Integrand size = 33, antiderivative size = 238 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=-\frac {b^3 (3 b d-4 a e) x \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)}+\frac {b^4 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x)}-\frac {(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^5 (a+b x) (d+e x)^2}+\frac {4 b (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) (d+e x)}+\frac {6 b^2 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^5 (a+b x)} \] Output:
-b^3*(-4*a*e+3*b*d)*x*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+1/2*b^4*x^2*((b*x+a)^2 )^(1/2)/e^3/(b*x+a)-1/2*(-a*e+b*d)^4*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d) ^2+4*b*(-a*e+b*d)^3*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)+6*b^2*(-a*e+b*d) ^2*((b*x+a)^2)^(1/2)*ln(e*x+d)/e^5/(b*x+a)
Time = 1.14 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=\frac {\sqrt {(a+b x)^2} \left (-a^4 e^4-4 a^3 b e^3 (d+2 e x)+6 a^2 b^2 d e^2 (3 d+4 e x)+4 a b^3 e \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )+b^4 \left (7 d^4+2 d^3 e x-11 d^2 e^2 x^2-4 d e^3 x^3+e^4 x^4\right )+12 b^2 (b d-a e)^2 (d+e x)^2 \log (d+e x)\right )}{2 e^5 (a+b x) (d+e x)^2} \] Input:
Integrate[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^3,x]
Output:
(Sqrt[(a + b*x)^2]*(-(a^4*e^4) - 4*a^3*b*e^3*(d + 2*e*x) + 6*a^2*b^2*d*e^2 *(3*d + 4*e*x) + 4*a*b^3*e*(-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3) + b^4*(7*d^4 + 2*d^3*e*x - 11*d^2*e^2*x^2 - 4*d*e^3*x^3 + e^4*x^4) + 12*b^ 2*(b*d - a*e)^2*(d + e*x)^2*Log[d + e*x]))/(2*e^5*(a + b*x)*(d + e*x)^2)
Time = 0.55 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.55, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1187, 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.
| \(\displaystyle \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^4}{(d+e x)^3}dx}{b^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^4}{(d+e x)^3}dx}{a+b x}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {x b^4}{e^3}-\frac {(3 b d-4 a e) b^3}{e^4}+\frac {6 (b d-a e)^2 b^2}{e^4 (d+e x)}-\frac {4 (b d-a e)^3 b}{e^4 (d+e x)^2}+\frac {(a e-b d)^4}{e^4 (d+e x)^3}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {b^3 x (3 b d-4 a e)}{e^4}+\frac {6 b^2 (b d-a e)^2 \log (d+e x)}{e^5}+\frac {4 b (b d-a e)^3}{e^5 (d+e x)}-\frac {(b d-a e)^4}{2 e^5 (d+e x)^2}+\frac {b^4 x^2}{2 e^3}\right )}{a+b x}\) |
Input:
Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^3,x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-((b^3*(3*b*d - 4*a*e)*x)/e^4) + (b^4*x^2) /(2*e^3) - (b*d - a*e)^4/(2*e^5*(d + e*x)^2) + (4*b*(b*d - a*e)^3)/(e^5*(d + e*x)) + (6*b^2*(b*d - a*e)^2*Log[d + e*x])/e^5))/(a + b*x)
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_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ IntPart[p]*(b/2 + c*x)^(2*FracPart[p])) Int[(d + e*x)^m*(f + g*x)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 1.41 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.92
| method | result | size |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{3} \left (\frac {1}{2} b e \,x^{2}+4 a e x -3 b d x \right )}{\left (b x +a \right ) e^{4}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-4 a^{3} b \,e^{3}+12 a^{2} b^{2} d \,e^{2}-12 d^{2} e a \,b^{3}+4 b^{4} d^{3}\right ) x -\frac {a^{4} e^{4}+4 a^{3} b d \,e^{3}-18 a^{2} b^{2} d^{2} e^{2}+20 a \,b^{3} d^{3} e -7 b^{4} d^{4}}{2 e}\right )}{\left (b x +a \right ) e^{4} \left (e x +d \right )^{2}}+\frac {6 \sqrt {\left (b x +a \right )^{2}}\, b^{2} \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \ln \left (e x +d \right )}{\left (b x +a \right ) e^{5}}\) | \(219\) |
