Integrand size = 20, antiderivative size = 90 \[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {c}{3 a^2 x^3}+\frac {2 b c-a d}{a^3 x}+\frac {b (b c-a d) x}{2 a^3 \left (a+b x^2\right )}+\frac {\sqrt {b} (5 b c-3 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}} \] Output:
-1/3*c/a^2/x^3+(-a*d+2*b*c)/a^3/x+1/2*b*(-a*d+b*c)*x/a^3/(b*x^2+a)+1/2*b^( 1/2)*(-3*a*d+5*b*c)*arctan(b^(1/2)*x/a^(1/2))/a^(7/2)
Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {c}{3 a^2 x^3}+\frac {2 b c-a d}{a^3 x}-\frac {b (-b c+a d) x}{2 a^3 \left (a+b x^2\right )}-\frac {\sqrt {b} (-5 b c+3 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}} \] Input:
Integrate[(c + d*x^2)/(x^4*(a + b*x^2)^2),x]
Output:
-1/3*c/(a^2*x^3) + (2*b*c - a*d)/(a^3*x) - (b*(-(b*c) + a*d)*x)/(2*a^3*(a + b*x^2)) - (Sqrt[b]*(-5*b*c + 3*a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7 /2))
Time = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {361, 25, 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.
\(\displaystyle \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 361 |
\(\displaystyle \frac {b x (b c-a d)}{2 a^3 \left (a+b x^2\right )}-\frac {1}{2} b \int -\frac {\frac {(b c-a d) x^4}{a^3}-\frac {2 (b c-a d) x^2}{a^2 b}+\frac {2 c}{a b}}{x^4 \left (b x^2+a\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} b \int \frac {\frac {(b c-a d) x^4}{a^3}-\frac {2 (b c-a d) x^2}{a^2 b}+\frac {2 c}{a b}}{x^4 \left (b x^2+a\right )}dx+\frac {b x (b c-a d)}{2 a^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 1584 |
\(\displaystyle \frac {1}{2} b \int \left (\frac {2 c}{a^2 b x^4}+\frac {5 b c-3 a d}{a^3 \left (b x^2+a\right )}+\frac {2 (a d-2 b c)}{a^3 b x^2}\right )dx+\frac {b x (b c-a d)}{2 a^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b x (b c-a d)}{2 a^3 \left (a+b x^2\right )}+\frac {1}{2} b \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (5 b c-3 a d)}{a^{7/2} \sqrt {b}}+\frac {2 (2 b c-a d)}{a^3 b x}-\frac {2 c}{3 a^2 b x^3}\right )\) |
Input:
Int[(c + d*x^2)/(x^4*(a + b*x^2)^2),x]
Output:
(b*(b*c - a*d)*x)/(2*a^3*(a + b*x^2)) + (b*((-2*c)/(3*a^2*b*x^3) + (2*(2*b *c - a*d))/(a^3*b*x) + ((5*b*c - 3*a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(7 /2)*Sqrt[b])))/2
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[x^m*(a + b*x^2)^(p + 1)*E xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
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]
Time = 0.43 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88
method | result | size |
default | \(-\frac {b \left (\frac {\left (\frac {a d}{2}-\frac {b c}{2}\right ) x}{b \,x^{2}+a}+\frac {\left (3 a d -5 b c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}-\frac {c}{3 a^{2} x^{3}}-\frac {a d -2 b c}{a^{3} x}\) | \(79\) |
risch | \(\frac {-\frac {b \left (3 a d -5 b c \right ) x^{4}}{2 a^{3}}-\frac {\left (3 a d -5 b c \right ) x^{2}}{3 a^{2}}-\frac {c}{3 a}}{x^{3} \left (b \,x^{2}+a \right )}+\frac {3 \sqrt {-a b}\, \ln \left (-b x +\sqrt {-a b}\right ) d}{4 a^{3}}-\frac {5 \sqrt {-a b}\, \ln \left (-b x +\sqrt {-a b}\right ) b c}{4 a^{4}}-\frac {3 \sqrt {-a b}\, \ln \left (-b x -\sqrt {-a b}\right ) d}{4 a^{3}}+\frac {5 \sqrt {-a b}\, \ln \left (-b x -\sqrt {-a b}\right ) b c}{4 a^{4}}\) | \(159\) |
Input:
int((d*x^2+c)/x^4/(b*x^2+a)^2,x,method=_RETURNVERBOSE)
Output:
-1/a^3*b*((1/2*a*d-1/2*b*c)*x/(b*x^2+a)+1/2*(3*a*d-5*b*c)/(a*b)^(1/2)*arct an(b*x/(a*b)^(1/2)))-1/3*c/a^2/x^3-(a*d-2*b*c)/a^3/x
Time = 0.08 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.78 \[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^2} \, dx=\left [\frac {6 \, {\left (5 \, b^{2} c - 3 \, a b d\right )} x^{4} - 4 \, a^{2} c + 4 \, {\left (5 \, a b c - 3 \, a^{2} d\right )} x^{2} - 3 \, {\left ({\left (5 \, b^{2} c - 3 \, a b d\right )} x^{5} + {\left (5 \, a b c - 3 \, a^{2} d\right )} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{12 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}, \frac {3 \, {\left (5 \, b^{2} c - 3 \, a b d\right )} x^{4} - 2 \, a^{2} c + 2 \, {\left (5 \, a b c - 3 \, a^{2} d\right )} x^{2} + 3 \, {\left ({\left (5 \, b^{2} c - 3 \, a b d\right )} x^{5} + {\left (5 \, a b c - 3 \, a^{2} d\right )} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{6 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}\right ] \] Input:
integrate((d*x^2+c)/x^4/(b*x^2+a)^2,x, algorithm="fricas")
Output:
[1/12*(6*(5*b^2*c - 3*a*b*d)*x^4 - 4*a^2*c + 4*(5*a*b*c - 3*a^2*d)*x^2 - 3 *((5*b^2*c - 3*a*b*d)*x^5 + (5*a*b*c - 3*a^2*d)*x^3)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^3*b*x^5 + a^4*x^3), 1/6*(3*(5*b^ 2*c - 3*a*b*d)*x^4 - 2*a^2*c + 2*(5*a*b*c - 3*a^2*d)*x^2 + 3*((5*b^2*c - 3 *a*b*d)*x^5 + (5*a*b*c - 3*a^2*d)*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a^3 *b*x^5 + a^4*x^3)]
Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (80) = 160\).
Time = 0.29 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.04 \[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {\sqrt {- \frac {b}{a^{7}}} \cdot \left (3 a d - 5 b c\right ) \log {\left (- \frac {a^{4} \sqrt {- \frac {b}{a^{7}}} \cdot \left (3 a d - 5 b c\right )}{3 a b d - 5 b^{2} c} + x \right )}}{4} - \frac {\sqrt {- \frac {b}{a^{7}}} \cdot \left (3 a d - 5 b c\right ) \log {\left (\frac {a^{4} \sqrt {- \frac {b}{a^{7}}} \cdot \left (3 a d - 5 b c\right )}{3 a b d - 5 b^{2} c} + x \right )}}{4} + \frac {- 2 a^{2} c + x^{4} \left (- 9 a b d + 15 b^{2} c\right ) + x^{2} \left (- 6 a^{2} d + 10 a b c\right )}{6 a^{4} x^{3} + 6 a^{3} b x^{5}} \] Input:
integrate((d*x**2+c)/x**4/(b*x**2+a)**2,x)
Output:
