Integrand size = 38, antiderivative size = 110 \[ \int \frac {1-(1-6 b)^{3/2}-9 b+54 b x-54 x^2+108 x^3}{(e+f x)^2} \, dx=-\frac {54 (4 e+f) x}{f^3}+\frac {54 x^2}{f^2}+\frac {108 e^3+54 e^2 f+54 b e f^2-\left (1-\sqrt {1-6 b}-\left (9-6 \sqrt {1-6 b}\right ) b\right ) f^3}{f^4 (e+f x)}+\frac {54 \left (6 e^2+2 e f+b f^2\right ) \log (e+f x)}{f^4} \] Output:
-54*(4*e+f)*x/f^3+54*x^2/f^2+(108*e^3+54*e^2*f+54*b*e*f^2-(1-(1-6*b)^(1/2) -(9-6*(1-6*b)^(1/2))*b)*f^3)/f^4/(f*x+e)+54*(b*f^2+6*e^2+2*e*f)*ln(f*x+e)/ f^4
Time = 0.07 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.02 \[ \int \frac {1-(1-6 b)^{3/2}-9 b+54 b x-54 x^2+108 x^3}{(e+f x)^2} \, dx=\frac {108 e^3-54 e^2 f (-1+4 x)+f^3 \left (-1+\sqrt {1-6 b}+\left (9-6 \sqrt {1-6 b}\right ) b-54 x^2+54 x^3\right )+54 e f^2 (b-x (1+3 x))+54 \left (6 e^2+2 e f+b f^2\right ) (e+f x) \log (e+f x)}{f^4 (e+f x)} \] Input:
Integrate[(1 - (1 - 6*b)^(3/2) - 9*b + 54*b*x - 54*x^2 + 108*x^3)/(e + f*x )^2,x]
Output:
(108*e^3 - 54*e^2*f*(-1 + 4*x) + f^3*(-1 + Sqrt[1 - 6*b] + (9 - 6*Sqrt[1 - 6*b])*b - 54*x^2 + 54*x^3) + 54*e*f^2*(b - x*(1 + 3*x)) + 54*(6*e^2 + 2*e *f + b*f^2)*(e + f*x)*Log[e + f*x])/(f^4*(e + f*x))
Time = 0.51 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.91, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2389, 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.
\(\displaystyle \int \frac {54 b x-(1-6 b)^{3/2}-9 b+108 x^3-54 x^2+1}{(e+f x)^2} \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {54 \left (b f^2+6 e^2+2 e f\right )}{f^3 (e+f x)}+\frac {-54 b e f^2-(1-6 b)^{3/2} f^3-9 b f^3-108 e^3-54 e^2 f+f^3}{f^3 (e+f x)^2}-\frac {54 (4 e+f)}{f^3}+\frac {108 x}{f^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {54 \left (b f^2+6 e^2+2 e f\right ) \log (e+f x)}{f^4}+\frac {54 b e f^2+(1-6 b)^{3/2} f^3+9 b f^3+108 e^3+54 e^2 f-f^3}{f^4 (e+f x)}-\frac {54 x (4 e+f)}{f^3}+\frac {54 x^2}{f^2}\) |
Input:
Int[(1 - (1 - 6*b)^(3/2) - 9*b + 54*b*x - 54*x^2 + 108*x^3)/(e + f*x)^2,x]
Output:
(-54*(4*e + f)*x)/f^3 + (54*x^2)/f^2 + (108*e^3 + 54*e^2*f + 54*b*e*f^2 - f^3 + (1 - 6*b)^(3/2)*f^3 + 9*b*f^3)/(f^4*(e + f*x)) + (54*(6*e^2 + 2*e*f + b*f^2)*Log[e + f*x])/f^4
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand [Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p , 0] || EqQ[n, 1])
Time = 0.43 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {54 \left (-f \,x^{2}+4 e x +f x \right )}{f^{3}}-\frac {-\left (1-6 b \right )^{\frac {3}{2}} f^{3}-54 b e \,f^{2}-9 b \,f^{3}-108 e^{3}-54 e^{2} f +f^{3}}{f^{4} \left (f x +e \right )}+\frac {\left (54 b \,f^{2}+324 e^{2}+108 e f \right ) \ln \left (f x +e \right )}{f^{4}}\) | \(99\) |
norman | \(\frac {\frac {54 x^{3}}{f}-\frac {54 \left (3 e +f \right ) x^{2}}{f^{2}}-\frac {\left (-6 \sqrt {1-6 b}\, f^{3} b +\sqrt {1-6 b}\, f^{3}+54 b e \,f^{2}+9 b \,f^{3}+324 e^{3}+108 e^{2} f -f^{3}\right ) x}{e \,f^{3}}}{f x +e}+\frac {54 \left (b \,f^{2}+6 e^{2}+2 e f \right ) \ln \left (f x +e \right )}{f^{4}}\) | \(121\) |
parallelrisch | \(\frac {54 \ln \left (f x +e \right ) x b \,f^{3}+54 x^{3} f^{3}+324 \ln \left (f x +e \right ) x \,e^{2} f +108 \ln \left (f x +e \right ) x e \,f^{2}+54 \ln \left (f x +e \right ) b e \,f^{2}-6 \sqrt {1-6 b}\, f^{3} b -162 x^{2} f^{2} e -54 x^{2} f^{3}+324 \ln \left (f x +e \right ) e^{3}+108 \ln \left (f x +e \right ) e^{2} f +\sqrt {1-6 b}\, f^{3}+54 b e \,f^{2}+9 b \,f^{3}+324 e^{3}+108 e^{2} f -f^{3}}{f^{4} \left (f x +e \right )}\) | \(166\) |
Input:
int((1-(1-6*b)^(3/2)-9*b+54*b*x-54*x^2+108*x^3)/(f*x+e)^2,x,method=_RETURN VERBOSE)
Output:
-54/f^3*(-f*x^2+4*e*x+f*x)-1/f^4*(-(1-6*b)^(3/2)*f^3-54*b*e*f^2-9*b*f^3-10 8*e^3-54*e^2*f+f^3)/(f*x+e)+(54*b*f^2+324*e^2+108*e*f)/f^4*ln(f*x+e)
Time = 0.07 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.30 \[ \int \frac {1-(1-6 b)^{3/2}-9 b+54 b x-54 x^2+108 x^3}{(e+f x)^2} \, dx=\frac {54 \, f^{3} x^{3} - {\left (6 \, b - 1\right )} \sqrt {-6 \, b + 1} f^{3} + 54 \, b e f^{2} + {\left (9 \, b - 1\right )} f^{3} + 108 \, e^{3} + 54 \, e^{2} f - 54 \, {\left (3 \, e f^{2} + f^{3}\right )} x^{2} - 54 \, {\left (4 \, e^{2} f + e f^{2}\right )} x + 54 \, {\left (b e f^{2} + 6 \, e^{3} + 2 \, e^{2} f + {\left (b f^{3} + 6 \, e^{2} f + 2 \, e f^{2}\right )} x\right )} \log \left (f x + e\right )}{f^{5} x + e f^{4}} \] Input:
integrate((1-(1-6*b)^(3/2)-9*b+54*b*x-54*x^2+108*x^3)/(f*x+e)^2,x, algorit hm="fricas")
Output:
(54*f^3*x^3 - (6*b - 1)*sqrt(-6*b + 1)*f^3 + 54*b*e*f^2 + (9*b - 1)*f^3 + 108*e^3 + 54*e^2*f - 54*(3*e*f^2 + f^3)*x^2 - 54*(4*e^2*f + e*f^2)*x + 54* (b*e*f^2 + 6*e^3 + 2*e^2*f + (b*f^3 + 6*e^2*f + 2*e*f^2)*x)*log(f*x + e))/ (f^5*x + e*f^4)