| default | \(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (b^{4} x^{4} e^{4}+12 \ln \left (e x +d \right ) a^{2} b^{2} e^{4} x^{2}-24 \ln \left (e x +d \right ) a \,b^{3} d \,e^{3} x^{2}+12 \ln \left (e x +d \right ) b^{4} d^{2} e^{2} x^{2}+8 x^{3} a \,b^{3} e^{4}-4 x^{3} b^{4} d \,e^{3}+24 \ln \left (e x +d \right ) a^{2} b^{2} d \,e^{3} x -48 \ln \left (e x +d \right ) a \,b^{3} d^{2} e^{2} x +24 \ln \left (e x +d \right ) b^{4} d^{3} e x +16 x^{2} a \,b^{3} d \,e^{3}-11 x^{2} b^{4} d^{2} e^{2}+12 \ln \left (e x +d \right ) a^{2} b^{2} d^{2} e^{2}-24 \ln \left (e x +d \right ) a \,b^{3} d^{3} e +12 \ln \left (e x +d \right ) b^{4} d^{4}-8 x \,a^{3} b \,e^{4}+24 x \,a^{2} b^{2} d \,e^{3}-16 x a \,b^{3} d^{2} e^{2}+2 x \,b^{4} d^{3} e -a^{4} e^{4}-4 a^{3} b d \,e^{3}+18 a^{2} b^{2} d^{2} e^{2}-20 a \,b^{3} d^{3} e +7 b^{4} d^{4}\right )}{2 \left (b x +a \right )^{3} e^{5} \left (e x +d \right )^{2}}\) | \(350\) |
Input:
int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^3,x,method=_RETURNVERBOSE)
Output:
((b*x+a)^2)^(1/2)/(b*x+a)*b^3/e^4*(1/2*b*e*x^2+4*a*e*x-3*b*d*x)+((b*x+a)^2 )^(1/2)/(b*x+a)*((-4*a^3*b*e^3+12*a^2*b^2*d*e^2-12*a*b^3*d^2*e+4*b^4*d^3)* x-1/2*(a^4*e^4+4*a^3*b*d*e^3-18*a^2*b^2*d^2*e^2+20*a*b^3*d^3*e-7*b^4*d^4)/ e)/e^4/(e*x+d)^2+6*((b*x+a)^2)^(1/2)/(b*x+a)*b^2/e^5*(a^2*e^2-2*a*b*d*e+b^ 2*d^2)*ln(e*x+d)
Time = 0.07 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.22 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=\frac {b^{4} e^{4} x^{4} + 7 \, b^{4} d^{4} - 20 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} - a^{4} e^{4} - 4 \, {\left (b^{4} d e^{3} - 2 \, a b^{3} e^{4}\right )} x^{3} - {\left (11 \, b^{4} d^{2} e^{2} - 16 \, a b^{3} d e^{3}\right )} x^{2} + 2 \, {\left (b^{4} d^{3} e - 8 \, a b^{3} d^{2} e^{2} + 12 \, a^{2} b^{2} d e^{3} - 4 \, a^{3} b e^{4}\right )} x + 12 \, {\left (b^{4} d^{4} - 2 \, a b^{3} d^{3} e + a^{2} b^{2} d^{2} e^{2} + {\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \, {\left (b^{4} d^{3} e - 2 \, a b^{3} d^{2} e^{2} + a^{2} b^{2} d e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^3,x, algorithm="fric as")
Output:
1/2*(b^4*e^4*x^4 + 7*b^4*d^4 - 20*a*b^3*d^3*e + 18*a^2*b^2*d^2*e^2 - 4*a^3 *b*d*e^3 - a^4*e^4 - 4*(b^4*d*e^3 - 2*a*b^3*e^4)*x^3 - (11*b^4*d^2*e^2 - 1 6*a*b^3*d*e^3)*x^2 + 2*(b^4*d^3*e - 8*a*b^3*d^2*e^2 + 12*a^2*b^2*d*e^3 - 4 *a^3*b*e^4)*x + 12*(b^4*d^4 - 2*a*b^3*d^3*e + a^2*b^2*d^2*e^2 + (b^4*d^2*e ^2 - 2*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 2*(b^4*d^3*e - 2*a*b^3*d^2*e^2 + a ^2*b^2*d*e^3)*x)*log(e*x + d))/(e^7*x^2 + 2*d*e^6*x + d^2*e^5)
\[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=\int \frac {\left (a + b x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{3}}\, dx \] Input:
integrate((b*x+a)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**3,x)
Output:
Integral((a + b*x)*((a + b*x)**2)**(3/2)/(d + e*x)**3, x)
Exception generated. \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^3,x, algorithm="maxi ma")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for m ore detail
