Integrand size = 23, antiderivative size = 77 \[ \int \frac {x^2 (d+e x)}{\left (a^2-c^2 x^2\right )^2} \, dx=\frac {x (d+e x)}{2 c^2 \left (a^2-c^2 x^2\right )}+\frac {(c d+2 a e) \log (a-c x)}{4 a c^4}-\frac {(c d-2 a e) \log (a+c x)}{4 a c^4} \]
1/2*x*(e*x+d)/c^2/(-c^2*x^2+a^2)+1/4*(2*a*e+c*d)*ln(-c*x+a)/a/c^4-1/4*(-2* a*e+c*d)*ln(c*x+a)/a/c^4
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 (d+e x)}{\left (a^2-c^2 x^2\right )^2} \, dx=\frac {\frac {a^2 e+c^2 d x}{a^2-c^2 x^2}-\frac {c d \text {arctanh}\left (\frac {c x}{a}\right )}{a}+e \log \left (a^2-c^2 x^2\right )}{2 c^4} \]
((a^2*e + c^2*d*x)/(a^2 - c^2*x^2) - (c*d*ArcTanh[(c*x)/a])/a + e*Log[a^2 - c^2*x^2])/(2*c^4)
Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {530, 27, 452, 221, 240}
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.
\(\displaystyle \int \frac {x^2 (d+e x)}{\left (a^2-c^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 530 |
\(\displaystyle \frac {a^2 e+c^2 d x}{2 c^4 \left (a^2-c^2 x^2\right )}-\frac {\int \frac {a^2 (d+2 e x)}{c^2 \left (a^2-c^2 x^2\right )}dx}{2 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 e+c^2 d x}{2 c^4 \left (a^2-c^2 x^2\right )}-\frac {\int \frac {d+2 e x}{a^2-c^2 x^2}dx}{2 c^2}\) |
\(\Big \downarrow \) 452 |
\(\displaystyle \frac {a^2 e+c^2 d x}{2 c^4 \left (a^2-c^2 x^2\right )}-\frac {d \int \frac {1}{a^2-c^2 x^2}dx+2 e \int \frac {x}{a^2-c^2 x^2}dx}{2 c^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {a^2 e+c^2 d x}{2 c^4 \left (a^2-c^2 x^2\right )}-\frac {2 e \int \frac {x}{a^2-c^2 x^2}dx+\frac {d \text {arctanh}\left (\frac {c x}{a}\right )}{a c}}{2 c^2}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {a^2 e+c^2 d x}{2 c^4 \left (a^2-c^2 x^2\right )}-\frac {\frac {d \text {arctanh}\left (\frac {c x}{a}\right )}{a c}-\frac {e \log \left (a^2-c^2 x^2\right )}{c^2}}{2 c^2}\) |
(a^2*e + c^2*d*x)/(2*c^4*(a^2 - c^2*x^2)) - ((d*ArcTanh[(c*x)/a])/(a*c) - (e*Log[a^2 - c^2*x^2])/c^2)/(2*c^2)
3.4.8.3.1 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)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c Int[1/ (a + b*x^2), x], x] + Simp[d Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c^2 + a*d^2, 0]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb ol] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Co eff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Po lynomialRemainder[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[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + e*(2*p + 3), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && EqQ[n, 1] && IntegerQ[2*p]
Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04
method | result | size |
norman | \(\frac {\frac {a^{2} e}{2 c^{4}}+\frac {d x}{2 c^{2}}}{-c^{2} x^{2}+a^{2}}+\frac {\left (2 a e -c d \right ) \ln \left (c x +a \right )}{4 a \,c^{4}}+\frac {\left (2 a e +c d \right ) \ln \left (-c x +a \right )}{4 a \,c^{4}}\) | \(80\) |
default | \(-\frac {-a e +c d}{4 c^{4} \left (c x +a \right )}+\frac {\left (2 a e -c d \right ) \ln \left (c x +a \right )}{4 a \,c^{4}}+\frac {a e +c d}{4 c^{4} \left (-c x +a \right )}+\frac {\left (2 a e +c d \right ) \ln \left (-c x +a \right )}{4 a \,c^{4}}\) | \(88\) |
risch | \(\frac {\frac {a^{2} e}{2 c^{4}}+\frac {d x}{2 c^{2}}}{-c^{2} x^{2}+a^{2}}+\frac {\ln \left (c x -a \right ) e}{2 c^{4}}+\frac {\ln \left (c x -a \right ) d}{4 a \,c^{3}}+\frac {\ln \left (-c x -a \right ) e}{2 c^{4}}-\frac {\ln \left (-c x -a \right ) d}{4 a \,c^{3}}\) | \(98\) |
