Integrand size = 20, antiderivative size = 96 \[ \int \frac {x^2}{(c+d x) \left (a+b x^2\right )} \, dx=-\frac {\sqrt {a} c \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b} \left (b c^2+a d^2\right )}+\frac {c^2 \log (c+d x)}{d \left (b c^2+a d^2\right )}+\frac {a d \log \left (a+b x^2\right )}{2 b \left (b c^2+a d^2\right )} \]
c^2*ln(d*x+c)/d/(a*d^2+b*c^2)+1/2*a*d*ln(b*x^2+a)/b/(a*d^2+b*c^2)-c*arctan (x*b^(1/2)/a^(1/2))*a^(1/2)/(a*d^2+b*c^2)/b^(1/2)
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.76 \[ \int \frac {x^2}{(c+d x) \left (a+b x^2\right )} \, dx=\frac {-2 \sqrt {a} \sqrt {b} c d \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+2 b c^2 \log (c+d x)+a d^2 \log \left (a+b x^2\right )}{2 b^2 c^2 d+2 a b d^3} \]
(-2*Sqrt[a]*Sqrt[b]*c*d*ArcTan[(Sqrt[b]*x)/Sqrt[a]] + 2*b*c^2*Log[c + d*x] + a*d^2*Log[a + b*x^2])/(2*b^2*c^2*d + 2*a*b*d^3)
Time = 0.26 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {615, 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.
\(\displaystyle \int \frac {x^2}{\left (a+b x^2\right ) (c+d x)} \, dx\) |
\(\Big \downarrow \) 615 |
\(\displaystyle \int \left (\frac {c^2}{(c+d x) \left (a d^2+b c^2\right )}-\frac {a (c-d x)}{\left (a+b x^2\right ) \left (a d^2+b c^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sqrt {a} c \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b} \left (a d^2+b c^2\right )}+\frac {a d \log \left (a+b x^2\right )}{2 b \left (a d^2+b c^2\right )}+\frac {c^2 \log (c+d x)}{d \left (a d^2+b c^2\right )}\) |
-((Sqrt[a]*c*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[b]*(b*c^2 + a*d^2))) + (c^ 2*Log[c + d*x])/(d*(b*c^2 + a*d^2)) + (a*d*Log[a + b*x^2])/(2*b*(b*c^2 + a *d^2))
3.4.40.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Time = 0.85 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.78
method | result | size |
default | \(-\frac {a \left (-\frac {d \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{a \,d^{2}+b \,c^{2}}+\frac {c^{2} \ln \left (d x +c \right )}{d \left (a \,d^{2}+b \,c^{2}\right )}\) | \(75\) |
risch | \(\frac {\ln \left (\left (-3 a^{2} b c \,d^{3}+5 a \,b^{2} c^{3} d +\sqrt {-a b}\, a^{2} d^{4}-5 \sqrt {-a b}\, a b \,c^{2} d^{2}+2 \sqrt {-a b}\, b^{2} c^{4}\right ) x -5 a^{2} b \,c^{2} d^{2}+3 \sqrt {-a b}\, a^{2} c \,d^{3}-5 \sqrt {-a b}\, a b \,c^{3} d +a^{3} d^{4}+2 a \,b^{2} c^{4}\right ) c \sqrt {-a b}}{2 \left (a \,d^{2}+b \,c^{2}\right ) b}+\frac {\ln \left (\left (-3 a^{2} b c \,d^{3}+5 a \,b^{2} c^{3} d +\sqrt {-a b}\, a^{2} d^{4}-5 \sqrt {-a b}\, a b \,c^{2} d^{2}+2 \sqrt {-a b}\, b^{2} c^{4}\right ) x -5 a^{2} b \,c^{2} d^{2}+3 \sqrt {-a b}\, a^{2} c \,d^{3}-5 \sqrt {-a b}\, a b \,c^{3} d +a^{3} d^{4}+2 a \,b^{2} c^{4}\right ) d a}{2 \left (a \,d^{2}+b \,c^{2}\right ) b}-\frac {\ln \left (\left (-3 a^{2} b c \,d^{3}+5 a \,b^{2} c^{3} d -\sqrt {-a b}\, a^{2} d^{4}+5 \sqrt {-a b}\, a b \,c^{2} d^{2}-2 \sqrt {-a b}\, b^{2} c^{4}\right ) x -5 a^{2} b \,c^{2} d^{2}-3 \sqrt {-a b}\, a^{2} c \,d^{3}+5 \sqrt {-a b}\, a b \,c^{3} d +a^{3} d^{4}+2 a \,b^{2} c^{4}\right ) c \sqrt {-a b}}{2 \left (a \,d^{2}+b \,c^{2}\right ) b}+\frac {\ln \left (\left (-3 a^{2} b c \,d^{3}+5 a \,b^{2} c^{3} d -\sqrt {-a b}\, a^{2} d^{4}+5 \sqrt {-a b}\, a b \,c^{2} d^{2}-2 \sqrt {-a b}\, b^{2} c^{4}\right ) x -5 a^{2} b \,c^{2} d^{2}-3 \sqrt {-a b}\, a^{2} c \,d^{3}+5 \sqrt {-a b}\, a b \,c^{3} d +a^{3} d^{4}+2 a \,b^{2} c^{4}\right ) d a}{2 \left (a \,d^{2}+b \,c^{2}\right ) b}+\frac {c^{2} \ln \left (d x +c \right )}{d \left (a \,d^{2}+b \,c^{2}\right )}\) | \(620\) |