Time = 0.17 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=\frac {6 \, {\left (b^{4} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{3} d e \mathrm {sgn}\left (b x + a\right ) + a^{2} b^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | e x + d \right |}\right )}{e^{5}} + \frac {b^{4} e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, b^{4} d e^{2} x \mathrm {sgn}\left (b x + a\right ) + 8 \, a b^{3} e^{3} x \mathrm {sgn}\left (b x + a\right )}{2 \, e^{6}} + \frac {7 \, b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 20 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 18 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) - a^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) + 8 \, {\left (b^{4} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{3} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b^{2} d e^{3} \mathrm {sgn}\left (b x + a\right ) - a^{3} b e^{4} \mathrm {sgn}\left (b x + a\right )\right )} x}{2 \, {\left (e x + d\right )}^{2} e^{5}} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^3,x, algorithm="giac ")
Output:
6*(b^4*d^2*sgn(b*x + a) - 2*a*b^3*d*e*sgn(b*x + a) + a^2*b^2*e^2*sgn(b*x + a))*log(abs(e*x + d))/e^5 + 1/2*(b^4*e^3*x^2*sgn(b*x + a) - 6*b^4*d*e^2*x *sgn(b*x + a) + 8*a*b^3*e^3*x*sgn(b*x + a))/e^6 + 1/2*(7*b^4*d^4*sgn(b*x + a) - 20*a*b^3*d^3*e*sgn(b*x + a) + 18*a^2*b^2*d^2*e^2*sgn(b*x + a) - 4*a^ 3*b*d*e^3*sgn(b*x + a) - a^4*e^4*sgn(b*x + a) + 8*(b^4*d^3*e*sgn(b*x + a) - 3*a*b^3*d^2*e^2*sgn(b*x + a) + 3*a^2*b^2*d*e^3*sgn(b*x + a) - a^3*b*e^4* sgn(b*x + a))*x)/((e*x + d)^2*e^5)
Timed out. \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=\int \frac {\left (a+b\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^3} \,d x \] Input:
int(((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^3,x)
Output:
int(((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^3, x)
Time = 0.20 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.39 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=\frac {12 \,\mathrm {log}\left (e x +d \right ) a^{2} b^{2} d^{3} e^{2}+24 \,\mathrm {log}\left (e x +d \right ) a^{2} b^{2} d^{2} e^{3} x +12 \,\mathrm {log}\left (e x +d \right ) a^{2} b^{2} d \,e^{4} x^{2}-24 \,\mathrm {log}\left (e x +d \right ) a \,b^{3} d^{4} e -48 \,\mathrm {log}\left (e x +d \right ) a \,b^{3} d^{3} e^{2} x -24 \,\mathrm {log}\left (e x +d \right ) a \,b^{3} d^{2} e^{3} x^{2}+12 \,\mathrm {log}\left (e x +d \right ) b^{4} d^{5}+24 \,\mathrm {log}\left (e x +d \right ) b^{4} d^{4} e x +12 \,\mathrm {log}\left (e x +d \right ) b^{4} d^{3} e^{2} x^{2}-a^{4} d \,e^{4}+4 a^{3} b \,e^{5} x^{2}+6 a^{2} b^{2} d^{3} e^{2}-12 a^{2} b^{2} d \,e^{4} x^{2}-12 a \,b^{3} d^{4} e +24 a \,b^{3} d^{2} e^{3} x^{2}+8 a \,b^{3} d \,e^{4} x^{3}+6 b^{4} d^{5}-12 b^{4} d^{3} e^{2} x^{2}-4 b^{4} d^{2} e^{3} x^{3}+b^{4} d \,e^{4} x^{4}}{2 d \,e^{5} \left (e^{2} x^{2}+2 d e x +d^{2}\right )} \] Input:
int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^3,x)
Output:
(12*log(d + e*x)*a**2*b**2*d**3*e**2 + 24*log(d + e*x)*a**2*b**2*d**2*e**3 *x + 12*log(d + e*x)*a**2*b**2*d*e**4*x**2 - 24*log(d + e*x)*a*b**3*d**4*e - 48*log(d + e*x)*a*b**3*d**3*e**2*x - 24*log(d + e*x)*a*b**3*d**2*e**3*x **2 + 12*log(d + e*x)*b**4*d**5 + 24*log(d + e*x)*b**4*d**4*e*x + 12*log(d + e*x)*b**4*d**3*e**2*x**2 - a**4*d*e**4 + 4*a**3*b*e**5*x**2 + 6*a**2*b* *2*d**3*e**2 - 12*a**2*b**2*d*e**4*x**2 - 12*a*b**3*d**4*e + 24*a*b**3*d** 2*e**3*x**2 + 8*a*b**3*d*e**4*x**3 + 6*b**4*d**5 - 12*b**4*d**3*e**2*x**2 - 4*b**4*d**2*e**3*x**3 + b**4*d*e**4*x**4)/(2*d*e**5*(d**2 + 2*d*e*x + e* *2*x**2))