Integrand size = 28, antiderivative size = 534 \[ \int \frac {A+B x+C x^2}{x^3 \left (a+b x^2+c x^4\right )^2} \, dx=-\frac {2 A b^2-6 a A c-a b C}{2 a^2 \left (b^2-4 a c\right ) x^2}-\frac {B \left (3 b^2-10 a c\right )}{2 a^2 \left (b^2-4 a c\right ) x}+\frac {B \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) x \left (a+b x^2+c x^4\right )}+\frac {A \left (b^2-2 a c\right )-a b C+c (A b-2 a C) x^2}{2 a \left (b^2-4 a c\right ) x^2 \left (a+b x^2+c x^4\right )}-\frac {B \sqrt {c} \left (3 b^3-16 a b c+\left (3 b^2-10 a c\right ) \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {B \sqrt {c} \left (3 b^3-16 a b c-\left (3 b^2-10 a c\right ) \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}}-\frac {\left (2 A \left (b^4-6 a b^2 c+6 a^2 c^2\right )-a b \left (b^2-6 a c\right ) C\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 a^3 \left (b^2-4 a c\right )^{3/2}}-\frac {(2 A b-a C) \log (x)}{a^3}+\frac {(2 A b-a C) \log \left (a+b x^2+c x^4\right )}{4 a^3} \]
1/2*(6*A*a*c-2*A*b^2+C*a*b)/a^2/(-4*a*c+b^2)/x^2-1/2*B*(-10*a*c+3*b^2)/a^2 /(-4*a*c+b^2)/x+1/2*B*(b*c*x^2-2*a*c+b^2)/a/(-4*a*c+b^2)/x/(c*x^4+b*x^2+a) +1/2*(A*(-2*a*c+b^2)-a*b*C+c*(A*b-2*C*a)*x^2)/a/(-4*a*c+b^2)/x^2/(c*x^4+b* x^2+a)-1/2*(2*A*(6*a^2*c^2-6*a*b^2*c+b^4)-a*b*(-6*a*c+b^2)*C)*arctanh((2*c *x^2+b)/(-4*a*c+b^2)^(1/2))/a^3/(-4*a*c+b^2)^(3/2)-(2*A*b-C*a)*ln(x)/a^3+1 /4*(2*A*b-C*a)*ln(c*x^4+b*x^2+a)/a^3-1/4*B*arctan(x*2^(1/2)*c^(1/2)/(b-(-4 *a*c+b^2)^(1/2))^(1/2))*c^(1/2)*(3*b^3-16*a*b*c+(-10*a*c+3*b^2)*(-4*a*c+b^ 2)^(1/2))/a^2/(-4*a*c+b^2)^(3/2)*2^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)+1/4* B*arctan(x*2^(1/2)*c^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2))*c^(1/2)*(3*b^3-16 *a*b*c-(-10*a*c+3*b^2)*(-4*a*c+b^2)^(1/2))/a^2/(-4*a*c+b^2)^(3/2)*2^(1/2)/ (b+(-4*a*c+b^2)^(1/2))^(1/2)
Time = 1.50 (sec) , antiderivative size = 655, normalized size of antiderivative = 1.23 \[ \int \frac {A+B x+C x^2}{x^3 \left (a+b x^2+c x^4\right )^2} \, dx=\frac {-\frac {2 a A}{x^2}-\frac {4 a B}{x}-\frac {2 a \left (2 a^2 c C+b^2 B x \left (b+c x^2\right )+A \left (b^3-3 a b c+b^2 c x^2-2 a c^2 x^2\right )-a \left (b^2 C+2 B c^2 x^3+b c x (3 B+C x)\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {\sqrt {2} a B \sqrt {c} \left (-3 b^3+16 a b c-3 b^2 \sqrt {b^2-4 a c}+10 a c \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\sqrt {2} a B \sqrt {c} \left (3 b^3-16 a b c-3 b^2 \sqrt {b^2-4 a c}+10 a c \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}}+4 (-2 A b+a C) \log (x)+\frac {\left (2 A \left (b^4-6 a b^2 c+6 a^2 c^2+b^3 \sqrt {b^2-4 a c}-4 a b c \sqrt {b^2-4 a c}\right )+a \left (-b^3+6 a b c-b^2 \sqrt {b^2-4 a c}+4 a c \sqrt {b^2-4 a c}\right ) C\right ) \log \left (-b+\sqrt {b^2-4 a c}-2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac {\left (2 A \left (-b^4+6 a b^2 c-6 a^2 c^2+b^3 \sqrt {b^2-4 a c}-4 a b c \sqrt {b^2-4 a c}\right )+a \left (b^3-6 a b c-b^2 \sqrt {b^2-4 a c}+4 a c \sqrt {b^2-4 a c}\right ) C\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}}{4 a^3} \]
