Integrand size = 20, antiderivative size = 126 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^4} \, dx=-\frac {a B-(A c-a C) x}{6 a c \left (a+c x^2\right )^3}+\frac {(5 A c+a C) x}{24 a^2 c \left (a+c x^2\right )^2}+\frac {(5 A c+a C) x}{16 a^3 c \left (a+c x^2\right )}+\frac {(5 A c+a C) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{3/2}} \]
1/6*(-a*B+(A*c-C*a)*x)/a/c/(c*x^2+a)^3+1/24*(5*A*c+C*a)*x/a^2/c/(c*x^2+a)^ 2+1/16*(5*A*c+C*a)*x/a^3/c/(c*x^2+a)+1/16*(5*A*c+C*a)*arctan(x*c^(1/2)/a^( 1/2))/a^(7/2)/c^(3/2)
Time = 0.06 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^4} \, dx=\frac {15 A c^3 x^5-a^3 (8 B+3 C x)+a c^2 x^3 \left (40 A+3 C x^2\right )+a^2 c x \left (33 A+8 C x^2\right )}{48 a^3 c \left (a+c x^2\right )^3}+\frac {(5 A c+a C) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{3/2}} \]
(15*A*c^3*x^5 - a^3*(8*B + 3*C*x) + a*c^2*x^3*(40*A + 3*C*x^2) + a^2*c*x*( 33*A + 8*C*x^2))/(48*a^3*c*(a + c*x^2)^3) + ((5*A*c + a*C)*ArcTan[(Sqrt[c] *x)/Sqrt[a]])/(16*a^(7/2)*c^(3/2))
Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2345, 25, 27, 215, 215, 218}
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.
\(\displaystyle \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^4} \, dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {\int -\frac {5 A c+a C}{c \left (c x^2+a\right )^3}dx}{6 a}-\frac {a B-x (A c-a C)}{6 a c \left (a+c x^2\right )^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {5 A c+a C}{c \left (c x^2+a\right )^3}dx}{6 a}-\frac {a B-x (A c-a C)}{6 a c \left (a+c x^2\right )^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a C+5 A c) \int \frac {1}{\left (c x^2+a\right )^3}dx}{6 a c}-\frac {a B-x (A c-a C)}{6 a c \left (a+c x^2\right )^3}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {(a C+5 A c) \left (\frac {3 \int \frac {1}{\left (c x^2+a\right )^2}dx}{4 a}+\frac {x}{4 a \left (a+c x^2\right )^2}\right )}{6 a c}-\frac {a B-x (A c-a C)}{6 a c \left (a+c x^2\right )^3}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {(a C+5 A c) \left (\frac {3 \left (\frac {\int \frac {1}{c x^2+a}dx}{2 a}+\frac {x}{2 a \left (a+c x^2\right )}\right )}{4 a}+\frac {x}{4 a \left (a+c x^2\right )^2}\right )}{6 a c}-\frac {a B-x (A c-a C)}{6 a c \left (a+c x^2\right )^3}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(a C+5 A c) \left (\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {c}}+\frac {x}{2 a \left (a+c x^2\right )}\right )}{4 a}+\frac {x}{4 a \left (a+c x^2\right )^2}\right )}{6 a c}-\frac {a B-x (A c-a C)}{6 a c \left (a+c x^2\right )^3}\) |
-1/6*(a*B - (A*c - a*C)*x)/(a*c*(a + c*x^2)^3) + ((5*A*c + a*C)*(x/(4*a*(a + c*x^2)^2) + (3*(x/(2*a*(a + c*x^2)) + ArcTan[(Sqrt[c]*x)/Sqrt[a]]/(2*a^ (3/2)*Sqrt[c])))/(4*a)))/(6*a*c)
3.1.68.3.1 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)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
