Integrand size = 30, antiderivative size = 254 \[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {35 e^2 (b d-a e) (a+b x) \sqrt {d+e x}}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^2 (a+b x) (d+e x)^{3/2}}{12 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (d+e x)^{5/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{7/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^2 (b d-a e)^{3/2} (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
35/12*e^2*(b*x+a)*(e*x+d)^(3/2)/b^3/((b*x+a)^2)^(1/2)-7/4*e*(e*x+d)^(5/2)/ b^2/((b*x+a)^2)^(1/2)-1/2*(e*x+d)^(7/2)/b/(b*x+a)/((b*x+a)^2)^(1/2)-35/4*e ^2*(-a*e+b*d)^(3/2)*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2) )/b^(9/2)/((b*x+a)^2)^(1/2)+35/4*e^2*(-a*e+b*d)*(b*x+a)*(e*x+d)^(1/2)/b^4/ ((b*x+a)^2)^(1/2)
Time = 0.55 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.73 \[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e^2 (a+b x)^3 \left (-\frac {\sqrt {b} \sqrt {d+e x} \left (105 a^3 e^3+35 a^2 b e^2 (-4 d+5 e x)+7 a b^2 e \left (3 d^2-34 d e x+8 e^2 x^2\right )+b^3 \left (6 d^3+39 d^2 e x-80 d e^2 x^2-8 e^3 x^3\right )\right )}{e^2 (a+b x)^2}+105 (-b d+a e)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )\right )}{12 b^{9/2} \left ((a+b x)^2\right )^{3/2}} \]
(e^2*(a + b*x)^3*(-((Sqrt[b]*Sqrt[d + e*x]*(105*a^3*e^3 + 35*a^2*b*e^2*(-4 *d + 5*e*x) + 7*a*b^2*e*(3*d^2 - 34*d*e*x + 8*e^2*x^2) + b^3*(6*d^3 + 39*d ^2*e*x - 80*d*e^2*x^2 - 8*e^3*x^3)))/(e^2*(a + b*x)^2)) + 105*(-(b*d) + a* e)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]]))/(12*b^(9/2)* ((a + b*x)^2)^(3/2))
Time = 0.28 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.70, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1102, 27, 51, 51, 60, 60, 73, 221}
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.
| \(\displaystyle \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1102 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {(d+e x)^{7/2}}{b^3 (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^{7/2}}{(a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {7 e \int \frac {(d+e x)^{5/2}}{(a+b x)^2}dx}{4 b}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {5 e \int \frac {(d+e x)^{3/2}}{a+b x}dx}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {5 e \left (\frac {(b d-a e) \int \frac {\sqrt {d+e x}}{a+b x}dx}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {5 e \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {5 e \left (\frac {(b d-a e) \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {5 e \left (\frac {(b d-a e) \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/2*(d + e*x)^(7/2)/(b*(a + b*x)^2) + (7*e*(-((d + e*x)^(5/2) /(b*(a + b*x))) + (5*e*((2*(d + e*x)^(3/2))/(3*b) + ((b*d - a*e)*((2*Sqrt[ d + e*x])/b - (2*Sqrt[b*d - a*e]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(3/2)))/b))/(2*b)))/(4*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.18.12.3.1 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] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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 = 2.34 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.69
| method | result | size |
| risch | \(-\frac {2 e^{2} \left (-b e x +9 a e -10 b d \right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{3 b^{4} \left (b x +a \right )}+\frac {\left (2 a^{2} e^{2}-4 a b d e +2 b^{2} d^{2}\right ) e^{2} \left (\frac {-\frac {13 \left (e x +d \right )^{\frac {3}{2}} b}{8}+\left (-\frac {11 a e}{8}+\frac {11 b d}{8}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {35 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{4} \left (b x +a \right )}\) | \(176\) |
| default | \(-\frac {\left (-8 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3} e^{2} x^{2}-105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b^{2} e^{4} x^{2}+210 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{3} d \,e^{3} x^{2}-105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{4} d^{2} e^{2} x^{2}-16 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} e^{2} x +72 a \,b^{2} e^{3} x^{2} \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}-72 b^{3} d \,e^{2} x^{2} \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}-210 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{3} b \,e^{4} x +420 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b^{2} d \,e^{3} x -210 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{3} d^{2} e^{2} x +31 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} a^{2} b \,e^{2}-78 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} a \,b^{2} d e +39 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} b^{3} d^{2}+144 