Integrand size = 35, antiderivative size = 262 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=-\frac {2 (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^5 (a+b x) (d+e x)^{9/2}}+\frac {8 b (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^5 (a+b x) (d+e x)^{7/2}}-\frac {12 b^2 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x) (d+e x)^{5/2}}+\frac {8 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x) (d+e x)^{3/2}}-\frac {2 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) \sqrt {d+e x}} \] Output:
-2/9*(-a*e+b*d)^4*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^(9/2)+8/7*b*(-a*e+ b*d)^3*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^(7/2)-12/5*b^2*(-a*e+b*d)^2*( (b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^(5/2)+8/3*b^3*(-a*e+b*d)*((b*x+a)^2)^ (1/2)/e^5/(b*x+a)/(e*x+d)^(3/2)-2*b^4*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d )^(1/2)
Time = 0.18 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.66 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (35 a^4 e^4+20 a^3 b e^3 (2 d+9 e x)+6 a^2 b^2 e^2 \left (8 d^2+36 d e x+63 e^2 x^2\right )+4 a b^3 e \left (16 d^3+72 d^2 e x+126 d e^2 x^2+105 e^3 x^3\right )+b^4 \left (128 d^4+576 d^3 e x+1008 d^2 e^2 x^2+840 d e^3 x^3+315 e^4 x^4\right )\right )}{315 e^5 (a+b x) (d+e x)^{9/2}} \] Input:
Integrate[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^(11/2),x]
Output:
(-2*Sqrt[(a + b*x)^2]*(35*a^4*e^4 + 20*a^3*b*e^3*(2*d + 9*e*x) + 6*a^2*b^2 *e^2*(8*d^2 + 36*d*e*x + 63*e^2*x^2) + 4*a*b^3*e*(16*d^3 + 72*d^2*e*x + 12 6*d*e^2*x^2 + 105*e^3*x^3) + b^4*(128*d^4 + 576*d^3*e*x + 1008*d^2*e^2*x^2 + 840*d*e^3*x^3 + 315*e^4*x^4)))/(315*e^5*(a + b*x)*(d + e*x)^(9/2))
Time = 0.50 (sec) , antiderivative size = 155, 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.
| \(\displaystyle \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^4}{(d+e x)^{11/2}}dx}{b^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^4}{(d+e x)^{11/2}}dx}{a+b x}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^4}{e^4 (d+e x)^{3/2}}-\frac {4 (b d-a e) b^3}{e^4 (d+e x)^{5/2}}+\frac {6 (b d-a e)^2 b^2}{e^4 (d+e x)^{7/2}}-\frac {4 (b d-a e)^3 b}{e^4 (d+e x)^{9/2}}+\frac {(a e-b d)^4}{e^4 (d+e x)^{11/2}}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {8 b^3 (b d-a e)}{3 e^5 (d+e x)^{3/2}}-\frac {12 b^2 (b d-a e)^2}{5 e^5 (d+e x)^{5/2}}+\frac {8 b (b d-a e)^3}{7 e^5 (d+e x)^{7/2}}-\frac {2 (b d-a e)^4}{9 e^5 (d+e x)^{9/2}}-\frac {2 b^4}{e^5 \sqrt {d+e x}}\right )}{a+b x}\) |
Input:
Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^(11/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*(b*d - a*e)^4)/(9*e^5*(d + e*x)^(9/2)) + (8*b*(b*d - a*e)^3)/(7*e^5*(d + e*x)^(7/2)) - (12*b^2*(b*d - a*e)^2)/(5 *e^5*(d + e*x)^(5/2)) + (8*b^3*(b*d - a*e))/(3*e^5*(d + e*x)^(3/2)) - (2*b ^4)/(e^5*Sqrt[d + e*x])))/(a + b*x)
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 = 1.29 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.77
| method | result | size |
| gosper | \(-\frac {2 \left (315 b^{4} x^{4} e^{4}+420 x^{3} a \,b^{3} e^{4}+840 x^{3} b^{4} d \,e^{3}+378 x^{2} a^{2} b^{2} e^{4}+504 x^{2} a \,b^{3} d \,e^{3}+1008 x^{2} b^{4} d^{2} e^{2}+180 x \,a^{3} b \,e^{4}+216 x \,a^{2} b^{2} d \,e^{3}+288 x a \,b^{3} d^{2} e^{2}+576 x \,b^{4} d^{3} e +35 a^{4} e^{4}+40 a^{3} b d \,e^{3}+48 a^{2} b^{2} d^{2} e^{2}+64 a \,b^{3} d^{3} e +128 b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{315 \left (e x +d \right )^{\frac {9}{2}} e^{5} \left (b x +a \right )^{3}}\) | \(202\) |
| default | \(-\frac {2 \left (315 b^{4} x^{4} e^{4}+420 x^{3} a \,b^{3} e^{4}+840 x^{3} b^{4} d \,e^{3}+378 x^{2} a^{2} b^{2} e^{4}+504 x^{2} a \,b^{3} d \,e^{3}+1008 x^{2} b^{4} d^{2} e^{2}+180 x \,a^{3} b \,e^{4}+216 x \,a^{2} b^{2} d \,e^{3}+288 x a \,b^{3} d^{2} e^{2}+576 x \,b^{4} d^{3} e +35 a^{4} e^{4}+40 a^{3} b d \,e^{3}+48 a^{2} b^{2} d^{2} e^{2}+64 a \,b^{3} d^{3} e +128 b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{315 \left (e x +d \right )^{\frac {9}{2}} e^{5} \left (b x +a \right )^{3}}\) | \(202\) |
| orering | \(-\frac {2 \left (315 b^{4} x^{4} e^{4}+420 x^{3} a \,b^{3} e^{4}+840 x^{3} b^{4} d \,e^{3}+378 x^{2} a^{2} b^{2} e^{4}+504 x^{2} a \,b^{3} d \,e^{3}+1008 x^{2} b^{4} d^{2} e^{2}+180 x \,a^{3} b \,e^{4}+216 x \,a^{2} b^{2} d \,e^{3}+288 x a \,b^{3} d^{2} e^{2}+576 x \,b^{4} d^{3} e +35 a^{4} e^{4}+40 a^{3} b d \,e^{3}+48 a^{2} b^{2} d^{2} e^{2}+64 a \,b^{3} d^{3} e +128 b^{4} d^{4}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{315 e^{5} \left (b x +a \right )^{3} \left (e x +d \right )^{\frac {9}{2}}}\) | \(211\) |
Input:
int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(11/2),x,method=_RETURNVER BOSE)
Output:
-2/315/(e*x+d)^(9/2)*(315*b^4*e^4*x^4+420*a*b^3*e^4*x^3+840*b^4*d*e^3*x^3+ 378*a^2*b^2*e^4*x^2+504*a*b^3*d*e^3*x^2+1008*b^4*d^2*e^2*x^2+180*a^3*b*e^4 *x+216*a^2*b^2*d*e^3*x+288*a*b^3*d^2*e^2*x+576*b^4*d^3*e*x+35*a^4*e^4+40*a ^3*b*d*e^3+48*a^2*b^2*d^2*e^2+64*a*b^3*d^3*e+128*b^4*d^4)*((b*x+a)^2)^(3/2 )/e^5/(b*x+a)^3
