Integrand size = 33, antiderivative size = 223 \[ \int \frac {a+b x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {b^2}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e (a+b x)}{2 (b d-a e)^2 (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b e (a+b x)}{(b d-a e)^3 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b^2 e (a+b x) \log (a+b x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b^2 e (a+b x) \log (d+e x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]
-b^2/(-a*e+b*d)^3/((b*x+a)^2)^(1/2)-1/2*e*(b*x+a)/(-a*e+b*d)^2/(e*x+d)^2/( (b*x+a)^2)^(1/2)-2*b*e*(b*x+a)/(-a*e+b*d)^3/(e*x+d)/((b*x+a)^2)^(1/2)-3*b^ 2*e*(b*x+a)*ln(b*x+a)/(-a*e+b*d)^4/((b*x+a)^2)^(1/2)+3*b^2*e*(b*x+a)*ln(e* x+d)/(-a*e+b*d)^4/((b*x+a)^2)^(1/2)
Time = 1.07 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.61 \[ \int \frac {a+b x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-\left ((b d-a e) \left (-a^2 e^2+a b e (5 d+3 e x)+b^2 \left (2 d^2+9 d e x+6 e^2 x^2\right )\right )\right )-6 b^2 e (a+b x) (d+e x)^2 \log (a+b x)+6 b^2 e (a+b x) (d+e x)^2 \log (d+e x)}{2 (b d-a e)^4 \sqrt {(a+b x)^2} (d+e x)^2} \]
(-((b*d - a*e)*(-(a^2*e^2) + a*b*e*(5*d + 3*e*x) + b^2*(2*d^2 + 9*d*e*x + 6*e^2*x^2))) - 6*b^2*e*(a + b*x)*(d + e*x)^2*Log[a + b*x] + 6*b^2*e*(a + b *x)*(d + e*x)^2*Log[d + e*x])/(2*(b*d - a*e)^4*Sqrt[(a + b*x)^2]*(d + e*x) ^2)
Time = 0.33 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.61, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1187, 27, 54, 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 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {1}{b^3 (a+b x)^2 (d+e x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {1}{(a+b x)^2 (d+e x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {(a+b x) \int \left (-\frac {3 e b^3}{(b d-a e)^4 (a+b x)}+\frac {b^3}{(b d-a e)^3 (a+b x)^2}+\frac {3 e^2 b^2}{(b d-a e)^4 (d+e x)}+\frac {2 e^2 b}{(b d-a e)^3 (d+e x)^2}+\frac {e^2}{(b d-a e)^2 (d+e x)^3}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {b^2}{(a+b x) (b d-a e)^3}-\frac {3 b^2 e \log (a+b x)}{(b d-a e)^4}+\frac {3 b^2 e \log (d+e x)}{(b d-a e)^4}-\frac {2 b e}{(d+e x) (b d-a e)^3}-\frac {e}{2 (d+e x)^2 (b d-a e)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-(b^2/((b*d - a*e)^3*(a + b*x))) - e/(2*(b*d - a*e)^2*(d + e*x )^2) - (2*b*e)/((b*d - a*e)^3*(d + e*x)) - (3*b^2*e*Log[a + b*x])/(b*d - a *e)^4 + (3*b^2*e*Log[d + e*x])/(b*d - a*e)^4))/Sqrt[a^2 + 2*a*b*x + b^2*x^ 2]
3.21.32.3.1 Defintions of rubi rules used
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[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[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 = 0.48 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.48
| method | result | size |
| default | \(-\frac {\left (6 \ln \left (b x +a \right ) b^{3} e^{3} x^{3}-6 \ln \left (e x +d \right ) x^{3} b^{3} e^{3}+6 \ln \left (b x +a \right ) x^{2} a \,b^{2} e^{3}+12 \ln \left (b x +a \right ) x^{2} b^{3} d \,e^{2}-6 \ln \left (e x +d \right ) x^{2} a \,b^{2} e^{3}-12 \ln \left (e x +d \right ) x^{2} b^{3} d \,e^{2}+12 \ln \left (b x +a \right ) x a \,b^{2} d \,e^{2}+6 \ln \left (b x +a \right ) b^{3} d^{2} e x -12 \ln \left (e x +d \right ) x a \,b^{2} d \,e^{2}-6 \ln \left (e x +d \right ) x \,b^{3} d^{2} e -6 x^{2} a \,b^{2} e^{3}+6 x^{2} b^{3} d \,e^{2}+6 \ln \left (b x +a \right ) a \,b^{2} d^{2} e -6 \ln \left (e x +d \right ) a \,b^{2} d^{2} e -3 x \,a^{2} b \,e^{3}-6 x a \,b^{2} d \,e^{2}+9 x \,b^{3} d^{2} e +a^{3} e^{3}-6 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +2 b^{3} d^{3}\right ) \left (b x +a \right )^{2}}{2 \left (e x +d \right )^{2} \left (a e -b d \right )^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(331\) |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {3 b^{2} e^{2} x^{2}}{a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}}+\frac {3 \left (a e +3 b d \right ) b e x}{2 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}-\frac {e^{2} a^{2}-5 a b d e -2 b^{2} d^{2}}{2 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}\right )}{\left (b x +a \right )^{2} \left (e x +d \right )^{2}}+\frac {3 \sqrt {\left (b x +a \right )^{2}}\, b^{2} e \ln \left (-e x -d \right )}{\left (b x +a \right ) \left (e^{4} a^{4}-4 b d \,e^{3} a^{3}+6 b^{2} d^{2} e^{2} a^{2}-4 b^{3} d^{3} e a +b^{4} d^{4}\right )}-\frac {3 \sqrt {\left (b x +a \right )^{2}}\, b^{2} e \ln \left (b x +a \right )}{\left (b x +a \right ) \left (e^{4} a^{4}-4 b d \,e^{3} a^{3}+6 b^{2} d^{2} e^{2} a^{2}-4 b^{3} d^{3} e a +b^{4} d^{4}\right )}\) | \(350\) |
-1/2*(6*ln(b*x+a)*b^3*e^3*x^3-6*ln(e*x+d)*x^3*b^3*e^3+6*ln(b*x+a)*x^2*a*b^ 2*e^3+12*ln(b*x+a)*x^2*b^3*d*e^2-6*ln(e*x+d)*x^2*a*b^2*e^3-12*ln(e*x+d)*x^ 2*b^3*d*e^2+12*ln(b*x+a)*x*a*b^2*d*e^2+6*ln(b*x+a)*b^3*d^2*e*x-12*ln(e*x+d )*x*a*b^2*d*e^2-6*ln(e*x+d)*x*b^3*d^2*e-6*x^2*a*b^2*e^3+6*x^2*b^3*d*e^2+6* ln(b*x+a)*a*b^2*d^2*e-6*ln(e*x+d)*a*b^2*d^2*e-3*x*a^2*b*e^3-6*x*a*b^2*d*e^ 2+9*x*b^3*d^2*e+a^3*e^3-6*a^2*b*d*e^2+3*a*b^2*d^2*e+2*b^3*d^3)*(b*x+a)^2/( e*x+d)^2/(a*e-b*d)^4/((b*x+a)^2)^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 495 vs. \(2 (166) = 332\).
Time = 0.35 (sec) , antiderivative size = 495, normalized size of antiderivative = 2.22 \[ \int \frac {a+b x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {2 \, b^{3} d^{3} + 3 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} + a^{3} e^{3} + 6 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 3 \, {\left (3 \, b^{3} d^{2} e - 2 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x + 6 \, {\left (b^{3} e^{3} x^{3} + a b^{2} d^{2} e + {\left (2 \, b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{2} + {\left (b^{3} d^{2} e + 2 \, a b^{2} d e^{2}\right )} x\right )} \log \left (b x + a\right ) - 6 \, {\left (b^{3} e^{3} x^{3} + a b^{2} d^{2} e + {\left (2 \, b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{2} + {\left (b^{3} d^{2} e + 2 \, a b^{2} d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (a b^{4} d^{6} - 4 \, a^{2} b^{3} d^{5} e + 6 \, a^{3} b^{2} d^{4} e^{2} - 4 \, a^{4} b d^{3} e^{3} + a^{5} d^{2} e^{4} + {\left (b^{5} d^{4} e^{2} - 4 \, a b^{4} d^{3} e^{3} + 6 \, a^{2} b^{3} d^{2} e^{4} - 4 \, a^{3} b^{2} d e^{5} + a^{4} b e^{6}\right )} x^{3} + {\left (2 \, b^{5} d^{5} e - 7 \, a b^{4} d^{4} e^{2} + 8 \, a^{2} b^{3} d^{3} e^{3} - 2 \, a^{3} b^{2} d^{2} e^{4} - 2 \, a^{4} b d e^{5} + a^{5} e^{6}\right )} x^{2} + {\left (b^{5} d^{6} - 2 \, a b^{4} d^{5} e - 2 \, a^{2} b^{3} d^{4} e^{2} + 8 \, a^{3} b^{2} d^{3} e^{3} - 7 \, a^{4} b d^{2} e^{4} + 2 \, a^{5} d e^{5}\right )} x\right )}} \]
-1/2*(2*b^3*d^3 + 3*a*b^2*d^2*e - 6*a^2*b*d*e^2 + a^3*e^3 + 6*(b^3*d*e^2 - a*b^2*e^3)*x^2 + 3*(3*b^3*d^2*e - 2*a*b^2*d*e^2 - a^2*b*e^3)*x + 6*(b^3*e ^3*x^3 + a*b^2*d^2*e + (2*b^3*d*e^2 + a*b^2*e^3)*x^2 + (b^3*d^2*e + 2*a*b^ 2*d*e^2)*x)*log(b*x + a) - 6*(b^3*e^3*x^3 + a*b^2*d^2*e + (2*b^3*d*e^2 + a *b^2*e^3)*x^2 + (b^3*d^2*e + 2*a*b^2*d*e^2)*x)*log(e*x + d))/(a*b^4*d^6 - 4*a^2*b^3*d^5*e + 6*a^3*b^2*d^4*e^2 - 4*a^4*b*d^3*e^3 + a^5*d^2*e^4 + (b^5 *d^4*e^2 - 4*a*b^4*d^3*e^3 + 6*a^2*b^3*d^2*e^4 - 4*a^3*b^2*d*e^5 + a^4*b*e ^6)*x^3 + (2*b^5*d^5*e - 7*a*b^4*d^4*e^2 + 8*a^2*b^3*d^3*e^3 - 2*a^3*b^2*d ^2*e^4 - 2*a^4*b*d*e^5 + a^5*e^6)*x^2 + (b^5*d^6 - 2*a*b^4*d^5*e - 2*a^2*b ^3*d^4*e^2 + 8*a^3*b^2*d^3*e^3 - 7*a^4*b*d^2*e^4 + 2*a^5*d*e^5)*x)
\[ \int \frac {a+b x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {a + b x}{\left (d + e x\right )^{3} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {a+b x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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 = 0.29 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.44 \[ \int \frac {a+b x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {3 \, b^{3} e \log \left ({\left | b x + a \right |}\right )}{b^{5} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{4} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{3} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b^{2} d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} b e^{4} \mathrm {sgn}\left (b x + a\right )} + \frac {3 \, b^{2} e^{2} \log \left ({\left | e x + d \right |}\right )}{b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{4} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{5} \mathrm {sgn}\left (b x + a\right )} - \frac {2 \, b^{3} d^{3} + 3 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} + a^{3} e^{3} + 6 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 3 \, {\left (3 \, b^{3} d^{2} e - 2 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x}{2 \, {\left (b d - a e\right )}^{4} {\left (b x + a\right )} {\left (e x + d\right )}^{2} \mathrm {sgn}\left (b x + a\right )} \]
-3*b^3*e*log(abs(b*x + a))/(b^5*d^4*sgn(b*x + a) - 4*a*b^4*d^3*e*sgn(b*x + a) + 6*a^2*b^3*d^2*e^2*sgn(b*x + a) - 4*a^3*b^2*d*e^3*sgn(b*x + a) + a^4* b*e^4*sgn(b*x + a)) + 3*b^2*e^2*log(abs(e*x + d))/(b^4*d^4*e*sgn(b*x + a) - 4*a*b^3*d^3*e^2*sgn(b*x + a) + 6*a^2*b^2*d^2*e^3*sgn(b*x + a) - 4*a^3*b* d*e^4*sgn(b*x + a) + a^4*e^5*sgn(b*x + a)) - 1/2*(2*b^3*d^3 + 3*a*b^2*d^2* e - 6*a^2*b*d*e^2 + a^3*e^3 + 6*(b^3*d*e^2 - a*b^2*e^3)*x^2 + 3*(3*b^3*d^2 *e - 2*a*b^2*d*e^2 - a^2*b*e^3)*x)/((b*d - a*e)^4*(b*x + a)*(e*x + d)^2*sg n(b*x + a))
Timed out. \[ \int \frac {a+b x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {a+b\,x}{{\left (d+e\,x\right )}^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]