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{2} b \,e^{3} x -144 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d \,e^{2} x -105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{4} e^{4}+210 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{3} b d \,e^{3}-105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b^{2} d^{2} e^{2}+105 \sqrt {e x +d}\, a^{3} e^{3} \sqrt {\left (a e -b d \right ) b}-171 \sqrt {e x +d}\, a^{2} d \,e^{2} b \sqrt {\left (a e -b d \right ) b}+99 \sqrt {e x +d}\, a \,d^{2} e \,b^{2} \sqrt {\left (a e -b d \right ) b}-33 \sqrt {e x +d}\, d^{3} b^{3} \sqrt {\left (a e -b d \right ) b}\right ) \left (b x +a \right )}{12 \sqrt {\left (a e -b d \right ) b}\, b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(714\) |
-2/3*e^2*(-b*e*x+9*a*e-10*b*d)*(e*x+d)^(1/2)/b^4*((b*x+a)^2)^(1/2)/(b*x+a) +1/b^4*(2*a^2*e^2-4*a*b*d*e+2*b^2*d^2)*e^2*((-13/8*(e*x+d)^(3/2)*b+(-11/8* a*e+11/8*b*d)*(e*x+d)^(1/2))/(b*(e*x+d)+a*e-b*d)^2+35/8/((a*e-b*d)*b)^(1/2 )*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)))*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.28 (sec) , antiderivative size = 520, normalized size of antiderivative = 2.05 \[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [-\frac {105 \, {\left (a^{2} b d e^{2} - a^{3} e^{3} + {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) - 2 \, {\left (8 \, b^{3} e^{3} x^{3} - 6 \, b^{3} d^{3} - 21 \, a b^{2} d^{2} e + 140 \, a^{2} b d e^{2} - 105 \, a^{3} e^{3} + 8 \, {\left (10 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} - {\left (39 \, b^{3} d^{2} e - 238 \, a b^{2} d e^{2} + 175 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, -\frac {105 \, {\left (a^{2} b d e^{2} - a^{3} e^{3} + {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (8 \, b^{3} e^{3} x^{3} - 6 \, b^{3} d^{3} - 21 \, a b^{2} d^{2} e + 140 \, a^{2} b d e^{2} - 105 \, a^{3} e^{3} + 8 \, {\left (10 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} - {\left (39 \, b^{3} d^{2} e - 238 \, a b^{2} d e^{2} + 175 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{12 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \]
[-1/24*(105*(a^2*b*d*e^2 - a^3*e^3 + (b^3*d*e^2 - a*b^2*e^3)*x^2 + 2*(a*b^ 2*d*e^2 - a^2*b*e^3)*x)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e + 2*s qrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) - 2*(8*b^3*e^3*x^3 - 6*b^3* d^3 - 21*a*b^2*d^2*e + 140*a^2*b*d*e^2 - 105*a^3*e^3 + 8*(10*b^3*d*e^2 - 7 *a*b^2*e^3)*x^2 - (39*b^3*d^2*e - 238*a*b^2*d*e^2 + 175*a^2*b*e^3)*x)*sqrt (e*x + d))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), -1/12*(105*(a^2*b*d*e^2 - a^3* e^3 + (b^3*d*e^2 - a*b^2*e^3)*x^2 + 2*(a*b^2*d*e^2 - a^2*b*e^3)*x)*sqrt(-( b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (8*b^3*e^3*x^3 - 6*b^3*d^3 - 21*a*b^2*d^2*e + 140*a^2*b*d*e^2 - 105*a^3*e^ 3 + 8*(10*b^3*d*e^2 - 7*a*b^2*e^3)*x^2 - (39*b^3*d^2*e - 238*a*b^2*d*e^2 + 175*a^2*b*e^3)*x)*sqrt(e*x + d))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)]
\[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{\frac {7}{2}}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {7}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.13 \[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {35 \, {\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, \sqrt {-b^{2} d + a b e} b^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {13 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{3} d^{2} e^{2} - 11 \, \sqrt {e x + d} b^{3} d^{3} e^{2} - 26 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{2} d e^{3} + 33 \, \sqrt {e x + d} a b^{2} d^{2} e^{3} + 13 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{2} b e^{4} - 33 \, \sqrt {e x + d} a^{2} b d e^{4} + 11 \, \sqrt {e x + d} a^{3} e^{5}}{4 \, {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2} b^{4} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} b^{6} e^{2} + 9 \, \sqrt {e x + d} b^{6} d e^{2} - 9 \, \sqrt {e x + d} a b^{5} e^{3}\right )}}{3 \, b^{9} \mathrm {sgn}\left (b x + a\right )} \]
35/4*(b^2*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*arctan(sqrt(e*x + d)*b/sqrt(-b^ 2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^4*sgn(b*x + a)) - 1/4*(13*(e*x + d)^ (3/2)*b^3*d^2*e^2 - 11*sqrt(e*x + d)*b^3*d^3*e^2 - 26*(e*x + d)^(3/2)*a*b^ 2*d*e^3 + 33*sqrt(e*x + d)*a*b^2*d^2*e^3 + 13*(e*x + d)^(3/2)*a^2*b*e^4 - 33*sqrt(e*x + d)*a^2*b*d*e^4 + 11*sqrt(e*x + d)*a^3*e^5)/(((e*x + d)*b - b *d + a*e)^2*b^4*sgn(b*x + a)) + 2/3*((e*x + d)^(3/2)*b^6*e^2 + 9*sqrt(e*x + d)*b^6*d*e^2 - 9*sqrt(e*x + d)*a*b^5*e^3)/(b^9*sgn(b*x + a))
Timed out. \[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^{7/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]