Integrand size = 30, antiderivative size = 208 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\frac {2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^4 (a+b x) (d+e x)^{9/2}}-\frac {6 b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^4 (a+b x) (d+e x)^{7/2}}+\frac {6 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^4 (a+b x) (d+e x)^{5/2}}-\frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^4 (a+b x) (d+e x)^{3/2}} \]
2/9*(-a*e+b*d)^3*((b*x+a)^2)^(1/2)/e^4/(b*x+a)/(e*x+d)^(9/2)-6/7*b*(-a*e+b *d)^2*((b*x+a)^2)^(1/2)/e^4/(b*x+a)/(e*x+d)^(7/2)+6/5*b^2*(-a*e+b*d)*((b*x +a)^2)^(1/2)/e^4/(b*x+a)/(e*x+d)^(5/2)-2/3*b^3*((b*x+a)^2)^(1/2)/e^4/(b*x+ a)/(e*x+d)^(3/2)
Time = 0.09 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.58 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (35 a^3 e^3+15 a^2 b e^2 (2 d+9 e x)+3 a b^2 e \left (8 d^2+36 d e x+63 e^2 x^2\right )+b^3 \left (16 d^3+72 d^2 e x+126 d e^2 x^2+105 e^3 x^3\right )\right )}{315 e^4 (a+b x) (d+e x)^{9/2}} \]
(-2*Sqrt[(a + b*x)^2]*(35*a^3*e^3 + 15*a^2*b*e^2*(2*d + 9*e*x) + 3*a*b^2*e *(8*d^2 + 36*d*e*x + 63*e^2*x^2) + b^3*(16*d^3 + 72*d^2*e*x + 126*d*e^2*x^ 2 + 105*e^3*x^3)))/(315*e^4*(a + b*x)*(d + e*x)^(9/2))
Time = 0.26 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.62, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3}{(d+e x)^{11/2}}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3}{(d+e x)^{11/2}}dx}{a+b x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^3}{e^3 (d+e x)^{5/2}}-\frac {3 (b d-a e) b^2}{e^3 (d+e x)^{7/2}}+\frac {3 (b d-a e)^2 b}{e^3 (d+e x)^{9/2}}+\frac {(a e-b d)^3}{e^3 (d+e x)^{11/2}}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {6 b^2 (b d-a e)}{5 e^4 (d+e x)^{5/2}}-\frac {6 b (b d-a e)^2}{7 e^4 (d+e x)^{7/2}}+\frac {2 (b d-a e)^3}{9 e^4 (d+e x)^{9/2}}-\frac {2 b^3}{3 e^4 (d+e x)^{3/2}}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((2*(b*d - a*e)^3)/(9*e^4*(d + e*x)^(9/2)) - (6*b*(b*d - a*e)^2)/(7*e^4*(d + e*x)^(7/2)) + (6*b^2*(b*d - a*e))/(5*e^4 *(d + e*x)^(5/2)) - (2*b^3)/(3*e^4*(d + e*x)^(3/2))))/(a + b*x)
3.17.91.3.1 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 = 2.35 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.63
method | result | size |
gosper | \(-\frac {2 \left (105 e^{3} x^{3} b^{3}+189 x^{2} a \,b^{2} e^{3}+126 x^{2} b^{3} d \,e^{2}+135 a^{2} b \,e^{3} x +108 x a \,b^{2} d \,e^{2}+72 b^{3} d^{2} e x +35 a^{3} e^{3}+30 a^{2} b d \,e^{2}+24 a \,b^{2} d^{2} e +16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{315 \left (e x +d \right )^{\frac {9}{2}} e^{4} \left (b x +a \right )^{3}}\) | \(132\) |
default | \(-\frac {2 \left (105 e^{3} x^{3} b^{3}+189 x^{2} a \,b^{2} e^{3}+126 x^{2} b^{3} d \,e^{2}+135 a^{2} b \,e^{3} x +108 x a \,b^{2} d \,e^{2}+72 b^{3} d^{2} e x +35 a^{3} e^{3}+30 a^{2} b d \,e^{2}+24 a \,b^{2} d^{2} e +16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{315 \left (e x +d \right )^{\frac {9}{2}} e^{4} \left (b x +a \right )^{3}}\) | \(132\) |
-2/315/(e*x+d)^(9/2)*(105*b^3*e^3*x^3+189*a*b^2*e^3*x^2+126*b^3*d*e^2*x^2+ 135*a^2*b*e^3*x+108*a*b^2*d*e^2*x+72*b^3*d^2*e*x+35*a^3*e^3+30*a^2*b*d*e^2 +24*a*b^2*d^2*e+16*b^3*d^3)*((b*x+a)^2)^(3/2)/e^4/(b*x+a)^3
Time = 0.40 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=-\frac {2 \, {\left (105 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} + 24 \, a b^{2} d^{2} e + 30 \, a^{2} b d e^{2} + 35 \, a^{3} e^{3} + 63 \, {\left (2 \, b^{3} d e^{2} + 3 \, a b^{2} e^{3}\right )} x^{2} + 9 \, {\left (8 \, b^{3} d^{2} e + 12 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{315 \, {\left (e^{9} x^{5} + 5 \, d e^{8} x^{4} + 10 \, d^{2} e^{7} x^{3} + 10 \, d^{3} e^{6} x^{2} + 5 \, d^{4} e^{5} x + d^{5} e^{4}\right )}} \]