Time = 0.09 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=-\frac {2 \, {\left (315 \, b^{4} e^{4} x^{4} + 128 \, b^{4} d^{4} + 64 \, a b^{3} d^{3} e + 48 \, a^{2} b^{2} d^{2} e^{2} + 40 \, a^{3} b d e^{3} + 35 \, a^{4} e^{4} + 420 \, {\left (2 \, b^{4} d e^{3} + a b^{3} e^{4}\right )} x^{3} + 126 \, {\left (8 \, b^{4} d^{2} e^{2} + 4 \, a b^{3} d e^{3} + 3 \, a^{2} b^{2} e^{4}\right )} x^{2} + 36 \, {\left (16 \, b^{4} d^{3} e + 8 \, a b^{3} d^{2} e^{2} + 6 \, a^{2} b^{2} d e^{3} + 5 \, a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{315 \, {\left (e^{10} x^{5} + 5 \, d e^{9} x^{4} + 10 \, d^{2} e^{8} x^{3} + 10 \, d^{3} e^{7} x^{2} + 5 \, d^{4} e^{6} x + d^{5} e^{5}\right )}} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(11/2),x, algorithm= "fricas")
Output:
-2/315*(315*b^4*e^4*x^4 + 128*b^4*d^4 + 64*a*b^3*d^3*e + 48*a^2*b^2*d^2*e^ 2 + 40*a^3*b*d*e^3 + 35*a^4*e^4 + 420*(2*b^4*d*e^3 + a*b^3*e^4)*x^3 + 126* (8*b^4*d^2*e^2 + 4*a*b^3*d*e^3 + 3*a^2*b^2*e^4)*x^2 + 36*(16*b^4*d^3*e + 8 *a*b^3*d^2*e^2 + 6*a^2*b^2*d*e^3 + 5*a^3*b*e^4)*x)*sqrt(e*x + d)/(e^10*x^5 + 5*d*e^9*x^4 + 10*d^2*e^8*x^3 + 10*d^3*e^7*x^2 + 5*d^4*e^6*x + d^5*e^5)
Timed out. \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\text {Timed out} \] Input:
integrate((b*x+a)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**(11/2),x)
Output:
Timed out
Time = 0.08 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.42 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=-\frac {2 \, {\left (105 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} + 24 \, a b^{2} d^{2} e + 30 \, a^{2} b d e^{2} + 35 \, a^{3} e^{3} + 63 \, {\left (2 \, b^{3} d e^{2} + 3 \, a b^{2} e^{3}\right )} x^{2} + 9 \, {\left (8 \, b^{3} d^{2} e + 12 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3}\right )} x\right )} a}{315 \, {\left (e^{8} x^{4} + 4 \, d e^{7} x^{3} + 6 \, d^{2} e^{6} x^{2} + 4 \, d^{3} e^{5} x + d^{4} e^{4}\right )} \sqrt {e x + d}} - \frac {2 \, {\left (315 \, b^{3} e^{4} x^{4} + 128 \, b^{3} d^{4} + 48 \, a b^{2} d^{3} e + 24 \, a^{2} b d^{2} e^{2} + 10 \, a^{3} d e^{3} + 105 \, {\left (8 \, b^{3} d e^{3} + 3 \, a b^{2} e^{4}\right )} x^{3} + 63 \, {\left (16 \, b^{3} d^{2} e^{2} + 6 \, a b^{2} d e^{3} + 3 \, a^{2} b e^{4}\right )} x^{2} + 9 \, {\left (64 \, b^{3} d^{3} e + 24 \, a b^{2} d^{2} e^{2} + 12 \, a^{2} b d e^{3} + 5 \, a^{3} e^{4}\right )} x\right )} b}{315 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )} \sqrt {e x + d}} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(11/2),x, algorithm= "maxima")
Output:
-2/315*(105*b^3*e^3*x^3 + 16*b^3*d^3 + 24*a*b^2*d^2*e + 30*a^2*b*d*e^2 + 3 5*a^3*e^3 + 63*(2*b^3*d*e^2 + 3*a*b^2*e^3)*x^2 + 9*(8*b^3*d^2*e + 12*a*b^2 *d*e^2 + 15*a^2*b*e^3)*x)*a/((e^8*x^4 + 4*d*e^7*x^3 + 6*d^2*e^6*x^2 + 4*d^ 3*e^5*x + d^4*e^4)*sqrt(e*x + d)) - 2/315*(315*b^3*e^4*x^4 + 128*b^3*d^4 + 48*a*b^2*d^3*e + 24*a^2*b*d^2*e^2 + 10*a^3*d*e^3 + 105*(8*b^3*d*e^3 + 3*a *b^2*e^4)*x^3 + 63*(16*b^3*d^2*e^2 + 6*a*b^2*d*e^3 + 3*a^2*b*e^4)*x^2 + 9* (64*b^3*d^3*e + 24*a*b^2*d^2*e^2 + 12*a^2*b*d*e^3 + 5*a^3*e^4)*x)*b/((e^9* x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*d^3*e^6*x + d^4*e^5)*sqrt(e*x + d))
Time = 0.18 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=-\frac {2 \, {\left (315 \, {\left (e x + d\right )}^{4} b^{4} \mathrm {sgn}\left (b x + a\right ) - 420 \, {\left (e x + d\right )}^{3} b^{4} d \mathrm {sgn}\left (b x + a\right ) + 378 \, {\left (e x + d\right )}^{2} b^{4} d^{2} \mathrm {sgn}\left (b x + a\right ) - 180 \, {\left (e x + d\right )} b^{4} d^{3} \mathrm {sgn}\left (b x + a\right ) + 35 \, b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) + 420 \, {\left (e x + d\right )}^{3} a b^{3} e \mathrm {sgn}\left (b x + a\right ) - 756 \, {\left (e x + d\right )}^{2} a b^{3} d e \mathrm {sgn}\left (b x + a\right ) + 540 \, {\left (e x + d\right )} a b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 140 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 378 \, {\left (e x + d\right )}^{2} a^{2} b^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 540 \, {\left (e x + d\right )} a^{2} b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) + 210 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 180 \, {\left (e x + d\right )} a^{3} b e^{3} \mathrm {sgn}\left (b x + a\right ) - 140 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 35 \, a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )}}{315 \, {\left (e x + d\right )}^{\frac {9}{2}} e^{5}} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(11/2),x, algorithm= "giac")
Output:
-2/315*(315*(e*x + d)^4*b^4*sgn(b*x + a) - 420*(e*x + d)^3*b^4*d*sgn(b*x + a) + 378*(e*x + d)^2*b^4*d^2*sgn(b*x + a) - 180*(e*x + d)*b^4*d^3*sgn(b*x + a) + 35*b^4*d^4*sgn(b*x + a) + 420*(e*x + d)^3*a*b^3*e*sgn(b*x + a) - 7 56*(e*x + d)^2*a*b^3*d*e*sgn(b*x + a) + 540*(e*x + d)*a*b^3*d^2*e*sgn(b*x + a) - 140*a*b^3*d^3*e*sgn(b*x + a) + 378*(e*x + d)^2*a^2*b^2*e^2*sgn(b*x + a) - 540*(e*x + d)*a^2*b^2*d*e^2*sgn(b*x + a) + 210*a^2*b^2*d^2*e^2*sgn( b*x + a) + 180*(e*x + d)*a^3*b*e^3*sgn(b*x + a) - 140*a^3*b*d*e^3*sgn(b*x + a) + 35*a^4*e^4*sgn(b*x + a))/((e*x + d)^(9/2)*e^5)
Time = 12.15 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {8\,x\,\left (5\,a^3\,e^3+6\,a^2\,b\,d\,e^2+8\,a\,b^2\,d^2\,e+16\,b^3\,d^3\right )}{35\,e^8}+\frac {2\,b^3\,x^4}{e^5}+\frac {70\,a^4\,e^4+80\,a^3\,b\,d\,e^3+96\,a^2\,b^2\,d^2\,e^2+128\,a\,b^3\,d^3\,e+256\,b^4\,d^4}{315\,b\,e^9}+\frac {8\,b^2\,x^3\,\left (a\,e+2\,b\,d\right )}{3\,e^6}+\frac {4\,b\,x^2\,\left (3\,a^2\,e^2+4\,a\,b\,d\,e+8\,b^2\,d^2\right )}{5\,e^7}\right )}{x^5\,\sqrt {d+e\,x}+\frac {a\,d^4\,\sqrt {d+e\,x}}{b\,e^4}+\frac {x^4\,\left (315\,a\,e^9+1260\,b\,d\,e^8\right )\,\sqrt {d+e\,x}}{315\,b\,e^9}+\frac {2\,d\,x^3\,\left (2\,a\,e+3\,b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}+\frac {d^3\,x\,\left (4\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^4}+\frac {2\,d^2\,x^2\,\left (3\,a\,e+2\,b\,d\right )\,\sqrt {d+e\,x}}{b\,e^3}} \] Input:
int(((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^(11/2),x)
Output:
-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((8*x*(5*a^3*e^3 + 16*b^3*d^3 + 8*a*b^2* d^2*e + 6*a^2*b*d*e^2))/(35*e^8) + (2*b^3*x^4)/e^5 + (70*a^4*e^4 + 256*b^4 *d^4 + 96*a^2*b^2*d^2*e^2 + 128*a*b^3*d^3*e + 80*a^3*b*d*e^3)/(315*b*e^9) + (8*b^2*x^3*(a*e + 2*b*d))/(3*e^6) + (4*b*x^2*(3*a^2*e^2 + 8*b^2*d^2 + 4* a*b*d*e))/(5*e^7)))/(x^5*(d + e*x)^(1/2) + (a*d^4*(d + e*x)^(1/2))/(b*e^4) + (x^4*(315*a*e^9 + 1260*b*d*e^8)*(d + e*x)^(1/2))/(315*b*e^9) + (2*d*x^3 *(2*a*e + 3*b*d)*(d + e*x)^(1/2))/(b*e^2) + (d^3*x*(4*a*e + b*d)*(d + e*x) ^(1/2))/(b*e^4) + (2*d^2*x^2*(3*a*e + 2*b*d)*(d + e*x)^(1/2))/(b*e^3))
Time = 0.26 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\frac {-2 b^{4} e^{4} x^{4}-\frac {8}{3} a \,b^{3} e^{4} x^{3}-\frac {16}{3} b^{4} d \,e^{3} x^{3}-\frac {12}{5} a^{2} b^{2} e^{4} x^{2}-\frac {16}{5} a \,b^{3} d \,e^{3} x^{2}-\frac {32}{5} b^{4} d^{2} e^{2} x^{2}-\frac {8}{7} a^{3} b \,e^{4} x -\frac {48}{35} a^{2} b^{2} d \,e^{3} x -\frac {64}{35} a \,b^{3} d^{2} e^{2} x -\frac {128}{35} b^{4} d^{3} e x -\frac {2}{9} a^{4} e^{4}-\frac {16}{63} a^{3} b d \,e^{3}-\frac {32}{105} a^{2} b^{2} d^{2} e^{2}-\frac {128}{315} a \,b^{3} d^{3} e -\frac {256}{315} b^{4} d^{4}}{\sqrt {e x +d}\, e^{5} \left (e^{4} x^{4}+4 d \,e^{3} x^{3}+6 d^{2} e^{2} x^{2}+4 d^{3} e x +d^{4}\right )} \] Input:
int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(11/2),x)
Output:
(2*( - 35*a**4*e**4 - 40*a**3*b*d*e**3 - 180*a**3*b*e**4*x - 48*a**2*b**2* d**2*e**2 - 216*a**2*b**2*d*e**3*x - 378*a**2*b**2*e**4*x**2 - 64*a*b**3*d **3*e - 288*a*b**3*d**2*e**2*x - 504*a*b**3*d*e**3*x**2 - 420*a*b**3*e**4* x**3 - 128*b**4*d**4 - 576*b**4*d**3*e*x - 1008*b**4*d**2*e**2*x**2 - 840* b**4*d*e**3*x**3 - 315*b**4*e**4*x**4))/(315*sqrt(d + e*x)*e**5*(d**4 + 4* d**3*e*x + 6*d**2*e**2*x**2 + 4*d*e**3*x**3 + e**4*x**4))