-a/(a*d^2+b*c^2)*(-1/2*d*ln(b*x^2+a)/b+c/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2 )))+c^2*ln(d*x+c)/d/(a*d^2+b*c^2)
Time = 0.29 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.69 \[ \int \frac {x^2}{(c+d x) \left (a+b x^2\right )} \, dx=\left [\frac {b c d \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + a d^{2} \log \left (b x^{2} + a\right ) + 2 \, b c^{2} \log \left (d x + c\right )}{2 \, {\left (b^{2} c^{2} d + a b d^{3}\right )}}, -\frac {2 \, b c d \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - a d^{2} \log \left (b x^{2} + a\right ) - 2 \, b c^{2} \log \left (d x + c\right )}{2 \, {\left (b^{2} c^{2} d + a b d^{3}\right )}}\right ] \]
[1/2*(b*c*d*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) + a *d^2*log(b*x^2 + a) + 2*b*c^2*log(d*x + c))/(b^2*c^2*d + a*b*d^3), -1/2*(2 *b*c*d*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) - a*d^2*log(b*x^2 + a) - 2*b*c^2* log(d*x + c))/(b^2*c^2*d + a*b*d^3)]
Timed out. \[ \int \frac {x^2}{(c+d x) \left (a+b x^2\right )} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int \frac {x^2}{(c+d x) \left (a+b x^2\right )} \, dx=\frac {a d \log \left (b x^{2} + a\right )}{2 \, {\left (b^{2} c^{2} + a b d^{2}\right )}} + \frac {c^{2} \log \left (d x + c\right )}{b c^{2} d + a d^{3}} - \frac {a c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b c^{2} + a d^{2}\right )} \sqrt {a b}} \]
1/2*a*d*log(b*x^2 + a)/(b^2*c^2 + a*b*d^2) + c^2*log(d*x + c)/(b*c^2*d + a *d^3) - a*c*arctan(b*x/sqrt(a*b))/((b*c^2 + a*d^2)*sqrt(a*b))
Time = 0.28 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int \frac {x^2}{(c+d x) \left (a+b x^2\right )} \, dx=\frac {a d \log \left (b x^{2} + a\right )}{2 \, {\left (b^{2} c^{2} + a b d^{2}\right )}} + \frac {c^{2} \log \left ({\left | d x + c \right |}\right )}{b c^{2} d + a d^{3}} - \frac {a c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b c^{2} + a d^{2}\right )} \sqrt {a b}} \]
1/2*a*d*log(b*x^2 + a)/(b^2*c^2 + a*b*d^2) + c^2*log(abs(d*x + c))/(b*c^2* d + a*d^3) - a*c*arctan(b*x/sqrt(a*b))/((b*c^2 + a*d^2)*sqrt(a*b))
Time = 0.65 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.61 \[ \int \frac {x^2}{(c+d x) \left (a+b x^2\right )} \, dx=\frac {\ln \left (a\,c+a\,d\,x+\frac {\left (c\,\sqrt {-a\,b^3}+a\,b\,d\right )\,\left (x\,\left (2\,b^2\,c^2-5\,a\,b\,d^2\right )-5\,a\,b\,c\,d+\frac {2\,b^2\,d\,\left (c\,\sqrt {-a\,b^3}+a\,b\,d\right )\,\left (-b\,x\,c^2+4\,a\,c\,d+3\,a\,x\,d^2\right )}{2\,b^3\,c^2+2\,a\,b^2\,d^2}\right )}{2\,b^3\,c^2+2\,a\,b^2\,d^2}\right )\,\left (c\,\sqrt {-a\,b^3}+a\,b\,d\right )}{2\,b^3\,c^2+2\,a\,b^2\,d^2}-\frac {\ln \left (a\,c+a\,d\,x+\frac {\left (c\,\sqrt {-a\,b^3}-a\,b\,d\right )\,\left (b\,x\,\left (5\,a\,d^2-2\,b\,c^2\right )+5\,a\,b\,c\,d+\frac {d\,\left (c\,\sqrt {-a\,b^3}-a\,b\,d\right )\,\left (-b\,x\,c^2+4\,a\,c\,d+3\,a\,x\,d^2\right )}{b\,c^2+a\,d^2}\right )}{2\,b^2\,\left (b\,c^2+a\,d^2\right )}\right )\,\left (c\,\sqrt {-a\,b^3}-a\,b\,d\right )}{2\,\left (b^3\,c^2+a\,b^2\,d^2\right )}+\frac {c^2\,\ln \left (c+d\,x\right )}{b\,c^2\,d+a\,d^3} \]
(log(a*c + a*d*x + ((c*(-a*b^3)^(1/2) + a*b*d)*(x*(2*b^2*c^2 - 5*a*b*d^2) - 5*a*b*c*d + (2*b^2*d*(c*(-a*b^3)^(1/2) + a*b*d)*(4*a*c*d + 3*a*d^2*x - b *c^2*x))/(2*b^3*c^2 + 2*a*b^2*d^2)))/(2*b^3*c^2 + 2*a*b^2*d^2))*(c*(-a*b^3 )^(1/2) + a*b*d))/(2*b^3*c^2 + 2*a*b^2*d^2) - (log(a*c + a*d*x + ((c*(-a*b ^3)^(1/2) - a*b*d)*(b*x*(5*a*d^2 - 2*b*c^2) + 5*a*b*c*d + (d*(c*(-a*b^3)^( 1/2) - a*b*d)*(4*a*c*d + 3*a*d^2*x - b*c^2*x))/(a*d^2 + b*c^2)))/(2*b^2*(a *d^2 + b*c^2)))*(c*(-a*b^3)^(1/2) - a*b*d))/(2*(b^3*c^2 + a*b^2*d^2)) + (c ^2*log(c + d*x))/(a*d^3 + b*c^2*d)