((-2*a*A)/x^2 - (4*a*B)/x - (2*a*(2*a^2*c*C + b^2*B*x*(b + c*x^2) + A*(b^3 - 3*a*b*c + b^2*c*x^2 - 2*a*c^2*x^2) - a*(b^2*C + 2*B*c^2*x^3 + b*c*x*(3* B + C*x))))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (Sqrt[2]*a*B*Sqrt[c]*(-3 *b^3 + 16*a*b*c - 3*b^2*Sqrt[b^2 - 4*a*c] + 10*a*c*Sqrt[b^2 - 4*a*c])*ArcT an[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/((b^2 - 4*a*c)^(3/2)* Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (Sqrt[2]*a*B*Sqrt[c]*(3*b^3 - 16*a*b*c - 3* b^2*Sqrt[b^2 - 4*a*c] + 10*a*c*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]* x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/((b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) + 4*(-2*A*b + a*C)*Log[x] + ((2*A*(b^4 - 6*a*b^2*c + 6*a^2*c^2 + b^3*Sqrt[b^2 - 4*a*c] - 4*a*b*c*Sqrt[b^2 - 4*a*c]) + a*(-b^3 + 6*a*b*c - b ^2*Sqrt[b^2 - 4*a*c] + 4*a*c*Sqrt[b^2 - 4*a*c])*C)*Log[-b + Sqrt[b^2 - 4*a *c] - 2*c*x^2])/(b^2 - 4*a*c)^(3/2) + ((2*A*(-b^4 + 6*a*b^2*c - 6*a^2*c^2 + b^3*Sqrt[b^2 - 4*a*c] - 4*a*b*c*Sqrt[b^2 - 4*a*c]) + a*(b^3 - 6*a*b*c - b^2*Sqrt[b^2 - 4*a*c] + 4*a*c*Sqrt[b^2 - 4*a*c])*C)*Log[b + Sqrt[b^2 - 4*a *c] + 2*c*x^2])/(b^2 - 4*a*c)^(3/2))/(4*a^3)
Time = 1.25 (sec) , antiderivative size = 553, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2193, 27, 1441, 25, 1578, 1235, 25, 1200, 1604, 1480, 218, 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.
| \(\displaystyle \int \frac {A+B x+C x^2}{x^3 \left (a+b x^2+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 2193 |
| \(\displaystyle \int \frac {C x^2+A}{x^3 \left (c x^4+b x^2+a\right )^2}dx+\int \frac {B}{x^2 \left (c x^4+b x^2+a\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {C x^2+A}{x^3 \left (c x^4+b x^2+a\right )^2}dx+B \int \frac {1}{x^2 \left (c x^4+b x^2+a\right )^2}dx\) |
| \(\Big \downarrow \) 1441 |
| \(\displaystyle \int \frac {C x^2+A}{x^3 \left (c x^4+b x^2+a\right )^2}dx+B \left (\frac {-2 a c+b^2+b c x^2}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {3 b^2+3 c x^2 b-10 a c}{x^2 \left (c x^4+b x^2+a\right )}dx}{2 a \left (b^2-4 a c\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {C x^2+A}{x^3 \left (c x^4+b x^2+a\right )^2}dx+B \left (\frac {\int \frac {3 b^2+3 c x^2 b-10 a c}{x^2 \left (c x^4+b x^2+a\right )}dx}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle \frac {1}{2} \int \frac {C x^2+A}{x^4 \left (c x^4+b x^2+a\right )^2}dx^2+B \left (\frac {\int \frac {3 b^2+3 c x^2 b-10 a c}{x^2 \left (c x^4+b x^2+a\right )}dx}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle \frac {1}{2} \left (\frac {c x^2 (A b-2 a C)-2 a A c-a b C+A b^2}{a x^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {2 A b^2-a C b+2 c (A b-2 a C) x^2-6 a A c}{x^4 \left (c x^4+b x^2+a\right )}dx^2}{a \left (b^2-4 a c\right )}\right )+B \left (\frac {\int \frac {3 b^2+3 c x^2 b-10 a c}{x^2 \left (c x^4+b x^2+a\right )}dx}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {2 c (A b-2 a C) x^2+2 A \left (b^2-3 a c\right )-a b C}{x^4 \left (c x^4+b x^2+a\right )}dx^2}{a \left (b^2-4 a c\right )}+\frac {c x^2 (A b-2 a C)-2 a A c-a b C+A b^2}{a x^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )+B \left (\frac {\int \frac {3 b^2+3 c x^2 b-10 a c}{x^2 \left (c x^4+b x^2+a\right )}dx}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 1200 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \left (-\frac {\left (4 a c-b^2\right ) (a C-2 A b)}{a^2 x^2}+\frac {c \left (b^2-4 a c\right ) (2 A b-a C) x^2+2 A \left (b^4-5 a c b^2+3 a^2 c^2\right )-a b \left (b^2-5 a c\right ) C}{a^2 \left (c x^4+b x^2+a\right )}+\frac {2 A \left (b^2-3 a c\right )-a b C}{a x^4}\right )dx^2}{a \left (b^2-4 a c\right )}+\frac {c x^2 (A b-2 a C)-2 a A c-a b C+A b^2}{a x^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )+B \left (\frac {\int \frac {3 b^2+3 c x^2 b-10 a c}{x^2 \left (c x^4+b x^2+a\right )}dx}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 1604 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \left (-\frac {\left (4 a c-b^2\right ) (a C-2 A b)}{a^2 x^2}+\frac {c \left (b^2-4 a c\right ) (2 A b-a C) x^2+2 A \left (b^4-5 a c b^2+3 a^2 c^2\right )-a b \left (b^2-5 a c\right ) C}{a^2 \left (c x^4+b x^2+a\right )}+\frac {2 A \left (b^2-3 a c\right )-a b C}{a x^4}\right )dx^2}{a \left (b^2-4 a c\right )}+\frac {c x^2 (A b-2 a C)-2 a A c-a b C+A b^2}{a x^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )+B \left (\frac {-\frac {\int \frac {c \left (3 b^2-10 a c\right ) x^2+b \left (3 b^2-13 a c\right )}{c x^4+b x^2+a}dx}{a}-\frac {3 b^2-10 a c}{a x}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \left (-\frac {\left (4 a c-b^2\right ) (a C-2 A b)}{a^2 x^2}+\frac {c \left (b^2-4 a c\right ) (2 A b-a C) x^2+2 A \left (b^4-5 a c b^2+3 a^2 c^2\right )-a b \left (b^2-5 a c\right ) C}{a^2 \left (c x^4+b x^2+a\right )}+\frac {2 A \left (b^2-3 a c\right )-a b C}{a x^4}\right )dx^2}{a \left (b^2-4 a c\right )}+\frac {c x^2 (A b-2 a C)-2 a A c-a b C+A b^2}{a x^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )+B \left (\frac {-\frac {\frac {1}{2} c \left (-\frac {16 a b c}{\sqrt {b^2-4 a c}}+\frac {3 b^3}{\sqrt {b^2-4 a c}}-10 a c+3 b^2\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx-\frac {c \left (-\left (3 b^2-10 a c\right ) \sqrt {b^2-4 a c}-16 