Integrand size = 30, antiderivative size = 202 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\frac {2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) \sqrt {d+e x}}+\frac {6 b (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)}-\frac {2 b^2 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)}+\frac {2 b^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^4 (a+b x)} \]
-2*b^2*(-a*e+b*d)*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+2/5*b^3*(e*x +d)^(5/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+2*(-a*e+b*d)^3*((b*x+a)^2)^(1/2)/e ^4/(b*x+a)/(e*x+d)^(1/2)+6*b*(-a*e+b*d)^2*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2)/ e^4/(b*x+a)
Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.59 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (5 a^3 e^3-15 a^2 b e^2 (2 d+e x)+5 a b^2 e \left (8 d^2+4 d e x-e^2 x^2\right )-b^3 \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )\right )}{5 e^4 (a+b x) \sqrt {d+e x}} \]
(-2*Sqrt[(a + b*x)^2]*(5*a^3*e^3 - 15*a^2*b*e^2*(2*d + e*x) + 5*a*b^2*e*(8 *d^2 + 4*d*e*x - e^2*x^2) - b^3*(16*d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^3*x^ 3)))/(5*e^4*(a + b*x)*Sqrt[d + e*x])
Time = 0.26 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.60, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3}{(d+e x)^{3/2}}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3}{(d+e x)^{3/2}}dx}{a+b x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {(d+e x)^{3/2} b^3}{e^3}-\frac {3 (b d-a e) \sqrt {d+e x} b^2}{e^3}+\frac {3 (b d-a e)^2 b}{e^3 \sqrt {d+e x}}+\frac {(a e-b d)^3}{e^3 (d+e x)^{3/2}}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {2 b^2 (d+e x)^{3/2} (b d-a e)}{e^4}+\frac {6 b \sqrt {d+e x} (b d-a e)^2}{e^4}+\frac {2 (b d-a e)^3}{e^4 \sqrt {d+e x}}+\frac {2 b^3 (d+e x)^{5/2}}{5 e^4}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((2*(b*d - a*e)^3)/(e^4*Sqrt[d + e*x]) + (6 *b*(b*d - a*e)^2*Sqrt[d + e*x])/e^4 - (2*b^2*(b*d - a*e)*(d + e*x)^(3/2))/ e^4 + (2*b^3*(d + e*x)^(5/2))/(5*e^4)))/(a + b*x)
3.17.87.3.1 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 = 2.27 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.65
method | result | size |
gosper | \(-\frac {2 \left (-e^{3} x^{3} b^{3}-5 x^{2} a \,b^{2} e^{3}+2 x^{2} b^{3} d \,e^{2}-15 a^{2} b \,e^{3} x +20 x a \,b^{2} d \,e^{2}-8 b^{3} d^{2} e x +5 a^{3} e^{3}-30 a^{2} b d \,e^{2}+40 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{5 \sqrt {e x +d}\, e^{4} \left (b x +a \right )^{3}}\) | \(132\) |
default | \(-\frac {2 \left (-e^{3} x^{3} b^{3}-5 x^{2} a \,b^{2} e^{3}+2 x^{2} b^{3} d \,e^{2}-15 a^{2} b \,e^{3} x +20 x a \,b^{2} d \,e^{2}-8 b^{3} d^{2} e x +5 a^{3} e^{3}-30 a^{2} b d \,e^{2}+40 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{5 \sqrt {e x +d}\, e^{4} \left (b x +a \right )^{3}}\) | \(132\) |
risch | \(\frac {2 b \left (x^{2} b^{2} e^{2}+5 x a b \,e^{2}-3 b^{2} d e x +15 a^{2} e^{2}-25 a b d e +11 b^{2} d^{2}\right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{5 e^{4} \left (b x +a \right )}-\frac {2 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {\left (b x +a \right )^{2}}}{e^{4} \sqrt {e x +d}\, \left (b x +a \right )}\) | \(144\) |
