Integrand size = 23, antiderivative size = 83 \[ \int \frac {x^2 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=\frac {a^2 d+b^2 c x}{2 b^4 \left (a^2-b^2 x^2\right )}+\frac {(b c+2 a d) \log (a-b x)}{4 a b^4}-\frac {(b c-2 a d) \log (a+b x)}{4 a b^4} \] Output:
1/2*(b^2*c*x+a^2*d)/b^4/(-b^2*x^2+a^2)+1/4*(2*a*d+b*c)*ln(-b*x+a)/a/b^4-1/ 4*(-2*a*d+b*c)*ln(b*x+a)/a/b^4
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.77 \[ \int \frac {x^2 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=\frac {\frac {a^2 d+b^2 c x}{a^2-b^2 x^2}-\frac {b c \text {arctanh}\left (\frac {b x}{a}\right )}{a}+d \log \left (a^2-b^2 x^2\right )}{2 b^4} \] Input:
Integrate[(x^2*(c + d*x))/(a^2 - b^2*x^2)^2,x]
Output:
((a^2*d + b^2*c*x)/(a^2 - b^2*x^2) - (b*c*ArcTanh[(b*x)/a])/a + d*Log[a^2 - b^2*x^2])/(2*b^4)
Time = 0.34 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.92, 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.
\(\displaystyle \int \frac {x^2 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 530 |
\(\displaystyle \frac {a^2 d+b^2 c x}{2 b^4 \left (a^2-b^2 x^2\right )}-\frac {\int \frac {a^2 (c+2 d x)}{b^2 \left (a^2-b^2 x^2\right )}dx}{2 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 d+b^2 c x}{2 b^4 \left (a^2-b^2 x^2\right )}-\frac {\int \frac {c+2 d x}{a^2-b^2 x^2}dx}{2 b^2}\) |
\(\Big \downarrow \) 452 |
\(\displaystyle \frac {a^2 d+b^2 c x}{2 b^4 \left (a^2-b^2 x^2\right )}-\frac {c \int \frac {1}{a^2-b^2 x^2}dx+2 d \int \frac {x}{a^2-b^2 x^2}dx}{2 b^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {a^2 d+b^2 c x}{2 b^4 \left (a^2-b^2 x^2\right )}-\frac {2 d \int \frac {x}{a^2-b^2 x^2}dx+\frac {c \text {arctanh}\left (\frac {b x}{a}\right )}{a b}}{2 b^2}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {a^2 d+b^2 c x}{2 b^4 \left (a^2-b^2 x^2\right )}-\frac {\frac {c \text {arctanh}\left (\frac {b x}{a}\right )}{a b}-\frac {d \log \left (a^2-b^2 x^2\right )}{b^2}}{2 b^2}\) |
Input:
Int[(x^2*(c + d*x))/(a^2 - b^2*x^2)^2,x]
Output:
(a^2*d + b^2*c*x)/(2*b^4*(a^2 - b^2*x^2)) - ((c*ArcTanh[(b*x)/a])/(a*b) - (d*Log[a^2 - b^2*x^2])/b^2)/(2*b^2)
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 = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96
method | result | size |
norman | \(\frac {\frac {a^{2} d}{2 b^{4}}+\frac {c x}{2 b^{2}}}{-b^{2} x^{2}+a^{2}}+\frac {\left (2 a d -b c \right ) \ln \left (b x +a \right )}{4 a \,b^{4}}+\frac {\left (2 a d +b c \right ) \ln \left (-b x +a \right )}{4 a \,b^{4}}\) | \(80\) |
default | \(-\frac {-a d +b c}{4 b^{4} \left (b x +a \right )}+\frac {\left (2 a d -b c \right ) \ln \left (b x +a \right )}{4 a \,b^{4}}+\frac {a d +b c}{4 b^{4} \left (-b x +a \right )}+\frac {\left (2 a d +b c \right ) \ln \left (-b x +a \right )}{4 a \,b^{4}}\) | \(88\) |