Time = 0.89 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.11 \[ \int \frac {1-(1-6 b)^{3/2}-9 b+54 b x-54 x^2+108 x^3}{(e+f x)^2} \, dx=x \left (- \frac {216 e}{f^{3}} - \frac {54}{f^{2}}\right ) + \frac {54 b e f^{6} - 6 b f^{7} \sqrt {1 - 6 b} + 9 b f^{7} + 108 e^{3} f^{4} + 54 e^{2} f^{5} + f^{7} \sqrt {1 - 6 b} - f^{7}}{e f^{8} + f^{9} x} + \frac {54 x^{2}}{f^{2}} + \frac {54 \left (b f^{2} + 6 e^{2} + 2 e f\right ) \log {\left (e + f x \right )}}{f^{4}} \] Input:
integrate((1-(1-6*b)**(3/2)-9*b+54*b*x-54*x**2+108*x**3)/(f*x+e)**2,x)
Output:
x*(-216*e/f**3 - 54/f**2) + (54*b*e*f**6 - 6*b*f**7*sqrt(1 - 6*b) + 9*b*f* *7 + 108*e**3*f**4 + 54*e**2*f**5 + f**7*sqrt(1 - 6*b) - f**7)/(e*f**8 + f **9*x) + 54*x**2/f**2 + 54*(b*f**2 + 6*e**2 + 2*e*f)*log(e + f*x)/f**4
Time = 0.04 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.86 \[ \int \frac {1-(1-6 b)^{3/2}-9 b+54 b x-54 x^2+108 x^3}{(e+f x)^2} \, dx=\frac {54 \, b e f^{2} + {\left ({\left (-6 \, b + 1\right )}^{\frac {3}{2}} + 9 \, b - 1\right )} f^{3} + 108 \, e^{3} + 54 \, e^{2} f}{f^{5} x + e f^{4}} + \frac {54 \, {\left (f x^{2} - {\left (4 \, e + f\right )} x\right )}}{f^{3}} + \frac {54 \, {\left (b f^{2} + 6 \, e^{2} + 2 \, e f\right )} \log \left (f x + e\right )}{f^{4}} \] Input:
integrate((1-(1-6*b)^(3/2)-9*b+54*b*x-54*x^2+108*x^3)/(f*x+e)^2,x, algorit hm="maxima")
Output:
(54*b*e*f^2 + ((-6*b + 1)^(3/2) + 9*b - 1)*f^3 + 108*e^3 + 54*e^2*f)/(f^5* x + e*f^4) + 54*(f*x^2 - (4*e + f)*x)/f^3 + 54*(b*f^2 + 6*e^2 + 2*e*f)*log (f*x + e)/f^4
Time = 0.11 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.85 \[ \int \frac {1-(1-6 b)^{3/2}-9 b+54 b x-54 x^2+108 x^3}{(e+f x)^2} \, dx=-\frac {54 \, b {\left (\frac {\log \left (\frac {{\left | f x + e \right |}}{{\left (f x + e\right )}^{2} {\left | f \right |}}\right )}{f} - \frac {e}{{\left (f x + e\right )} f}\right )}}{f} + \frac {{\left (-6 \, b + 1\right )}^{\frac {3}{2}}}{{\left (f x + e\right )} f} + \frac {9 \, b}{{\left (f x + e\right )} f} - \frac {54 \, {\left (f x + e\right )}^{2} {\left (\frac {6 \, e}{f x + e} - 1\right )}}{f^{4}} - \frac {324 \, e^{2} \log \left (\frac {{\left | f x + e \right |}}{{\left (f x + e\right )}^{2} {\left | f \right |}}\right )}{f^{4}} - \frac {108 \, e \log \left (\frac {{\left | f x + e \right |}}{{\left (f x + e\right )}^{2} {\left | f \right |}}\right )}{f^{3}} + \frac {108 \, e^{3}}{{\left (f x + e\right )} f^{4}} - \frac {54 \, {\left (f x + e\right )}}{f^{3}} + \frac {54 \, e^{2}}{{\left (f x + e\right )} f^{3}} - \frac {1}{{\left (f x + e\right )} f} \] Input:
integrate((1-(1-6*b)^(3/2)-9*b+54*b*x-54*x^2+108*x^3)/(f*x+e)^2,x, algorit hm="giac")
Output:
-54*b*(log(abs(f*x + e)/((f*x + e)^2*abs(f)))/f - e/((f*x + e)*f))/f + (-6 *b + 1)^(3/2)/((f*x + e)*f) + 9*b/((f*x + e)*f) - 54*(f*x + e)^2*(6*e/(f*x + e) - 1)/f^4 - 324*e^2*log(abs(f*x + e)/((f*x + e)^2*abs(f)))/f^4 - 108* e*log(abs(f*x + e)/((f*x + e)^2*abs(f)))/f^3 + 108*e^3/((f*x + e)*f^4) - 5 4*(f*x + e)/f^3 + 54*e^2/((f*x + e)*f^3) - 1/((f*x + e)*f)
Time = 12.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.98 \[ \int \frac {1-(1-6 b)^{3/2}-9 b+54 b x-54 x^2+108 x^3}{(e+f x)^2} \, dx=\frac {\frac {{\left (1-6\,b\right )}^{3/2}}{f^2}+\frac {9\,b\,f^3+54\,e^2\,f+108\,e^3-f^3+54\,b\,e\,f^2}{f^5}}{x+\frac {e}{f}}-x\,\left (\frac {216\,e}{f^3}+\frac {54}{f^2}\right )+\frac {54\,x^2}{f^2}+\frac {\ln \left (x+\frac {e}{f}\right )\,\left (324\,e^2+108\,e\,f+54\,b\,f^2\right )}{f^4} \] Input:
int(-(9*b - 54*b*x + (1 - 6*b)^(3/2) + 54*x^2 - 108*x^3 - 1)/(e + f*x)^2,x )
Output:
((1 - 6*b)^(3/2)/f^2 + (9*b*f^3 + 54*e^2*f + 108*e^3 - f^3 + 54*b*e*f^2)/f ^5)/(x + e/f) - x*((216*e)/f^3 + 54/f^2) + (54*x^2)/f^2 + (log(x + e/f)*(1 08*e*f + 54*b*f^2 + 324*e^2))/f^4
Time = 0.16 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.68 \[ \int \frac {1-(1-6 b)^{3/2}-9 b+54 b x-54 x^2+108 x^3}{(e+f x)^2} \, dx=\frac {6 \sqrt {-6 b +1}\, b \,f^{4} x -\sqrt {-6 b +1}\, f^{4} x +54 \,\mathrm {log}\left (f x +e \right ) b \,e^{2} f^{2}+54 \,\mathrm {log}\left (f x +e \right ) b e \,f^{3} x +324 \,\mathrm {log}\left (f x +e \right ) e^{4}+324 \,\mathrm {log}\left (f x +e \right ) e^{3} f x +108 \,\mathrm {log}\left (f x +e \right ) e^{3} f +108 \,\mathrm {log}\left (f x +e \right ) e^{2} f^{2} x -54 b e \,f^{3} x -9 b \,f^{4} x -324 e^{3} f x -162 e^{2} f^{2} x^{2}-108 e^{2} f^{2} x +54 e \,f^{3} x^{3}-54 e \,f^{3} x^{2}+f^{4} x}{e \,f^{4} \left (f x +e \right )} \] Input:
int((1-(1-6*b)^(3/2)-9*b+54*b*x-54*x^2+108*x^3)/(f*x+e)^2,x)
Output:
(6*sqrt( - 6*b + 1)*b*f**4*x - sqrt( - 6*b + 1)*f**4*x + 54*log(e + f*x)*b *e**2*f**2 + 54*log(e + f*x)*b*e*f**3*x + 324*log(e + f*x)*e**4 + 324*log( e + f*x)*e**3*f*x + 108*log(e + f*x)*e**3*f + 108*log(e + f*x)*e**2*f**2*x - 54*b*e*f**3*x - 9*b*f**4*x - 324*e**3*f*x - 162*e**2*f**2*x**2 - 108*e* *2*f**2*x + 54*e*f**3*x**3 - 54*e*f**3*x**2 + f**4*x)/(e*f**4*(e + f*x))