Integrand size = 28, antiderivative size = 143 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^7} \, dx=\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{6 (b d-a e) (d+e x)^6}+\frac {b (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{15 (b d-a e)^2 (d+e x)^5}+\frac {b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{60 (b d-a e)^3 (d+e x)^4} \]
1/6*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(-a*e+b*d)/(e*x+d)^6+1/15*b*(b*x+a )*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(-a*e+b*d)^2/(e*x+d)^5+1/60*b^2*(b*x+a)*(b^2 *x^2+2*a*b*x+a^2)^(3/2)/(-a*e+b*d)^3/(e*x+d)^4
Time = 1.03 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^7} \, dx=-\frac {\sqrt {(a+b x)^2} \left (10 a^3 e^3+6 a^2 b e^2 (d+6 e x)+3 a b^2 e \left (d^2+6 d e x+15 e^2 x^2\right )+b^3 \left (d^3+6 d^2 e x+15 d e^2 x^2+20 e^3 x^3\right )\right )}{60 e^4 (a+b x) (d+e x)^6} \]
-1/60*(Sqrt[(a + b*x)^2]*(10*a^3*e^3 + 6*a^2*b*e^2*(d + 6*e*x) + 3*a*b^2*e *(d^2 + 6*d*e*x + 15*e^2*x^2) + b^3*(d^3 + 6*d^2*e*x + 15*d*e^2*x^2 + 20*e ^3*x^3)))/(e^4*(a + b*x)*(d + e*x)^6)
Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.84, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^7} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3}{(d+e x)^7}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3}{(d+e x)^7}dx}{a+b x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^3}{e^3 (d+e x)^4}-\frac {3 (b d-a e) b^2}{e^3 (d+e x)^5}+\frac {3 (b d-a e)^2 b}{e^3 (d+e x)^6}+\frac {(a e-b d)^3}{e^3 (d+e x)^7}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {3 b^2 (b d-a e)}{4 e^4 (d+e x)^4}-\frac {3 b (b d-a e)^2}{5 e^4 (d+e x)^5}+\frac {(b d-a e)^3}{6 e^4 (d+e x)^6}-\frac {b^3}{3 e^4 (d+e x)^3}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((b*d - a*e)^3/(6*e^4*(d + e*x)^6) - (3*b*( b*d - a*e)^2)/(5*e^4*(d + e*x)^5) + (3*b^2*(b*d - a*e))/(4*e^4*(d + e*x)^4 ) - b^3/(3*e^4*(d + e*x)^3)))/(a + b*x)
3.16.66.3.1 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 = 3.48 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {b^{3} x^{3}}{3 e}-\frac {b^{2} \left (3 a e +b d \right ) x^{2}}{4 e^{2}}-\frac {b \left (6 a^{2} e^{2}+3 a b d e +b^{2} d^{2}\right ) x}{10 e^{3}}-\frac {10 a^{3} e^{3}+6 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +b^{3} d^{3}}{60 e^{4}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{6}}\) | \(126\) |
gosper | \(-\frac {\left (20 e^{3} x^{3} b^{3}+45 x^{2} a \,b^{2} e^{3}+15 x^{2} b^{3} d \,e^{2}+36 a^{2} b \,e^{3} x +18 x a \,b^{2} d \,e^{2}+6 b^{3} d^{2} e x +10 a^{3} e^{3}+6 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 e^{4} \left (e x +d \right )^{6} \left (b x +a \right )^{3}}\) | \(131\) |
default | \(-\frac {\left (20 e^{3} x^{3} b^{3}+45 x^{2} a \,b^{2} e^{3}+15 x^{2} b^{3} d \,e^{2}+36 a^{2} b \,e^{3} x +18 x a \,b^{2} d \,e^{2}+6 b^{3} d^{2} e x +10 a^{3} e^{3}+6 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 e^{4} \left (e x +d \right )^{6} \left (b x +a \right )^{3}}\) | \(131\) |
((b*x+a)^2)^(1/2)/(b*x+a)*(-1/3*b^3/e*x^3-1/4*b^2/e^2*(3*a*e+b*d)*x^2-1/10 /e^3*b*(6*a^2*e^2+3*a*b*d*e+b^2*d^2)*x-1/60/e^4*(10*a^3*e^3+6*a^2*b*d*e^2+ 3*a*b^2*d^2*e+b^3*d^3))/(e*x+d)^6
