Integrand size = 28, antiderivative size = 130 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx=\frac {A b-a C+(b B-a D) x}{4 a b \left (a+b x^2\right )^2}+\frac {4 A b+(3 b B+a D) x}{8 a^2 b \left (a+b x^2\right )}+\frac {(3 b B+a D) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2}}+\frac {A \log (x)}{a^3}-\frac {A \log \left (a+b x^2\right )}{2 a^3} \]
1/4*(A*b-C*a+(B*b-D*a)*x)/a/b/(b*x^2+a)^2+1/8*(4*A*b+(3*B*b+D*a)*x)/a^2/b/ (b*x^2+a)+1/8*(3*B*b+D*a)*arctan(x*b^(1/2)/a^(1/2))/a^(5/2)/b^(3/2)+A*ln(x )/a^3-1/2*A*ln(b*x^2+a)/a^3
Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx=\frac {\frac {a (4 A b+3 b B x+a D x)}{b \left (a+b x^2\right )}+\frac {2 a^2 (A b+b B x-a (C+D x))}{b \left (a+b x^2\right )^2}+\frac {\sqrt {a} (3 b B+a D) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}+8 A \log (x)-4 A \log \left (a+b x^2\right )}{8 a^3} \]
((a*(4*A*b + 3*b*B*x + a*D*x))/(b*(a + b*x^2)) + (2*a^2*(A*b + b*B*x - a*( C + D*x)))/(b*(a + b*x^2)^2) + (Sqrt[a]*(3*b*B + a*D)*ArcTan[(Sqrt[b]*x)/S qrt[a]])/b^(3/2) + 8*A*Log[x] - 4*A*Log[a + b*x^2])/(8*a^3)
Time = 0.38 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2336, 25, 27, 532, 25, 523, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 2336 |
\(\displaystyle \frac {x (b B-a D)-a C+A b}{4 a b \left (a+b x^2\right )^2}-\frac {\int -\frac {4 A b+(3 b B+a D) x}{b x \left (b x^2+a\right )^2}dx}{4 a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {4 A b+(3 b B+a D) x}{b x \left (b x^2+a\right )^2}dx}{4 a}+\frac {x (b B-a D)-a C+A b}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {4 A b+(3 b B+a D) x}{x \left (b x^2+a\right )^2}dx}{4 a b}+\frac {x (b B-a D)-a C+A b}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 532 |
\(\displaystyle \frac {\frac {x (a D+3 b B)+4 A b}{2 a \left (a+b x^2\right )}-\frac {\int -\frac {8 A b+(3 b B+a D) x}{x \left (b x^2+a\right )}dx}{2 a}}{4 a b}+\frac {x (b B-a D)-a C+A b}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {8 A b+(3 b B+a D) x}{x \left (b x^2+a\right )}dx}{2 a}+\frac {x (a D+3 b B)+4 A b}{2 a \left (a+b x^2\right )}}{4 a b}+\frac {x (b B-a D)-a C+A b}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 523 |
\(\displaystyle \frac {\frac {\int \left (\frac {8 A b}{a x}+\frac {D a^2+3 b B a-8 A b^2 x}{a \left (b x^2+a\right )}\right )dx}{2 a}+\frac {x (a D+3 b B)+4 A b}{2 a \left (a+b x^2\right )}}{4 a b}+\frac {x (b B-a D)-a C+A b}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {-\frac {4 A b \log \left (a+b x^2\right )}{a}+\frac {8 A b \log (x)}{a}+\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (a D+3 b B)}{\sqrt {a} \sqrt {b}}}{2 a}+\frac {x (a D+3 b B)+4 A b}{2 a \left (a+b x^2\right )}}{4 a b}+\frac {x (b B-a D)-a C+A b}{4 a b \left (a+b x^2\right )^2}\) |
(A*b - a*C + (b*B - a*D)*x)/(4*a*b*(a + b*x^2)^2) + ((4*A*b + (3*b*B + a*D )*x)/(2*a*(a + b*x^2)) + (((3*b*B + a*D)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqr t[a]*Sqrt[b]) + (8*A*b*Log[x])/a - (4*A*b*Log[a + b*x^2])/a)/(2*a))/(4*a*b )
3.2.7.3.1 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[((x_)^(m_.)*((c_) + (d_.)*(x_)))/((a_) + (b_.)*(x_)^2), x_Symbol] :> In t[ExpandIntegrand[x^m*((c + d*x)/(a + b*x^2)), x], x] /; FreeQ[{a, b, c, d} , x] && IntegerQ[m]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[x^m *(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*p]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) ^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
Time = 3.45 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {A \ln \left (x \right )}{a^{3}}-\frac {\frac {\left (-\frac {3}{8} a b B -\frac {1}{8} D a^{2}\right ) x^{3}-\frac {a A b \,x^{2}}{2}-\frac {a^{2} \left (5 B b -D a \right ) x}{8 b}-\frac {a^{2} \left (3 A b -C a \right )}{4 b}}{\left (b \,x^{2}+a \right )^{2}}+\frac {4 b A \ln \left (b \,x^{2}+a \right )+\frac {\left (-3 a b B -D a^{2}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{8 b}}{a^{3}}\) | \(130\) |
A*ln(x)/a^3-1/a^3*(((-3/8*a*b*B-1/8*D*a^2)*x^3-1/2*a*A*b*x^2-1/8*a^2*(5*B* b-D*a)/b*x-1/4*a^2*(3*A*b-C*a)/b)/(b*x^2+a)^2+1/8/b*(4*b*A*ln(b*x^2+a)+(-3 *B*a*b-D*a^2)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (114) = 228\).