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[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[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)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Time = 0.52 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.79
method | result | size |
default | \(\frac {\frac {\left (5 A c +C a \right ) c \,x^{5}}{16 a^{3}}+\frac {\left (5 A c +C a \right ) x^{3}}{6 a^{2}}+\frac {\left (11 A c -C a \right ) x}{16 a c}-\frac {B}{6 c}}{\left (c \,x^{2}+a \right )^{3}}+\frac {\left (5 A c +C a \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 a^{3} c \sqrt {a c}}\) | \(100\) |
risch | \(\frac {\frac {\left (5 A c +C a \right ) c \,x^{5}}{16 a^{3}}+\frac {\left (5 A c +C a \right ) x^{3}}{6 a^{2}}+\frac {\left (11 A c -C a \right ) x}{16 a c}-\frac {B}{6 c}}{\left (c \,x^{2}+a \right )^{3}}-\frac {5 \ln \left (c x +\sqrt {-a c}\right ) A}{32 \sqrt {-a c}\, a^{3}}-\frac {\ln \left (c x +\sqrt {-a c}\right ) C}{32 \sqrt {-a c}\, c \,a^{2}}+\frac {5 \ln \left (-c x +\sqrt {-a c}\right ) A}{32 \sqrt {-a c}\, a^{3}}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) C}{32 \sqrt {-a c}\, c \,a^{2}}\) | \(170\) |
(1/16*(5*A*c+C*a)/a^3*c*x^5+1/6/a^2*(5*A*c+C*a)*x^3+1/16*(11*A*c-C*a)/a/c* x-1/6*B/c)/(c*x^2+a)^3+1/16*(5*A*c+C*a)/a^3/c/(a*c)^(1/2)*arctan(c*x/(a*c) ^(1/2))
Time = 0.29 (sec) , antiderivative size = 430, normalized size of antiderivative = 3.41 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^4} \, dx=\left [-\frac {16 \, B a^{4} c - 6 \, {\left (C a^{2} c^{3} + 5 \, A a c^{4}\right )} x^{5} - 16 \, {\left (C a^{3} c^{2} + 5 \, A a^{2} c^{3}\right )} x^{3} + 3 \, {\left ({\left (C a c^{3} + 5 \, A c^{4}\right )} x^{6} + C a^{4} + 5 \, A a^{3} c + 3 \, {\left (C a^{2} c^{2} + 5 \, A a c^{3}\right )} x^{4} + 3 \, {\left (C a^{3} c + 5 \, A a^{2} c^{2}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 6 \, {\left (C a^{4} c - 11 \, A a^{3} c^{2}\right )} x}{96 \, {\left (a^{4} c^{5} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{6} c^{3} x^{2} + a^{7} c^{2}\right )}}, -\frac {8 \, B a^{4} c - 3 \, {\left (C a^{2} c^{3} + 5 \, A a c^{4}\right )} x^{5} - 8 \, {\left (C a^{3} c^{2} + 5 \, A a^{2} c^{3}\right )} x^{3} - 3 \, {\left ({\left (C a c^{3} + 5 \, A c^{4}\right )} x^{6} + C a^{4} + 5 \, A a^{3} c + 3 \, {\left (C a^{2} c^{2} + 5 \, A a c^{3}\right )} x^{4} + 3 \, {\left (C a^{3} c + 5 \, A a^{2} c^{2}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + 3 \, {\left (C a^{4} c - 11 \, A a^{3} c^{2}\right )} x}{48 \, {\left (a^{4} c^{5} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{6} c^{3} x^{2} + a^{7} c^{2}\right )}}\right ] \]
[-1/96*(16*B*a^4*c - 6*(C*a^2*c^3 + 5*A*a*c^4)*x^5 - 16*(C*a^3*c^2 + 5*A*a ^2*c^3)*x^3 + 3*((C*a*c^3 + 5*A*c^4)*x^6 + C*a^4 + 5*A*a^3*c + 3*(C*a^2*c^ 2 + 5*A*a*c^3)*x^4 + 3*(C*a^3*c + 5*A*a^2*c^2)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 6*(C*a^4*c - 11*A*a^3*c^2)*x)/(a^4*c^ 5*x^6 + 3*a^5*c^4*x^4 + 3*a^6*c^3*x^2 + a^7*c^2), -1/48*(8*B*a^4*c - 3*(C* a^2*c^3 + 5*A*a*c^4)*x^5 - 8*(C*a^3*c^2 + 5*A*a^2*c^3)*x^3 - 3*((C*a*c^3 + 5*A*c^4)*x^6 + C*a^4 + 5*A*a^3*c + 3*(C*a^2*c^2 + 5*A*a*c^3)*x^4 + 3*(C*a ^3*c + 5*A*a^2*c^2)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + 3*(C*a^4*c - 11 *A*a^3*c^2)*x)/(a^4*c^5*x^6 + 3*a^5*c^4*x^4 + 3*a^6*c^3*x^2 + a^7*c^2)]