a b c+3 b^3\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx}{2 \sqrt {b^2-4 a c}}}{a}-\frac {3 b^2-10 a c}{a x}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \left (-\frac {\left (4 a c-b^2\right ) (a C-2 A b)}{a^2 x^2}+\frac {c \left (b^2-4 a c\right ) (2 A b-a C) x^2+2 A \left (b^4-5 a c b^2+3 a^2 c^2\right )-a b \left (b^2-5 a c\right ) C}{a^2 \left (c x^4+b x^2+a\right )}+\frac {2 A \left (b^2-3 a c\right )-a b C}{a x^4}\right )dx^2}{a \left (b^2-4 a c\right )}+\frac {c x^2 (A b-2 a C)-2 a A c-a b C+A b^2}{a x^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )+B \left (\frac {-\frac {\frac {\sqrt {c} \left (-\frac {16 a b c}{\sqrt {b^2-4 a c}}+\frac {3 b^3}{\sqrt {b^2-4 a c}}-10 a c+3 b^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \left (-\left (3 b^2-10 a c\right ) \sqrt {b^2-4 a c}-16 a b c+3 b^3\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {\sqrt {b^2-4 a c}+b}}}{a}-\frac {3 b^2-10 a c}{a x}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {-\frac {\left (2 A \left (6 a^2 c^2-6 a b^2 c+b^4\right )-a b C \left (b^2-6 a c\right )\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c}}-\frac {\log \left (x^2\right ) \left (b^2-4 a c\right ) (2 A b-a C)}{a^2}+\frac {\left (b^2-4 a c\right ) (2 A b-a C) \log \left (a+b x^2+c x^4\right )}{2 a^2}-\frac {-6 a A c-a b C+2 A b^2}{a x^2}}{a \left (b^2-4 a c\right )}+\frac {c x^2 (A b-2 a C)-2 a A c-a b C+A b^2}{a x^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )+B \left (\frac {-\frac {\frac {\sqrt {c} \left (-\frac {16 a b c}{\sqrt {b^2-4 a c}}+\frac {3 b^3}{\sqrt {b^2-4 a c}}-10 a c+3 b^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \left (-\left (3 b^2-10 a c\right ) \sqrt {b^2-4 a c}-16 a b c+3 b^3\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {\sqrt {b^2-4 a c}+b}}}{a}-\frac {3 b^2-10 a c}{a x}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
B*((b^2 - 2*a*c + b*c*x^2)/(2*a*(b^2 - 4*a*c)*x*(a + b*x^2 + c*x^4)) + (-( (3*b^2 - 10*a*c)/(a*x)) - ((Sqrt[c]*(3*b^2 - 10*a*c + (3*b^3)/Sqrt[b^2 - 4 *a*c] - (16*a*b*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[b - Sqrt[b^2 - 4*a*c]]) - (Sqrt[c]*(3*b ^3 - 16*a*b*c - (3*b^2 - 10*a*c)*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c ]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[b^2 - 4*a*c]*Sqrt[b + Sqr t[b^2 - 4*a*c]]))/a)/(2*a*(b^2 - 4*a*c))) + ((A*b^2 - 2*a*A*c - a*b*C + c* (A*b - 2*a*C)*x^2)/(a*(b^2 - 4*a*c)*x^2*(a + b*x^2 + c*x^4)) + (-((2*A*b^2 - 6*a*A*c - a*b*C)/(a*x^2)) - ((2*A*(b^4 - 6*a*b^2*c + 6*a^2*c^2) - a*b*( b^2 - 6*a*c)*C)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(a^2*Sqrt[b^2 - 4*a*c]) - ((b^2 - 4*a*c)*(2*A*b - a*C)*Log[x^2])/a^2 + ((b^2 - 4*a*c)*(2*A *b - a*C)*Log[a + b*x^2 + c*x^4])/(2*a^2))/(a*(b^2 - 4*a*c)))/2
