Integrand size = 23, antiderivative size = 81 \[ \int \frac {x^3 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=\frac {d x}{b^4}+\frac {a^2 (c+d x)}{2 b^4 \left (a^2-b^2 x^2\right )}+\frac {(2 b c+3 a d) \log (a-b x)}{4 b^5}+\frac {(2 b c-3 a d) \log (a+b x)}{4 b^5} \] Output:
d*x/b^4+1/2*a^2*(d*x+c)/b^4/(-b^2*x^2+a^2)+1/4*(3*a*d+2*b*c)*ln(-b*x+a)/b^ 5+1/4*(-3*a*d+2*b*c)*ln(b*x+a)/b^5
Time = 0.06 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=\frac {2 b d x+\frac {a^2 b (c+d x)}{a^2-b^2 x^2}-3 a d \text {arctanh}\left (\frac {b x}{a}\right )+b c \log \left (a^2-b^2 x^2\right )}{2 b^5} \] Input:
Integrate[(x^3*(c + d*x))/(a^2 - b^2*x^2)^2,x]
Output:
(2*b*d*x + (a^2*b*(c + d*x))/(a^2 - b^2*x^2) - 3*a*d*ArcTanh[(b*x)/a] + b* c*Log[a^2 - b^2*x^2])/(2*b^5)
Time = 0.49 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {530, 2341, 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.
\(\displaystyle \int \frac {x^3 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 530 |
\(\displaystyle \frac {a^2 (c+d x)}{2 b^4 \left (a^2-b^2 x^2\right )}-\frac {\int \frac {\frac {d a^4}{b^4}+\frac {2 d x^2 a^2}{b^2}+\frac {2 c x a^2}{b^2}}{a^2-b^2 x^2}dx}{2 a^2}\) |
\(\Big \downarrow \) 2341 |
\(\displaystyle \frac {a^2 (c+d x)}{2 b^4 \left (a^2-b^2 x^2\right )}-\frac {\int \left (\frac {3 d a^4+2 b^2 c x a^2}{b^4 \left (a^2-b^2 x^2\right )}-\frac {2 a^2 d}{b^4}\right )dx}{2 a^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 (c+d x)}{2 b^4 \left (a^2-b^2 x^2\right )}-\frac {\frac {3 a^3 d \text {arctanh}\left (\frac {b x}{a}\right )}{b^5}-\frac {2 a^2 d x}{b^4}-\frac {a^2 c \log \left (a^2-b^2 x^2\right )}{b^4}}{2 a^2}\) |
Input:
Int[(x^3*(c + d*x))/(a^2 - b^2*x^2)^2,x]
Output:
(a^2*(c + d*x))/(2*b^4*(a^2 - b^2*x^2)) - ((-2*a^2*d*x)/b^4 + (3*a^3*d*Arc Tanh[(b*x)/a])/b^5 - (a^2*c*Log[a^2 - b^2*x^2])/b^4)/(2*a^2)
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]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.07
method | result | size |
norman | \(\frac {\frac {a^{2} c}{2 b^{4}}-\frac {d \,x^{3}}{b^{2}}+\frac {3 a^{2} d x}{2 b^{4}}}{-b^{2} x^{2}+a^{2}}+\frac {\left (3 a d +2 b c \right ) \ln \left (-b x +a \right )}{4 b^{5}}-\frac {\left (3 a d -2 b c \right ) \ln \left (b x +a \right )}{4 b^{5}}\) | \(87\) |
default | \(\frac {d x}{b^{4}}+\frac {\left (-3 a d +2 b c \right ) \ln \left (b x +a \right )}{4 b^{5}}-\frac {a \left (a d -b c \right )}{4 b^{5} \left (b x +a \right )}+\frac {\left (3 a d +2 b c \right ) \ln \left (-b x +a \right )}{4 b^{5}}+\frac {a \left (a d +b c \right )}{4 b^{5} \left (-b x +a \right )}\) | \(91\) |