Time = 0.31 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^7} \, dx=-\frac {20 \, b^{3} e^{3} x^{3} + b^{3} d^{3} + 3 \, a b^{2} d^{2} e + 6 \, a^{2} b d e^{2} + 10 \, a^{3} e^{3} + 15 \, {\left (b^{3} d e^{2} + 3 \, a b^{2} e^{3}\right )} x^{2} + 6 \, {\left (b^{3} d^{2} e + 3 \, a b^{2} d e^{2} + 6 \, a^{2} b e^{3}\right )} x}{60 \, {\left (e^{10} x^{6} + 6 \, d e^{9} x^{5} + 15 \, d^{2} e^{8} x^{4} + 20 \, d^{3} e^{7} x^{3} + 15 \, d^{4} e^{6} x^{2} + 6 \, d^{5} e^{5} x + d^{6} e^{4}\right )}} \]
-1/60*(20*b^3*e^3*x^3 + b^3*d^3 + 3*a*b^2*d^2*e + 6*a^2*b*d*e^2 + 10*a^3*e ^3 + 15*(b^3*d*e^2 + 3*a*b^2*e^3)*x^2 + 6*(b^3*d^2*e + 3*a*b^2*d*e^2 + 6*a ^2*b*e^3)*x)/(e^10*x^6 + 6*d*e^9*x^5 + 15*d^2*e^8*x^4 + 20*d^3*e^7*x^3 + 1 5*d^4*e^6*x^2 + 6*d^5*e^5*x + d^6*e^4)
\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^7} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{7}}\, dx \]
Exception generated. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^7} \, dx=\text {Exception raised: ValueError} \]
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 = 0.27 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.60 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^7} \, dx=\frac {b^{6} \mathrm {sgn}\left (b x + a\right )}{60 \, {\left (b^{3} d^{3} e^{4} - 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} b d e^{6} - a^{3} e^{7}\right )}} - \frac {20 \, b^{3} e^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 15 \, b^{3} d e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 45 \, a b^{2} e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, b^{3} d^{2} e x \mathrm {sgn}\left (b x + a\right ) + 18 \, a b^{2} d e^{2} x \mathrm {sgn}\left (b x + a\right ) + 36 \, a^{2} b e^{3} x \mathrm {sgn}\left (b x + a\right ) + b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )}{60 \, {\left (e x + d\right )}^{6} e^{4}} \]
1/60*b^6*sgn(b*x + a)/(b^3*d^3*e^4 - 3*a*b^2*d^2*e^5 + 3*a^2*b*d*e^6 - a^3 *e^7) - 1/60*(20*b^3*e^3*x^3*sgn(b*x + a) + 15*b^3*d*e^2*x^2*sgn(b*x + a) + 45*a*b^2*e^3*x^2*sgn(b*x + a) + 6*b^3*d^2*e*x*sgn(b*x + a) + 18*a*b^2*d* e^2*x*sgn(b*x + a) + 36*a^2*b*e^3*x*sgn(b*x + a) + b^3*d^3*sgn(b*x + a) + 3*a*b^2*d^2*e*sgn(b*x + a) + 6*a^2*b*d*e^2*sgn(b*x + a) + 10*a^3*e^3*sgn(b *x + a))/((e*x + d)^6*e^4)
Time = 9.50 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.99 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^7} \, dx=\frac {\left (\frac {2\,b^3\,d-3\,a\,b^2\,e}{4\,e^4}+\frac {b^3\,d}{4\,e^4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}-\frac {\left (\frac {3\,a^2\,b\,e^2-3\,a\,b^2\,d\,e+b^3\,d^2}{5\,e^4}+\frac {d\,\left (\frac {b^3\,d}{5\,e^3}-\frac {b^2\,\left (3\,a\,e-b\,d\right )}{5\,e^3}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}-\frac {\left (\frac {a^3}{6\,e}-\frac {d\,\left (\frac {a^2\,b}{2\,e}-\frac {d\,\left (\frac {a\,b^2}{2\,e}-\frac {b^3\,d}{6\,e^2}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,e^4\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3} \]
(((2*b^3*d - 3*a*b^2*e)/(4*e^4) + (b^3*d)/(4*e^4))*(a^2 + b^2*x^2 + 2*a*b* x)^(1/2))/((a + b*x)*(d + e*x)^4) - (((b^3*d^2 + 3*a^2*b*e^2 - 3*a*b^2*d*e )/(5*e^4) + (d*((b^3*d)/(5*e^3) - (b^2*(3*a*e - b*d))/(5*e^3)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^5) - ((a^3/(6*e) - (d*((a^2 *b)/(2*e) - (d*((a*b^2)/(2*e) - (b^3*d)/(6*e^2)))/e))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^6) - (b^3*(a^2 + b^2*x^2 + 2*a*b*x)^( 1/2))/(3*e^4*(a + b*x)*(d + e*x)^3)