3.1.36.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 *a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m *(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p] )
Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-(d*x)^(m + 1))*(b^2 - 2*a*c + b*c*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*a*d*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[(d*x)^m*(a + b*x^2 + c*x^4)^(p + 1)*Simp[b^2*(m + 2*p + 3) - 2*a*c*(m + 4*p + 5) + b*c*(m + 4*p + 7)*x^2, x], x], x] /; FreeQ[{a, b, c, d, m}, x ] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int egerQ[(m - 1)/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_), x_Symbol] :> Simp[d*(f*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1) /(a*f*(m + 1))), x] + Simp[1/(a*f^2*(m + 1)) Int[(f*x)^(m + 2)*(a + b*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x ], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[ m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_S ymbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[Pq, x, 2*k]*x^(2*k), {k, 0, q/2 + 1}]*(d*x)^m*(a + b*x^2 + c*x^4)^p, x] + Simp[1/d Int[Sum[Coe ff[Pq, x, 2*k + 1]*x^(2*k), {k, 0, (q + 1)/2}]*(d*x)^(m + 1)*(a + b*x^2 + c *x^4)^p, x], x]] /; FreeQ[{a, b, c, d, m, p}, x] && PolyQ[Pq, x] && !PolyQ [Pq, x^2]
Time = 0.25 (sec) , antiderivative size = 778, normalized size of antiderivative = 1.46
| method | result | size |
| default | \(-\frac {A}{2 a^{2} x^{2}}-\frac {B}{a^{2} x}+\frac {\left (-2 A b +C a \right ) \ln \left (x \right )}{a^{3}}-\frac {\frac {\frac {B a c \left (2 a c -b^{2}\right ) x^{3}}{8 a c -2 b^{2}}+\frac {a c \left (2 A a c -A \,b^{2}+a b C \right ) x^{2}}{8 a c -2 b^{2}}+\frac {B a b \left (3 a c -b^{2}\right ) x}{8 a c -2 b^{2}}+\frac {a \left (3 A a b c -A \,b^{3}-2 a^{2} c C +C a \,b^{2}\right )}{8 a c -2 b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {2 c \left (\frac {\frac {\left (24 A \sqrt {-4 a c +b^{2}}\, a^{2} c^{2}-24 A \sqrt {-4 a c +b^{2}}\, a \,b^{2} c +4 A \sqrt {-4 a c +b^{2}}\, b^{4}-64 A \,a^{2} b \,c^{2}+32 A a \,b^{3} c -4 A \,b^{5}+12 C \sqrt {-4 a c +b^{2}}\, a^{2} b c -2 C \sqrt {-4 a c +b^{2}}\, a \,b^{3}+32 C \,a^{3} c^{2}-16 C \,a^{2} b^{2} c +2 C a \,b^{4}\right ) \ln \left (2 c \,x^{2}+\sqrt {-4 a c +b^{2}}+b \right )}{4 c}+\frac {\left (16 B \sqrt {-4 a c +b^{2}}\, a^{2} b c -3 B \sqrt {-4 a c +b^{2}}\, a \,b^{3}+40 a^{3} B \,c^{2}-22 B \,a^{2} b^{2} c +3 B a \,b^{4}\right ) \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}}{16 a c -4 b^{2}}+\frac {-\frac {\left (24 A \sqrt {-4 a c +b^{2}}\, a^{2} c^{2}-24 A \sqrt {-4 a c +b^{2}}\, a \,b^{2} c +4 A \sqrt {-4 a c +b^{2}}\, b^{4}+64 A \,a^{2} b \,c^{2}-32 A a \,b^{3} c +4 A \,b^{5}+12 C \sqrt {-4 a c +b^{2}}\, a^{2} b c -2 C \sqrt {-4 a c +b^{2}}\, a \,b^{3}-32 C \,a^{3} c^{2}+16 C \,a^{2} b^{2} c -2 C a \,b^{4}\right ) \ln \left (-2 c \,x^{2}+\sqrt {-4 a c +b^{2}}-b \right )}{4 c}+\frac {\left (16 B \sqrt {-4 a c +b^{2}}\, a^{2} b c -3 B \sqrt {-4 a c +b^{2}}\, a \,b^{3}-40 a^{3} B \,c^{2}+22 B \,a^{2} b^{2} c -3 B a \,b^{4}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}}{16 a c -4 b^{2}}\right )}{4 a c -b^{2}}}{a^{3}}\) | \(778\) |
