Integrand size = 28, antiderivative size = 292 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {b^4 (4 b d-5 a e) x \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac {b^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^4 (a+b x)}+\frac {(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x) (d+e x)^3}-\frac {5 b (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^6 (a+b x) (d+e x)^2}+\frac {10 b^2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) (d+e x)}+\frac {10 b^3 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^6 (a+b x)} \]
-b^4*(-5*a*e+4*b*d)*x*((b*x+a)^2)^(1/2)/e^5/(b*x+a)+1/2*b^5*x^2*((b*x+a)^2 )^(1/2)/e^4/(b*x+a)+1/3*(-a*e+b*d)^5*((b*x+a)^2)^(1/2)/e^6/(b*x+a)/(e*x+d) ^3-5/2*b*(-a*e+b*d)^4*((b*x+a)^2)^(1/2)/e^6/(b*x+a)/(e*x+d)^2+10*b^2*(-a*e +b*d)^3*((b*x+a)^2)^(1/2)/e^6/(b*x+a)/(e*x+d)+10*b^3*(-a*e+b*d)^2*ln(e*x+d )*((b*x+a)^2)^(1/2)/e^6/(b*x+a)
Time = 1.07 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {\sqrt {(a+b x)^2} \left (-2 a^5 e^5-5 a^4 b e^4 (d+3 e x)-20 a^3 b^2 e^3 \left (d^2+3 d e x+3 e^2 x^2\right )+10 a^2 b^3 d e^2 \left (11 d^2+27 d e x+18 e^2 x^2\right )+10 a b^4 e \left (-13 d^4-27 d^3 e x-9 d^2 e^2 x^2+9 d e^3 x^3+3 e^4 x^4\right )+b^5 \left (47 d^5+81 d^4 e x-9 d^3 e^2 x^2-63 d^2 e^3 x^3-15 d e^4 x^4+3 e^5 x^5\right )+60 b^3 (b d-a e)^2 (d+e x)^3 \log (d+e x)\right )}{6 e^6 (a+b x) (d+e x)^3} \]
(Sqrt[(a + b*x)^2]*(-2*a^5*e^5 - 5*a^4*b*e^4*(d + 3*e*x) - 20*a^3*b^2*e^3* (d^2 + 3*d*e*x + 3*e^2*x^2) + 10*a^2*b^3*d*e^2*(11*d^2 + 27*d*e*x + 18*e^2 *x^2) + 10*a*b^4*e*(-13*d^4 - 27*d^3*e*x - 9*d^2*e^2*x^2 + 9*d*e^3*x^3 + 3 *e^4*x^4) + b^5*(47*d^5 + 81*d^4*e*x - 9*d^3*e^2*x^2 - 63*d^2*e^3*x^3 - 15 *d*e^4*x^4 + 3*e^5*x^5) + 60*b^3*(b*d - a*e)^2*(d + e*x)^3*Log[d + e*x]))/ (6*e^6*(a + b*x)*(d + e*x)^3)
Time = 0.35 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.54, 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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 1102 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5}{(d+e x)^4}dx}{b^5 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5}{(d+e x)^4}dx}{a+b x}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {x b^5}{e^4}-\frac {(4 b d-5 a e) b^4}{e^5}+\frac {10 (b d-a e)^2 b^3}{e^5 (d+e x)}-\frac {10 (b d-a e)^3 b^2}{e^5 (d+e x)^2}+\frac {5 (b d-a e)^4 b}{e^5 (d+e x)^3}+\frac {(a e-b d)^5}{e^5 (d+e x)^4}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {b^4 x (4 b d-5 a e)}{e^5}+\frac {10 b^3 (b d-a e)^2 \log (d+e x)}{e^6}+\frac {10 b^2 (b d-a e)^3}{e^6 (d+e x)}-\frac {5 b (b d-a e)^4}{2 e^6 (d+e x)^2}+\frac {(b d-a e)^5}{3 e^6 (d+e x)^3}+\frac {b^5 x^2}{2 e^4}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-((b^4*(4*b*d - 5*a*e)*x)/e^5) + (b^5*x^2) /(2*e^4) + (b*d - a*e)^5/(3*e^6*(d + e*x)^3) - (5*b*(b*d - a*e)^4)/(2*e^6* (d + e*x)^2) + (10*b^2*(b*d - a*e)^3)/(e^6*(d + e*x)) + (10*b^3*(b*d - a*e )^2*Log[d + e*x])/e^6))/(a + b*x)
3.16.78.3.1 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 = 2.84 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.01
