Integrand size = 22, antiderivative size = 246 \[ \int \frac {x^2 (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=-\frac {\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^4 d}-\frac {\left (10 a c+\frac {b c^2}{d}-\frac {35 a^2 d}{b}\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^2 (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 d}-\frac {(b c-a d) \left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{9/2} d^{3/2}} \]
-1/8*(-a*d+b*c)*(-35*a^2*d^2+10*a*b*c*d+b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^( 1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(9/2)/d^(3/2)-2*a^2*(d*x+c)^(5/2)/b^2/(-a*d+ b*c)/(b*x+a)^(1/2)-1/12*(10*a*c+b*c^2/d-35*a^2*d/b)*(d*x+c)^(3/2)*(b*x+a)^ (1/2)/b^2/(-a*d+b*c)+1/3*(d*x+c)^(5/2)*(b*x+a)^(1/2)/b^2/d-1/8*(-35*a^2*d^ 2+10*a*b*c*d+b^2*c^2)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^4/d
Time = 10.33 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.79 \[ \int \frac {x^2 (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {c+d x} \left (\frac {\sqrt {d} \left (105 a^3 d^2+5 a^2 b d (-20 c+7 d x)+a b^2 \left (3 c^2-38 c d x-14 d^2 x^2\right )+b^3 x \left (3 c^2+14 c d x+8 d^2 x^2\right )\right )}{\sqrt {a+b x}}-\frac {3 \sqrt {b c-a d} \left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {\frac {b (c+d x)}{b c-a d}}}\right )}{24 b^4 d^{3/2}} \]
(Sqrt[c + d*x]*((Sqrt[d]*(105*a^3*d^2 + 5*a^2*b*d*(-20*c + 7*d*x) + a*b^2* (3*c^2 - 38*c*d*x - 14*d^2*x^2) + b^3*x*(3*c^2 + 14*c*d*x + 8*d^2*x^2)))/S qrt[a + b*x] - (3*Sqrt[b*c - a*d]*(b^2*c^2 + 10*a*b*c*d - 35*a^2*d^2)*ArcS inh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/Sqrt[(b*(c + d*x))/(b*c - a* d)]))/(24*b^4*d^(3/2))
Time = 0.30 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {100, 27, 90, 60, 60, 66, 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.
| \(\displaystyle \int \frac {x^2 (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 100 |
| \(\displaystyle \frac {2 \int -\frac {(c+d x)^{3/2} (a (b c-5 a d)-b (b c-a d) x)}{2 \sqrt {a+b x}}dx}{b^2 (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 \sqrt {a+b x} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {(c+d x)^{3/2} (a (b c-5 a d)-b (b c-a d) x)}{\sqrt {a+b x}}dx}{b^2 (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 \sqrt {a+b x} (b c-a d)}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle -\frac {\frac {\left (-35 a^2 d^2+10 a b c d+b^2 c^2\right ) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}}dx}{6 d}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-a d)}{3 d}}{b^2 (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 \sqrt {a+b x} (b c-a d)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle -\frac {\frac {\left (-35 a^2 d^2+10 a b c d+b^2 c^2\right ) \left (\frac {3 (b c-a d) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}}dx}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 d}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-a d)}{3 d}}{b^2 (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 \sqrt {a+b x} (b c-a d)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle -\frac {\frac {\left (-35 a^2 d^2+10 a b c d+b^2 c^2\right ) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 d}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-a d)}{3 d}}{b^2 (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 \sqrt {a+b x} (b c-a d)}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle -\frac {\frac {\left (-35 a^2 d^2+10 a b c d+b^2 c^2\right ) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 d}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-a d)}{3 d}}{b^2 (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 \sqrt {a+b x} (b c-a d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {\frac {\left (-35 a^2 d^2+10 a b c d+b^2 c^2\right ) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 d}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-a d)}{3 d}}{b^2 (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 \sqrt {a+b x} (b c-a d)}\) |
(-2*a^2*(c + d*x)^(5/2))/(b^2*(b*c - a*d)*Sqrt[a + b*x]) - (-1/3*((b*c - a *d)*Sqrt[a + b*x]*(c + d*x)^(5/2))/d + ((b^2*c^2 + 10*a*b*c*d - 35*a^2*d^2 )*((Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*b) + (3*(b*c - a*d)*((Sqrt[a + b*x]* Sqrt[c + d*x])/b + ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*S qrt[c + d*x])])/(b^(3/2)*Sqrt[d])))/(4*b)))/(6*d))/(b^2*(b*c - a*d))
3.8.60.3.1 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 + 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[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
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]
Leaf count of result is larger than twice the leaf count of optimal. \(691\) vs. \(2(210)=420\).