| risch | \(\text {Expression too large to display}\) | \(3877\) |
-1/2*A/a^2/x^2-1/a^2*B/x+(-2*A*b+C*a)/a^3*ln(x)-1/a^3*((1/2*B*a*c*(2*a*c-b ^2)/(4*a*c-b^2)*x^3+1/2*a*c*(2*A*a*c-A*b^2+C*a*b)/(4*a*c-b^2)*x^2+1/2*B*a* b*(3*a*c-b^2)/(4*a*c-b^2)*x+1/2*a*(3*A*a*b*c-A*b^3-2*C*a^2*c+C*a*b^2)/(4*a *c-b^2))/(c*x^4+b*x^2+a)+2/(4*a*c-b^2)*c*(1/(16*a*c-4*b^2)*(1/4*(24*A*(-4* a*c+b^2)^(1/2)*a^2*c^2-24*A*(-4*a*c+b^2)^(1/2)*a*b^2*c+4*A*(-4*a*c+b^2)^(1 /2)*b^4-64*A*a^2*b*c^2+32*A*a*b^3*c-4*A*b^5+12*C*(-4*a*c+b^2)^(1/2)*a^2*b* c-2*C*(-4*a*c+b^2)^(1/2)*a*b^3+32*C*a^3*c^2-16*C*a^2*b^2*c+2*C*a*b^4)/c*ln (2*c*x^2+(-4*a*c+b^2)^(1/2)+b)+1/2*(16*B*(-4*a*c+b^2)^(1/2)*a^2*b*c-3*B*(- 4*a*c+b^2)^(1/2)*a*b^3+40*a^3*B*c^2-22*B*a^2*b^2*c+3*B*a*b^4)*2^(1/2)/((b+ (-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c) ^(1/2)))+1/(16*a*c-4*b^2)*(-1/4*(24*A*(-4*a*c+b^2)^(1/2)*a^2*c^2-24*A*(-4* a*c+b^2)^(1/2)*a*b^2*c+4*A*(-4*a*c+b^2)^(1/2)*b^4+64*A*a^2*b*c^2-32*A*a*b^ 3*c+4*A*b^5+12*C*(-4*a*c+b^2)^(1/2)*a^2*b*c-2*C*(-4*a*c+b^2)^(1/2)*a*b^3-3 2*C*a^3*c^2+16*C*a^2*b^2*c-2*C*a*b^4)/c*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)+ 1/2*(16*B*(-4*a*c+b^2)^(1/2)*a^2*b*c-3*B*(-4*a*c+b^2)^(1/2)*a*b^3-40*a^3*B *c^2+22*B*a^2*b^2*c-3*B*a*b^4)*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*a rctanh(c*x*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2)))))
Timed out. \[ \int \frac {A+B x+C x^2}{x^3 \left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B x+C x^2}{x^3 \left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {A+B x+C x^2}{x^3 \left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {C x^{2} + B x + A}{{\left (c x^{4} + b x^{2} + a\right )}^{2} x^{3}} \,d x } \]
-1/2*((3*B*b^2*c - 10*B*a*c^2)*x^5 - (6*A*a*c^2 + (C*a*b - 2*A*b^2)*c)*x^4 + A*a*b^2 - 4*A*a^2*c + (3*B*b^3 - 11*B*a*b*c)*x^3 - (C*a*b^2 - 2*A*b^3 - (2*C*a^2 - 7*A*a*b)*c)*x^2 + 2*(B*a*b^2 - 4*B*a^2*c)*x)/((a^2*b^2*c - 4*a ^3*c^2)*x^6 + (a^2*b^3 - 4*a^3*b*c)*x^4 + (a^3*b^2 - 4*a^4*c)*x^2) - 1/2*i ntegrate((3*B*a*b^3 - 13*B*a^2*b*c - 2*(4*(C*a^2 - 2*A*a*b)*c^2 - (C*a*b^2 - 2*A*b^3)*c)*x^3 + (3*B*a*b^2*c - 10*B*a^2*c^2)*x^2 + 2*(C*a*b^3 - 2*A*b ^4 - 6*A*a^2*c^2 - 5*(C*a^2*b - 2*A*a*b^2)*c)*x)/(c*x^4 + b*x^2 + a), x)/( a^3*b^2 - 4*a^4*c) + (C*a - 2*A*b)*log(x)/a^3
Leaf count of result is larger than twice the leaf count of optimal. 6939 vs. \(2 (470) = 940\).
