Integrand size = 23, antiderivative size = 76 \[ \int \frac {(c+d x)^3}{(a-b x) (a+b x)} \, dx=-\frac {3 c d^2 x}{b^2}-\frac {d^3 x^2}{2 b^2}-\frac {(b c+a d)^3 \log (a-b x)}{2 a b^4}+\frac {(b c-a d)^3 \log (a+b x)}{2 a b^4} \] Output:
-3*c*d^2*x/b^2-1/2*d^3*x^2/b^2-1/2*(a*d+b*c)^3*ln(-b*x+a)/a/b^4+1/2*(-a*d+ b*c)^3*ln(b*x+a)/a/b^4
Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.82 \[ \int \frac {(c+d x)^3}{(a-b x) (a+b x)} \, dx=-\frac {a b^2 d^2 x (6 c+d x)+(b c+a d)^3 \log (a-b x)-(b c-a d)^3 \log (a+b x)}{2 a b^4} \] Input:
Integrate[(c + d*x)^3/((a - b*x)*(a + b*x)),x]
Output:
-1/2*(a*b^2*d^2*x*(6*c + d*x) + (b*c + a*d)^3*Log[a - b*x] - (b*c - a*d)^3 *Log[a + b*x])/(a*b^4)
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {82, 477, 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.
\(\displaystyle \int \frac {(c+d x)^3}{(a-b x) (a+b x)} \, dx\) |
\(\Big \downarrow \) 82 |
\(\displaystyle \int \frac {(c+d x)^3}{a^2-b^2 x^2}dx\) |
\(\Big \downarrow \) 477 |
\(\displaystyle \frac {\int \left (-\frac {a^2 x d^3}{b^2}-\frac {3 a^2 c d^2}{b^2}+\frac {a (b c+a d)^3}{2 b^3 (a-b x)}+\frac {a (b c-a d)^3}{2 b^3 (a+b x)}\right )dx}{a^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {3 a^2 c d^2 x}{b^2}-\frac {a^2 d^3 x^2}{2 b^2}-\frac {a (a d+b c)^3 \log (a-b x)}{2 b^4}+\frac {a (b c-a d)^3 \log (a+b x)}{2 b^4}}{a^2}\) |
Input:
Int[(c + d*x)^3/((a - b*x)*(a + b*x)),x]
Output:
((-3*a^2*c*d^2*x)/b^2 - (a^2*d^3*x^2)/(2*b^2) - (a*(b*c + a*d)^3*Log[a - b *x])/(2*b^4) + (a*(b*c - a*d)^3*Log[a + b*x])/(2*b^4))/a^2
Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) )^(p_.), x_] :> Int[(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && IntegerQ[m]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Time = 0.31 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.62
method | result | size |
default | \(-\frac {d^{2} \left (\frac {1}{2} x^{2} d +3 c x \right )}{b^{2}}+\frac {\left (-a^{3} d^{3}-3 a^{2} b c \,d^{2}-3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (-b x +a \right )}{2 b^{4} a}+\frac {\left (-a^{3} d^{3}+3 a^{2} b c \,d^{2}-3 a \,b^{2} c^{2} d +b^{3} c^{3}\right ) \ln \left (b x +a \right )}{2 b^{4} a}\) | \(123\) |
norman | \(-\frac {d^{3} x^{2}}{2 b^{2}}-\frac {3 c \,d^{2} x}{b^{2}}-\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (b x +a \right )}{2 a \,b^{4}}-\frac {\left (a^{3} d^{3}+3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d +b^{3} c^{3}\right ) \ln \left (-b x +a \right )}{2 a \,b^{4}}\) | \(123\) |
