Integrand size = 28, antiderivative size = 210 \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {4 e (b d-a e)^3}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^4}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^3 (4 b d-3 a e) x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^4 x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {6 e^2 (b d-a e)^2 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:
-4*e*(-a*e+b*d)^3/b^5/((b*x+a)^2)^(1/2)-1/2*(-a*e+b*d)^4/b^5/(b*x+a)/((b*x +a)^2)^(1/2)+e^3*(-3*a*e+4*b*d)*x*(b*x+a)/b^4/((b*x+a)^2)^(1/2)+1/2*e^4*x^ 2*(b*x+a)/b^3/((b*x+a)^2)^(1/2)+6*e^2*(-a*e+b*d)^2*(b*x+a)*ln(b*x+a)/b^5/( (b*x+a)^2)^(1/2)
Time = 1.06 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.83 \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {7 a^4 e^4+2 a^3 b e^3 (-10 d+e x)+a^2 b^2 e^2 \left (18 d^2-16 d e x-11 e^2 x^2\right )-4 a b^3 e \left (d^3-6 d^2 e x-4 d e^2 x^2+e^3 x^3\right )+b^4 \left (-d^4-8 d^3 e x+8 d e^3 x^3+e^4 x^4\right )+12 e^2 (b d-a e)^2 (a+b x)^2 \log (a+b x)}{2 b^5 (a+b x) \sqrt {(a+b x)^2}} \] Input:
Integrate[(d + e*x)^4/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
Output:
(7*a^4*e^4 + 2*a^3*b*e^3*(-10*d + e*x) + a^2*b^2*e^2*(18*d^2 - 16*d*e*x - 11*e^2*x^2) - 4*a*b^3*e*(d^3 - 6*d^2*e*x - 4*d*e^2*x^2 + e^3*x^3) + b^4*(- d^4 - 8*d^3*e*x + 8*d*e^3*x^3 + e^4*x^4) + 12*e^2*(b*d - a*e)^2*(a + b*x)^ 2*Log[a + b*x])/(2*b^5*(a + b*x)*Sqrt[(a + b*x)^2])
Time = 0.54 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.61, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1102, 27, 49, 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.
| \(\displaystyle \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1102 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {(d+e x)^4}{b^3 (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^4}{(a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {(a+b x) \int \left (\frac {x e^4}{b^3}+\frac {(4 b d-3 a e) e^3}{b^4}+\frac {6 (b d-a e)^2 e^2}{b^4 (a+b x)}+\frac {4 (b d-a e)^3 e}{b^4 (a+b x)^2}+\frac {(b d-a e)^4}{b^4 (a+b x)^3}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(a+b x) \left (\frac {6 e^2 (b d-a e)^2 \log (a+b x)}{b^5}-\frac {4 e (b d-a e)^3}{b^5 (a+b x)}-\frac {(b d-a e)^4}{2 b^5 (a+b x)^2}+\frac {e^3 x (4 b d-3 a e)}{b^4}+\frac {e^4 x^2}{2 b^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
Input:
Int[(d + e*x)^4/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
Output:
((a + b*x)*((e^3*(4*b*d - 3*a*e)*x)/b^4 + (e^4*x^2)/(2*b^3) - (b*d - a*e)^ 4/(2*b^5*(a + b*x)^2) - (4*e*(b*d - a*e)^3)/(b^5*(a + b*x)) + (6*e^2*(b*d - a*e)^2*Log[a + b*x])/b^5))/Sqrt[a^2 + 2*a*b*x + b^2*x^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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 1.60 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.02
| method | result | size |
| risch | \(-\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{3} \left (-\frac {1}{2} b e \,x^{2}+3 a e x -4 b d x \right )}{\left (b x +a \right ) b^{4}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (4 e^{4} a^{3}-12 a^{2} d \,e^{3} b +12 a \,b^{2} d^{2} e^{2}-4 d^{3} e \,b^{3}\right ) x +\frac {7 a^{4} e^{4}-20 a^{3} b d \,e^{3}+18 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e -b^{4} d^{4}}{2 b}\right )}{\left (b x +a \right )^{3} b^{4}}+\frac {6 \sqrt {\left (b x +a \right )^{2}}\, e^{2} \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\) | \(214\) |
