Integrand size = 33, antiderivative size = 252 \[ \int \frac {(a+b x) (d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {(d+e x)^5}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {20 e^2 (b d-a e)^3}{3 b^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (b d-a e)^4}{6 b^6 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e^4 (4 b d-3 a e) x (a+b x)}{3 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e^5 x^2 (a+b x)}{6 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {10 e^3 (b d-a e)^2 (a+b x) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}} \]
-1/3*(e*x+d)^5/b/(b^2*x^2+2*a*b*x+a^2)^(3/2)-20/3*e^2*(-a*e+b*d)^3/b^6/((b *x+a)^2)^(1/2)-5/6*e*(-a*e+b*d)^4/b^6/(b*x+a)/((b*x+a)^2)^(1/2)+5/3*e^4*(- 3*a*e+4*b*d)*x*(b*x+a)/b^5/((b*x+a)^2)^(1/2)+5/6*e^5*x^2*(b*x+a)/b^4/((b*x +a)^2)^(1/2)+10*e^3*(-a*e+b*d)^2*(b*x+a)*ln(b*x+a)/b^6/((b*x+a)^2)^(1/2)
Time = 1.08 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b x) (d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {47 a^5 e^5+a^4 b e^4 (-130 d+81 e x)+a^3 b^2 e^3 \left (110 d^2-270 d e x-9 e^2 x^2\right )-a^2 b^3 e^2 \left (20 d^3-270 d^2 e x+90 d e^2 x^2+63 e^3 x^3\right )-5 a b^4 e \left (d^4+12 d^3 e x-36 d^2 e^2 x^2-18 d e^3 x^3+3 e^4 x^4\right )+b^5 \left (-2 d^5-15 d^4 e x-60 d^3 e^2 x^2+30 d e^4 x^4+3 e^5 x^5\right )+60 e^3 (b d-a e)^2 (a+b x)^3 \log (a+b x)}{6 b^6 \left ((a+b x)^2\right )^{3/2}} \]
(47*a^5*e^5 + a^4*b*e^4*(-130*d + 81*e*x) + a^3*b^2*e^3*(110*d^2 - 270*d*e *x - 9*e^2*x^2) - a^2*b^3*e^2*(20*d^3 - 270*d^2*e*x + 90*d*e^2*x^2 + 63*e^ 3*x^3) - 5*a*b^4*e*(d^4 + 12*d^3*e*x - 36*d^2*e^2*x^2 - 18*d*e^3*x^3 + 3*e ^4*x^4) + b^5*(-2*d^5 - 15*d^4*e*x - 60*d^3*e^2*x^2 + 30*d*e^4*x^4 + 3*e^5 *x^5) + 60*e^3*(b*d - a*e)^2*(a + b*x)^3*Log[a + b*x])/(6*b^6*((a + b*x)^2 )^(3/2))
Time = 0.39 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.62, 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.
| \(\displaystyle \int \frac {(a+b x) (d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^5 (a+b x) \int \frac {(d+e x)^5}{b^5 (a+b x)^4}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^5}{(a+b x)^4}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {(a+b x) \int \left (\frac {x e^5}{b^4}+\frac {(5 b d-4 a e) e^4}{b^5}+\frac {10 (b d-a e)^2 e^3}{b^5 (a+b x)}+\frac {10 (b d-a e)^3 e^2}{b^5 (a+b x)^2}+\frac {5 (b d-a e)^4 e}{b^5 (a+b x)^3}+\frac {(b d-a e)^5}{b^5 (a+b x)^4}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(a+b x) \left (\frac {10 e^3 (b d-a e)^2 \log (a+b x)}{b^6}-\frac {10 e^2 (b d-a e)^3}{b^6 (a+b x)}-\frac {5 e (b d-a e)^4}{2 b^6 (a+b x)^2}-\frac {(b d-a e)^5}{3 b^6 (a+b x)^3}+\frac {e^4 x (5 b d-4 a e)}{b^5}+\frac {e^5 x^2}{2 b^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((e^4*(5*b*d - 4*a*e)*x)/b^5 + (e^5*x^2)/(2*b^4) - (b*d - a*e)^ 5/(3*b^6*(a + b*x)^3) - (5*e*(b*d - a*e)^4)/(2*b^6*(a + b*x)^2) - (10*e^2* (b*d - a*e)^3)/(b^6*(a + b*x)) + (10*e^3*(b*d - a*e)^2*Log[a + b*x])/b^6)) /Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.21.33.3.1 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_.)*((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 = 0.74 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.15
