Integrand size = 33, antiderivative size = 201 \[ \int \frac {(a+b x) (d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {(d+e x)^4}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {4 e^2 (b d-a e)^2}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 e (b d-a e)^3}{3 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 e^4 x (a+b x)}{3 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 e^3 (b d-a e) (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]
-1/3*(e*x+d)^4/b/(b^2*x^2+2*a*b*x+a^2)^(3/2)-4*e^2*(-a*e+b*d)^2/b^5/((b*x+ a)^2)^(1/2)-2/3*e*(-a*e+b*d)^3/b^5/(b*x+a)/((b*x+a)^2)^(1/2)+4/3*e^4*x*(b* x+a)/b^4/((b*x+a)^2)^(1/2)+4*e^3*(-a*e+b*d)*(b*x+a)*ln(b*x+a)/b^5/((b*x+a) ^2)^(1/2)
Time = 1.06 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b x) (d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-13 a^4 e^4+a^3 b e^3 (22 d-27 e x)-3 a^2 b^2 e^2 \left (2 d^2-18 d e x+3 e^2 x^2\right )+a b^3 e \left (-2 d^3-18 d^2 e x+36 d e^2 x^2+9 e^3 x^3\right )-b^4 \left (d^4+6 d^3 e x+18 d^2 e^2 x^2-3 e^4 x^4\right )-12 e^3 (-b d+a e) (a+b x)^3 \log (a+b x)}{3 b^5 \left ((a+b x)^2\right )^{3/2}} \]
(-13*a^4*e^4 + a^3*b*e^3*(22*d - 27*e*x) - 3*a^2*b^2*e^2*(2*d^2 - 18*d*e*x + 3*e^2*x^2) + a*b^3*e*(-2*d^3 - 18*d^2*e*x + 36*d*e^2*x^2 + 9*e^3*x^3) - b^4*(d^4 + 6*d^3*e*x + 18*d^2*e^2*x^2 - 3*e^4*x^4) - 12*e^3*(-(b*d) + a*e )*(a + b*x)^3*Log[a + b*x])/(3*b^5*((a + b*x)^2)^(3/2))
Time = 0.32 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.64, 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.
| \(\displaystyle \int \frac {(a+b x) (d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^5 (a+b x) \int \frac {(d+e x)^4}{b^5 (a+b x)^4}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^4}{(a+b x)^4}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {(a+b x) \int \left (\frac {e^4}{b^4}+\frac {4 (b d-a e) e^3}{b^4 (a+b x)}+\frac {6 (b d-a e)^2 e^2}{b^4 (a+b x)^2}+\frac {4 (b d-a e)^3 e}{b^4 (a+b x)^3}+\frac {(b d-a e)^4}{b^4 (a+b x)^4}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(a+b x) \left (\frac {4 e^3 (b d-a e) \log (a+b x)}{b^5}-\frac {6 e^2 (b d-a e)^2}{b^5 (a+b x)}-\frac {2 e (b d-a e)^3}{b^5 (a+b x)^2}-\frac {(b d-a e)^4}{3 b^5 (a+b x)^3}+\frac {e^4 x}{b^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((e^4*x)/b^4 - (b*d - a*e)^4/(3*b^5*(a + b*x)^3) - (2*e*(b*d - a*e)^3)/(b^5*(a + b*x)^2) - (6*e^2*(b*d - a*e)^2)/(b^5*(a + b*x)) + (4*e^3 *(b*d - a*e)*Log[a + b*x])/b^5))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.21.34.3.1 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 = 0.64 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.07
| method | result | size |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{4} x}{\left (b x +a \right ) b^{4}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-6 a^{2} b \,e^{4}+12 a \,b^{2} d \,e^{3}-6 b^{3} d^{2} e^{2}\right ) x^{2}-2 e \left (5 a^{3} e^{3}-9 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) x -\frac {13 e^{4} a^{4}-22 b d \,e^{3} a^{3}+6 b^{2} d^{2} e^{2} a^{2}+2 b^{3} d^{3} e a +b^{4} d^{4}}{3 b}\right )}{\left (b x +a \right )^{4} b^{4}}-\frac {4 \sqrt {\left (b x +a \right )^{2}}\, e^{3} \left (a e -b d \right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\) | \(215\) |
| default | \(-\frac {\left (12 \ln \left (b x +a \right ) x^{3} a \,b^{3} e^{4}-12 \ln \left (b x +a \right ) b^{4} d \,e^{3} x^{3}-3 e^{4} x^{4} b^{4}+36 \ln \left (b x +a \right ) x^{2} a^{2} b^{2} e^{4}-36 \ln \left (b x +a \right ) x^{2} a \,b^{3} d \,e^{3}-9 x^{3} a \,b^{3} e^{4}+36 \ln \left (b x +a \right ) x \,a^{3} b \,e^{4}-36 \ln \left (b x +a \right ) x \,a^{2} b^{2} d \,e^{3}+9 x^{2} a^{2} b^{2} e^{4}-36 x^{2} a \,b^{3} d \,e^{3}+18 x^{2} b^{4} d^{2} e^{2}+12 \ln \left (b x +a \right ) a^{4} e^{4}-12 \ln \left (b x +a \right ) a^{3} b d \,e^{3}+27 x \,a^{3} b \,e^{4}-54 x \,a^{2} b^{2} d \,e^{3}+18 x a \,b^{3} d^{2} e^{2}+6 x \,b^{4} d^{3} e +13 e^{4} a^{4}-22 b d \,e^{3} a^{3}+6 b^{2} d^{2} e^{2} a^{2}+2 b^{3} d^{3} e a +b^{4} d^{4}\right ) \left (b x +a \right )^{2}}{3 b^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(322\) |
((b*x+a)^2)^(1/2)/(b*x+a)*e^4/b^4*x+((b*x+a)^2)^(1/2)/(b*x+a)^4*((-6*a^2*b *e^4+12*a*b^2*d*e^3-6*b^3*d^2*e^2)*x^2-2*e*(5*a^3*e^3-9*a^2*b*d*e^2+3*a*b^ 2*d^2*e+b^3*d^3)*x-1/3*(13*a^4*e^4-22*a^3*b*d*e^3+6*a^2*b^2*d^2*e^2+2*a*b^ 3*d^3*e+b^4*d^4)/b)/b^4-4*((b*x+a)^2)^(1/2)/(b*x+a)/b^5*e^3*(a*e-b*d)*ln(b *x+a)
