Integrand size = 20, antiderivative size = 87 \[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {(b c-2 a d) x}{b^3}+\frac {d x^3}{3 b^2}+\frac {a (b c-a d) x}{2 b^3 \left (a+b x^2\right )}-\frac {\sqrt {a} (3 b c-5 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}} \]
(-2*a*d+b*c)*x/b^3+1/3*d*x^3/b^2+1/2*a*(-a*d+b*c)*x/b^3/(b*x^2+a)-1/2*(-5* a*d+3*b*c)*arctan(x*b^(1/2)/a^(1/2))*a^(1/2)/b^(7/2)
Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02 \[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {(b c-2 a d) x}{b^3}+\frac {d x^3}{3 b^2}+\frac {\left (a b c-a^2 d\right ) x}{2 b^3 \left (a+b x^2\right )}+\frac {\sqrt {a} (-3 b c+5 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}} \]
((b*c - 2*a*d)*x)/b^3 + (d*x^3)/(3*b^2) + ((a*b*c - a^2*d)*x)/(2*b^3*(a + b*x^2)) + (Sqrt[a]*(-3*b*c + 5*a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*b^(7/2 ))
Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {360, 1467, 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.
\(\displaystyle \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 360 |
\(\displaystyle \frac {a x (b c-a d)}{2 b^3 \left (a+b x^2\right )}-\frac {\int \frac {-2 b^2 d x^4-2 b (b c-a d) x^2+a (b c-a d)}{b x^2+a}dx}{2 b^3}\) |
\(\Big \downarrow \) 1467 |
\(\displaystyle \frac {a x (b c-a d)}{2 b^3 \left (a+b x^2\right )}-\frac {\int \left (-2 b d x^2-2 (b c-2 a d)+\frac {3 a b c-5 a^2 d}{b x^2+a}\right )dx}{2 b^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a x (b c-a d)}{2 b^3 \left (a+b x^2\right )}-\frac {\frac {\sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (3 b c-5 a d)}{\sqrt {b}}-2 x (b c-2 a d)-\frac {2}{3} b d x^3}{2 b^3}\) |
(a*(b*c - a*d)*x)/(2*b^3*(a + b*x^2)) - (-2*(b*c - 2*a*d)*x - (2*b*d*x^3)/ 3 + (Sqrt[a]*(3*b*c - 5*a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[b])/(2*b^3)
3.3.62.3.1 Defintions of rubi rules used
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Time = 2.67 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87
method | result | size |
default | \(-\frac {-\frac {1}{3} b d \,x^{3}+2 a d x -b c x}{b^{3}}+\frac {a \left (\frac {\left (-\frac {a d}{2}+\frac {b c}{2}\right ) x}{b \,x^{2}+a}+\frac {\left (5 a d -3 b c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{3}}\) | \(76\) |
risch | \(\frac {d \,x^{3}}{3 b^{2}}-\frac {2 a d x}{b^{3}}+\frac {c x}{b^{2}}+\frac {\left (-\frac {1}{2} a^{2} d +\frac {1}{2} a b c \right ) x}{b^{3} \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x +a \right ) a d}{4 b^{4}}-\frac {3 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x +a \right ) c}{4 b^{3}}-\frac {5 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x +a \right ) a d}{4 b^{4}}+\frac {3 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x +a \right ) c}{4 b^{3}}\) | \(147\) |