| method | result | size |
| risch | \(-\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{4} \left (-\frac {1}{2} b e \,x^{2}+4 a e x -5 b d x \right )}{\left (b x +a \right ) b^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (10 a^{3} b \,e^{5}-30 a^{2} d \,e^{4} b^{2}+30 a \,b^{3} d^{2} e^{3}-10 d^{3} e^{2} b^{4}\right ) x^{2}+\frac {5 e \left (7 e^{4} a^{4}-20 b d \,e^{3} a^{3}+18 b^{2} d^{2} e^{2} a^{2}-4 b^{3} d^{3} e a -b^{4} d^{4}\right ) x}{2}+\frac {47 e^{5} a^{5}-130 b d \,e^{4} a^{4}+110 b^{2} d^{2} e^{3} a^{3}-20 b^{3} d^{3} e^{2} a^{2}-5 b^{4} d^{4} e a -2 b^{5} d^{5}}{6 b}\right )}{\left (b x +a \right )^{4} b^{5}}+\frac {10 \sqrt {\left (b x +a \right )^{2}}\, e^{3} \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{6}}\) | \(290\) |
| default | \(\frac {\left (60 \ln \left (b x +a \right ) a^{2} b^{3} e^{5} x^{3}+60 \ln \left (b x +a \right ) b^{5} d^{2} e^{3} x^{3}+180 \ln \left (b x +a \right ) a^{3} b^{2} e^{5} x^{2}+180 \ln \left (b x +a \right ) a^{4} b \,e^{5} x -120 \ln \left (b x +a \right ) a^{4} b d \,e^{4}+60 \ln \left (b x +a \right ) a^{3} b^{2} d^{2} e^{3}-130 b d \,e^{4} a^{4}+110 b^{2} d^{2} e^{3} a^{3}-20 b^{3} d^{3} e^{2} a^{2}-5 b^{4} d^{4} e a -15 x^{4} a \,b^{4} e^{5}+30 x^{4} b^{5} d \,e^{4}-63 x^{3} a^{2} b^{3} e^{5}-9 x^{2} a^{3} b^{2} e^{5}-60 x^{2} b^{5} d^{3} e^{2}+81 x \,a^{4} b \,e^{5}-15 x \,b^{5} d^{4} e -120 \ln \left (b x +a \right ) a \,b^{4} d \,e^{4} x^{3}-360 \ln \left (b x +a \right ) a^{2} b^{3} d \,e^{4} x^{2}+180 \ln \left (b x +a \right ) a \,b^{4} d^{2} e^{3} x^{2}-360 \ln \left (b x +a \right ) a^{3} b^{2} d \,e^{4} x +180 \ln \left (b x +a \right ) a^{2} b^{3} d^{2} e^{3} x -270 x \,a^{3} b^{2} d \,e^{4}+270 x \,a^{2} b^{3} d^{2} e^{3}-60 x a \,b^{4} d^{3} e^{2}+90 x^{3} a \,b^{4} d \,e^{4}-90 x^{2} a^{2} b^{3} d \,e^{4}+180 x^{2} a \,b^{4} d^{2} e^{3}-2 b^{5} d^{5}+3 x^{5} b^{5} e^{5}+60 \ln \left (b x +a \right ) a^{5} e^{5}+47 e^{5} a^{5}\right ) \left (b x +a \right )^{2}}{6 b^{6} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(495\) |
-((b*x+a)^2)^(1/2)/(b*x+a)*e^4/b^5*(-1/2*b*e*x^2+4*a*e*x-5*b*d*x)+((b*x+a) ^2)^(1/2)/(b*x+a)^4*((10*a^3*b*e^5-30*a^2*b^2*d*e^4+30*a*b^3*d^2*e^3-10*b^ 4*d^3*e^2)*x^2+5/2*e*(7*a^4*e^4-20*a^3*b*d*e^3+18*a^2*b^2*d^2*e^2-4*a*b^3* d^3*e-b^4*d^4)*x+1/6*(47*a^5*e^5-130*a^4*b*d*e^4+110*a^3*b^2*d^2*e^3-20*a^ 2*b^3*d^3*e^2-5*a*b^4*d^4*e-2*b^5*d^5)/b)/b^5+10*((b*x+a)^2)^(1/2)/(b*x+a) /b^6*e^3*(a^2*e^2-2*a*b*d*e+b^2*d^2)*ln(b*x+a)
Leaf count of result is larger than twice the leaf count of optimal. 426 vs. \(2 (185) = 370\).
