Integrand size = 33, antiderivative size = 169 \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {b}{(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e (a+b x)}{(b d-a e)^2 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b e (a+b x) \log (a+b x)}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b e (a+b x) \log (d+e x)}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]
-b/(-a*e+b*d)^2/((b*x+a)^2)^(1/2)-e*(b*x+a)/(-a*e+b*d)^2/(e*x+d)/((b*x+a)^ 2)^(1/2)-2*b*e*(b*x+a)*ln(b*x+a)/(-a*e+b*d)^3/((b*x+a)^2)^(1/2)+2*b*e*(b*x +a)*ln(e*x+d)/(-a*e+b*d)^3/((b*x+a)^2)^(1/2)
Time = 1.05 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.54 \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-((b d-a e) (a e+b (d+2 e x)))-2 b e (a+b x) (d+e x) \log (a+b x)+2 b e (a+b x) (d+e x) \log (d+e x)}{(b d-a e)^3 \sqrt {(a+b x)^2} (d+e x)} \]
(-((b*d - a*e)*(a*e + b*(d + 2*e*x))) - 2*b*e*(a + b*x)*(d + e*x)*Log[a + b*x] + 2*b*e*(a + b*x)*(d + e*x)*Log[d + e*x])/((b*d - a*e)^3*Sqrt[(a + b* x)^2]*(d + e*x))
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.63, 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.
| \(\displaystyle \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {1}{b^3 (a+b x)^2 (d+e x)^2}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {1}{(a+b x)^2 (d+e x)^2}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {(a+b x) \int \left (-\frac {2 e b^2}{(b d-a e)^3 (a+b x)}+\frac {b^2}{(b d-a e)^2 (a+b x)^2}+\frac {2 e^2 b}{(b d-a e)^3 (d+e x)}+\frac {e^2}{(b d-a e)^2 (d+e x)^2}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {b}{(a+b x) (b d-a e)^2}-\frac {e}{(d+e x) (b d-a e)^2}-\frac {2 b e \log (a+b x)}{(b d-a e)^3}+\frac {2 b e \log (d+e x)}{(b d-a e)^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-(b/((b*d - a*e)^2*(a + b*x))) - e/((b*d - a*e)^2*(d + e*x)) - (2*b*e*Log[a + b*x])/(b*d - a*e)^3 + (2*b*e*Log[d + e*x])/(b*d - a*e)^3)) /Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.21.31.3.1 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 = 0.40 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.07
| method | result | size |
| default | \(\frac {\left (2 \ln \left (b x +a \right ) x^{2} b^{2} e^{2}-2 \ln \left (e x +d \right ) b^{2} e^{2} x^{2}+2 \ln \left (b x +a \right ) x a b \,e^{2}+2 \ln \left (b x +a \right ) b^{2} d e x -2 \ln \left (e x +d \right ) a b \,e^{2} x -2 \ln \left (e x +d \right ) b^{2} d e x +2 \ln \left (b x +a \right ) a b d e -2 \ln \left (e x +d \right ) a b d e -2 a b \,e^{2} x +2 b^{2} d e x -e^{2} a^{2}+b^{2} d^{2}\right ) \left (b x +a \right )^{2}}{\left (e x +d \right ) \left (a e -b d \right )^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(181\) |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {2 b e x}{e^{2} a^{2}-2 a b d e +b^{2} d^{2}}-\frac {a e +b d}{e^{2} a^{2}-2 a b d e +b^{2} d^{2}}\right )}{\left (b x +a \right )^{2} \left (e x +d \right )}+\frac {2 \sqrt {\left (b x +a \right )^{2}}\, b e \ln \left (-b x -a \right )}{\left (b x +a \right ) \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}-\frac {2 \sqrt {\left (b x +a \right )^{2}}\, b e \ln \left (e x +d \right )}{\left (b x +a \right ) \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}\) | \(218\) |
(2*ln(b*x+a)*x^2*b^2*e^2-2*ln(e*x+d)*b^2*e^2*x^2+2*ln(b*x+a)*x*a*b*e^2+2*l n(b*x+a)*b^2*d*e*x-2*ln(e*x+d)*a*b*e^2*x-2*ln(e*x+d)*b^2*d*e*x+2*ln(b*x+a) *a*b*d*e-2*ln(e*x+d)*a*b*d*e-2*a*b*e^2*x+2*b^2*d*e*x-e^2*a^2+b^2*d^2)*(b*x +a)^2/(e*x+d)/(a*e-b*d)^3/((b*x+a)^2)^(3/2)
Time = 0.46 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.43 \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {b^{2} d^{2} - a^{2} e^{2} + 2 \, {\left (b^{2} d e - a b e^{2}\right )} x + 2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )} \log \left (b x + a\right ) - 2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )} \log \left (e x + d\right )}{a b^{3} d^{4} - 3 \, a^{2} b^{2} d^{3} e + 3 \, a^{3} b d^{2} e^{2} - a^{4} d e^{3} + {\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x^{2} + {\left (b^{4} d^{4} - 2 \, a b^{3} d^{3} e + 2 \, a^{3} b d e^{3} - a^{4} e^{4}\right )} x} \]
-(b^2*d^2 - a^2*e^2 + 2*(b^2*d*e - a*b*e^2)*x + 2*(b^2*e^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)*log(b*x + a) - 2*(b^2*e^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)*log(e*x + d))/(a*b^3*d^4 - 3*a^2*b^2*d^3*e + 3*a^3*b*d^2*e^ 2 - a^4*d*e^3 + (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^2 + (b^4*d^4 - 2*a*b^3*d^3*e + 2*a^3*b*d*e^3 - a^4*e^4)*x)
\[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {a + b x}{\left (d + e x\right )^{2} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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 = 0.27 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.37 \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {2 \, b^{2} e \log \left ({\left | b x + a \right |}\right )}{b^{4} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} b e^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, b e^{2} \log \left ({\left | e x + d \right |}\right )}{b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {2 \, b e x + b d + a e}{{\left (b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b d e \mathrm {sgn}\left (b x + a\right ) + a^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} {\left (b e x^{2} + b d x + a e x + a d\right )}} \]
-2*b^2*e*log(abs(b*x + a))/(b^4*d^3*sgn(b*x + a) - 3*a*b^3*d^2*e*sgn(b*x + a) + 3*a^2*b^2*d*e^2*sgn(b*x + a) - a^3*b*e^3*sgn(b*x + a)) + 2*b*e^2*log (abs(e*x + d))/(b^3*d^3*e*sgn(b*x + a) - 3*a*b^2*d^2*e^2*sgn(b*x + a) + 3* a^2*b*d*e^3*sgn(b*x + a) - a^3*e^4*sgn(b*x + a)) - (2*b*e*x + b*d + a*e)/( (b^2*d^2*sgn(b*x + a) - 2*a*b*d*e*sgn(b*x + a) + a^2*e^2*sgn(b*x + a))*(b* e*x^2 + b*d*x + a*e*x + a*d))
Timed out. \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {a+b\,x}{{\left (d+e\,x\right )}^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]