frac {\arctan \left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {-a +b}}\right )}{\left (a -b \right )^{2} \sqrt {-a +b}}+\frac {1}{3 \left (a -b \right ) \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )\) | \(148\) |
int(x*(b^2*x^4+b*x^4+2*a*b*x^2+a*x^2+b*x^2+a^2+x^2+a+1)/(x^2+1)/(b*x^2+a)^ (5/2),x,method=_RETURNVERBOSE)
1/(-a+b)^(1/2)*(arctan((b*x^2+a)^(1/2)/(-a+b)^(1/2))*b*(b*x^2+a)^(3/2)-(-a +b)^(1/2)*(b*x^2+a+1/3))/(b*x^2+a)^(3/2)/b
Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (56) = 112\).
Time = 0.32 (sec) , antiderivative size = 382, normalized size of antiderivative = 5.62 \[ \int \frac {x \left (1+a+a^2+x^2+a x^2+b x^2+2 a b x^2+b x^4+b^2 x^4\right )}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{2} + a^{2} b\right )} \sqrt {a - b} \log \left (\frac {b^{2} x^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} x^{2} - 4 \, {\left (b x^{2} + 2 \, a - b\right )} \sqrt {b x^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{x^{4} + 2 \, x^{2} + 1}\right ) - 4 \, {\left (3 \, {\left (a b - b^{2}\right )} x^{2} + 3 \, a^{2} - {\left (3 \, a + 1\right )} b + a\right )} \sqrt {b x^{2} + a}}{12 \, {\left ({\left (a b^{3} - b^{4}\right )} x^{4} + a^{3} b - a^{2} b^{2} + 2 \, {\left (a^{2} b^{2} - a b^{3}\right )} x^{2}\right )}}, -\frac {3 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{2} + a^{2} b\right )} \sqrt {-a + b} \arctan \left (-\frac {{\left (b x^{2} + 2 \, a - b\right )} \sqrt {b x^{2} + a} \sqrt {-a + b}}{2 \, {\left ({\left (a b - b^{2}\right )} x^{2} + a^{2} - a b\right )}}\right ) + 2 \, {\left (3 \, {\left (a b - b^{2}\right )} x^{2} + 3 \, a^{2} - {\left (3 \, a + 1\right )} b + a\right )} \sqrt {b x^{2} + a}}{6 \, {\left ({\left (a b^{3} - b^{4}\right )} x^{4} + a^{3} b - a^{2} b^{2} + 2 \, {\left (a^{2} b^{2} - a b^{3}\right )} x^{2}\right )}}\right ] \]
integrate(x*(b^2*x^4+b*x^4+2*a*b*x^2+a*x^2+b*x^2+a^2+x^2+a+1)/(x^2+1)/(b*x ^2+a)^(5/2),x, algorithm="fricas")
[1/12*(3*(b^3*x^4 + 2*a*b^2*x^2 + a^2*b)*sqrt(a - b)*log((b^2*x^4 + 2*(4*a *b - 3*b^2)*x^2 - 4*(b*x^2 + 2*a - b)*sqrt(b*x^2 + a)*sqrt(a - b) + 8*a^2 - 8*a*b + b^2)/(x^4 + 2*x^2 + 1)) - 4*(3*(a*b - b^2)*x^2 + 3*a^2 - (3*a + 1)*b + a)*sqrt(b*x^2 + a))/((a*b^3 - b^4)*x^4 + a^3*b - a^2*b^2 + 2*(a^2*b ^2 - a*b^3)*x^2), -1/6*(3*(b^3*x^4 + 2*a*b^2*x^2 + a^2*b)*sqrt(-a + b)*arc tan(-1/2*(b*x^2 + 2*a - b)*sqrt(b*x^2 + a)*sqrt(-a + b)/((a*b - b^2)*x^2 + a^2 - a*b)) + 2*(3*(a*b - b^2)*x^2 + 3*a^2 - (3*a + 1)*b + a)*sqrt(b*x^2 + a))/((a*b^3 - b^4)*x^4 + a^3*b - a^2*b^2 + 2*(a^2*b^2 - a*b^3)*x^2)]
\[ \int \frac {x \left (1+a+a^2+x^2+a x^2+b x^2+2 a b x^2+b x^4+b^2 x^4\right )}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx=\int \frac {x \left (a^{2} + 2 a b x^{2} + a x^{2} + a + b^{2} x^{4} + b x^{4} + b x^{2} + x^{2} + 1\right )}{\left (a + b x^{2}\right )^{\frac {5}{2}} \left (x^{2} + 1\right )}\, dx \]
integrate(x*(b**2*x**4+b*x**4+2*a*b*x**2+a*x**2+b*x**2+a**2+x**2+a+1)/(x** 2+1)/(b*x**2+a)**(5/2),x)
Integral(x*(a**2 + 2*a*b*x**2 + a*x**2 + a + b**2*x**4 + b*x**4 + b*x**2 + x**2 + 1)/((a + b*x**2)**(5/2)*(x**2 + 1)), x)
Exception generated. \[ \int \frac {x \left (1+a+a^2+x^2+a x^2+b x^2+2 a b x^2+b x^4+b^2 x^4\right )}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
integrate(x*(b^2*x^4+b*x^4+2*a*b*x^2+a*x^2+b*x^2+a^2+x^2+a+1)/(x^2+1)/(b*x ^2+a)^(5/2),x, algorithm="maxima")
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for m ore detail
Time = 0.42 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {x \left (1+a+a^2+x^2+a x^2+b x^2+2 a b x^2+b x^4+b^2 x^4\right )}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a + b}}\right )}{\sqrt {-a + b}} - \frac {3 \, b x^{2} + 3 \, a + 1}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} \]
integrate(x*(b^2*x^4+b*x^4+2*a*b*x^2+a*x^2+b*x^2+a^2+x^2+a+1)/(x^2+1)/(b*x ^2+a)^(5/2),x, algorithm="giac")
arctan(sqrt(b*x^2 + a)/sqrt(-a + b))/sqrt(-a + b) - 1/3*(3*b*x^2 + 3*a + 1 )/((b*x^2 + a)^(3/2)*b)
Time = 18.90 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.74 \[ \int \frac {x \left (1+a+a^2+x^2+a x^2+b x^2+2 a b x^2+b x^4+b^2 x^4\right )}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx=-\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}-\frac {b\,x^2+a+\frac {1}{3}}{b\,{\left (b\,x^2+a\right )}^{3/2}} \]
int((x*(a + a*x^2 + b*x^2 + b*x^4 + a^2 + x^2 + b^2*x^4 + 2*a*b*x^2 + 1))/ ((x^2 + 1)*(a + b*x^2)^(5/2)),x)