-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)*sqrt(e*x + d)/(e^9*x^5 + 5*d*e^8*x^4 + 10*d^2*e^ 7*x^3 + 10*d^3*e^6*x^2 + 5*d^4*e^5*x + d^5*e^4)
\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {11}{2}}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=-\frac {2 \, {\left (105 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} + 24 \, a b^{2} d^{2} e + 30 \, a^{2} b d e^{2} + 35 \, a^{3} e^{3} + 63 \, {\left (2 \, b^{3} d e^{2} + 3 \, a b^{2} e^{3}\right )} x^{2} + 9 \, {\left (8 \, b^{3} d^{2} e + 12 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3}\right )} x\right )}}{315 \, {\left (e^{8} x^{4} + 4 \, d e^{7} x^{3} + 6 \, d^{2} e^{6} x^{2} + 4 \, d^{3} e^{5} x + d^{4} e^{4}\right )} \sqrt {e x + d}} \]
-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)/((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))
Time = 0.35 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=-\frac {2 \, {\left (105 \, {\left (e x + d\right )}^{3} b^{3} \mathrm {sgn}\left (b x + a\right ) - 189 \, {\left (e x + d\right )}^{2} b^{3} d \mathrm {sgn}\left (b x + a\right ) + 135 \, {\left (e x + d\right )} b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 35 \, b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) + 189 \, {\left (e x + d\right )}^{2} a b^{2} e \mathrm {sgn}\left (b x + a\right ) - 270 \, {\left (e x + d\right )} a b^{2} d e \mathrm {sgn}\left (b x + a\right ) + 105 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 135 \, {\left (e x + d\right )} a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right ) - 105 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + 35 \, a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )}}{315 \, {\left (e x + d\right )}^{\frac {9}{2}} e^{4}} \]
-2/315*(105*(e*x + d)^3*b^3*sgn(b*x + a) - 189*(e*x + d)^2*b^3*d*sgn(b*x + a) + 135*(e*x + d)*b^3*d^2*sgn(b*x + a) - 35*b^3*d^3*sgn(b*x + a) + 189*( e*x + d)^2*a*b^2*e*sgn(b*x + a) - 270*(e*x + d)*a*b^2*d*e*sgn(b*x + a) + 1 05*a*b^2*d^2*e*sgn(b*x + a) + 135*(e*x + d)*a^2*b*e^2*sgn(b*x + a) - 105*a ^2*b*d*e^2*sgn(b*x + a) + 35*a^3*e^3*sgn(b*x + a))/((e*x + d)^(9/2)*e^4)
Time = 10.30 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {\frac {2\,a^3\,e^3}{9}+\frac {4\,a^2\,b\,d\,e^2}{21}+\frac {16\,a\,b^2\,d^2\,e}{105}+\frac {32\,b^3\,d^3}{315}}{b\,e^8}+\frac {2\,x\,\left (15\,a^2\,e^2+12\,a\,b\,d\,e+8\,b^2\,d^2\right )}{35\,e^7}+\frac {2\,b^2\,x^3}{3\,e^5}+\frac {2\,b\,x^2\,\left (3\,a\,e+2\,b\,d\right )}{5\,e^6}\right )}{x^5\,\sqrt {d+e\,x}+\frac {a\,d^4\,\sqrt {d+e\,x}}{b\,e^4}+\frac {x^4\,\left (a\,e^8+4\,b\,d\,e^7\right )\,\sqrt {d+e\,x}}{b\,e^8}+\frac {2\,d\,x^3\,\left (2\,a\,e+3\,b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}+\frac {d^3\,x\,\left (4\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^4}+\frac {d^2\,x^2\,\left (6\,a\,e+4\,b\,d\right )\,\sqrt {d+e\,x}}{b\,e^3}} \]
-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(((2*a^3*e^3)/9 + (32*b^3*d^3)/315 + (16 *a*b^2*d^2*e)/105 + (4*a^2*b*d*e^2)/21)/(b*e^8) + (2*x*(15*a^2*e^2 + 8*b^2 *d^2 + 12*a*b*d*e))/(35*e^7) + (2*b^2*x^3)/(3*e^5) + (2*b*x^2*(3*a*e + 2*b *d))/(5*e^6)))/(x^5*(d + e*x)^(1/2) + (a*d^4*(d + e*x)^(1/2))/(b*e^4) + (x ^4*(a*e^8 + 4*b*d*e^7)*(d + e*x)^(1/2))/(b*e^8) + (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) + (d^2*x^2*(6*a*e + 4*b*d)*(d + e*x)^(1/2))/(b*e^3))