-2/5/(e*x+d)^(1/2)*(-b^3*e^3*x^3-5*a*b^2*e^3*x^2+2*b^3*d*e^2*x^2-15*a^2*b* e^3*x+20*a*b^2*d*e^2*x-8*b^3*d^2*e*x+5*a^3*e^3-30*a^2*b*d*e^2+40*a*b^2*d^2 *e-16*b^3*d^3)*((b*x+a)^2)^(3/2)/e^4/(b*x+a)^3
Time = 0.28 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.61 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 40 \, a b^{2} d^{2} e + 30 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3} - {\left (2 \, b^{3} d e^{2} - 5 \, a b^{2} e^{3}\right )} x^{2} + {\left (8 \, b^{3} d^{2} e - 20 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{5 \, {\left (e^{5} x + d e^{4}\right )}} \]
2/5*(b^3*e^3*x^3 + 16*b^3*d^3 - 40*a*b^2*d^2*e + 30*a^2*b*d*e^2 - 5*a^3*e^ 3 - (2*b^3*d*e^2 - 5*a*b^2*e^3)*x^2 + (8*b^3*d^2*e - 20*a*b^2*d*e^2 + 15*a ^2*b*e^3)*x)*sqrt(e*x + d)/(e^5*x + d*e^4)
\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {3}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.56 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 40 \, a b^{2} d^{2} e + 30 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3} - {\left (2 \, b^{3} d e^{2} - 5 \, a b^{2} e^{3}\right )} x^{2} + {\left (8 \, b^{3} d^{2} e - 20 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3}\right )} x\right )}}{5 \, \sqrt {e x + d} e^{4}} \]
2/5*(b^3*e^3*x^3 + 16*b^3*d^3 - 40*a*b^2*d^2*e + 30*a^2*b*d*e^2 - 5*a^3*e^ 3 - (2*b^3*d*e^2 - 5*a*b^2*e^3)*x^2 + (8*b^3*d^2*e - 20*a*b^2*d*e^2 + 15*a ^2*b*e^3)*x)/(sqrt(e*x + d)*e^4)
Time = 0.35 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )}}{\sqrt {e x + d} e^{4}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {5}{2}} b^{3} e^{16} \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{3} d e^{16} \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {e x + d} b^{3} d^{2} e^{16} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{2} e^{17} \mathrm {sgn}\left (b x + a\right ) - 30 \, \sqrt {e x + d} a b^{2} d e^{17} \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {e x + d} a^{2} b e^{18} \mathrm {sgn}\left (b x + a\right )\right )}}{5 \, e^{20}} \]
2*(b^3*d^3*sgn(b*x + a) - 3*a*b^2*d^2*e*sgn(b*x + a) + 3*a^2*b*d*e^2*sgn(b *x + a) - a^3*e^3*sgn(b*x + a))/(sqrt(e*x + d)*e^4) + 2/5*((e*x + d)^(5/2) *b^3*e^16*sgn(b*x + a) - 5*(e*x + d)^(3/2)*b^3*d*e^16*sgn(b*x + a) + 15*sq rt(e*x + d)*b^3*d^2*e^16*sgn(b*x + a) + 5*(e*x + d)^(3/2)*a*b^2*e^17*sgn(b *x + a) - 30*sqrt(e*x + d)*a*b^2*d*e^17*sgn(b*x + a) + 15*sqrt(e*x + d)*a^ 2*b*e^18*sgn(b*x + a))/e^20
Time = 10.07 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.73 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {2\,x\,\left (15\,a^2\,e^2-20\,a\,b\,d\,e+8\,b^2\,d^2\right )}{5\,e^3}-\frac {2\,a^3\,e^3-12\,a^2\,b\,d\,e^2+16\,a\,b^2\,d^2\,e-\frac {32\,b^3\,d^3}{5}}{b\,e^4}+\frac {2\,b^2\,x^3}{5\,e}+\frac {2\,b\,x^2\,\left (5\,a\,e-2\,b\,d\right )}{5\,e^2}\right )}{x\,\sqrt {d+e\,x}+\frac {a\,\sqrt {d+e\,x}}{b}} \]