Time = 1.72 (sec) , antiderivative size = 6939, normalized size of antiderivative = 12.99 \[ \int \frac {A+B x+C x^2}{x^3 \left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
-1/16*((a^6*b^4*c - 8*a^7*b^2*c^2 + 16*a^8*c^3)^2*(6*b^4*c^2 - 44*a*b^2*c^ 3 + 80*a^2*c^4 - 3*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)* c)*b^4 + 22*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^ 2*c + 6*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c - 40*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*c^2 - 20* sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^2 - 3*sqrt (2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2*c^2 + 10*sqrt(2) *sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*c^3 - 6*(b^2 - 4*a*c) *b^2*c^2 + 20*(b^2 - 4*a*c)*a*c^3)*B + 2*(3*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^6*b^9*c - 49*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^7*b^7*c ^2 - 6*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^6*b^8*c^2 - 6*a^6*b^9*c^2 + 300*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^8*b^5*c^3 + 74*sqrt(2)*sq rt(b*c + sqrt(b^2 - 4*a*c)*c)*a^7*b^6*c^3 + 3*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^6*b^7*c^3 + 98*a^7*b^7*c^3 - 816*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^9*b^3*c^4 - 304*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^8 *b^4*c^4 - 37*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^7*b^5*c^4 - 600*a^ 8*b^5*c^4 + 832*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^10*b*c^5 + 416*s qrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^9*b^2*c^5 + 152*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^8*b^3*c^5 + 1632*a^9*b^3*c^5 - 208*sqrt(2)*sqrt(b *c + sqrt(b^2 - 4*a*c)*c)*a^9*b*c^6 - 1664*a^10*b*c^6 + 6*(b^2 - 4*a*c)...
Time = 9.01 (sec) , antiderivative size = 10595, normalized size of antiderivative = 19.84 \[ \int \frac {A+B x+C x^2}{x^3 \left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
symsum(log(root(1572864*a^11*b^2*c^5*z^4 - 983040*a^10*b^4*c^4*z^4 + 32768 0*a^9*b^6*c^3*z^4 - 61440*a^8*b^8*c^2*z^4 + 6144*a^7*b^10*c*z^4 - 1048576* a^12*c^6*z^4 - 256*a^6*b^12*z^4 + 1572864*C*a^9*b^2*c^5*z^3 - 983040*C*a^8 *b^4*c^4*z^3 + 327680*C*a^7*b^6*c^3*z^3 - 61440*C*a^6*b^8*c^2*z^3 - 314572 8*A*a^8*b^3*c^5*z^3 + 1966080*A*a^7*b^5*c^4*z^3 - 655360*A*a^6*b^7*c^3*z^3 + 122880*A*a^5*b^9*c^2*z^3 + 6144*C*a^5*b^10*c*z^3 + 2097152*A*a^9*b*c^6* z^3 - 12288*A*a^4*b^11*c*z^3 - 1048576*C*a^10*c^6*z^3 - 256*C*a^4*b^12*z^3 + 512*A*a^3*b^13*z^3 + 1277952*A*C*a^7*b*c^6*z^2 - 6144*A*C*a^2*b^11*c*z^ 2 - 1794048*A*C*a^6*b^3*c^5*z^2 + 1062912*A*C*a^5*b^5*c^4*z^2 - 340480*A*C *a^4*b^7*c^3*z^2 + 62208*A*C*a^3*b^9*c^2*z^2 + 256*A*C*a*b^13*z^2 + 1536*C ^2*a^3*b^10*c*z^2 - 430080*B^2*a^7*b*c^6*z^2 + 3408*B^2*a^2*b^11*c*z^2 + 6 144*A^2*a*b^12*c*z^2 + 516096*C^2*a^7*b^2*c^5*z^2 - 288768*C^2*a^6*b^4*c^4 *z^2 + 88576*C^2*a^5*b^6*c^3*z^2 - 15744*C^2*a^4*b^8*c^2*z^2 + 716800*B^2* a^6*b^3*c^5*z^2 - 483840*B^2*a^5*b^5*c^4*z^2 + 170496*B^2*a^4*b^7*c^3*z^2 - 33232*B^2*a^3*b^9*c^2*z^2 + 1468416*A^2*a^5*b^4*c^5*z^2 - 966144*A^2*a^4 *b^6*c^4*z^2 - 761856*A^2*a^6*b^2*c^6*z^2 + 326656*A^2*a^3*b^8*c^3*z^2 - 6 1440*A^2*a^2*b^10*c^2*z^2 - 144*B^2*a*b^13*z^2 - 393216*C^2*a^8*c^6*z^2 - 64*C^2*a^2*b^12*z^2 - 294912*A^2*a^7*c^7*z^2 - 256*A^2*b^14*z^2 - 138240*B ^2*C*a^5*b*c^6*z - 432*B^2*C*a*b^9*c^2*z + 245760*A*C^2*a^5*b*c^6*z + 1228 8*A^2*C*a*b^8*c^3*z + 768*A*C^2*a*b^9*c^2*z + 576*A*B^2*a*b^8*c^3*z + 1...