Time = 0.90 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.56 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^4} \, dx=- \frac {\sqrt {- \frac {1}{a^{7} c^{3}}} \cdot \left (5 A c + C a\right ) \log {\left (- a^{4} c \sqrt {- \frac {1}{a^{7} c^{3}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{a^{7} c^{3}}} \cdot \left (5 A c + C a\right ) \log {\left (a^{4} c \sqrt {- \frac {1}{a^{7} c^{3}}} + x \right )}}{32} + \frac {- 8 B a^{3} + x^{5} \cdot \left (15 A c^{3} + 3 C a c^{2}\right ) + x^{3} \cdot \left (40 A a c^{2} + 8 C a^{2} c\right ) + x \left (33 A a^{2} c - 3 C a^{3}\right )}{48 a^{6} c + 144 a^{5} c^{2} x^{2} + 144 a^{4} c^{3} x^{4} + 48 a^{3} c^{4} x^{6}} \]
-sqrt(-1/(a**7*c**3))*(5*A*c + C*a)*log(-a**4*c*sqrt(-1/(a**7*c**3)) + x)/ 32 + sqrt(-1/(a**7*c**3))*(5*A*c + C*a)*log(a**4*c*sqrt(-1/(a**7*c**3)) + x)/32 + (-8*B*a**3 + x**5*(15*A*c**3 + 3*C*a*c**2) + x**3*(40*A*a*c**2 + 8 *C*a**2*c) + x*(33*A*a**2*c - 3*C*a**3))/(48*a**6*c + 144*a**5*c**2*x**2 + 144*a**4*c**3*x**4 + 48*a**3*c**4*x**6)
Time = 0.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^4} \, dx=\frac {3 \, {\left (C a c^{2} + 5 \, A c^{3}\right )} x^{5} - 8 \, B a^{3} + 8 \, {\left (C a^{2} c + 5 \, A a c^{2}\right )} x^{3} - 3 \, {\left (C a^{3} - 11 \, A a^{2} c\right )} x}{48 \, {\left (a^{3} c^{4} x^{6} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{5} c^{2} x^{2} + a^{6} c\right )}} + \frac {{\left (C a + 5 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c} \]
1/48*(3*(C*a*c^2 + 5*A*c^3)*x^5 - 8*B*a^3 + 8*(C*a^2*c + 5*A*a*c^2)*x^3 - 3*(C*a^3 - 11*A*a^2*c)*x)/(a^3*c^4*x^6 + 3*a^4*c^3*x^4 + 3*a^5*c^2*x^2 + a ^6*c) + 1/16*(C*a + 5*A*c)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3*c)
Time = 0.26 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^4} \, dx=\frac {{\left (C a + 5 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c} + \frac {3 \, C a c^{2} x^{5} + 15 \, A c^{3} x^{5} + 8 \, C a^{2} c x^{3} + 40 \, A a c^{2} x^{3} - 3 \, C a^{3} x + 33 \, A a^{2} c x - 8 \, B a^{3}}{48 \, {\left (c x^{2} + a\right )}^{3} a^{3} c} \]
1/16*(C*a + 5*A*c)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3*c) + 1/48*(3*C*a*c ^2*x^5 + 15*A*c^3*x^5 + 8*C*a^2*c*x^3 + 40*A*a*c^2*x^3 - 3*C*a^3*x + 33*A* a^2*c*x - 8*B*a^3)/((c*x^2 + a)^3*a^3*c)
Time = 12.91 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^4} \, dx=\frac {\frac {x^3\,\left (5\,A\,c+C\,a\right )}{6\,a^2}-\frac {B}{6\,c}+\frac {c\,x^5\,\left (5\,A\,c+C\,a\right )}{16\,a^3}+\frac {x\,\left (11\,A\,c-C\,a\right )}{16\,a\,c}}{a^3+3\,a^2\,c\,x^2+3\,a\,c^2\,x^4+c^3\,x^6}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (5\,A\,c+C\,a\right )}{16\,a^{7/2}\,c^{3/2}} \]