| method | result | size |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{4} \left (\frac {1}{2} b e \,x^{2}+5 a e x -4 b d x \right )}{\left (b x +a \right ) e^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-10 a^{3} b^{2} e^{4}+30 d \,e^{3} a^{2} b^{3}-30 d^{2} e^{2} a \,b^{4}+10 b^{5} d^{3} e \right ) x^{2}-\frac {5 b \left (e^{4} a^{4}+4 b \,e^{3} d \,a^{3}-18 b^{2} e^{2} d^{2} a^{2}+20 a \,b^{3} d^{3} e -7 b^{4} d^{4}\right ) x}{2}-\frac {2 a^{5} e^{5}+5 a^{4} b d \,e^{4}+20 a^{3} b^{2} d^{2} e^{3}-110 a^{2} b^{3} d^{3} e^{2}+130 a \,b^{4} d^{4} e -47 b^{5} d^{5}}{6 e}\right )}{\left (b x +a \right ) e^{5} \left (e x +d \right )^{3}}+\frac {10 \sqrt {\left (b x +a \right )^{2}}\, b^{3} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \ln \left (e x +d \right )}{\left (b x +a \right ) e^{6}}\) | \(295\) |
| default | \(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (180 \ln \left (e x +d \right ) b^{5} d^{4} e x +47 b^{5} d^{5}-2 a^{5} e^{5}-5 a^{4} b d \,e^{4}-20 a^{3} b^{2} d^{2} e^{3}+110 a^{2} b^{3} d^{3} e^{2}-130 a \,b^{4} d^{4} e -60 a^{3} b^{2} d \,e^{4} x -15 a^{4} b \,e^{5} x -360 \ln \left (e x +d \right ) a \,b^{4} d^{3} e^{2} x +180 \ln \left (e x +d \right ) b^{5} d^{3} e^{2} x^{2}+60 \ln \left (e x +d \right ) a^{2} b^{3} d^{3} e^{2}-120 \ln \left (e x +d \right ) a \,b^{4} d^{4} e +60 \ln \left (e x +d \right ) a^{2} b^{3} e^{5} x^{3}+180 \ln \left (e x +d \right ) a^{2} b^{3} d \,e^{4} x^{2}-360 \ln \left (e x +d \right ) a \,b^{4} d^{2} e^{3} x^{2}+90 x^{3} a \,b^{4} d \,e^{4}+180 x^{2} a^{2} b^{3} d \,e^{4}-90 x^{2} a \,b^{4} d^{2} e^{3}+180 \ln \left (e x +d \right ) a^{2} b^{3} d^{2} e^{3} x +60 \ln \left (e x +d \right ) b^{5} d^{2} e^{3} x^{3}-120 \ln \left (e x +d \right ) a \,b^{4} d \,e^{4} x^{3}-15 x^{4} b^{5} d \,e^{4}-63 x^{3} b^{5} d^{2} e^{3}+60 \ln \left (e x +d \right ) b^{5} d^{5}+81 b^{5} d^{4} e x +3 x^{5} e^{5} b^{5}+30 x^{4} a \,b^{4} e^{5}-60 x^{2} a^{3} b^{2} e^{5}-9 x^{2} b^{5} d^{3} e^{2}+270 x \,a^{2} b^{3} d^{2} e^{3}-270 x a \,b^{4} d^{3} e^{2}\right )}{6 \left (b x +a \right )^{5} e^{6} \left (e x +d \right )^{3}}\) | \(502\) |
((b*x+a)^2)^(1/2)/(b*x+a)*b^4/e^5*(1/2*b*e*x^2+5*a*e*x-4*b*d*x)+((b*x+a)^2 )^(1/2)/(b*x+a)*((-10*a^3*b^2*e^4+30*a^2*b^3*d*e^3-30*a*b^4*d^2*e^2+10*b^5 *d^3*e)*x^2-5/2*b*(a^4*e^4+4*a^3*b*d*e^3-18*a^2*b^2*d^2*e^2+20*a*b^3*d^3*e -7*b^4*d^4)*x-1/6*(2*a^5*e^5+5*a^4*b*d*e^4+20*a^3*b^2*d^2*e^3-110*a^2*b^3* d^3*e^2+130*a*b^4*d^4*e-47*b^5*d^5)/e)/e^5/(e*x+d)^3+10*((b*x+a)^2)^(1/2)/ (b*x+a)*b^3/e^6*(a^2*e^2-2*a*b*d*e+b^2*d^2)*ln(e*x+d)
Time = 0.28 (sec) , antiderivative size = 425, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {3 \, b^{5} e^{5} x^{5} + 47 \, b^{5} d^{5} - 130 \, a b^{4} d^{4} e + 110 \, a^{2} b^{3} d^{3} e^{2} - 20 \, a^{3} b^{2} d^{2} e^{3} - 5 \, a^{4} b d e^{4} - 2 \, a^{5} e^{5} - 15 \, {\left (b^{5} d e^{4} - 2 \, a b^{4} e^{5}\right )} x^{4} - 9 \, {\left (7 \, b^{5} d^{2} e^{3} - 10 \, a b^{4} d e^{4}\right )} x^{3} - 3 \, {\left (3 \, b^{5} d^{3} e^{2} + 30 \, a b^{4} d^{2} e^{3} - 60 \, a^{2} b^{3} d e^{4} + 20 \, a^{3} b^{2} e^{5}\right )} x^{2} + 3 \, {\left (27 \, b^{5} d^{4} e - 90 \, a b^{4} d^{3} e^{2} + 90 \, a^{2} b^{3} d^{2} e^{3} - 20 \, a^{3} b^{2} d e^{4} - 5 \, a^{4} b e^{5}\right )} x + 60 \, {\left (b^{5} d^{5} - 2 \, a b^{4} d^{4} e + a^{2} b^{3} d^{3} e^{2} + {\left (b^{5} d^{2} e^{3} - 2 \, a b^{4} d e^{4} + a^{2} b^{3} e^{5}\right )} x^{3} + 3 \, {\left (b^{5} d^{3} e^{2} - 2 \, a b^{4} d^{2} e^{3} + a^{2} b^{3} d e^{4}\right )} x^{2} + 3 \, {\left (b^{5} d^{4} e - 2 \, a b^{4} d^{3} e^{2} + a^{2} b^{3} d^{2} e^{3}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\right )}} \]
