Integrand size = 35, antiderivative size = 260 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=-\frac {2 (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x) (d+e x)^{3/2}}+\frac {8 b (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) \sqrt {d+e x}}+\frac {12 b^2 (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}-\frac {8 b^3 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x)}+\frac {2 b^4 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x)} \]
-2/3*(-a*e+b*d)^4*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^(3/2)-8/3*b^3*(-a* e+b*d)*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)+2/5*b^4*(e*x+d)^(5/2)*( (b*x+a)^2)^(1/2)/e^5/(b*x+a)+8*b*(-a*e+b*d)^3*((b*x+a)^2)^(1/2)/e^5/(b*x+a )/(e*x+d)^(1/2)+12*b^2*(-a*e+b*d)^2*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2)/e^5/(b *x+a)
Time = 0.04 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.67 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (5 a^4 e^4+20 a^3 b e^3 (2 d+3 e x)-30 a^2 b^2 e^2 \left (8 d^2+12 d e x+3 e^2 x^2\right )+20 a b^3 e \left (16 d^3+24 d^2 e x+6 d e^2 x^2-e^3 x^3\right )-b^4 \left (128 d^4+192 d^3 e x+48 d^2 e^2 x^2-8 d e^3 x^3+3 e^4 x^4\right )\right )}{15 e^5 (a+b x) (d+e x)^{3/2}} \]
(-2*Sqrt[(a + b*x)^2]*(5*a^4*e^4 + 20*a^3*b*e^3*(2*d + 3*e*x) - 30*a^2*b^2 *e^2*(8*d^2 + 12*d*e*x + 3*e^2*x^2) + 20*a*b^3*e*(16*d^3 + 24*d^2*e*x + 6* d*e^2*x^2 - e^3*x^3) - b^4*(128*d^4 + 192*d^3*e*x + 48*d^2*e^2*x^2 - 8*d*e ^3*x^3 + 3*e^4*x^4)))/(15*e^5*(a + b*x)*(d + e*x)^(3/2))
Time = 0.29 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.59, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1187, 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.
| \(\displaystyle \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^4}{(d+e x)^{5/2}}dx}{b^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^4}{(d+e x)^{5/2}}dx}{a+b x}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {(d+e x)^{3/2} b^4}{e^4}-\frac {4 (b d-a e) \sqrt {d+e x} b^3}{e^4}+\frac {6 (b d-a e)^2 b^2}{e^4 \sqrt {d+e x}}-\frac {4 (b d-a e)^3 b}{e^4 (d+e x)^{3/2}}+\frac {(a e-b d)^4}{e^4 (d+e x)^{5/2}}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {8 b^3 (d+e x)^{3/2} (b d-a e)}{3 e^5}+\frac {12 b^2 \sqrt {d+e x} (b d-a e)^2}{e^5}+\frac {8 b (b d-a e)^3}{e^5 \sqrt {d+e x}}-\frac {2 (b d-a e)^4}{3 e^5 (d+e x)^{3/2}}+\frac {2 b^4 (d+e x)^{5/2}}{5 e^5}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*(b*d - a*e)^4)/(3*e^5*(d + e*x)^(3/2)) + (8*b*(b*d - a*e)^3)/(e^5*Sqrt[d + e*x]) + (12*b^2*(b*d - a*e)^2*Sqrt[d + e*x])/e^5 - (8*b^3*(b*d - a*e)*(d + e*x)^(3/2))/(3*e^5) + (2*b^4*(d + e* x)^(5/2))/(5*e^5)))/(a + b*x)
3.22.5.3.1 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_.)*((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 = 0.31 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.62
| method | result | size |
| risch | \(\frac {2 b^{2} \left (3 b^{2} e^{2} x^{2}+20 a b \,e^{2} x -14 b^{2} d e x +90 e^{2} a^{2}-160 a b d e +73 b^{2} d^{2}\right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{15 e^{5} \left (b x +a \right )}-\frac {2 \left (12 b e x +a e +11 b d \right ) \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {\left (b x +a \right )^{2}}}{3 e^{5} \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right )}\) | \(160\) |
| gosper | \(-\frac {2 \left (-3 e^{4} x^{4} b^{4}-20 x^{3} a \,b^{3} e^{4}+8 x^{3} b^{4} d \,e^{3}-90 x^{2} a^{2} b^{2} e^{4}+120 x^{2} a \,b^{3} d \,e^{3}-48 x^{2} b^{4} d^{2} e^{2}+60 x \,a^{3} b \,e^{4}-360 x \,a^{2} b^{2} d \,e^{3}+480 x a \,b^{3} d^{2} e^{2}-192 x \,b^{4} d^{3} e +5 e^{4} a^{4}+40 b d \,e^{3} a^{3}-240 b^{2} d^{2} e^{2} a^{2}+320 b^{3} d^{3} e a -128 b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{15 \left (e x +d \right )^{\frac {3}{2}} e^{5} \left (b x +a \right )^{3}}\) | \(202\) |
| default | \(-\frac {2 \left (-3 e^{4} x^{4} b^{4}-20 x^{3} a \,b^{3} e^{4}+8 x^{3} b^{4} d \,e^{3}-90 x^{2} a^{2} b^{2} e^{4}+120 x^{2} a \,b^{3} d \,e^{3}-48 x^{2} b^{4} d^{2} e^{2}+60 x \,a^{3} b \,e^{4}-360 x \,a^{2} b^{2} d \,e^{3}+480 x a \,b^{3} d^{2} e^{2}-192 x \,b^{4} d^{3} e +5 e^{4} a^{4}+40 b d \,e^{3} a^{3}-240 b^{2} d^{2} e^{2} a^{2}+320 b^{3} d^{3} e a -128 b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{15 \left (e x +d \right )^{\frac {3}{2}} e^{5} \left (b x +a \right )^{3}}\) | \(202\) |
2/15*b^2*(3*b^2*e^2*x^2+20*a*b*e^2*x-14*b^2*d*e*x+90*a^2*e^2-160*a*b*d*e+7 3*b^2*d^2)*(e*x+d)^(1/2)/e^5*((b*x+a)^2)^(1/2)/(b*x+a)-2/3*(12*b*e*x+a*e+1 1*b*d)*(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)/e^5/(e*x+d)^(3/2)*((b *x+a)^2)^(1/2)/(b*x+a)
Time = 0.38 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, b^{4} e^{4} x^{4} + 128 \, b^{4} d^{4} - 320 \, a b^{3} d^{3} e + 240 \, a^{2} b^{2} d^{2} e^{2} - 40 \, a^{3} b d e^{3} - 5 \, a^{4} e^{4} - 4 \, {\left (2 \, b^{4} d e^{3} - 5 \, a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (8 \, b^{4} d^{2} e^{2} - 20 \, a b^{3} d e^{3} + 15 \, a^{2} b^{2} e^{4}\right )} x^{2} + 12 \, {\left (16 \, b^{4} d^{3} e - 40 \, a b^{3} d^{2} e^{2} + 30 \, a^{2} b^{2} d e^{3} - 5 \, a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \]
2/15*(3*b^4*e^4*x^4 + 128*b^4*d^4 - 320*a*b^3*d^3*e + 240*a^2*b^2*d^2*e^2 - 40*a^3*b*d*e^3 - 5*a^4*e^4 - 4*(2*b^4*d*e^3 - 5*a*b^3*e^4)*x^3 + 6*(8*b^ 4*d^2*e^2 - 20*a*b^3*d*e^3 + 15*a^2*b^2*e^4)*x^2 + 12*(16*b^4*d^3*e - 40*a *b^3*d^2*e^2 + 30*a^2*b^2*d*e^3 - 5*a^3*b*e^4)*x)*sqrt(e*x + d)/(e^7*x^2 + 2*d*e^6*x + d^2*e^5)
\[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\int \frac {\left (a + b x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (b^{3} e^{3} x^{3} - 16 \, b^{3} d^{3} + 24 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} - a^{3} e^{3} - 3 \, {\left (2 \, b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} - 3 \, {\left (8 \, b^{3} d^{2} e - 12 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )} a}{3 \, {\left (e^{5} x + d e^{4}\right )} \sqrt {e x + d}} + \frac {2 \, {\left (3 \, b^{3} e^{4} x^{4} + 128 \, b^{3} d^{4} - 240 \, a b^{2} d^{3} e + 120 \, a^{2} b d^{2} e^{2} - 10 \, a^{3} d e^{3} - {\left (8 \, b^{3} d e^{3} - 15 \, a b^{2} e^{4}\right )} x^{3} + 3 \, {\left (16 \, b^{3} d^{2} e^{2} - 30 \, a b^{2} d e^{3} + 15 \, a^{2} b e^{4}\right )} x^{2} + 3 \, {\left (64 \, b^{3} d^{3} e - 120 \, a b^{2} d^{2} e^{2} + 60 \, a^{2} b d e^{3} - 5 \, a^{3} e^{4}\right )} x\right )} b}{15 \, {\left (e^{6} x + d e^{5}\right )} \sqrt {e x + d}} \]
