Integrand size = 28, antiderivative size = 200 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\frac {(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{8 e^4 (a+b x) (d+e x)^8}-\frac {3 b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^4 (a+b x) (d+e x)^7}+\frac {b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^4 (a+b x) (d+e x)^6}-\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^4 (a+b x) (d+e x)^5} \]
1/8*(-a*e+b*d)^3*((b*x+a)^2)^(1/2)/e^4/(b*x+a)/(e*x+d)^8-3/7*b*(-a*e+b*d)^ 2*((b*x+a)^2)^(1/2)/e^4/(b*x+a)/(e*x+d)^7+1/2*b^2*(-a*e+b*d)*((b*x+a)^2)^( 1/2)/e^4/(b*x+a)/(e*x+d)^6-1/5*b^3*((b*x+a)^2)^(1/2)/e^4/(b*x+a)/(e*x+d)^5
Time = 1.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.56 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=-\frac {\sqrt {(a+b x)^2} \left (35 a^3 e^3+15 a^2 b e^2 (d+8 e x)+5 a b^2 e \left (d^2+8 d e x+28 e^2 x^2\right )+b^3 \left (d^3+8 d^2 e x+28 d e^2 x^2+56 e^3 x^3\right )\right )}{280 e^4 (a+b x) (d+e x)^8} \]
-1/280*(Sqrt[(a + b*x)^2]*(35*a^3*e^3 + 15*a^2*b*e^2*(d + 8*e*x) + 5*a*b^2 *e*(d^2 + 8*d*e*x + 28*e^2*x^2) + b^3*(d^3 + 8*d^2*e*x + 28*d*e^2*x^2 + 56 *e^3*x^3)))/(e^4*(a + b*x)*(d + e*x)^8)
Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.60, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1102, 27, 53, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3}{(d+e x)^9}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3}{(d+e x)^9}dx}{a+b x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^3}{e^3 (d+e x)^6}-\frac {3 (b d-a e) b^2}{e^3 (d+e x)^7}+\frac {3 (b d-a e)^2 b}{e^3 (d+e x)^8}+\frac {(a e-b d)^3}{e^3 (d+e x)^9}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b^2 (b d-a e)}{2 e^4 (d+e x)^6}-\frac {3 b (b d-a e)^2}{7 e^4 (d+e x)^7}+\frac {(b d-a e)^3}{8 e^4 (d+e x)^8}-\frac {b^3}{5 e^4 (d+e x)^5}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((b*d - a*e)^3/(8*e^4*(d + e*x)^8) - (3*b*( b*d - a*e)^2)/(7*e^4*(d + e*x)^7) + (b^2*(b*d - a*e))/(2*e^4*(d + e*x)^6) - b^3/(5*e^4*(d + e*x)^5)))/(a + b*x)
3.16.68.3.1 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, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 = 3.80 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.63
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {b^{3} x^{3}}{5 e}-\frac {b^{2} \left (5 a e +b d \right ) x^{2}}{10 e^{2}}-\frac {b \left (15 a^{2} e^{2}+5 a b d e +b^{2} d^{2}\right ) x}{35 e^{3}}-\frac {35 a^{3} e^{3}+15 a^{2} b d \,e^{2}+5 a \,b^{2} d^{2} e +b^{3} d^{3}}{280 e^{4}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{8}}\) | \(126\) |
gosper | \(-\frac {\left (56 e^{3} x^{3} b^{3}+140 x^{2} a \,b^{2} e^{3}+28 x^{2} b^{3} d \,e^{2}+120 a^{2} b \,e^{3} x +40 x a \,b^{2} d \,e^{2}+8 b^{3} d^{2} e x +35 a^{3} e^{3}+15 a^{2} b d \,e^{2}+5 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 e^{4} \left (e x +d \right )^{8} \left (b x +a \right )^{3}}\) | \(131\) |
default | \(-\frac {\left (56 e^{3} x^{3} b^{3}+140 x^{2} a \,b^{2} e^{3}+28 x^{2} b^{3} d \,e^{2}+120 a^{2} b \,e^{3} x +40 x a \,b^{2} d \,e^{2}+8 b^{3} d^{2} e x +35 a^{3} e^{3}+15 a^{2} b d \,e^{2}+5 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 e^{4} \left (e x +d \right )^{8} \left (b x +a \right )^{3}}\) | \(131\) |
((b*x+a)^2)^(1/2)/(b*x+a)*(-1/5*b^3/e*x^3-1/10*b^2/e^2*(5*a*e+b*d)*x^2-1/3 5*b/e^3*(15*a^2*e^2+5*a*b*d*e+b^2*d^2)*x-1/280/e^4*(35*a^3*e^3+15*a^2*b*d* e^2+5*a*b^2*d^2*e+b^3*d^3))/(e*x+d)^8
Time = 0.33 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=-\frac {56 \, b^{3} e^{3} x^{3} + b^{3} d^{3} + 5 \, a b^{2} d^{2} e + 15 \, a^{2} b d e^{2} + 35 \, a^{3} e^{3} + 28 \, {\left (b^{3} d e^{2} + 5 \, a b^{2} e^{3}\right )} x^{2} + 8 \, {\left (b^{3} d^{2} e + 5 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3}\right )} x}{280 \, {\left (e^{12} x^{8} + 8 \, d e^{11} x^{7} + 28 \, d^{2} e^{10} x^{6} + 56 \, d^{3} e^{9} x^{5} + 70 \, d^{4} e^{8} x^{4} + 56 \, d^{5} e^{7} x^{3} + 28 \, d^{6} e^{6} x^{2} + 8 \, d^{7} e^{5} x + d^{8} e^{4}\right )}} \]
-1/280*(56*b^3*e^3*x^3 + b^3*d^3 + 5*a*b^2*d^2*e + 15*a^2*b*d*e^2 + 35*a^3 *e^3 + 28*(b^3*d*e^2 + 5*a*b^2*e^3)*x^2 + 8*(b^3*d^2*e + 5*a*b^2*d*e^2 + 1 5*a^2*b*e^3)*x)/(e^12*x^8 + 8*d*e^11*x^7 + 28*d^2*e^10*x^6 + 56*d^3*e^9*x^ 5 + 70*d^4*e^8*x^4 + 56*d^5*e^7*x^3 + 28*d^6*e^6*x^2 + 8*d^7*e^5*x + d^8*e ^4)
Timed out. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\text {Exception raised: ValueError} \]
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 = 257, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\frac {b^{8} \mathrm {sgn}\left (b x + a\right )}{280 \, {\left (b^{5} d^{5} e^{4} - 5 \, a b^{4} d^{4} e^{5} + 10 \, a^{2} b^{3} d^{3} e^{6} - 10 \, a^{3} b^{2} d^{2} e^{7} + 5 \, a^{4} b d e^{8} - a^{5} e^{9}\right )}} - \frac {56 \, b^{3} e^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 28 \, b^{3} d e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 140 \, a b^{2} e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 8 \, b^{3} d^{2} e x \mathrm {sgn}\left (b x + a\right ) + 40 \, a b^{2} d e^{2} x \mathrm {sgn}\left (b x + a\right ) + 120 \, a^{2} b e^{3} x \mathrm {sgn}\left (b x + a\right ) + b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 15 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + 35 \, a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )}{280 \, {\left (e x + d\right )}^{8} e^{4}} \]
1/280*b^8*sgn(b*x + a)/(b^5*d^5*e^4 - 5*a*b^4*d^4*e^5 + 10*a^2*b^3*d^3*e^6 - 10*a^3*b^2*d^2*e^7 + 5*a^4*b*d*e^8 - a^5*e^9) - 1/280*(56*b^3*e^3*x^3*s gn(b*x + a) + 28*b^3*d*e^2*x^2*sgn(b*x + a) + 140*a*b^2*e^3*x^2*sgn(b*x + a) + 8*b^3*d^2*e*x*sgn(b*x + a) + 40*a*b^2*d*e^2*x*sgn(b*x + a) + 120*a^2* b*e^3*x*sgn(b*x + a) + b^3*d^3*sgn(b*x + a) + 5*a*b^2*d^2*e*sgn(b*x + a) + 15*a^2*b*d*e^2*sgn(b*x + a) + 35*a^3*e^3*sgn(b*x + a))/((e*x + d)^8*e^4)
Time = 9.68 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.42 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\frac {\left (\frac {2\,b^3\,d-3\,a\,b^2\,e}{6\,e^4}+\frac {b^3\,d}{6\,e^4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6}-\frac {\left (\frac {3\,a^2\,b\,e^2-3\,a\,b^2\,d\,e+b^3\,d^2}{7\,e^4}+\frac {d\,\left (\frac {b^3\,d}{7\,e^3}-\frac {b^2\,\left (3\,a\,e-b\,d\right )}{7\,e^3}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^7}-\frac {\left (\frac {a^3}{8\,e}-\frac {d\,\left (\frac {3\,a^2\,b}{8\,e}-\frac {d\,\left (\frac {3\,a\,b^2}{8\,e}-\frac {b^3\,d}{8\,e^2}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^8}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,e^4\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5} \]
(((2*b^3*d - 3*a*b^2*e)/(6*e^4) + (b^3*d)/(6*e^4))*(a^2 + b^2*x^2 + 2*a*b* x)^(1/2))/((a + b*x)*(d + e*x)^6) - (((b^3*d^2 + 3*a^2*b*e^2 - 3*a*b^2*d*e )/(7*e^4) + (d*((b^3*d)/(7*e^3) - (b^2*(3*a*e - b*d))/(7*e^3)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^7) - ((a^3/(8*e) - (d*((3*a ^2*b)/(8*e) - (d*((3*a*b^2)/(8*e) - (b^3*d)/(8*e^2)))/e))/e)*(a^2 + b^2*x^ 2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^8) - (b^3*(a^2 + b^2*x^2 + 2*a*b* x)^(1/2))/(5*e^4*(a + b*x)*(d + e*x)^5)