risch | \(\frac {\frac {a^{2} d}{2 b^{4}}+\frac {c x}{2 b^{2}}}{-b^{2} x^{2}+a^{2}}+\frac {\ln \left (b x -a \right ) d}{2 b^{4}}+\frac {\ln \left (b x -a \right ) c}{4 a \,b^{3}}+\frac {\ln \left (-b x -a \right ) d}{2 b^{4}}-\frac {\ln \left (-b x -a \right ) c}{4 a \,b^{3}}\) | \(98\) |
parallelrisch | \(\frac {2 \ln \left (b x -a \right ) x^{2} a \,b^{2} d +\ln \left (b x -a \right ) x^{2} b^{3} c +2 \ln \left (b x +a \right ) x^{2} a \,b^{2} d -\ln \left (b x +a \right ) x^{2} b^{3} c -2 \ln \left (b x -a \right ) a^{3} d -\ln \left (b x -a \right ) a^{2} b c -2 \ln \left (b x +a \right ) a^{3} d +\ln \left (b x +a \right ) a^{2} b c -2 a \,b^{2} c x -2 a^{3} d}{4 b^{4} \left (b^{2} x^{2}-a^{2}\right ) a}\) | \(157\) |
Input:
int(x^2*(d*x+c)/(-b^2*x^2+a^2)^2,x,method=_RETURNVERBOSE)
Output:
(1/2*a^2*d/b^4+1/2/b^2*c*x)/(-b^2*x^2+a^2)+1/4/a/b^4*(2*a*d-b*c)*ln(b*x+a) +1/4*(2*a*d+b*c)*ln(-b*x+a)/a/b^4
Time = 0.24 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.39 \[ \int \frac {x^2 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=-\frac {2 \, a b^{2} c x + 2 \, a^{3} d - {\left (a^{2} b c - 2 \, a^{3} d - {\left (b^{3} c - 2 \, a b^{2} d\right )} x^{2}\right )} \log \left (b x + a\right ) + {\left (a^{2} b c + 2 \, a^{3} d - {\left (b^{3} c + 2 \, a b^{2} d\right )} x^{2}\right )} \log \left (b x - a\right )}{4 \, {\left (a b^{6} x^{2} - a^{3} b^{4}\right )}} \] Input:
integrate(x^2*(d*x+c)/(-b^2*x^2+a^2)^2,x, algorithm="fricas")
Output:
-1/4*(2*a*b^2*c*x + 2*a^3*d - (a^2*b*c - 2*a^3*d - (b^3*c - 2*a*b^2*d)*x^2 )*log(b*x + a) + (a^2*b*c + 2*a^3*d - (b^3*c + 2*a*b^2*d)*x^2)*log(b*x - a ))/(a*b^6*x^2 - a^3*b^4)
Time = 0.31 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.33 \[ \int \frac {x^2 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=\frac {- a^{2} d - b^{2} c x}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} + \frac {\left (2 a d - b c\right ) \log {\left (x + \frac {2 a^{2} d - a \left (2 a d - b c\right )}{b^{2} c} \right )}}{4 a b^{4}} + \frac {\left (2 a d + b c\right ) \log {\left (x + \frac {2 a^{2} d - a \left (2 a d + b c\right )}{b^{2} c} \right )}}{4 a b^{4}} \] Input:
integrate(x**2*(d*x+c)/(-b**2*x**2+a**2)**2,x)
Output:
(-a**2*d - b**2*c*x)/(-2*a**2*b**4 + 2*b**6*x**2) + (2*a*d - b*c)*log(x + (2*a**2*d - a*(2*a*d - b*c))/(b**2*c))/(4*a*b**4) + (2*a*d + b*c)*log(x + (2*a**2*d - a*(2*a*d + b*c))/(b**2*c))/(4*a*b**4)
Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {x^2 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=-\frac {b^{2} c x + a^{2} d}{2 \, {\left (b^{6} x^{2} - a^{2} b^{4}\right )}} - \frac {{\left (b c - 2 \, a d\right )} \log \left (b x + a\right )}{4 \, a b^{4}} + \frac {{\left (b c + 2 \, a d\right )} \log \left (b x - a\right )}{4 \, a b^{4}} \] Input:
integrate(x^2*(d*x+c)/(-b^2*x^2+a^2)^2,x, algorithm="maxima")
Output:
-1/2*(b^2*c*x + a^2*d)/(b^6*x^2 - a^2*b^4) - 1/4*(b*c - 2*a*d)*log(b*x + a )/(a*b^4) + 1/4*(b*c + 2*a*d)*log(b*x - a)/(a*b^4)
Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int \frac {x^2 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=-\frac {c x + \frac {a^{2} d}{b^{2}}}{2 \, {\left (b x + a\right )} {\left (b x - a\right )} b^{2}} - \frac {{\left (b c - 2 \, a d\right )} \log \left ({\left | b x + a \right |}\right )}{4 \, a b^{4}} + \frac {{\left (b c + 2 \, a d\right )} \log \left ({\left | b x - a \right |}\right )}{4 \, a b^{4}} \] Input:
integrate(x^2*(d*x+c)/(-b^2*x^2+a^2)^2,x, algorithm="giac")
Output:
-1/2*(c*x + a^2*d/b^2)/((b*x + a)*(b*x - a)*b^2) - 1/4*(b*c - 2*a*d)*log(a bs(b*x + a))/(a*b^4) + 1/4*(b*c + 2*a*d)*log(abs(b*x - a))/(a*b^4)
Time = 0.14 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.24 \[ \int \frac {x^2 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=\frac {a^2\,d}{2\,\left (a^2\,b^4-b^6\,x^2\right )}+\frac {c\,x}{2\,\left (a^2\,b^2-b^4\,x^2\right )}+\frac {d\,\ln \left (a+b\,x\right )}{2\,b^4}+\frac {d\,\ln \left (a-b\,x\right )}{2\,b^4}-\frac {c\,\ln \left (a+b\,x\right )}{4\,a\,b^3}+\frac {c\,\ln \left (a-b\,x\right )}{4\,a\,b^3} \] Input:
int((x^2*(c + d*x))/(a^2 - b^2*x^2)^2,x)
Output:
(a^2*d)/(2*(a^2*b^4 - b^6*x^2)) + (c*x)/(2*(a^2*b^2 - b^4*x^2)) + (d*log(a + b*x))/(2*b^4) + (d*log(a - b*x))/(2*b^4) - (c*log(a + b*x))/(4*a*b^3) + (c*log(a - b*x))/(4*a*b^3)
Time = 0.19 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.01 \[ \int \frac {x^2 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=\frac {2 \,\mathrm {log}\left (-b x -a \right ) a^{3} d -\mathrm {log}\left (-b x -a \right ) a^{2} b c -2 \,\mathrm {log}\left (-b x -a \right ) a \,b^{2} d \,x^{2}+\mathrm {log}\left (-b x -a \right ) b^{3} c \,x^{2}+2 \,\mathrm {log}\left (-b x +a \right ) a^{3} d +\mathrm {log}\left (-b x +a \right ) a^{2} b c -2 \,\mathrm {log}\left (-b x +a \right ) a \,b^{2} d \,x^{2}-\mathrm {log}\left (-b x +a \right ) b^{3} c \,x^{2}+2 a \,b^{2} c x +2 a \,b^{2} d \,x^{2}}{4 a \,b^{4} \left (-b^{2} x^{2}+a^{2}\right )} \] Input:
int(x^2*(d*x+c)/(-b^2*x^2+a^2)^2,x)
Output:
(2*log( - a - b*x)*a**3*d - log( - a - b*x)*a**2*b*c - 2*log( - a - b*x)*a *b**2*d*x**2 + log( - a - b*x)*b**3*c*x**2 + 2*log(a - b*x)*a**3*d + log(a - b*x)*a**2*b*c - 2*log(a - b*x)*a*b**2*d*x**2 - log(a - b*x)*b**3*c*x**2 + 2*a*b**2*c*x + 2*a*b**2*d*x**2)/(4*a*b**4*(a**2 - b**2*x**2))