parallelrisch | \(\frac {2 \ln \left (c x -a \right ) x^{2} a \,c^{2} e +\ln \left (c x -a \right ) x^{2} c^{3} d +2 \ln \left (c x +a \right ) x^{2} a \,c^{2} e -\ln \left (c x +a \right ) x^{2} c^{3} d -2 \ln \left (c x -a \right ) a^{3} e -\ln \left (c x -a \right ) a^{2} c d -2 \ln \left (c x +a \right ) a^{3} e +\ln \left (c x +a \right ) a^{2} c d -2 x a \,c^{2} d -2 a^{3} e}{4 c^{4} \left (c^{2} x^{2}-a^{2}\right ) a}\) | \(157\) |
(1/2*a^2*e/c^4+1/2*d*x/c^2)/(-c^2*x^2+a^2)+1/4/a/c^4*(2*a*e-c*d)*ln(c*x+a) +1/4*(2*a*e+c*d)*ln(-c*x+a)/a/c^4
Time = 0.39 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.49 \[ \int \frac {x^2 (d+e x)}{\left (a^2-c^2 x^2\right )^2} \, dx=-\frac {2 \, a c^{2} d x + 2 \, a^{3} e - {\left (a^{2} c d - 2 \, a^{3} e - {\left (c^{3} d - 2 \, a c^{2} e\right )} x^{2}\right )} \log \left (c x + a\right ) + {\left (a^{2} c d + 2 \, a^{3} e - {\left (c^{3} d + 2 \, a c^{2} e\right )} x^{2}\right )} \log \left (c x - a\right )}{4 \, {\left (a c^{6} x^{2} - a^{3} c^{4}\right )}} \]
-1/4*(2*a*c^2*d*x + 2*a^3*e - (a^2*c*d - 2*a^3*e - (c^3*d - 2*a*c^2*e)*x^2 )*log(c*x + a) + (a^2*c*d + 2*a^3*e - (c^3*d + 2*a*c^2*e)*x^2)*log(c*x - a ))/(a*c^6*x^2 - a^3*c^4)
Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.43 \[ \int \frac {x^2 (d+e x)}{\left (a^2-c^2 x^2\right )^2} \, dx=\frac {- a^{2} e - c^{2} d x}{- 2 a^{2} c^{4} + 2 c^{6} x^{2}} + \frac {\left (2 a e - c d\right ) \log {\left (x + \frac {2 a^{2} e - a \left (2 a e - c d\right )}{c^{2} d} \right )}}{4 a c^{4}} + \frac {\left (2 a e + c d\right ) \log {\left (x + \frac {2 a^{2} e - a \left (2 a e + c d\right )}{c^{2} d} \right )}}{4 a c^{4}} \]
(-a**2*e - c**2*d*x)/(-2*a**2*c**4 + 2*c**6*x**2) + (2*a*e - c*d)*log(x + (2*a**2*e - a*(2*a*e - c*d))/(c**2*d))/(4*a*c**4) + (2*a*e + c*d)*log(x + (2*a**2*e - a*(2*a*e + c*d))/(c**2*d))/(4*a*c**4)
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03 \[ \int \frac {x^2 (d+e x)}{\left (a^2-c^2 x^2\right )^2} \, dx=-\frac {c^{2} d x + a^{2} e}{2 \, {\left (c^{6} x^{2} - a^{2} c^{4}\right )}} - \frac {{\left (c d - 2 \, a e\right )} \log \left (c x + a\right )}{4 \, a c^{4}} + \frac {{\left (c d + 2 \, a e\right )} \log \left (c x - a\right )}{4 \, a c^{4}} \]
-1/2*(c^2*d*x + a^2*e)/(c^6*x^2 - a^2*c^4) - 1/4*(c*d - 2*a*e)*log(c*x + a )/(a*c^4) + 1/4*(c*d + 2*a*e)*log(c*x - a)/(a*c^4)
Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.06 \[ \int \frac {x^2 (d+e x)}{\left (a^2-c^2 x^2\right )^2} \, dx=-\frac {d x + \frac {a^{2} e}{c^{2}}}{2 \, {\left (c x + a\right )} {\left (c x - a\right )} c^{2}} - \frac {{\left (c d - 2 \, a e\right )} \log \left ({\left | c x + a \right |}\right )}{4 \, a c^{4}} + \frac {{\left (c d + 2 \, a e\right )} \log \left ({\left | c x - a \right |}\right )}{4 \, a c^{4}} \]
-1/2*(d*x + a^2*e/c^2)/((c*x + a)*(c*x - a)*c^2) - 1/4*(c*d - 2*a*e)*log(a bs(c*x + a))/(a*c^4) + 1/4*(c*d + 2*a*e)*log(abs(c*x - a))/(a*c^4)
Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.34 \[ \int \frac {x^2 (d+e x)}{\left (a^2-c^2 x^2\right )^2} \, dx=\frac {a^2\,e}{2\,\left (a^2\,c^4-c^6\,x^2\right )}+\frac {d\,x}{2\,\left (a^2\,c^2-c^4\,x^2\right )}+\frac {e\,\ln \left (a+c\,x\right )}{2\,c^4}+\frac {e\,\ln \left (a-c\,x\right )}{2\,c^4}-\frac {d\,\ln \left (a+c\,x\right )}{4\,a\,c^3}+\frac {d\,\ln \left (a-c\,x\right )}{4\,a\,c^3} \]