Time = 0.32 (sec) , antiderivative size = 488, normalized size of antiderivative = 3.75 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx=\left [\frac {8 \, A a b^{3} x^{2} - 4 \, C a^{3} b + 12 \, A a^{2} b^{2} + 2 \, {\left (D a^{2} b^{2} + 3 \, B a b^{3}\right )} x^{3} - {\left ({\left (D a b^{2} + 3 \, B b^{3}\right )} x^{4} + D a^{3} + 3 \, B a^{2} b + 2 \, {\left (D a^{2} b + 3 \, B a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 2 \, {\left (D a^{3} b - 5 \, B a^{2} b^{2}\right )} x - 8 \, {\left (A b^{4} x^{4} + 2 \, A a b^{3} x^{2} + A a^{2} b^{2}\right )} \log \left (b x^{2} + a\right ) + 16 \, {\left (A b^{4} x^{4} + 2 \, A a b^{3} x^{2} + A a^{2} b^{2}\right )} \log \left (x\right )}{16 \, {\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}, \frac {4 \, A a b^{3} x^{2} - 2 \, C a^{3} b + 6 \, A a^{2} b^{2} + {\left (D a^{2} b^{2} + 3 \, B a b^{3}\right )} x^{3} + {\left ({\left (D a b^{2} + 3 \, B b^{3}\right )} x^{4} + D a^{3} + 3 \, B a^{2} b + 2 \, {\left (D a^{2} b + 3 \, B a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - {\left (D a^{3} b - 5 \, B a^{2} b^{2}\right )} x - 4 \, {\left (A b^{4} x^{4} + 2 \, A a b^{3} x^{2} + A a^{2} b^{2}\right )} \log \left (b x^{2} + a\right ) + 8 \, {\left (A b^{4} x^{4} + 2 \, A a b^{3} x^{2} + A a^{2} b^{2}\right )} \log \left (x\right )}{8 \, {\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}\right ] \]
[1/16*(8*A*a*b^3*x^2 - 4*C*a^3*b + 12*A*a^2*b^2 + 2*(D*a^2*b^2 + 3*B*a*b^3 )*x^3 - ((D*a*b^2 + 3*B*b^3)*x^4 + D*a^3 + 3*B*a^2*b + 2*(D*a^2*b + 3*B*a* b^2)*x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 2*(D* a^3*b - 5*B*a^2*b^2)*x - 8*(A*b^4*x^4 + 2*A*a*b^3*x^2 + A*a^2*b^2)*log(b*x ^2 + a) + 16*(A*b^4*x^4 + 2*A*a*b^3*x^2 + A*a^2*b^2)*log(x))/(a^3*b^4*x^4 + 2*a^4*b^3*x^2 + a^5*b^2), 1/8*(4*A*a*b^3*x^2 - 2*C*a^3*b + 6*A*a^2*b^2 + (D*a^2*b^2 + 3*B*a*b^3)*x^3 + ((D*a*b^2 + 3*B*b^3)*x^4 + D*a^3 + 3*B*a^2* b + 2*(D*a^2*b + 3*B*a*b^2)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) - (D*a^3* b - 5*B*a^2*b^2)*x - 4*(A*b^4*x^4 + 2*A*a*b^3*x^2 + A*a^2*b^2)*log(b*x^2 + a) + 8*(A*b^4*x^4 + 2*A*a*b^3*x^2 + A*a^2*b^2)*log(x))/(a^3*b^4*x^4 + 2*a ^4*b^3*x^2 + a^5*b^2)]
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.02 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx=\frac {4 \, A b^{2} x^{2} + {\left (D a b + 3 \, B b^{2}\right )} x^{3} - 2 \, C a^{2} + 6 \, A a b - {\left (D a^{2} - 5 \, B a b\right )} x}{8 \, {\left (a^{2} b^{3} x^{4} + 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )}} - \frac {A \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {A \log \left (x\right )}{a^{3}} + \frac {{\left (D a + 3 \, B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b} \]
1/8*(4*A*b^2*x^2 + (D*a*b + 3*B*b^2)*x^3 - 2*C*a^2 + 6*A*a*b - (D*a^2 - 5* B*a*b)*x)/(a^2*b^3*x^4 + 2*a^3*b^2*x^2 + a^4*b) - 1/2*A*log(b*x^2 + a)/a^3 + A*log(x)/a^3 + 1/8*(D*a + 3*B*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2*b )
Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx=-\frac {A \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {A \log \left ({\left | x \right |}\right )}{a^{3}} + \frac {{\left (D a + 3 \, B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b} + \frac {4 \, A a b^{2} x^{2} - 2 \, C a^{3} + 6 \, A a^{2} b + {\left (D a^{2} b + 3 \, B a b^{2}\right )} x^{3} - {\left (D a^{3} - 5 \, B a^{2} b\right )} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3} b} \]
-1/2*A*log(b*x^2 + a)/a^3 + A*log(abs(x))/a^3 + 1/8*(D*a + 3*B*b)*arctan(b *x/sqrt(a*b))/(sqrt(a*b)*a^2*b) + 1/8*(4*A*a*b^2*x^2 - 2*C*a^3 + 6*A*a^2*b + (D*a^2*b + 3*B*a*b^2)*x^3 - (D*a^3 - 5*B*a^2*b)*x)/((b*x^2 + a)^2*a^3*b )
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{x\,{\left (b\,x^2+a\right )}^3} \,d x \]