Time = 0.62 (sec) , antiderivative size = 692, normalized size of antiderivative = 2.81
| method | result | size |
| default | \(-\frac {\sqrt {d x +c}\, \left (-16 b^{3} d^{2} x^{3} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+105 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} b \,d^{3} x -135 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b^{2} c \,d^{2} x +27 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{3} c^{2} d x +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{4} c^{3} x +28 a \,b^{2} d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-28 b^{3} c d \,x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+105 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{4} d^{3}-135 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} b c \,d^{2}+27 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b^{2} c^{2} d +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{3} c^{3}-70 a^{2} b \,d^{2} x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+76 a \,b^{2} c d x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-6 b^{3} c^{2} x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-210 a^{3} d^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+200 a^{2} b c d \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-6 a \,b^{2} c^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {b x +a}\, b^{4} d}\) | \(692\) |
-1/48*(d*x+c)^(1/2)*(-16*b^3*d^2*x^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+1 05*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1 /2))*a^3*b*d^3*x-135*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2) +a*d+b*c)/(b*d)^(1/2))*a^2*b^2*c*d^2*x+27*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+ c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b^3*c^2*d*x+3*ln(1/2*(2*b*d* x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^4*c^3*x+28 *a*b^2*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-28*b^3*c*d*x^2*((b*x+a) *(d*x+c))^(1/2)*(b*d)^(1/2)+105*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)* (b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^4*d^3-135*ln(1/2*(2*b*d*x+2*((b*x+a)*( d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3*b*c*d^2+27*ln(1/2*(2*b *d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b^2*c ^2*d+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d )^(1/2))*a*b^3*c^3-70*a^2*b*d^2*x*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+76*a *b^2*c*d*x*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-6*b^3*c^2*x*((b*x+a)*(d*x+c ))^(1/2)*(b*d)^(1/2)-210*a^3*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+200*a ^2*b*c*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-6*a*b^2*c^2*((b*x+a)*(d*x+c)) ^(1/2)*(b*d)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(b*x+a)^(1/2)/b^4/ d
Time = 0.33 (sec) , antiderivative size = 600, normalized size of antiderivative = 2.44 \[ \int \frac {x^2 (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\left [\frac {3 \, {\left (a b^{3} c^{3} + 9 \, a^{2} b^{2} c^{2} d - 45 \, a^{3} b c d^{2} + 35 \, a^{4} d^{3} + {\left (b^{4} c^{3} + 9 \, a b^{3} c^{2} d - 45 \, a^{2} b^{2} c d^{2} + 35 \, a^{3} b d^{3}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (8 \, b^{4} d^{3} x^{3} + 3 \, a b^{3} c^{2} d - 100 \, a^{2} b^{2} c d^{2} + 105 \, a^{3} b d^{3} + 14 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + {\left (3 \, b^{4} c^{2} d - 38 \, a b^{3} c d^{2} + 35 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, {\left (b^{6} d^{2} x + a b^{5} d^{2}\right )}}, \frac {3 \, {\left (a b^{3} c^{3} + 9 \, a^{2} b^{2} c^{2} d - 45 \, a^{3} b c d^{2} + 35 \, a^{4} d^{3} + {\left (b^{4} c^{3} + 9 \, a b^{3} c^{2} d - 45 \, a^{2} b^{2} c d^{2} + 35 \, a^{3} b d^{3}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{4} d^{3} x^{3} + 3 \, a b^{3} c^{2} d - 100 \, a^{2} b^{2} c d^{2} + 105 \, a^{3} b d^{3} + 14 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + {\left (3 \, b^{4} c^{2} d - 38 \, a b^{3} c d^{2} + 35 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (b^{6} d^{2} x + a b^{5} d^{2}\right )}}\right ] \]
[1/96*(3*(a*b^3*c^3 + 9*a^2*b^2*c^2*d - 45*a^3*b*c*d^2 + 35*a^4*d^3 + (b^4 *c^3 + 9*a*b^3*c^2*d - 45*a^2*b^2*c*d^2 + 35*a^3*b*d^3)*x)*sqrt(b*d)*log(8 *b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b*d*x + b*c + a*d)*sqr t(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(8*b^4*d ^3*x^3 + 3*a*b^3*c^2*d - 100*a^2*b^2*c*d^2 + 105*a^3*b*d^3 + 14*(b^4*c*d^2 - a*b^3*d^3)*x^2 + (3*b^4*c^2*d - 38*a*b^3*c*d^2 + 35*a^2*b^2*d^3)*x)*sqr t(b*x + a)*sqrt(d*x + c))/(b^6*d^2*x + a*b^5*d^2), 1/48*(3*(a*b^3*c^3 + 9* a^2*b^2*c^2*d - 45*a^3*b*c*d^2 + 35*a^4*d^3 + (b^4*c^3 + 9*a*b^3*c^2*d - 4 5*a^2*b^2*c*d^2 + 35*a^3*b*d^3)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2* c*d + a*b*d^2)*x)) + 2*(8*b^4*d^3*x^3 + 3*a*b^3*c^2*d - 100*a^2*b^2*c*d^2 + 105*a^3*b*d^3 + 14*(b^4*c*d^2 - a*b^3*d^3)*x^2 + (3*b^4*c^2*d - 38*a*b^3 *c*d^2 + 35*a^2*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^6*d^2*x + a*b^ 5*d^2)]
\[ \int \frac {x^2 (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\int \frac {x^{2} \left (c + d x\right )^{\frac {3}{2}}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {x^2 (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.45 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.32 \[ \int \frac {x^2 (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\frac {1}{24} \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} d {\left | b \right |}}{b^{6}} + \frac {7 \, b^{18} c d^{4} {\left | b \right |} - 19 \, a b^{17} d^{5} {\left | b \right |}}{b^{23} d^{4}}\right )} + \frac {3 \, {\left (b^{19} c^{2} d^{3} {\left | b \right |} - 22 \, a b^{18} c d^{4} {\left | b \right |} + 29 \, a^{2} b^{17} d^{5} {\left | b \right |}\right )}}{b^{23} d^{4}}\right )} - \frac {4 \, {\left (a^{2} b^{2} c^{2} d {\left | b \right |} - 2 \, a^{3} b c d^{2} {\left | b \right |} + a^{4} d^{3} {\left | b \right |}\right )}}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} \sqrt {b d} b^{4}} + \frac {{\left (b^{3} c^{3} {\left | b \right |} + 9 \, a b^{2} c^{2} d {\left | b \right |} - 45 \, a^{2} b c d^{2} {\left | b \right |} + 35 \, a^{3} d^{3} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{16 \, \sqrt {b d} b^{5} d} \]
1/24*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b* x + a)*d*abs(b)/b^6 + (7*b^18*c*d^4*abs(b) - 19*a*b^17*d^5*abs(b))/(b^23*d ^4)) + 3*(b^19*c^2*d^3*abs(b) - 22*a*b^18*c*d^4*abs(b) + 29*a^2*b^17*d^5*a bs(b))/(b^23*d^4)) - 4*(a^2*b^2*c^2*d*abs(b) - 2*a^3*b*c*d^2*abs(b) + a^4* d^3*abs(b))/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)*sqrt(b*d)*b^4) + 1/16*(b^3*c^3*abs(b) + 9*a*b^2*c^2 *d*abs(b) - 45*a^2*b*c*d^2*abs(b) + 35*a^3*d^3*abs(b))*log((sqrt(b*d)*sqrt (b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(b*d)*b^5*d)
Timed out. \[ \int \frac {x^2 (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\int \frac {x^2\,{\left (c+d\,x\right )}^{3/2}}{{\left (a+b\,x\right )}^{3/2}} \,d x \]