1/6*(3*b^5*e^5*x^5 + 47*b^5*d^5 - 130*a*b^4*d^4*e + 110*a^2*b^3*d^3*e^2 - 20*a^3*b^2*d^2*e^3 - 5*a^4*b*d*e^4 - 2*a^5*e^5 - 15*(b^5*d*e^4 - 2*a*b^4*e ^5)*x^4 - 9*(7*b^5*d^2*e^3 - 10*a*b^4*d*e^4)*x^3 - 3*(3*b^5*d^3*e^2 + 30*a *b^4*d^2*e^3 - 60*a^2*b^3*d*e^4 + 20*a^3*b^2*e^5)*x^2 + 3*(27*b^5*d^4*e - 90*a*b^4*d^3*e^2 + 90*a^2*b^3*d^2*e^3 - 20*a^3*b^2*d*e^4 - 5*a^4*b*e^5)*x + 60*(b^5*d^5 - 2*a*b^4*d^4*e + a^2*b^3*d^3*e^2 + (b^5*d^2*e^3 - 2*a*b^4*d *e^4 + a^2*b^3*e^5)*x^3 + 3*(b^5*d^3*e^2 - 2*a*b^4*d^2*e^3 + a^2*b^3*d*e^4 )*x^2 + 3*(b^5*d^4*e - 2*a*b^4*d^3*e^2 + a^2*b^3*d^2*e^3)*x)*log(e*x + d)) /(e^9*x^3 + 3*d*e^8*x^2 + 3*d^2*e^7*x + d^3*e^6)
\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \]
Exception generated. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\text {Exception raised: ValueError} \]
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 = 0.28 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {10 \, {\left (b^{5} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{4} d e \mathrm {sgn}\left (b x + a\right ) + a^{2} b^{3} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | e x + d \right |}\right )}{e^{6}} + \frac {b^{5} e^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) - 8 \, b^{5} d e^{3} x \mathrm {sgn}\left (b x + a\right ) + 10 \, a b^{4} e^{4} x \mathrm {sgn}\left (b x + a\right )}{2 \, e^{8}} + \frac {47 \, b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 130 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 110 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 20 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 5 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) - 2 \, a^{5} e^{5} \mathrm {sgn}\left (b x + a\right ) + 60 \, {\left (b^{5} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{4} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b^{3} d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{3} b^{2} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 15 \, {\left (7 \, b^{5} d^{4} e \mathrm {sgn}\left (b x + a\right ) - 20 \, a b^{4} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 18 \, a^{2} b^{3} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b^{2} d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{4} b e^{5} \mathrm {sgn}\left (b x + a\right )\right )} x}{6 \, {\left (e x + d\right )}^{3} e^{6}} \]
10*(b^5*d^2*sgn(b*x + a) - 2*a*b^4*d*e*sgn(b*x + a) + a^2*b^3*e^2*sgn(b*x + a))*log(abs(e*x + d))/e^6 + 1/2*(b^5*e^4*x^2*sgn(b*x + a) - 8*b^5*d*e^3* x*sgn(b*x + a) + 10*a*b^4*e^4*x*sgn(b*x + a))/e^8 + 1/6*(47*b^5*d^5*sgn(b* x + a) - 130*a*b^4*d^4*e*sgn(b*x + a) + 110*a^2*b^3*d^3*e^2*sgn(b*x + a) - 20*a^3*b^2*d^2*e^3*sgn(b*x + a) - 5*a^4*b*d*e^4*sgn(b*x + a) - 2*a^5*e^5* sgn(b*x + a) + 60*(b^5*d^3*e^2*sgn(b*x + a) - 3*a*b^4*d^2*e^3*sgn(b*x + a) + 3*a^2*b^3*d*e^4*sgn(b*x + a) - a^3*b^2*e^5*sgn(b*x + a))*x^2 + 15*(7*b^ 5*d^4*e*sgn(b*x + a) - 20*a*b^4*d^3*e^2*sgn(b*x + a) + 18*a^2*b^3*d^2*e^3* sgn(b*x + a) - 4*a^3*b^2*d*e^4*sgn(b*x + a) - a^4*b*e^5*sgn(b*x + a))*x)/( (e*x + d)^3*e^6)
Timed out. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \]