Integrand size = 28, antiderivative size = 161 \[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {3 e (b d-a e)^2}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^3}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^3 x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e^2 (b d-a e) (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]
-3*e*(-a*e+b*d)^2/b^4/((b*x+a)^2)^(1/2)-1/2*(-a*e+b*d)^3/b^4/(b*x+a)/((b*x +a)^2)^(1/2)+e^3*x*(b*x+a)/b^3/((b*x+a)^2)^(1/2)+3*e^2*(-a*e+b*d)*(b*x+a)* ln(b*x+a)/b^4/((b*x+a)^2)^(1/2)
Time = 1.04 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.78 \[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-5 a^3 e^3+a^2 b e^2 (9 d-4 e x)+a b^2 e \left (-3 d^2+12 d e x+4 e^2 x^2\right )-b^3 \left (d^3+6 d^2 e x-2 e^3 x^3\right )-6 e^2 (-b d+a e) (a+b x)^2 \log (a+b x)}{2 b^4 (a+b x) \sqrt {(a+b x)^2}} \]
(-5*a^3*e^3 + a^2*b*e^2*(9*d - 4*e*x) + a*b^2*e*(-3*d^2 + 12*d*e*x + 4*e^2 *x^2) - b^3*(d^3 + 6*d^2*e*x - 2*e^3*x^3) - 6*e^2*(-(b*d) + a*e)*(a + b*x) ^2*Log[a + b*x])/(2*b^4*(a + b*x)*Sqrt[(a + b*x)^2])
Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.65, 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.
\(\displaystyle \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {b^3 (a+b x) \int \frac {(d+e x)^3}{b^3 (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^3}{(a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {(a+b x) \int \left (\frac {e^3}{b^3}+\frac {3 (b d-a e) e^2}{b^3 (a+b x)}+\frac {3 (b d-a e)^2 e}{b^3 (a+b x)^2}+\frac {(b d-a e)^3}{b^3 (a+b x)^3}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (\frac {3 e^2 (b d-a e) \log (a+b x)}{b^4}-\frac {3 e (b d-a e)^2}{b^4 (a+b x)}-\frac {(b d-a e)^3}{2 b^4 (a+b x)^2}+\frac {e^3 x}{b^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((e^3*x)/b^3 - (b*d - a*e)^3/(2*b^4*(a + b*x)^2) - (3*e*(b*d - a*e)^2)/(b^4*(a + b*x)) + (3*e^2*(b*d - a*e)*Log[a + b*x])/b^4))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.16.97.3.1 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 [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 = 2.48 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{3} x}{\left (b x +a \right ) b^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-3 a^{2} e^{3}+6 a b d \,e^{2}-3 b^{2} d^{2} e \right ) x -\frac {5 a^{3} e^{3}-9 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +b^{3} d^{3}}{2 b}\right )}{\left (b x +a \right )^{3} b^{3}}-\frac {3 \sqrt {\left (b x +a \right )^{2}}\, e^{2} \left (a e -b d \right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{4}}\) | \(154\) |
default | \(-\frac {\left (6 \ln \left (b x +a \right ) x^{2} a \,b^{2} e^{3}-6 \ln \left (b x +a \right ) b^{3} d \,e^{2} x^{2}-2 e^{3} x^{3} b^{3}+12 \ln \left (b x +a \right ) x \,a^{2} b \,e^{3}-12 \ln \left (b x +a \right ) x a \,b^{2} d \,e^{2}-4 x^{2} a \,b^{2} e^{3}+6 \ln \left (b x +a \right ) a^{3} e^{3}-6 \ln \left (b x +a \right ) a^{2} b d \,e^{2}+4 a^{2} b \,e^{3} x -12 x a \,b^{2} d \,e^{2}+6 b^{3} d^{2} e x +5 a^{3} e^{3}-9 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (b x +a \right )}{2 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(209\) |
((b*x+a)^2)^(1/2)/(b*x+a)*e^3/b^3*x+((b*x+a)^2)^(1/2)/(b*x+a)^3*((-3*a^2*e ^3+6*a*b*d*e^2-3*b^2*d^2*e)*x-1/2*(5*a^3*e^3-9*a^2*b*d*e^2+3*a*b^2*d^2*e+b ^3*d^3)/b)/b^3-3*((b*x+a)^2)^(1/2)/(b*x+a)/b^4*e^2*(a*e-b*d)*ln(b*x+a)
Time = 0.26 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.17 \[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {2 \, b^{3} e^{3} x^{3} + 4 \, a b^{2} e^{3} x^{2} - b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 9 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3} - 2 \, {\left (3 \, b^{3} d^{2} e - 6 \, a b^{2} d e^{2} + 2 \, a^{2} b e^{3}\right )} x + 6 \, {\left (a^{2} b d e^{2} - a^{3} e^{3} + {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]
1/2*(2*b^3*e^3*x^3 + 4*a*b^2*e^3*x^2 - b^3*d^3 - 3*a*b^2*d^2*e + 9*a^2*b*d *e^2 - 5*a^3*e^3 - 2*(3*b^3*d^2*e - 6*a*b^2*d*e^2 + 2*a^2*b*e^3)*x + 6*(a^ 2*b*d*e^2 - a^3*e^3 + (b^3*d*e^2 - a*b^2*e^3)*x^2 + 2*(a*b^2*d*e^2 - a^2*b *e^3)*x)*log(b*x + a))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)
\[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (115) = 230\).
Time = 0.21 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.47 \[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e^{3} x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {3 \, d e^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {3 \, a e^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} - \frac {3 \, d^{2} e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {2 \, a^{2} e^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} + \frac {6 \, a d e^{2} x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {6 \, a^{2} e^{3} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {d^{3}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, a d^{2} e}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {9 \, a^{2} d e^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, a^{3} e^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} \]
e^3*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 3*d*e^2*log(x + a/b)/b^3 - 3 *a*e^3*log(x + a/b)/b^4 - 3*d^2*e/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 2* a^2*e^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) + 6*a*d*e^2*x/(b^4*(x + a/b)^2 ) - 6*a^2*e^3*x/(b^5*(x + a/b)^2) - 1/2*d^3/(b^3*(x + a/b)^2) + 3/2*a*d^2* e/(b^4*(x + a/b)^2) + 9/2*a^2*d*e^2/(b^5*(x + a/b)^2) - 11/2*a^3*e^3/(b^6* (x + a/b)^2)
Time = 0.28 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.84 \[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e^{3} x}{b^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {3 \, {\left (b d e^{2} - a e^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {b^{3} d^{3} + 3 \, a b^{2} d^{2} e - 9 \, a^{2} b d e^{2} + 5 \, a^{3} e^{3} + 6 \, {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{4} \mathrm {sgn}\left (b x + a\right )} \]
e^3*x/(b^3*sgn(b*x + a)) + 3*(b*d*e^2 - a*e^3)*log(abs(b*x + a))/(b^4*sgn( b*x + a)) - 1/2*(b^3*d^3 + 3*a*b^2*d^2*e - 9*a^2*b*d*e^2 + 5*a^3*e^3 + 6*( b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x)/((b*x + a)^2*b^4*sgn(b*x + a))
Timed out. \[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^3}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]