Time = 0.31 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.45 \[ \int \frac {(a+b x) (d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {3 \, b^{4} e^{4} x^{4} + 9 \, a b^{3} e^{4} x^{3} - b^{4} d^{4} - 2 \, a b^{3} d^{3} e - 6 \, a^{2} b^{2} d^{2} e^{2} + 22 \, a^{3} b d e^{3} - 13 \, a^{4} e^{4} - 9 \, {\left (2 \, b^{4} d^{2} e^{2} - 4 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} - 3 \, {\left (2 \, b^{4} d^{3} e + 6 \, a b^{3} d^{2} e^{2} - 18 \, a^{2} b^{2} d e^{3} + 9 \, a^{3} b e^{4}\right )} x + 12 \, {\left (a^{3} b d e^{3} - a^{4} e^{4} + {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 3 \, {\left (a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x\right )} \log \left (b x + a\right )}{3 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}} \]
1/3*(3*b^4*e^4*x^4 + 9*a*b^3*e^4*x^3 - b^4*d^4 - 2*a*b^3*d^3*e - 6*a^2*b^2 *d^2*e^2 + 22*a^3*b*d*e^3 - 13*a^4*e^4 - 9*(2*b^4*d^2*e^2 - 4*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 - 3*(2*b^4*d^3*e + 6*a*b^3*d^2*e^2 - 18*a^2*b^2*d*e^3 + 9*a^3*b*e^4)*x + 12*(a^3*b*d*e^3 - a^4*e^4 + (b^4*d*e^3 - a*b^3*e^4)*x^3 + 3*(a*b^3*d*e^3 - a^2*b^2*e^4)*x^2 + 3*(a^2*b^2*d*e^3 - a^3*b*e^4)*x)*log (b*x + a))/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5)
\[ \int \frac {(a+b x) (d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (a + b x\right ) \left (d + e x\right )^{4}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 755 vs. \(2 (149) = 298\).
Time = 0.24 (sec) , antiderivative size = 755, normalized size of antiderivative = 3.76 \[ \int \frac {(a+b x) (d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {1}{12} \, b e^{4} {\left (\frac {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}} - \frac {60 \, a \log \left (b x + a\right )}{b^{6}}\right )} + \frac {1}{3} \, b d e^{3} {\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 e^{4} {\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}{2} \, b d^{2} e^{2} {\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 )} - \frac {1}{3} \, a d e^{3} {\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 )} - \frac {1}{12} \, b d^{4} {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{3} \, a d^{3} e {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{3} \, b d^{3} e {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{2} \, a d^{2} e^{2} {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {a d^{4}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} \]
1/12*b*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) + 1/3*b*d*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) + 1/12*a*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*log(b*x + a)/b^5) - 1/2*b*d^2*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)) - 1/3*a*d*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^4*(4/((b ^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/(b^6*(x + a/b)^4)) - 1/3*a*d^3*e* (4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/(b^6*(x + a/b)^4)) - 1/3*b* d^3*e*(6/(b^5*(x + a/b)^2) - 8*a/(b^6*(x + a/b)^3) + 3*a^2/(b^7*(x + a/b)^ 4)) - 1/2*a*d^2*e^2*(6/(b^5*(x + a/b)^2) - 8*a/(b^6*(x + a/b)^3) + 3*a^2/( b^7*(x + a/b)^4)) - 1/4*a*d^4/(b^5*(x + a/b)^4)
Time = 0.27 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x) (d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^{4} x}{b^{4} \mathrm {sgn}\left (b x + a\right )} + \frac {4 \, {\left (b d e^{3} - a e^{4}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5} \mathrm {sgn}\left (b x + a\right )} - \frac {b^{4} d^{4} + 2 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 22 \, a^{3} b d e^{3} + 13 \, a^{4} e^{4} + 18 \, {\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 6 \, {\left (b^{4} d^{3} e + 3 \, a b^{3} d^{2} e^{2} - 9 \, a^{2} b^{2} d e^{3} + 5 \, a^{3} b e^{4}\right )} x}{3 \, {\left (b x + a\right )}^{3} b^{5} \mathrm {sgn}\left (b x + a\right )} \]
e^4*x/(b^4*sgn(b*x + a)) + 4*(b*d*e^3 - a*e^4)*log(abs(b*x + a))/(b^5*sgn( b*x + a)) - 1/3*(b^4*d^4 + 2*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 22*a^3*b*d* e^3 + 13*a^4*e^4 + 18*(b^4*d^2*e^2 - 2*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 6* (b^4*d^3*e + 3*a*b^3*d^2*e^2 - 9*a^2*b^2*d*e^3 + 5*a^3*b*e^4)*x)/((b*x + a )^3*b^5*sgn(b*x + a))
Timed out. \[ \int \frac {(a+b x) (d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]