2/3*(b^3*e^3*x^3 - 16*b^3*d^3 + 24*a*b^2*d^2*e - 6*a^2*b*d*e^2 - a^3*e^3 - 3*(2*b^3*d*e^2 - 3*a*b^2*e^3)*x^2 - 3*(8*b^3*d^2*e - 12*a*b^2*d*e^2 + 3*a ^2*b*e^3)*x)*a/((e^5*x + d*e^4)*sqrt(e*x + d)) + 2/15*(3*b^3*e^4*x^4 + 128 *b^3*d^4 - 240*a*b^2*d^3*e + 120*a^2*b*d^2*e^2 - 10*a^3*d*e^3 - (8*b^3*d*e ^3 - 15*a*b^2*e^4)*x^3 + 3*(16*b^3*d^2*e^2 - 30*a*b^2*d*e^3 + 15*a^2*b*e^4 )*x^2 + 3*(64*b^3*d^3*e - 120*a*b^2*d^2*e^2 + 60*a^2*b*d*e^3 - 5*a^3*e^4)* x)*b/((e^6*x + d*e^5)*sqrt(e*x + d))
Time = 0.28 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (12 \, {\left (e x + d\right )} b^{4} d^{3} \mathrm {sgn}\left (b x + a\right ) - b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 36 \, {\left (e x + d\right )} a b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 36 \, {\left (e x + d\right )} a^{2} b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 12 \, {\left (e x + d\right )} a^{3} b e^{3} \mathrm {sgn}\left (b x + a\right ) + 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) - a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{5}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{4} e^{20} \mathrm {sgn}\left (b x + a\right ) - 20 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{4} d e^{20} \mathrm {sgn}\left (b x + a\right ) + 90 \, \sqrt {e x + d} b^{4} d^{2} e^{20} \mathrm {sgn}\left (b x + a\right ) + 20 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{3} e^{21} \mathrm {sgn}\left (b x + a\right ) - 180 \, \sqrt {e x + d} a b^{3} d e^{21} \mathrm {sgn}\left (b x + a\right ) + 90 \, \sqrt {e x + d} a^{2} b^{2} e^{22} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, e^{25}} \]
2/3*(12*(e*x + d)*b^4*d^3*sgn(b*x + a) - b^4*d^4*sgn(b*x + a) - 36*(e*x + d)*a*b^3*d^2*e*sgn(b*x + a) + 4*a*b^3*d^3*e*sgn(b*x + a) + 36*(e*x + d)*a^ 2*b^2*d*e^2*sgn(b*x + a) - 6*a^2*b^2*d^2*e^2*sgn(b*x + a) - 12*(e*x + d)*a ^3*b*e^3*sgn(b*x + a) + 4*a^3*b*d*e^3*sgn(b*x + a) - a^4*e^4*sgn(b*x + a)) /((e*x + d)^(3/2)*e^5) + 2/15*(3*(e*x + d)^(5/2)*b^4*e^20*sgn(b*x + a) - 2 0*(e*x + d)^(3/2)*b^4*d*e^20*sgn(b*x + a) + 90*sqrt(e*x + d)*b^4*d^2*e^20* sgn(b*x + a) + 20*(e*x + d)^(3/2)*a*b^3*e^21*sgn(b*x + a) - 180*sqrt(e*x + d)*a*b^3*d*e^21*sgn(b*x + a) + 90*sqrt(e*x + d)*a^2*b^2*e^22*sgn(b*x + a) )/e^25
Time = 11.79 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {2\,b^3\,x^4}{5\,e^2}-\frac {10\,a^4\,e^4+80\,a^3\,b\,d\,e^3-480\,a^2\,b^2\,d^2\,e^2+640\,a\,b^3\,d^3\,e-256\,b^4\,d^4}{15\,b\,e^6}-\frac {x\,\left (120\,a^3\,b\,e^4-720\,a^2\,b^2\,d\,e^3+960\,a\,b^3\,d^2\,e^2-384\,b^4\,d^3\,e\right )}{15\,b\,e^6}+\frac {8\,b^2\,x^3\,\left (5\,a\,e-2\,b\,d\right )}{15\,e^3}+\frac {4\,b\,x^2\,\left (15\,a^2\,e^2-20\,a\,b\,d\,e+8\,b^2\,d^2\right )}{5\,e^4}\right )}{x^2\,\sqrt {d+e\,x}+\frac {a\,d\,\sqrt {d+e\,x}}{b\,e}+\frac {x\,\left (15\,a\,e^6+15\,b\,d\,e^5\right )\,\sqrt {d+e\,x}}{15\,b\,e^6}} \]
((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((2*b^3*x^4)/(5*e^2) - (10*a^4*e^4 - 256* b^4*d^4 - 480*a^2*b^2*d^2*e^2 + 640*a*b^3*d^3*e + 80*a^3*b*d*e^3)/(15*b*e^ 6) - (x*(120*a^3*b*e^4 - 384*b^4*d^3*e + 960*a*b^3*d^2*e^2 - 720*a^2*b^2*d *e^3))/(15*b*e^6) + (8*b^2*x^3*(5*a*e - 2*b*d))/(15*e^3) + (4*b*x^2*(15*a^ 2*e^2 + 8*b^2*d^2 - 20*a*b*d*e))/(5*e^4)))/(x^2*(d + e*x)^(1/2) + (a*d*(d + e*x)^(1/2))/(b*e) + (x*(15*a*e^6 + 15*b*d*e^5)*(d + e*x)^(1/2))/(15*b*e^ 6))