Integrand size = 35, antiderivative size = 258 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=-\frac {2 (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) \sqrt {d+e x}}-\frac {8 b (b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac {4 b^2 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}-\frac {8 b^3 (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x)}+\frac {2 b^4 (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^5 (a+b x)} \]
4*b^2*(-a*e+b*d)^2*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)-8/5*b^3*(-a *e+b*d)*(e*x+d)^(5/2)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)+2/7*b^4*(e*x+d)^(7/2)* ((b*x+a)^2)^(1/2)/e^5/(b*x+a)-2*(-a*e+b*d)^4*((b*x+a)^2)^(1/2)/e^5/(b*x+a) /(e*x+d)^(1/2)-8*b*(-a*e+b*d)^3*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2)/e^5/(b*x+a )
Time = 0.04 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.66 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (35 a^4 e^4-140 a^3 b e^3 (2 d+e x)+70 a^2 b^2 e^2 \left (8 d^2+4 d e x-e^2 x^2\right )-28 a b^3 e \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )+b^4 \left (128 d^4+64 d^3 e x-16 d^2 e^2 x^2+8 d e^3 x^3-5 e^4 x^4\right )\right )}{35 e^5 (a+b x) \sqrt {d+e x}} \]
(-2*Sqrt[(a + b*x)^2]*(35*a^4*e^4 - 140*a^3*b*e^3*(2*d + e*x) + 70*a^2*b^2 *e^2*(8*d^2 + 4*d*e*x - e^2*x^2) - 28*a*b^3*e*(16*d^3 + 8*d^2*e*x - 2*d*e^ 2*x^2 + e^3*x^3) + b^4*(128*d^4 + 64*d^3*e*x - 16*d^2*e^2*x^2 + 8*d*e^3*x^ 3 - 5*e^4*x^4)))/(35*e^5*(a + b*x)*Sqrt[d + e*x])
Time = 0.31 (sec) , antiderivative size = 151, 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.
| \(\displaystyle \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^4}{(d+e x)^{3/2}}dx}{b^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^4}{(d+e x)^{3/2}}dx}{a+b x}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {(d+e x)^{5/2} b^4}{e^4}-\frac {4 (b d-a e) (d+e x)^{3/2} b^3}{e^4}+\frac {6 (b d-a e)^2 \sqrt {d+e x} b^2}{e^4}-\frac {4 (b d-a e)^3 b}{e^4 \sqrt {d+e x}}+\frac {(a e-b d)^4}{e^4 (d+e x)^{3/2}}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {8 b^3 (d+e x)^{5/2} (b d-a e)}{5 e^5}+\frac {4 b^2 (d+e x)^{3/2} (b d-a e)^2}{e^5}-\frac {8 b \sqrt {d+e x} (b d-a e)^3}{e^5}-\frac {2 (b d-a e)^4}{e^5 \sqrt {d+e x}}+\frac {2 b^4 (d+e x)^{7/2}}{7 e^5}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*(b*d - a*e)^4)/(e^5*Sqrt[d + e*x]) - ( 8*b*(b*d - a*e)^3*Sqrt[d + e*x])/e^5 + (4*b^2*(b*d - a*e)^2*(d + e*x)^(3/2 ))/e^5 - (8*b^3*(b*d - a*e)*(d + e*x)^(5/2))/(5*e^5) + (2*b^4*(d + e*x)^(7 /2))/(7*e^5)))/(a + b*x)
3.22.4.3.1 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 = 0.29 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.78
| method | result | size |
| gosper | \(-\frac {2 \left (-5 e^{4} x^{4} b^{4}-28 x^{3} a \,b^{3} e^{4}+8 x^{3} b^{4} d \,e^{3}-70 x^{2} a^{2} b^{2} e^{4}+56 x^{2} a \,b^{3} d \,e^{3}-16 x^{2} b^{4} d^{2} e^{2}-140 x \,a^{3} b \,e^{4}+280 x \,a^{2} b^{2} d \,e^{3}-224 x a \,b^{3} d^{2} e^{2}+64 x \,b^{4} d^{3} e +35 e^{4} a^{4}-280 b d \,e^{3} a^{3}+560 b^{2} d^{2} e^{2} a^{2}-448 b^{3} d^{3} e a +128 b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{35 \sqrt {e x +d}\, e^{5} \left (b x +a \right )^{3}}\) | \(202\) |