-1/b^3*(-1/3*b*d*x^3+2*a*d*x-b*c*x)+a/b^3*((-1/2*a*d+1/2*b*c)*x/(b*x^2+a)+ 1/2*(5*a*d-3*b*c)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.25 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.76 \[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=\left [\frac {4 \, b^{2} d x^{5} + 4 \, {\left (3 \, b^{2} c - 5 \, a b d\right )} x^{3} - 3 \, {\left (3 \, a b c - 5 \, a^{2} d + {\left (3 \, b^{2} c - 5 \, a b d\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 6 \, {\left (3 \, a b c - 5 \, a^{2} d\right )} x}{12 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, \frac {2 \, b^{2} d x^{5} + 2 \, {\left (3 \, b^{2} c - 5 \, a b d\right )} x^{3} - 3 \, {\left (3 \, a b c - 5 \, a^{2} d + {\left (3 \, b^{2} c - 5 \, a b d\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 3 \, {\left (3 \, a b c - 5 \, a^{2} d\right )} x}{6 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \]
[1/12*(4*b^2*d*x^5 + 4*(3*b^2*c - 5*a*b*d)*x^3 - 3*(3*a*b*c - 5*a^2*d + (3 *b^2*c - 5*a*b*d)*x^2)*sqrt(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*x^ 2 + a)) + 6*(3*a*b*c - 5*a^2*d)*x)/(b^4*x^2 + a*b^3), 1/6*(2*b^2*d*x^5 + 2 *(3*b^2*c - 5*a*b*d)*x^3 - 3*(3*a*b*c - 5*a^2*d + (3*b^2*c - 5*a*b*d)*x^2) *sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) + 3*(3*a*b*c - 5*a^2*d)*x)/(b^4*x^2 + a *b^3)]
Time = 0.37 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.48 \[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=x \left (- \frac {2 a d}{b^{3}} + \frac {c}{b^{2}}\right ) + \frac {x \left (- a^{2} d + a b c\right )}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {\sqrt {- \frac {a}{b^{7}}} \cdot \left (5 a d - 3 b c\right ) \log {\left (- b^{3} \sqrt {- \frac {a}{b^{7}}} + x \right )}}{4} + \frac {\sqrt {- \frac {a}{b^{7}}} \cdot \left (5 a d - 3 b c\right ) \log {\left (b^{3} \sqrt {- \frac {a}{b^{7}}} + x \right )}}{4} + \frac {d x^{3}}{3 b^{2}} \]
x*(-2*a*d/b**3 + c/b**2) + x*(-a**2*d + a*b*c)/(2*a*b**3 + 2*b**4*x**2) - sqrt(-a/b**7)*(5*a*d - 3*b*c)*log(-b**3*sqrt(-a/b**7) + x)/4 + sqrt(-a/b** 7)*(5*a*d - 3*b*c)*log(b**3*sqrt(-a/b**7) + x)/4 + d*x**3/(3*b**2)
Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.97 \[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (a b c - a^{2} d\right )} x}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} - \frac {{\left (3 \, a b c - 5 \, a^{2} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {b d x^{3} + 3 \, {\left (b c - 2 \, a d\right )} x}{3 \, b^{3}} \]
1/2*(a*b*c - a^2*d)*x/(b^4*x^2 + a*b^3) - 1/2*(3*a*b*c - 5*a^2*d)*arctan(b *x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/3*(b*d*x^3 + 3*(b*c - 2*a*d)*x)/b^3
Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01 \[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=-\frac {{\left (3 \, a b c - 5 \, a^{2} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {a b c x - a^{2} d x}{2 \, {\left (b x^{2} + a\right )} b^{3}} + \frac {b^{4} d x^{3} + 3 \, b^{4} c x - 6 \, a b^{3} d x}{3 \, b^{6}} \]
-1/2*(3*a*b*c - 5*a^2*d)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/2*(a*b* c*x - a^2*d*x)/((b*x^2 + a)*b^3) + 1/3*(b^4*d*x^3 + 3*b^4*c*x - 6*a*b^3*d* x)/b^6
Time = 0.09 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.20 \[ \int \frac {x^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=x\,\left (\frac {c}{b^2}-\frac {2\,a\,d}{b^3}\right )+\frac {d\,x^3}{3\,b^2}-\frac {x\,\left (\frac {a^2\,d}{2}-\frac {a\,b\,c}{2}\right )}{b^4\,x^2+a\,b^3}+\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,x\,\left (5\,a\,d-3\,b\,c\right )}{5\,a^2\,d-3\,a\,b\,c}\right )\,\left (5\,a\,d-3\,b\,c\right )}{2\,b^{7/2}} \]