Time = 0.34 (sec) , antiderivative size = 426, normalized size of antiderivative = 1.69 \[ \int \frac {(a+b x) (d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {3 \, b^{5} e^{5} x^{5} - 2 \, b^{5} d^{5} - 5 \, a b^{4} d^{4} e - 20 \, a^{2} b^{3} d^{3} e^{2} + 110 \, a^{3} b^{2} d^{2} e^{3} - 130 \, a^{4} b d e^{4} + 47 \, a^{5} e^{5} + 15 \, {\left (2 \, b^{5} d e^{4} - a b^{4} e^{5}\right )} x^{4} + 9 \, {\left (10 \, a b^{4} d e^{4} - 7 \, a^{2} b^{3} e^{5}\right )} x^{3} - 3 \, {\left (20 \, b^{5} d^{3} e^{2} - 60 \, a b^{4} d^{2} e^{3} + 30 \, a^{2} b^{3} d e^{4} + 3 \, a^{3} b^{2} e^{5}\right )} x^{2} - 3 \, {\left (5 \, b^{5} d^{4} e + 20 \, a b^{4} d^{3} e^{2} - 90 \, a^{2} b^{3} d^{2} e^{3} + 90 \, a^{3} b^{2} d e^{4} - 27 \, a^{4} b e^{5}\right )} x + 60 \, {\left (a^{3} b^{2} d^{2} e^{3} - 2 \, a^{4} b d e^{4} + a^{5} e^{5} + {\left (b^{5} d^{2} e^{3} - 2 \, a b^{4} d e^{4} + a^{2} b^{3} e^{5}\right )} x^{3} + 3 \, {\left (a b^{4} d^{2} e^{3} - 2 \, a^{2} b^{3} d e^{4} + a^{3} b^{2} e^{5}\right )} x^{2} + 3 \, {\left (a^{2} b^{3} d^{2} e^{3} - 2 \, a^{3} b^{2} d e^{4} + a^{4} b e^{5}\right )} x\right )} \log \left (b x + a\right )}{6 \, {\left (b^{9} x^{3} + 3 \, a b^{8} x^{2} + 3 \, a^{2} b^{7} x + a^{3} b^{6}\right )}} \]
1/6*(3*b^5*e^5*x^5 - 2*b^5*d^5 - 5*a*b^4*d^4*e - 20*a^2*b^3*d^3*e^2 + 110* a^3*b^2*d^2*e^3 - 130*a^4*b*d*e^4 + 47*a^5*e^5 + 15*(2*b^5*d*e^4 - a*b^4*e ^5)*x^4 + 9*(10*a*b^4*d*e^4 - 7*a^2*b^3*e^5)*x^3 - 3*(20*b^5*d^3*e^2 - 60* a*b^4*d^2*e^3 + 30*a^2*b^3*d*e^4 + 3*a^3*b^2*e^5)*x^2 - 3*(5*b^5*d^4*e + 2 0*a*b^4*d^3*e^2 - 90*a^2*b^3*d^2*e^3 + 90*a^3*b^2*d*e^4 - 27*a^4*b*e^5)*x + 60*(a^3*b^2*d^2*e^3 - 2*a^4*b*d*e^4 + a^5*e^5 + (b^5*d^2*e^3 - 2*a*b^4*d *e^4 + a^2*b^3*e^5)*x^3 + 3*(a*b^4*d^2*e^3 - 2*a^2*b^3*d*e^4 + a^3*b^2*e^5 )*x^2 + 3*(a^2*b^3*d^2*e^3 - 2*a^3*b^2*d*e^4 + a^4*b*e^5)*x)*log(b*x + a)) /(b^9*x^3 + 3*a*b^8*x^2 + 3*a^2*b^7*x + a^3*b^6)
\[ \int \frac {(a+b x) (d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (a + b x\right ) \left (d + e x\right )^{5}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 1010 vs. \(2 (185) = 370\).