| default | \(-\frac {2 \left (-5 e^{4} x^{4} b^{4}-28 x^{3} a \,b^{3} e^{4}+8 x^{3} b^{4} d \,e^{3}-70 x^{2} a^{2} b^{2} e^{4}+56 x^{2} a \,b^{3} d \,e^{3}-16 x^{2} b^{4} d^{2} e^{2}-140 x \,a^{3} b \,e^{4}+280 x \,a^{2} b^{2} d \,e^{3}-224 x a \,b^{3} d^{2} e^{2}+64 x \,b^{4} d^{3} e +35 e^{4} a^{4}-280 b d \,e^{3} a^{3}+560 b^{2} d^{2} e^{2} a^{2}-448 b^{3} d^{3} e a +128 b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{35 \sqrt {e x +d}\, e^{5} \left (b x +a \right )^{3}}\) | \(202\) |
| risch | \(\frac {2 b \left (5 b^{3} x^{3} e^{3}+28 x^{2} a \,b^{2} e^{3}-13 x^{2} b^{3} d \,e^{2}+70 x \,a^{2} b \,e^{3}-84 x a \,b^{2} d \,e^{2}+29 x \,b^{3} d^{2} e +140 a^{3} e^{3}-350 a^{2} b d \,e^{2}+308 a \,b^{2} d^{2} e -93 b^{3} d^{3}\right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{35 e^{5} \left (b x +a \right )}-\frac {2 \left (e^{4} a^{4}-4 b d \,e^{3} a^{3}+6 b^{2} d^{2} e^{2} a^{2}-4 b^{3} d^{3} e a +b^{4} d^{4}\right ) \sqrt {\left (b x +a \right )^{2}}}{e^{5} \sqrt {e x +d}\, \left (b x +a \right )}\) | \(211\) |
-2/35/(e*x+d)^(1/2)*(-5*b^4*e^4*x^4-28*a*b^3*e^4*x^3+8*b^4*d*e^3*x^3-70*a^ 2*b^2*e^4*x^2+56*a*b^3*d*e^3*x^2-16*b^4*d^2*e^2*x^2-140*a^3*b*e^4*x+280*a^ 2*b^2*d*e^3*x-224*a*b^3*d^2*e^2*x+64*b^4*d^3*e*x+35*a^4*e^4-280*a^3*b*d*e^ 3+560*a^2*b^2*d^2*e^2-448*a*b^3*d^3*e+128*b^4*d^4)*((b*x+a)^2)^(3/2)/e^5/( b*x+a)^3
Time = 0.41 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (5 \, b^{4} e^{4} x^{4} - 128 \, b^{4} d^{4} + 448 \, a b^{3} d^{3} e - 560 \, a^{2} b^{2} d^{2} e^{2} + 280 \, a^{3} b d e^{3} - 35 \, a^{4} e^{4} - 4 \, {\left (2 \, b^{4} d e^{3} - 7 \, a b^{3} e^{4}\right )} x^{3} + 2 \, {\left (8 \, b^{4} d^{2} e^{2} - 28 \, a b^{3} d e^{3} + 35 \, a^{2} b^{2} e^{4}\right )} x^{2} - 4 \, {\left (16 \, b^{4} d^{3} e - 56 \, a b^{3} d^{2} e^{2} + 70 \, a^{2} b^{2} d e^{3} - 35 \, a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{35 \, {\left (e^{6} x + d e^{5}\right )}} \]
2/35*(5*b^4*e^4*x^4 - 128*b^4*d^4 + 448*a*b^3*d^3*e - 560*a^2*b^2*d^2*e^2 + 280*a^3*b*d*e^3 - 35*a^4*e^4 - 4*(2*b^4*d*e^3 - 7*a*b^3*e^4)*x^3 + 2*(8* b^4*d^2*e^2 - 28*a*b^3*d*e^3 + 35*a^2*b^2*e^4)*x^2 - 4*(16*b^4*d^3*e - 56* a*b^3*d^2*e^2 + 70*a^2*b^2*d*e^3 - 35*a^3*b*e^4)*x)*sqrt(e*x + d)/(e^6*x + d*e^5)
\[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\int \frac {\left (a + b x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {3}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 40 \, a b^{2} d^{2} e + 30 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3} - {\left (2 \, b^{3} d e^{2} - 5 \, a b^{2} e^{3}\right )} x^{2} + {\left (8 \, b^{3} d^{2} e - 20 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3}\right )} x\right )} a}{5 \, \sqrt {e x + d} e^{4}} + \frac {2 \, {\left (5 \, b^{3} e^{4} x^{4} - 128 \, b^{3} d^{4} + 336 \, a b^{2} d^{3} e - 280 \, a^{2} b d^{2} e^{2} + 70 \, a^{3} d e^{3} - {\left (8 \, b^{3} d e^{3} - 21 \, a b^{2} e^{4}\right )} x^{3} + {\left (16 \, b^{3} d^{2} e^{2} - 42 \, a b^{2} d e^{3} + 35 \, a^{2} b e^{4}\right )} x^{2} - {\left (64 \, b^{3} d^{3} e - 168 \, a b^{2} d^{2} e^{2} + 140 \, a^{2} b d e^{3} - 35 \, a^{3} e^{4}\right )} x\right )} b}{35 \, \sqrt {e x + d} e^{5}} \]
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)*a/(sqrt(e*x + d)*e^4) + 2/35*(5*b^3*e^4*x^4 - 128*b^3*d^4 + 3 36*a*b^2*d^3*e - 280*a^2*b*d^2*e^2 + 70*a^3*d*e^3 - (8*b^3*d*e^3 - 21*a*b^ 2*e^4)*x^3 + (16*b^3*d^2*e^2 - 42*a*b^2*d*e^3 + 35*a^2*b*e^4)*x^2 - (64*b^ 3*d^3*e - 168*a*b^2*d^2*e^2 + 140*a^2*b*d*e^3 - 35*a^3*e^4)*x)*b/(sqrt(e*x + d)*e^5)
Time = 0.27 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.28 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=-\frac {2 \, {\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )}}{\sqrt {e x + d} e^{5}} + \frac {2 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} b^{4} e^{30} \mathrm {sgn}\left (b x + a\right ) - 28 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{4} d e^{30} \mathrm {sgn}\left (b x + a\right ) + 70 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{4} d^{2} e^{30} \mathrm {sgn}\left (b x + a\right ) - 140 \, \sqrt {e x + d} b^{4} d^{3} e^{30} \mathrm {sgn}\left (b x + a\right ) + 28 \, {\left (e x + d\right )}^{\frac {5}{2}} a b^{3} e^{31} \mathrm {sgn}\left (b x + a\right ) - 140 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{3} d e^{31} \mathrm {sgn}\left (b x + a\right ) + 420 \, \sqrt {e x + d} a b^{3} d^{2} e^{31} \mathrm {sgn}\left (b x + a\right ) + 70 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{2} b^{2} e^{32} \mathrm {sgn}\left (b x + a\right ) - 420 \, \sqrt {e x + d} a^{2} b^{2} d e^{32} \mathrm {sgn}\left (b x + a\right ) + 140 \, \sqrt {e x + d} a^{3} b e^{33} \mathrm {sgn}\left (b x + a\right )\right )}}{35 \, e^{35}} \]
-2*(b^4*d^4*sgn(b*x + a) - 4*a*b^3*d^3*e*sgn(b*x + a) + 6*a^2*b^2*d^2*e^2* sgn(b*x + a) - 4*a^3*b*d*e^3*sgn(b*x + a) + a^4*e^4*sgn(b*x + a))/(sqrt(e* x + d)*e^5) + 2/35*(5*(e*x + d)^(7/2)*b^4*e^30*sgn(b*x + a) - 28*(e*x + d) ^(5/2)*b^4*d*e^30*sgn(b*x + a) + 70*(e*x + d)^(3/2)*b^4*d^2*e^30*sgn(b*x + a) - 140*sqrt(e*x + d)*b^4*d^3*e^30*sgn(b*x + a) + 28*(e*x + d)^(5/2)*a*b ^3*e^31*sgn(b*x + a) - 140*(e*x + d)^(3/2)*a*b^3*d*e^31*sgn(b*x + a) + 420 *sqrt(e*x + d)*a*b^3*d^2*e^31*sgn(b*x + a) + 70*(e*x + d)^(3/2)*a^2*b^2*e^ 32*sgn(b*x + a) - 420*sqrt(e*x + d)*a^2*b^2*d*e^32*sgn(b*x + a) + 140*sqrt (e*x + d)*a^3*b*e^33*sgn(b*x + a))/e^35
Time = 11.59 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {2\,b^3\,x^4}{7\,e}-\frac {70\,a^4\,e^4-560\,a^3\,b\,d\,e^3+1120\,a^2\,b^2\,d^2\,e^2-896\,a\,b^3\,d^3\,e+256\,b^4\,d^4}{35\,b\,e^5}+\frac {x\,\left (280\,a^3\,b\,e^4-560\,a^2\,b^2\,d\,e^3+448\,a\,b^3\,d^2\,e^2-128\,b^4\,d^3\,e\right )}{35\,b\,e^5}+\frac {8\,b^2\,x^3\,\left (7\,a\,e-2\,b\,d\right )}{35\,e^2}+\frac {4\,b\,x^2\,\left (35\,a^2\,e^2-28\,a\,b\,d\,e+8\,b^2\,d^2\right )}{35\,e^3}\right )}{x\,\sqrt {d+e\,x}+\frac {a\,\sqrt {d+e\,x}}{b}} \]
((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((2*b^3*x^4)/(7*e) - (70*a^4*e^4 + 256*b^ 4*d^4 + 1120*a^2*b^2*d^2*e^2 - 896*a*b^3*d^3*e - 560*a^3*b*d*e^3)/(35*b*e^ 5) + (x*(280*a^3*b*e^4 - 128*b^4*d^3*e + 448*a*b^3*d^2*e^2 - 560*a^2*b^2*d *e^3))/(35*b*e^5) + (8*b^2*x^3*(7*a*e - 2*b*d))/(35*e^2) + (4*b*x^2*(35*a^ 2*e^2 + 8*b^2*d^2 - 28*a*b*d*e))/(35*e^3)))/(x*(d + e*x)^(1/2) + (a*(d + e *x)^(1/2))/b)