Integrand size = 29, antiderivative size = 188 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {3 a B}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^4}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 B}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a^2 B}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]
3*a*B/b^5/((b*x+a)^2)^(1/2)+1/4*(A*b-B*a)*x^4/a/b/(b*x+a)^3/((b*x+a)^2)^(1 /2)+1/3*a^3*B/b^5/(b*x+a)^2/((b*x+a)^2)^(1/2)-3/2*a^2*B/b^5/(b*x+a)/((b*x+ a)^2)^(1/2)+B*(b*x+a)*ln(b*x+a)/b^5/((b*x+a)^2)^(1/2)
Time = 1.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.55 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {25 a^4 B-12 A b^4 x^3-12 a^2 b^2 x (A-9 B x)+6 a b^3 x^2 (-3 A+8 B x)+a^3 (-3 A b+88 b B x)+12 B (a+b x)^4 \log (a+b x)}{12 b^5 (a+b x)^3 \sqrt {(a+b x)^2}} \]
(25*a^4*B - 12*A*b^4*x^3 - 12*a^2*b^2*x*(A - 9*B*x) + 6*a*b^3*x^2*(-3*A + 8*B*x) + a^3*(-3*A*b + 88*b*B*x) + 12*B*(a + b*x)^4*Log[a + b*x])/(12*b^5* (a + b*x)^3*Sqrt[(a + b*x)^2])
Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.63, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1187, 27, 87, 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.
\(\displaystyle \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b^5 (a+b x) \int \frac {x^3 (A+B x)}{b^5 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {x^3 (A+B x)}{(a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(a+b x) \left (\frac {B \int \frac {x^3}{(a+b x)^4}dx}{b}+\frac {x^4 (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {(a+b x) \left (\frac {B \int \left (-\frac {a^3}{b^3 (a+b x)^4}+\frac {3 a^2}{b^3 (a+b x)^3}-\frac {3 a}{b^3 (a+b x)^2}+\frac {1}{b^3 (a+b x)}\right )dx}{b}+\frac {x^4 (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (\frac {B \left (\frac {a^3}{3 b^4 (a+b x)^3}-\frac {3 a^2}{2 b^4 (a+b x)^2}+\frac {3 a}{b^4 (a+b x)}+\frac {\log (a+b x)}{b^4}\right )}{b}+\frac {x^4 (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(((A*b - a*B)*x^4)/(4*a*b*(a + b*x)^4) + (B*(a^3/(3*b^4*(a + b* x)^3) - (3*a^2)/(2*b^4*(a + b*x)^2) + (3*a)/(b^4*(a + b*x)) + Log[a + b*x] /b^4))/b))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.8.23.3.1 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[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.34 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.61
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {\left (A b -4 B a \right ) x^{3}}{b^{2}}-\frac {3 a \left (A b -6 B a \right ) x^{2}}{2 b^{3}}-\frac {a^{2} \left (3 A b -22 B a \right ) x}{3 b^{4}}-\frac {a^{3} \left (3 A b -25 B a \right )}{12 b^{5}}\right )}{\left (b x +a \right )^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, B \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\) | \(115\) |
default | \(-\frac {\left (-12 B \ln \left (b x +a \right ) b^{4} x^{4}-48 B \ln \left (b x +a \right ) x^{3} a \,b^{3}+12 A \,b^{4} x^{3}-72 B \ln \left (b x +a \right ) x^{2} a^{2} b^{2}-48 B a \,b^{3} x^{3}+18 A a \,b^{3} x^{2}-48 B \ln \left (b x +a \right ) x \,a^{3} b -108 B \,a^{2} b^{2} x^{2}+12 A \,a^{2} b^{2} x -12 B \ln \left (b x +a \right ) a^{4}-88 B \,a^{3} b x +3 A \,a^{3} b -25 B \,a^{4}\right ) \left (b x +a \right )}{12 b^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(168\) |
((b*x+a)^2)^(1/2)/(b*x+a)^5*(-(A*b-4*B*a)/b^2*x^3-3/2*a*(A*b-6*B*a)/b^3*x^ 2-1/3*a^2*(3*A*b-22*B*a)/b^4*x-1/12*a^3*(3*A*b-25*B*a)/b^5)+((b*x+a)^2)^(1 /2)/(b*x+a)*B/b^5*ln(b*x+a)
Time = 0.43 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.93 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {25 \, B a^{4} - 3 \, A a^{3} b + 12 \, {\left (4 \, B a b^{3} - A b^{4}\right )} x^{3} + 18 \, {\left (6 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 4 \, {\left (22 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x + 12 \, {\left (B b^{4} x^{4} + 4 \, B a b^{3} x^{3} + 6 \, B a^{2} b^{2} x^{2} + 4 \, B a^{3} b x + B a^{4}\right )} \log \left (b x + a\right )}{12 \, {\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}\right )}} \]
1/12*(25*B*a^4 - 3*A*a^3*b + 12*(4*B*a*b^3 - A*b^4)*x^3 + 18*(6*B*a^2*b^2 - A*a*b^3)*x^2 + 4*(22*B*a^3*b - 3*A*a^2*b^2)*x + 12*(B*b^4*x^4 + 4*B*a*b^ 3*x^3 + 6*B*a^2*b^2*x^2 + 4*B*a^3*b*x + B*a^4)*log(b*x + a))/(b^9*x^4 + 4* a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5)
\[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{3} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.26 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.07 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {1}{12} \, B {\left (\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} - \frac {1}{12} \, A {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} \]
1/12*B*((48*a*b^3*x^3 + 108*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5) + 12*log(b*x + a)/b^5 ) - 1/12*A*(12*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 8*a^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) + 6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7*(x + a/b )^3) - 3*a^3/(b^8*(x + a/b)^4))
Time = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.62 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {B \log \left ({\left | b x + a \right |}\right )}{b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {12 \, {\left (4 \, B a b^{2} - A b^{3}\right )} x^{3} + 18 \, {\left (6 \, B a^{2} b - A a b^{2}\right )} x^{2} + 4 \, {\left (22 \, B a^{3} - 3 \, A a^{2} b\right )} x + \frac {25 \, B a^{4} - 3 \, A a^{3} b}{b}}{12 \, {\left (b x + a\right )}^{4} b^{4} \mathrm {sgn}\left (b x + a\right )} \]
B*log(abs(b*x + a))/(b^5*sgn(b*x + a)) + 1/12*(12*(4*B*a*b^2 - A*b^3)*x^3 + 18*(6*B*a^2*b - A*a*b^2)*x^2 + 4*(22*B*a^3 - 3*A*a^2*b)*x + (25*B*a^4 - 3*A*a^3*b)/b)/((b*x + a)^4*b^4*sgn(b*x + a))
Timed out. \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]