risch | \(\frac {d x}{b^{4}}+\frac {\frac {1}{2} a^{2} d x +\frac {1}{2} a^{2} c}{b^{4} \left (-b^{2} x^{2}+a^{2}\right )}+\frac {3 \ln \left (b x -a \right ) a d}{4 b^{5}}+\frac {\ln \left (b x -a \right ) c}{2 b^{4}}-\frac {3 \ln \left (-b x -a \right ) a d}{4 b^{5}}+\frac {\ln \left (-b x -a \right ) c}{2 b^{4}}\) | \(100\) |
parallelrisch | \(\frac {3 \ln \left (b x -a \right ) x^{2} a \,b^{2} d +2 \ln \left (b x -a \right ) x^{2} b^{3} c -3 \ln \left (b x +a \right ) x^{2} a \,b^{2} d +2 \ln \left (b x +a \right ) x^{2} b^{3} c +4 b^{3} d \,x^{3}-3 \ln \left (b x -a \right ) a^{3} d -2 \ln \left (b x -a \right ) a^{2} b c +3 \ln \left (b x +a \right ) a^{3} d -2 \ln \left (b x +a \right ) a^{2} b c -6 a^{2} b d x -2 c \,a^{2} b}{4 b^{5} \left (b^{2} x^{2}-a^{2}\right )}\) | \(166\) |
Input:
int(x^3*(d*x+c)/(-b^2*x^2+a^2)^2,x,method=_RETURNVERBOSE)
Output:
(1/2*a^2*c/b^4-d/b^2*x^3+3/2*a^2*d*x/b^4)/(-b^2*x^2+a^2)+1/4*(3*a*d+2*b*c) *ln(-b*x+a)/b^5-1/4*(3*a*d-2*b*c)/b^5*ln(b*x+a)
Time = 0.16 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.59 \[ \int \frac {x^3 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=\frac {4 \, b^{3} d x^{3} - 6 \, a^{2} b d x - 2 \, a^{2} b c - {\left (2 \, a^{2} b c - 3 \, a^{3} d - {\left (2 \, b^{3} c - 3 \, a b^{2} d\right )} x^{2}\right )} \log \left (b x + a\right ) - {\left (2 \, a^{2} b c + 3 \, a^{3} d - {\left (2 \, b^{3} c + 3 \, a b^{2} d\right )} x^{2}\right )} \log \left (b x - a\right )}{4 \, {\left (b^{7} x^{2} - a^{2} b^{5}\right )}} \] Input:
integrate(x^3*(d*x+c)/(-b^2*x^2+a^2)^2,x, algorithm="fricas")
Output:
1/4*(4*b^3*d*x^3 - 6*a^2*b*d*x - 2*a^2*b*c - (2*a^2*b*c - 3*a^3*d - (2*b^3 *c - 3*a*b^2*d)*x^2)*log(b*x + a) - (2*a^2*b*c + 3*a^3*d - (2*b^3*c + 3*a* b^2*d)*x^2)*log(b*x - a))/(b^7*x^2 - a^2*b^5)
Time = 0.38 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.36 \[ \int \frac {x^3 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=\frac {- a^{2} c - a^{2} d x}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} + \frac {d x}{b^{4}} - \frac {\left (3 a d - 2 b c\right ) \log {\left (x + \frac {2 c + \frac {3 a d - 2 b c}{b}}{3 d} \right )}}{4 b^{5}} + \frac {\left (3 a d + 2 b c\right ) \log {\left (x + \frac {2 c - \frac {3 a d + 2 b c}{b}}{3 d} \right )}}{4 b^{5}} \] Input:
integrate(x**3*(d*x+c)/(-b**2*x**2+a**2)**2,x)
Output:
(-a**2*c - a**2*d*x)/(-2*a**2*b**4 + 2*b**6*x**2) + d*x/b**4 - (3*a*d - 2* b*c)*log(x + (2*c + (3*a*d - 2*b*c)/b)/(3*d))/(4*b**5) + (3*a*d + 2*b*c)*l og(x + (2*c - (3*a*d + 2*b*c)/b)/(3*d))/(4*b**5)
Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {x^3 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=-\frac {a^{2} d x + a^{2} c}{2 \, {\left (b^{6} x^{2} - a^{2} b^{4}\right )}} + \frac {d x}{b^{4}} + \frac {{\left (2 \, b c - 3 \, a d\right )} \log \left (b x + a\right )}{4 \, b^{5}} + \frac {{\left (2 \, b c + 3 \, a d\right )} \log \left (b x - a\right )}{4 \, b^{5}} \] Input:
integrate(x^3*(d*x+c)/(-b^2*x^2+a^2)^2,x, algorithm="maxima")
Output:
-1/2*(a^2*d*x + a^2*c)/(b^6*x^2 - a^2*b^4) + d*x/b^4 + 1/4*(2*b*c - 3*a*d) *log(b*x + a)/b^5 + 1/4*(2*b*c + 3*a*d)*log(b*x - a)/b^5
Time = 0.13 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.04 \[ \int \frac {x^3 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=\frac {d x}{b^{4}} + \frac {{\left (2 \, b c - 3 \, a d\right )} \log \left ({\left | b x + a \right |}\right )}{4 \, b^{5}} + \frac {{\left (2 \, b c + 3 \, a d\right )} \log \left ({\left | b x - a \right |}\right )}{4 \, b^{5}} - \frac {a^{2} d x + a^{2} c}{2 \, {\left (b x + a\right )} {\left (b x - a\right )} b^{4}} \] Input:
integrate(x^3*(d*x+c)/(-b^2*x^2+a^2)^2,x, algorithm="giac")
Output:
d*x/b^4 + 1/4*(2*b*c - 3*a*d)*log(abs(b*x + a))/b^5 + 1/4*(2*b*c + 3*a*d)* log(abs(b*x - a))/b^5 - 1/2*(a^2*d*x + a^2*c)/((b*x + a)*(b*x - a)*b^4)
Time = 7.43 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {x^3 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=\frac {\frac {a^2\,c}{2}+\frac {a^2\,d\,x}{2}}{a^2\,b^4-b^6\,x^2}-\frac {\ln \left (a+b\,x\right )\,\left (3\,a\,d-2\,b\,c\right )}{4\,b^5}+\frac {\ln \left (a-b\,x\right )\,\left (3\,a\,d+2\,b\,c\right )}{4\,b^5}+\frac {d\,x}{b^4} \] Input:
int((x^3*(c + d*x))/(a^2 - b^2*x^2)^2,x)
Output:
((a^2*c)/2 + (a^2*d*x)/2)/(a^2*b^4 - b^6*x^2) - (log(a + b*x)*(3*a*d - 2*b *c))/(4*b^5) + (log(a - b*x)*(3*a*d + 2*b*c))/(4*b^5) + (d*x)/b^4
Time = 0.21 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.15 \[ \int \frac {x^3 (c+d x)}{\left (a^2-b^2 x^2\right )^2} \, dx=\frac {-3 \,\mathrm {log}\left (-b x -a \right ) a^{3} d +2 \,\mathrm {log}\left (-b x -a \right ) a^{2} b c +3 \,\mathrm {log}\left (-b x -a \right ) a \,b^{2} d \,x^{2}-2 \,\mathrm {log}\left (-b x -a \right ) b^{3} c \,x^{2}+3 \,\mathrm {log}\left (-b x +a \right ) a^{3} d +2 \,\mathrm {log}\left (-b x +a \right ) a^{2} b c -3 \,\mathrm {log}\left (-b x +a \right ) a \,b^{2} d \,x^{2}-2 \,\mathrm {log}\left (-b x +a \right ) b^{3} c \,x^{2}+6 a^{2} b d x +2 b^{3} c \,x^{2}-4 b^{3} d \,x^{3}}{4 b^{5} \left (-b^{2} x^{2}+a^{2}\right )} \] Input:
int(x^3*(d*x+c)/(-b^2*x^2+a^2)^2,x)
Output:
( - 3*log( - a - b*x)*a**3*d + 2*log( - a - b*x)*a**2*b*c + 3*log( - a - b *x)*a*b**2*d*x**2 - 2*log( - a - b*x)*b**3*c*x**2 + 3*log(a - b*x)*a**3*d + 2*log(a - b*x)*a**2*b*c - 3*log(a - b*x)*a*b**2*d*x**2 - 2*log(a - b*x)* b**3*c*x**2 + 6*a**2*b*d*x + 2*b**3*c*x**2 - 4*b**3*d*x**3)/(4*b**5*(a**2 - b**2*x**2))