| default | \(\frac {\left (b^{4} x^{4} e^{4}+12 \ln \left (b x +a \right ) x^{2} a^{2} b^{2} e^{4}-24 \ln \left (b x +a \right ) x^{2} a \,b^{3} d \,e^{3}+12 \ln \left (b x +a \right ) b^{4} d^{2} e^{2} x^{2}-4 x^{3} a \,b^{3} e^{4}+8 x^{3} b^{4} d \,e^{3}+24 \ln \left (b x +a \right ) x \,a^{3} b \,e^{4}-48 \ln \left (b x +a \right ) x \,a^{2} b^{2} d \,e^{3}+24 \ln \left (b x +a \right ) x a \,b^{3} d^{2} e^{2}-11 x^{2} a^{2} b^{2} e^{4}+16 x^{2} a \,b^{3} d \,e^{3}+12 \ln \left (b x +a \right ) a^{4} e^{4}-24 \ln \left (b x +a \right ) a^{3} b d \,e^{3}+12 \ln \left (b x +a \right ) a^{2} b^{2} d^{2} e^{2}+2 x \,a^{3} b \,e^{4}-16 x \,a^{2} b^{2} d \,e^{3}+24 x a \,b^{3} d^{2} e^{2}-8 x \,b^{4} d^{3} e +7 a^{4} e^{4}-20 a^{3} b d \,e^{3}+18 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e -b^{4} d^{4}\right ) \left (b x +a \right )}{2 b^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(341\) |
Input:
int((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-((b*x+a)^2)^(1/2)/(b*x+a)*e^3/b^4*(-1/2*b*e*x^2+3*a*e*x-4*b*d*x)+((b*x+a) ^2)^(1/2)/(b*x+a)^3*((4*a^3*e^4-12*a^2*b*d*e^3+12*a*b^2*d^2*e^2-4*b^3*d^3* e)*x+1/2*(7*a^4*e^4-20*a^3*b*d*e^3+18*a^2*b^2*d^2*e^2-4*a*b^3*d^3*e-b^4*d^ 4)/b)/b^4+6*((b*x+a)^2)^(1/2)/(b*x+a)/b^5*e^2*(a^2*e^2-2*a*b*d*e+b^2*d^2)* ln(b*x+a)
Time = 0.08 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.39 \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {b^{4} e^{4} x^{4} - b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 20 \, a^{3} b d e^{3} + 7 \, a^{4} e^{4} + 4 \, {\left (2 \, b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + {\left (16 \, a b^{3} d e^{3} - 11 \, a^{2} b^{2} e^{4}\right )} x^{2} - 2 \, {\left (4 \, b^{4} d^{3} e - 12 \, a b^{3} d^{2} e^{2} + 8 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x + 12 \, {\left (a^{2} b^{2} d^{2} e^{2} - 2 \, a^{3} b d e^{3} + a^{4} e^{4} + {\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \, {\left (a b^{3} d^{2} e^{2} - 2 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \] Input:
integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")
Output:
1/2*(b^4*e^4*x^4 - b^4*d^4 - 4*a*b^3*d^3*e + 18*a^2*b^2*d^2*e^2 - 20*a^3*b *d*e^3 + 7*a^4*e^4 + 4*(2*b^4*d*e^3 - a*b^3*e^4)*x^3 + (16*a*b^3*d*e^3 - 1 1*a^2*b^2*e^4)*x^2 - 2*(4*b^4*d^3*e - 12*a*b^3*d^2*e^2 + 8*a^2*b^2*d*e^3 - a^3*b*e^4)*x + 12*(a^2*b^2*d^2*e^2 - 2*a^3*b*d*e^3 + a^4*e^4 + (b^4*d^2*e ^2 - 2*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 2*(a*b^3*d^2*e^2 - 2*a^2*b^2*d*e^3 + a^3*b*e^4)*x)*log(b*x + a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)
\[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{4}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((e*x+d)**4/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
Output:
Integral((d + e*x)**4/((a + b*x)**2)**(3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 397 vs. \(2 (151) = 302\).
Time = 0.04 (sec) , antiderivative size = 397, normalized size of antiderivative = 1.89 \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e^{4} x^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {4 \, d e^{3} x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {5 \, a e^{4} x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac {6 \, d^{2} e^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {12 \, a d e^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {6 \, a^{2} e^{4} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {4 \, d^{3} e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {8 \, a^{2} d e^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {5 \, a^{3} e^{4}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} + \frac {12 \, a d^{2} e^{2} x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {24 \, a^{2} d e^{3} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {12 \, a^{3} e^{4} x}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {d^{4}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, a d^{3} e}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {9 \, a^{2} d^{2} e^{2}}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {22 \, a^{3} d e^{3}}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, a^{4} e^{4}}{2 \, b^{7} {\left (x + \frac {a}{b}\right )}^{2}} \] Input:
integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")
Output:
1/2*e^4*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 4*d*e^3*x^2/(sqrt(b^2*x^ 2 + 2*a*b*x + a^2)*b^2) - 5/2*a*e^4*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3 ) + 6*d^2*e^2*log(x + a/b)/b^3 - 12*a*d*e^3*log(x + a/b)/b^4 + 6*a^2*e^4*l og(x + a/b)/b^5 - 4*d^3*e/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 8*a^2*d*e^ 3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - 5*a^3*e^4/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^5) + 12*a*d^2*e^2*x/(b^4*(x + a/b)^2) - 24*a^2*d*e^3*x/(b^5*(x + a/b)^2) + 12*a^3*e^4*x/(b^6*(x + a/b)^2) - 1/2*d^4/(b^3*(x + a/b)^2) + 2*a *d^3*e/(b^4*(x + a/b)^2) + 9*a^2*d^2*e^2/(b^5*(x + a/b)^2) - 22*a^3*d*e^3/ (b^6*(x + a/b)^2) + 23/2*a^4*e^4/(b^7*(x + a/b)^2)
Time = 0.17 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.03 \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {6 \, {\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {b^{3} e^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + 8 \, b^{3} d e^{3} x \mathrm {sgn}\left (b x + a\right ) - 6 \, a b^{2} e^{4} x \mathrm {sgn}\left (b x + a\right )}{2 \, b^{6}} - \frac {b^{4} d^{4} + 4 \, a b^{3} d^{3} e - 18 \, a^{2} b^{2} d^{2} e^{2} + 20 \, a^{3} b d e^{3} - 7 \, a^{4} e^{4} + 8 \, {\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{5} \mathrm {sgn}\left (b x + a\right )} \] Input:
integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")
Output:
6*(b^2*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*log(abs(b*x + a))/(b^5*sgn(b*x + a )) + 1/2*(b^3*e^4*x^2*sgn(b*x + a) + 8*b^3*d*e^3*x*sgn(b*x + a) - 6*a*b^2* e^4*x*sgn(b*x + a))/b^6 - 1/2*(b^4*d^4 + 4*a*b^3*d^3*e - 18*a^2*b^2*d^2*e^ 2 + 20*a^3*b*d*e^3 - 7*a^4*e^4 + 8*(b^4*d^3*e - 3*a*b^3*d^2*e^2 + 3*a^2*b^ 2*d*e^3 - a^3*b*e^4)*x)/((b*x + a)^2*b^5*sgn(b*x + a))
Timed out. \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^4}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \] Input:
int((d + e*x)^4/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)
Output:
int((d + e*x)^4/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)
Time = 0.21 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.57 \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {12 \,\mathrm {log}\left (b x +a \right ) a^{5} e^{4}-24 \,\mathrm {log}\left (b x +a \right ) a^{4} b d \,e^{3}+24 \,\mathrm {log}\left (b x +a \right ) a^{4} b \,e^{4} x +12 \,\mathrm {log}\left (b x +a \right ) a^{3} b^{2} d^{2} e^{2}-48 \,\mathrm {log}\left (b x +a \right ) a^{3} b^{2} d \,e^{3} x +12 \,\mathrm {log}\left (b x +a \right ) a^{3} b^{2} e^{4} x^{2}+24 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{3} d^{2} e^{2} x -24 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{3} d \,e^{3} x^{2}+12 \,\mathrm {log}\left (b x +a \right ) a \,b^{4} d^{2} e^{2} x^{2}+6 a^{5} e^{4}-12 a^{4} b d \,e^{3}+6 a^{3} b^{2} d^{2} e^{2}-12 a^{3} b^{2} e^{4} x^{2}+24 a^{2} b^{3} d \,e^{3} x^{2}-4 a^{2} b^{3} e^{4} x^{3}-a \,b^{4} d^{4}-12 a \,b^{4} d^{2} e^{2} x^{2}+8 a \,b^{4} d \,e^{3} x^{3}+a \,b^{4} e^{4} x^{4}+4 b^{5} d^{3} e \,x^{2}}{2 a \,b^{5} \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:
int((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)
Output:
(12*log(a + b*x)*a**5*e**4 - 24*log(a + b*x)*a**4*b*d*e**3 + 24*log(a + b* x)*a**4*b*e**4*x + 12*log(a + b*x)*a**3*b**2*d**2*e**2 - 48*log(a + b*x)*a **3*b**2*d*e**3*x + 12*log(a + b*x)*a**3*b**2*e**4*x**2 + 24*log(a + b*x)* a**2*b**3*d**2*e**2*x - 24*log(a + b*x)*a**2*b**3*d*e**3*x**2 + 12*log(a + b*x)*a*b**4*d**2*e**2*x**2 + 6*a**5*e**4 - 12*a**4*b*d*e**3 + 6*a**3*b**2 *d**2*e**2 - 12*a**3*b**2*e**4*x**2 + 24*a**2*b**3*d*e**3*x**2 - 4*a**2*b* *3*e**4*x**3 - a*b**4*d**4 - 12*a*b**4*d**2*e**2*x**2 + 8*a*b**4*d*e**3*x* *3 + a*b**4*e**4*x**4 + 4*b**5*d**3*e*x**2)/(2*a*b**5*(a**2 + 2*a*b*x + b* *2*x**2))