sqrt(-b/a**7)*(3*a*d - 5*b*c)*log(-a**4*sqrt(-b/a**7)*(3*a*d - 5*b*c)/(3*a *b*d - 5*b**2*c) + x)/4 - sqrt(-b/a**7)*(3*a*d - 5*b*c)*log(a**4*sqrt(-b/a **7)*(3*a*d - 5*b*c)/(3*a*b*d - 5*b**2*c) + x)/4 + (-2*a**2*c + x**4*(-9*a *b*d + 15*b**2*c) + x**2*(-6*a**2*d + 10*a*b*c))/(6*a**4*x**3 + 6*a**3*b*x **5)
Time = 0.12 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {3 \, {\left (5 \, b^{2} c - 3 \, a b d\right )} x^{4} - 2 \, a^{2} c + 2 \, {\left (5 \, a b c - 3 \, a^{2} d\right )} x^{2}}{6 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}} + \frac {{\left (5 \, b^{2} c - 3 \, a b d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} \] Input:
integrate((d*x^2+c)/x^4/(b*x^2+a)^2,x, algorithm="maxima")
Output:
1/6*(3*(5*b^2*c - 3*a*b*d)*x^4 - 2*a^2*c + 2*(5*a*b*c - 3*a^2*d)*x^2)/(a^3 *b*x^5 + a^4*x^3) + 1/2*(5*b^2*c - 3*a*b*d)*arctan(b*x/sqrt(a*b))/(sqrt(a* b)*a^3)
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {{\left (5 \, b^{2} c - 3 \, a b d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} + \frac {b^{2} c x - a b d x}{2 \, {\left (b x^{2} + a\right )} a^{3}} + \frac {6 \, b c x^{2} - 3 \, a d x^{2} - a c}{3 \, a^{3} x^{3}} \] Input:
integrate((d*x^2+c)/x^4/(b*x^2+a)^2,x, algorithm="giac")
Output:
1/2*(5*b^2*c - 3*a*b*d)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) + 1/2*(b^2*c *x - a*b*d*x)/((b*x^2 + a)*a^3) + 1/3*(6*b*c*x^2 - 3*a*d*x^2 - a*c)/(a^3*x ^3)
Time = 0.54 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.93 \[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {\frac {c}{3\,a}+\frac {x^2\,\left (3\,a\,d-5\,b\,c\right )}{3\,a^2}+\frac {b\,x^4\,\left (3\,a\,d-5\,b\,c\right )}{2\,a^3}}{b\,x^5+a\,x^3}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (3\,a\,d-5\,b\,c\right )}{2\,a^{7/2}} \] Input:
int((c + d*x^2)/(x^4*(a + b*x^2)^2),x)
Output:
- (c/(3*a) + (x^2*(3*a*d - 5*b*c))/(3*a^2) + (b*x^4*(3*a*d - 5*b*c))/(2*a^ 3))/(a*x^3 + b*x^5) - (b^(1/2)*atan((b^(1/2)*x)/a^(1/2))*(3*a*d - 5*b*c))/ (2*a^(7/2))
Time = 0.20 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.79 \[ \int \frac {c+d x^2}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {-9 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} d \,x^{3}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b c \,x^{3}-9 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b d \,x^{5}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} c \,x^{5}-2 a^{3} c -6 a^{3} d \,x^{2}+10 a^{2} b c \,x^{2}-9 a^{2} b d \,x^{4}+15 a \,b^{2} c \,x^{4}}{6 a^{4} x^{3} \left (b \,x^{2}+a \right )} \] Input:
int((d*x^2+c)/x^4/(b*x^2+a)^2,x)
Output:
( - 9*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*d*x**3 + 15*sqrt( b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*c*x**3 - 9*sqrt(b)*sqrt(a)*at an((b*x)/(sqrt(b)*sqrt(a)))*a*b*d*x**5 + 15*sqrt(b)*sqrt(a)*atan((b*x)/(sq rt(b)*sqrt(a)))*b**2*c*x**5 - 2*a**3*c - 6*a**3*d*x**2 + 10*a**2*b*c*x**2 - 9*a**2*b*d*x**4 + 15*a*b**2*c*x**4)/(6*a**4*x**3*(a + b*x**2))