risch | \(-\frac {d^{3} x^{2}}{2 b^{2}}-\frac {3 c \,d^{2} x}{b^{2}}-\frac {a^{2} \ln \left (b x +a \right ) d^{3}}{2 b^{4}}+\frac {3 a \ln \left (b x +a \right ) c \,d^{2}}{2 b^{3}}-\frac {3 \ln \left (b x +a \right ) c^{2} d}{2 b^{2}}+\frac {\ln \left (b x +a \right ) c^{3}}{2 a b}-\frac {a^{2} \ln \left (-b x +a \right ) d^{3}}{2 b^{4}}-\frac {3 a \ln \left (-b x +a \right ) c \,d^{2}}{2 b^{3}}-\frac {3 \ln \left (-b x +a \right ) c^{2} d}{2 b^{2}}-\frac {\ln \left (-b x +a \right ) c^{3}}{2 a b}\) | \(157\) |
parallelrisch | \(-\frac {x^{2} a \,b^{2} d^{3}+\ln \left (b x -a \right ) a^{3} d^{3}+3 \ln \left (b x -a \right ) a^{2} b c \,d^{2}+3 \ln \left (b x -a \right ) a \,b^{2} c^{2} d +\ln \left (b x -a \right ) b^{3} c^{3}+\ln \left (b x +a \right ) a^{3} d^{3}-3 \ln \left (b x +a \right ) a^{2} b c \,d^{2}+3 \ln \left (b x +a \right ) a \,b^{2} c^{2} d -\ln \left (b x +a \right ) b^{3} c^{3}+6 x a \,b^{2} c \,d^{2}}{2 a \,b^{4}}\) | \(157\) |
Input:
int((d*x+c)^3/(-b*x+a)/(b*x+a),x,method=_RETURNVERBOSE)
Output:
-d^2/b^2*(1/2*x^2*d+3*c*x)+1/2*(-a^3*d^3-3*a^2*b*c*d^2-3*a*b^2*c^2*d-b^3*c ^3)/b^4/a*ln(-b*x+a)+1/2*(-a^3*d^3+3*a^2*b*c*d^2-3*a*b^2*c^2*d+b^3*c^3)/b^ 4/a*ln(b*x+a)
Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.57 \[ \int \frac {(c+d x)^3}{(a-b x) (a+b x)} \, dx=-\frac {a b^{2} d^{3} x^{2} + 6 \, a b^{2} c d^{2} x - {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (b x + a\right ) + {\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left (b x - a\right )}{2 \, a b^{4}} \] Input:
integrate((d*x+c)^3/(-b*x+a)/(b*x+a),x, algorithm="fricas")
Output:
-1/2*(a*b^2*d^3*x^2 + 6*a*b^2*c*d^2*x - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b *c*d^2 - a^3*d^3)*log(b*x + a) + (b^3*c^3 + 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 + a^3*d^3)*log(b*x - a))/(a*b^4)
Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (66) = 132\).
Time = 0.50 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.14 \[ \int \frac {(c+d x)^3}{(a-b x) (a+b x)} \, dx=- \frac {3 c d^{2} x}{b^{2}} - \frac {d^{3} x^{2}}{2 b^{2}} - \frac {\left (a d - b c\right )^{3} \log {\left (x + \frac {a^{4} d^{3} + 3 a^{2} b^{2} c^{2} d - a \left (a d - b c\right )^{3}}{3 a^{2} b^{2} c d^{2} + b^{4} c^{3}} \right )}}{2 a b^{4}} - \frac {\left (a d + b c\right )^{3} \log {\left (x + \frac {a^{4} d^{3} + 3 a^{2} b^{2} c^{2} d - a \left (a d + b c\right )^{3}}{3 a^{2} b^{2} c d^{2} + b^{4} c^{3}} \right )}}{2 a b^{4}} \] Input:
integrate((d*x+c)**3/(-b*x+a)/(b*x+a),x)
Output:
-3*c*d**2*x/b**2 - d**3*x**2/(2*b**2) - (a*d - b*c)**3*log(x + (a**4*d**3 + 3*a**2*b**2*c**2*d - a*(a*d - b*c)**3)/(3*a**2*b**2*c*d**2 + b**4*c**3)) /(2*a*b**4) - (a*d + b*c)**3*log(x + (a**4*d**3 + 3*a**2*b**2*c**2*d - a*( a*d + b*c)**3)/(3*a**2*b**2*c*d**2 + b**4*c**3))/(2*a*b**4)
Time = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.61 \[ \int \frac {(c+d x)^3}{(a-b x) (a+b x)} \, dx=-\frac {d^{3} x^{2} + 6 \, c d^{2} x}{2 \, b^{2}} + \frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (b x + a\right )}{2 \, a b^{4}} - \frac {{\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left (b x - a\right )}{2 \, a b^{4}} \] Input:
integrate((d*x+c)^3/(-b*x+a)/(b*x+a),x, algorithm="maxima")
Output:
-1/2*(d^3*x^2 + 6*c*d^2*x)/b^2 + 1/2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c* d^2 - a^3*d^3)*log(b*x + a)/(a*b^4) - 1/2*(b^3*c^3 + 3*a*b^2*c^2*d + 3*a^2 *b*c*d^2 + a^3*d^3)*log(b*x - a)/(a*b^4)
Time = 0.13 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.71 \[ \int \frac {(c+d x)^3}{(a-b x) (a+b x)} \, dx=-\frac {b^{2} d^{3} x^{2} + 6 \, b^{2} c d^{2} x}{2 \, b^{4}} + \frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{2 \, a b^{4}} - \frac {{\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left ({\left | b x - a \right |}\right )}{2 \, a b^{4}} \] Input:
integrate((d*x+c)^3/(-b*x+a)/(b*x+a),x, algorithm="giac")
Output:
-1/2*(b^2*d^3*x^2 + 6*b^2*c*d^2*x)/b^4 + 1/2*(b^3*c^3 - 3*a*b^2*c^2*d + 3* a^2*b*c*d^2 - a^3*d^3)*log(abs(b*x + a))/(a*b^4) - 1/2*(b^3*c^3 + 3*a*b^2* c^2*d + 3*a^2*b*c*d^2 + a^3*d^3)*log(abs(b*x - a))/(a*b^4)
Time = 1.08 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.61 \[ \int \frac {(c+d x)^3}{(a-b x) (a+b x)} \, dx=-\frac {d^3\,x^2}{2\,b^2}-\frac {\ln \left (a+b\,x\right )\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{2\,a\,b^4}-\frac {\ln \left (a-b\,x\right )\,\left (a^3\,d^3+3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d+b^3\,c^3\right )}{2\,a\,b^4}-\frac {3\,c\,d^2\,x}{b^2} \] Input:
int((c + d*x)^3/((a + b*x)*(a - b*x)),x)
Output:
- (d^3*x^2)/(2*b^2) - (log(a + b*x)*(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2*d - 3 *a^2*b*c*d^2))/(2*a*b^4) - (log(a - b*x)*(a^3*d^3 + b^3*c^3 + 3*a*b^2*c^2* d + 3*a^2*b*c*d^2))/(2*a*b^4) - (3*c*d^2*x)/b^2
Time = 0.16 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.20 \[ \int \frac {(c+d x)^3}{(a-b x) (a+b x)} \, dx=\frac {-\mathrm {log}\left (-b x -a \right ) a^{3} d^{3}+3 \,\mathrm {log}\left (-b x -a \right ) a^{2} b c \,d^{2}-3 \,\mathrm {log}\left (-b x -a \right ) a \,b^{2} c^{2} d +\mathrm {log}\left (-b x -a \right ) b^{3} c^{3}-\mathrm {log}\left (-b x +a \right ) a^{3} d^{3}-3 \,\mathrm {log}\left (-b x +a \right ) a^{2} b c \,d^{2}-3 \,\mathrm {log}\left (-b x +a \right ) a \,b^{2} c^{2} d -\mathrm {log}\left (-b x +a \right ) b^{3} c^{3}-6 a \,b^{2} c \,d^{2} x -a \,b^{2} d^{3} x^{2}}{2 a \,b^{4}} \] Input:
int((d*x+c)^3/(-b*x+a)/(b*x+a),x)
Output:
( - log( - a - b*x)*a**3*d**3 + 3*log( - a - b*x)*a**2*b*c*d**2 - 3*log( - a - b*x)*a*b**2*c**2*d + log( - a - b*x)*b**3*c**3 - log(a - b*x)*a**3*d* *3 - 3*log(a - b*x)*a**2*b*c*d**2 - 3*log(a - b*x)*a*b**2*c**2*d - log(a - b*x)*b**3*c**3 - 6*a*b**2*c*d**2*x - a*b**2*d**3*x**2)/(2*a*b**4)