Time = 0.29 (sec) , antiderivative size = 1010, normalized size of antiderivative = 4.01 \[ \int \frac {(a+b x) (d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Too large to display} \]
1/4*b*e^5*((2*b^6*x^6 - 12*a*b^5*x^5 - 68*a^2*b^4*x^4 - 32*a^3*b^3*x^3 + 1 32*a^4*b^2*x^2 + 168*a^5*b*x + 57*a^6)/(b^11*x^4 + 4*a*b^10*x^3 + 6*a^2*b^ 9*x^2 + 4*a^3*b^8*x + a^4*b^7) + 60*a^2*log(b*x + a)/b^7) + 5/12*b*d*e^4*( (12*b^5*x^5 + 48*a*b^4*x^4 - 48*a^2*b^3*x^3 - 252*a^3*b^2*x^2 - 248*a^4*b* x - 77*a^5)/(b^10*x^4 + 4*a*b^9*x^3 + 6*a^2*b^8*x^2 + 4*a^3*b^7*x + a^4*b^ 6) - 60*a*log(b*x + a)/b^6) + 1/12*a*e^5*((12*b^5*x^5 + 48*a*b^4*x^4 - 48* a^2*b^3*x^3 - 252*a^3*b^2*x^2 - 248*a^4*b*x - 77*a^5)/(b^10*x^4 + 4*a*b^9* x^3 + 6*a^2*b^8*x^2 + 4*a^3*b^7*x + a^4*b^6) - 60*a*log(b*x + a)/b^6) + 5/ 6*b*d^2*e^3*((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) + 5/12*a*d*e^4*((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*l og(b*x + a)/b^5) - 5/6*b*d^3*e^2*(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)) - 5/6*a*d^2*e^3*(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)) - 1/12*b*d^5*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3 *a/(b^6*(x + a/b)^4)) - 5/12*a*d^4*e*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b ^2) - 3*a/(b^6*(x + a/b)^4)) - 5/12*b*d^4*e*(6/(b^5*(x + a/b)^2) - 8*a/...
Time = 0.27 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b x) (d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {10 \, {\left (b^{2} d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{6} \mathrm {sgn}\left (b x + a\right )} + \frac {b^{4} e^{5} x^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, b^{4} d e^{4} x \mathrm {sgn}\left (b x + a\right ) - 8 \, a b^{3} e^{5} x \mathrm {sgn}\left (b x + a\right )}{2 \, b^{8}} - \frac {2 \, b^{5} d^{5} + 5 \, a b^{4} d^{4} e + 20 \, a^{2} b^{3} d^{3} e^{2} - 110 \, a^{3} b^{2} d^{2} e^{3} + 130 \, a^{4} b d e^{4} - 47 \, a^{5} e^{5} + 60 \, {\left (b^{5} d^{3} e^{2} - 3 \, a b^{4} d^{2} e^{3} + 3 \, a^{2} b^{3} d e^{4} - a^{3} b^{2} e^{5}\right )} x^{2} + 15 \, {\left (b^{5} d^{4} e + 4 \, a b^{4} d^{3} e^{2} - 18 \, a^{2} b^{3} d^{2} e^{3} + 20 \, a^{3} b^{2} d e^{4} - 7 \, a^{4} b e^{5}\right )} x}{6 \, {\left (b x + a\right )}^{3} b^{6} \mathrm {sgn}\left (b x + a\right )} \]
10*(b^2*d^2*e^3 - 2*a*b*d*e^4 + a^2*e^5)*log(abs(b*x + a))/(b^6*sgn(b*x + a)) + 1/2*(b^4*e^5*x^2*sgn(b*x + a) + 10*b^4*d*e^4*x*sgn(b*x + a) - 8*a*b^ 3*e^5*x*sgn(b*x + a))/b^8 - 1/6*(2*b^5*d^5 + 5*a*b^4*d^4*e + 20*a^2*b^3*d^ 3*e^2 - 110*a^3*b^2*d^2*e^3 + 130*a^4*b*d*e^4 - 47*a^5*e^5 + 60*(b^5*d^3*e ^2 - 3*a*b^4*d^2*e^3 + 3*a^2*b^3*d*e^4 - a^3*b^2*e^5)*x^2 + 15*(b^5*d^4*e + 4*a*b^4*d^3*e^2 - 18*a^2*b^3*d^2*e^3 + 20*a^3*b^2*d*e^4 - 7*a^4*b*e^5)*x )/((b*x + a)^3*b^6*sgn(b*x + a))
Timed out. \[ \int \frac {(a+b x) (d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]