Integrand size = 35, antiderivative size = 260 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=-\frac {2 (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x) (d+e x)^{5/2}}+\frac {8 b (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x) (d+e x)^{3/2}}-\frac {12 b^2 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) \sqrt {d+e x}}-\frac {8 b^3 (b d-a e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac {2 b^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x)} \]
-2/5*(-a*e+b*d)^4*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^(5/2)+8/3*b*(-a*e+ b*d)^3*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^(3/2)+2/3*b^4*(e*x+d)^(3/2)*( (b*x+a)^2)^(1/2)/e^5/(b*x+a)-12*b^2*(-a*e+b*d)^2*((b*x+a)^2)^(1/2)/e^5/(b* x+a)/(e*x+d)^(1/2)-8*b^3*(-a*e+b*d)*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2)/e^5/(b *x+a)
Time = 0.04 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.66 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (3 a^4 e^4+4 a^3 b e^3 (2 d+5 e x)+6 a^2 b^2 e^2 \left (8 d^2+20 d e x+15 e^2 x^2\right )-12 a b^3 e \left (16 d^3+40 d^2 e x+30 d e^2 x^2+5 e^3 x^3\right )+b^4 \left (128 d^4+320 d^3 e x+240 d^2 e^2 x^2+40 d e^3 x^3-5 e^4 x^4\right )\right )}{15 e^5 (a+b x) (d+e x)^{5/2}} \]
(-2*Sqrt[(a + b*x)^2]*(3*a^4*e^4 + 4*a^3*b*e^3*(2*d + 5*e*x) + 6*a^2*b^2*e ^2*(8*d^2 + 20*d*e*x + 15*e^2*x^2) - 12*a*b^3*e*(16*d^3 + 40*d^2*e*x + 30* d*e^2*x^2 + 5*e^3*x^3) + b^4*(128*d^4 + 320*d^3*e*x + 240*d^2*e^2*x^2 + 40 *d*e^3*x^3 - 5*e^4*x^4)))/(15*e^5*(a + b*x)*(d + e*x)^(5/2))
Time = 0.30 (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.
| \(\displaystyle \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^4}{(d+e x)^{7/2}}dx}{b^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^4}{(d+e x)^{7/2}}dx}{a+b x}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {\sqrt {d+e x} b^4}{e^4}-\frac {4 (b d-a e) b^3}{e^4 \sqrt {d+e x}}+\frac {6 (b d-a e)^2 b^2}{e^4 (d+e x)^{3/2}}-\frac {4 (b d-a e)^3 b}{e^4 (d+e x)^{5/2}}+\frac {(a e-b d)^4}{e^4 (d+e x)^{7/2}}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {8 b^3 \sqrt {d+e x} (b d-a e)}{e^5}-\frac {12 b^2 (b d-a e)^2}{e^5 \sqrt {d+e x}}+\frac {8 b (b d-a e)^3}{3 e^5 (d+e x)^{3/2}}-\frac {2 (b d-a e)^4}{5 e^5 (d+e x)^{5/2}}+\frac {2 b^4 (d+e x)^{3/2}}{3 e^5}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*(b*d - a*e)^4)/(5*e^5*(d + e*x)^(5/2)) + (8*b*(b*d - a*e)^3)/(3*e^5*(d + e*x)^(3/2)) - (12*b^2*(b*d - a*e)^2)/(e ^5*Sqrt[d + e*x]) - (8*b^3*(b*d - a*e)*Sqrt[d + e*x])/e^5 + (2*b^4*(d + e* x)^(3/2))/(3*e^5)))/(a + b*x)
3.22.6.3.1 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 = 163, normalized size of antiderivative = 0.63
| method | result | size |
| risch | \(\frac {2 b^{3} \left (b e x +12 a e -11 b d \right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{3 e^{5} \left (b x +a \right )}-\frac {2 \left (90 b^{2} e^{2} x^{2}+20 a b \,e^{2} x +160 b^{2} d e x +3 e^{2} a^{2}+14 a b d e +73 b^{2} d^{2}\right ) \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{5} \sqrt {e x +d}\, \left (e^{2} x^{2}+2 d e x +d^{2}\right ) \left (b x +a \right )}\) | \(163\) |
| gosper | \(-\frac {2 \left (-5 e^{4} x^{4} b^{4}-60 x^{3} a \,b^{3} e^{4}+40 x^{3} b^{4} d \,e^{3}+90 x^{2} a^{2} b^{2} e^{4}-360 x^{2} a \,b^{3} d \,e^{3}+240 x^{2} b^{4} d^{2} e^{2}+20 x \,a^{3} b \,e^{4}+120 x \,a^{2} b^{2} d \,e^{3}-480 x a \,b^{3} d^{2} e^{2}+320 x \,b^{4} d^{3} e +3 e^{4} a^{4}+8 b d \,e^{3} a^{3}+48 b^{2} d^{2} e^{2} a^{2}-192 b^{3} d^{3} e a +128 b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{15 \left (e x +d \right )^{\frac {5}{2}} e^{5} \left (b x +a \right )^{3}}\) | \(202\) |
| default | \(-\frac {2 \left (-5 e^{4} x^{4} b^{4}-60 x^{3} a \,b^{3} e^{4}+40 x^{3} b^{4} d \,e^{3}+90 x^{2} a^{2} b^{2} e^{4}-360 x^{2} a \,b^{3} d \,e^{3}+240 x^{2} b^{4} d^{2} e^{2}+20 x \,a^{3} b \,e^{4}+120 x \,a^{2} b^{2} d \,e^{3}-480 x a \,b^{3} d^{2} e^{2}+320 x \,b^{4} d^{3} e +3 e^{4} a^{4}+8 b d \,e^{3} a^{3}+48 b^{2} d^{2} e^{2} a^{2}-192 b^{3} d^{3} e a +128 b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{15 \left (e x +d \right )^{\frac {5}{2}} e^{5} \left (b x +a \right )^{3}}\) | \(202\) |
2/3*b^3*(b*e*x+12*a*e-11*b*d)*(e*x+d)^(1/2)/e^5*((b*x+a)^2)^(1/2)/(b*x+a)- 2/15*(90*b^2*e^2*x^2+20*a*b*e^2*x+160*b^2*d*e*x+3*a^2*e^2+14*a*b*d*e+73*b^ 2*d^2)*(a^2*e^2-2*a*b*d*e+b^2*d^2)/e^5/(e*x+d)^(1/2)/(e^2*x^2+2*d*e*x+d^2) *((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.43 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.82 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (5 \, b^{4} e^{4} x^{4} - 128 \, b^{4} d^{4} + 192 \, a b^{3} d^{3} e - 48 \, a^{2} b^{2} d^{2} e^{2} - 8 \, a^{3} b d e^{3} - 3 \, a^{4} e^{4} - 20 \, {\left (2 \, b^{4} d e^{3} - 3 \, a b^{3} e^{4}\right )} x^{3} - 30 \, {\left (8 \, b^{4} d^{2} e^{2} - 12 \, a b^{3} d e^{3} + 3 \, a^{2} b^{2} e^{4}\right )} x^{2} - 20 \, {\left (16 \, b^{4} d^{3} e - 24 \, a b^{3} d^{2} e^{2} + 6 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} \]
2/15*(5*b^4*e^4*x^4 - 128*b^4*d^4 + 192*a*b^3*d^3*e - 48*a^2*b^2*d^2*e^2 - 8*a^3*b*d*e^3 - 3*a^4*e^4 - 20*(2*b^4*d*e^3 - 3*a*b^3*e^4)*x^3 - 30*(8*b^ 4*d^2*e^2 - 12*a*b^3*d*e^3 + 3*a^2*b^2*e^4)*x^2 - 20*(16*b^4*d^3*e - 24*a* b^3*d^2*e^2 + 6*a^2*b^2*d*e^3 + a^3*b*e^4)*x)*sqrt(e*x + d)/(e^8*x^3 + 3*d *e^7*x^2 + 3*d^2*e^6*x + d^3*e^5)
\[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=\int \frac {\left (a + b x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {7}{2}}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.25 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (5 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 8 \, a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} - a^{3} e^{3} + 15 \, {\left (2 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 5 \, {\left (8 \, b^{3} d^{2} e - 4 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )} a}{5 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )} \sqrt {e x + d}} + \frac {2 \, {\left (5 \, b^{3} e^{4} x^{4} - 128 \, b^{3} d^{4} + 144 \, a b^{2} d^{3} e - 24 \, a^{2} b d^{2} e^{2} - 2 \, a^{3} d e^{3} - 5 \, {\left (8 \, b^{3} d e^{3} - 9 \, a b^{2} e^{4}\right )} x^{3} - 15 \, {\left (16 \, b^{3} d^{2} e^{2} - 18 \, a b^{2} d e^{3} + 3 \, a^{2} b e^{4}\right )} x^{2} - 5 \, {\left (64 \, b^{3} d^{3} e - 72 \, a b^{2} d^{2} e^{2} + 12 \, a^{2} b d e^{3} + a^{3} e^{4}\right )} x\right )} b}{15 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )} \sqrt {e x + d}} \]
2/5*(5*b^3*e^3*x^3 + 16*b^3*d^3 - 8*a*b^2*d^2*e - 2*a^2*b*d*e^2 - a^3*e^3 + 15*(2*b^3*d*e^2 - a*b^2*e^3)*x^2 + 5*(8*b^3*d^2*e - 4*a*b^2*d*e^2 - a^2* b*e^3)*x)*a/((e^6*x^2 + 2*d*e^5*x + d^2*e^4)*sqrt(e*x + d)) + 2/15*(5*b^3* e^4*x^4 - 128*b^3*d^4 + 144*a*b^2*d^3*e - 24*a^2*b*d^2*e^2 - 2*a^3*d*e^3 - 5*(8*b^3*d*e^3 - 9*a*b^2*e^4)*x^3 - 15*(16*b^3*d^2*e^2 - 18*a*b^2*d*e^3 + 3*a^2*b*e^4)*x^2 - 5*(64*b^3*d^3*e - 72*a*b^2*d^2*e^2 + 12*a^2*b*d*e^3 + a^3*e^4)*x)*b/((e^7*x^2 + 2*d*e^6*x + d^2*e^5)*sqrt(e*x + d))
Time = 0.28 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (90 \, {\left (e x + d\right )}^{2} b^{4} d^{2} \mathrm {sgn}\left (b x + a\right ) - 20 \, {\left (e x + d\right )} b^{4} d^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 180 \, {\left (e x + d\right )}^{2} a b^{3} d e \mathrm {sgn}\left (b x + a\right ) + 60 \, {\left (e x + d\right )} a b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 12 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 90 \, {\left (e x + d\right )}^{2} a^{2} b^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 60 \, {\left (e x + d\right )} a^{2} b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) + 18 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 20 \, {\left (e x + d\right )} a^{3} b e^{3} \mathrm {sgn}\left (b x + a\right ) - 12 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{5}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} b^{4} e^{10} \mathrm {sgn}\left (b x + a\right ) - 12 \, \sqrt {e x + d} b^{4} d e^{10} \mathrm {sgn}\left (b x + a\right ) + 12 \, \sqrt {e x + d} a b^{3} e^{11} \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, e^{15}} \]
-2/15*(90*(e*x + d)^2*b^4*d^2*sgn(b*x + a) - 20*(e*x + d)*b^4*d^3*sgn(b*x + a) + 3*b^4*d^4*sgn(b*x + a) - 180*(e*x + d)^2*a*b^3*d*e*sgn(b*x + a) + 6 0*(e*x + d)*a*b^3*d^2*e*sgn(b*x + a) - 12*a*b^3*d^3*e*sgn(b*x + a) + 90*(e *x + d)^2*a^2*b^2*e^2*sgn(b*x + a) - 60*(e*x + d)*a^2*b^2*d*e^2*sgn(b*x + a) + 18*a^2*b^2*d^2*e^2*sgn(b*x + a) + 20*(e*x + d)*a^3*b*e^3*sgn(b*x + a) - 12*a^3*b*d*e^3*sgn(b*x + a) + 3*a^4*e^4*sgn(b*x + a))/((e*x + d)^(5/2)* e^5) + 2/3*((e*x + d)^(3/2)*b^4*e^10*sgn(b*x + a) - 12*sqrt(e*x + d)*b^4*d *e^10*sgn(b*x + a) + 12*sqrt(e*x + d)*a*b^3*e^11*sgn(b*x + a))/e^15
Time = 11.79 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {6\,a^4\,e^4+16\,a^3\,b\,d\,e^3+96\,a^2\,b^2\,d^2\,e^2-384\,a\,b^3\,d^3\,e+256\,b^4\,d^4}{15\,b\,e^7}-\frac {2\,b^3\,x^4}{3\,e^3}+\frac {x\,\left (40\,a^3\,b\,e^4+240\,a^2\,b^2\,d\,e^3-960\,a\,b^3\,d^2\,e^2+640\,b^4\,d^3\,e\right )}{15\,b\,e^7}-\frac {8\,b^2\,x^3\,\left (3\,a\,e-2\,b\,d\right )}{3\,e^4}+\frac {4\,b\,x^2\,\left (3\,a^2\,e^2-12\,a\,b\,d\,e+8\,b^2\,d^2\right )}{e^5}\right )}{x^3\,\sqrt {d+e\,x}+\frac {a\,d^2\,\sqrt {d+e\,x}}{b\,e^2}+\frac {x^2\,\left (15\,a\,e^7+30\,b\,d\,e^6\right )\,\sqrt {d+e\,x}}{15\,b\,e^7}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}} \]
-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((6*a^4*e^4 + 256*b^4*d^4 + 96*a^2*b^2*d ^2*e^2 - 384*a*b^3*d^3*e + 16*a^3*b*d*e^3)/(15*b*e^7) - (2*b^3*x^4)/(3*e^3 ) + (x*(40*a^3*b*e^4 + 640*b^4*d^3*e - 960*a*b^3*d^2*e^2 + 240*a^2*b^2*d*e ^3))/(15*b*e^7) - (8*b^2*x^3*(3*a*e - 2*b*d))/(3*e^4) + (4*b*x^2*(3*a^2*e^ 2 + 8*b^2*d^2 - 12*a*b*d*e))/e^5))/(x^3*(d + e*x)^(1/2) + (a*d^2*(d + e*x) ^(1/2))/(b*e^2) + (x^2*(15*a*e^7 + 30*b*d*e^6)*(d + e*x)^(1/2))/(15*b*e^7) + (d*x*(2*a*e + b*d)*(d + e*x)^(1/2))/(b*e^2))