Integrand size = 22, antiderivative size = 175 \[ \int \frac {x^3}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=-\frac {2 c x^2 \sqrt {a+b x}}{d (b c-a d) \sqrt {c+d x}}-\frac {\sqrt {a+b x} \sqrt {c+d x} ((5 b c-3 a d) (3 b c+a d)-2 b d (5 b c-a d) x)}{4 b^2 d^3 (b c-a d)}+\frac {3 \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{7/2}} \]
3/4*(a^2*d^2+2*a*b*c*d+5*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d *x+c)^(1/2))/b^(5/2)/d^(7/2)-2*c*x^2*(b*x+a)^(1/2)/d/(-a*d+b*c)/(d*x+c)^(1 /2)-1/4*((-3*a*d+5*b*c)*(a*d+3*b*c)-2*b*d*(-a*d+5*b*c)*x)*(b*x+a)^(1/2)*(d *x+c)^(1/2)/b^2/d^3/(-a*d+b*c)
Time = 0.43 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.11 \[ \int \frac {x^3}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\frac {-\sqrt {b} \sqrt {d} \sqrt {a+b x} \left (3 a^2 d^2 (c+d x)+2 a b d \left (2 c^2+c d x-d^2 x^2\right )+b^2 c \left (-15 c^2-5 c d x+2 d^2 x^2\right )\right )-3 \left (5 b^3 c^3-3 a b^2 c^2 d-a^2 b c d^2-a^3 d^3\right ) \sqrt {c+d x} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{7/2} (-b c+a d) \sqrt {c+d x}} \]
(-(Sqrt[b]*Sqrt[d]*Sqrt[a + b*x]*(3*a^2*d^2*(c + d*x) + 2*a*b*d*(2*c^2 + c *d*x - d^2*x^2) + b^2*c*(-15*c^2 - 5*c*d*x + 2*d^2*x^2))) - 3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*Sqrt[c + d*x]*ArcTanh[(Sqrt[d]*Sqr t[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(5/2)*d^(7/2)*(-(b*c) + a*d)*Sq rt[c + d*x])
Time = 0.29 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {109, 27, 164, 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.
| \(\displaystyle \int \frac {x^3}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {2 \int \frac {x (4 a c+(5 b c-a d) x)}{2 \sqrt {a+b x} \sqrt {c+d x}}dx}{d (b c-a d)}-\frac {2 c x^2 \sqrt {a+b x}}{d \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {x (4 a c+(5 b c-a d) x)}{\sqrt {a+b x} \sqrt {c+d x}}dx}{d (b c-a d)}-\frac {2 c x^2 \sqrt {a+b x}}{d \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {\frac {3 (b c-a d) \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{8 b^2 d^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} ((5 b c-3 a d) (a d+3 b c)-2 b d x (5 b c-a d))}{4 b^2 d^2}}{d (b c-a d)}-\frac {2 c x^2 \sqrt {a+b x}}{d \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {\frac {3 (b c-a d) \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{4 b^2 d^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} ((5 b c-3 a d) (a d+3 b c)-2 b d x (5 b c-a d))}{4 b^2 d^2}}{d (b c-a d)}-\frac {2 c x^2 \sqrt {a+b x}}{d \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {3 (b c-a d) \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{5/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} ((5 b c-3 a d) (a d+3 b c)-2 b d x (5 b c-a d))}{4 b^2 d^2}}{d (b c-a d)}-\frac {2 c x^2 \sqrt {a+b x}}{d \sqrt {c+d x} (b c-a d)}\) |
(-2*c*x^2*Sqrt[a + b*x])/(d*(b*c - a*d)*Sqrt[c + d*x]) + (-1/4*(Sqrt[a + b *x]*Sqrt[c + d*x]*((5*b*c - 3*a*d)*(3*b*c + a*d) - 2*b*d*(5*b*c - a*d)*x)) /(b^2*d^2) + (3*(b*c - a*d)*(5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*ArcTanh[(Sqr t[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(5/2)*d^(5/2)))/(d*(b*c - a*d))
3.8.41.3.1 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[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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f *x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
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. \(672\) vs. \(2(151)=302\).
Time = 1.67 (sec) , antiderivative size = 673, normalized size of antiderivative = 3.85
| method | result | size |
| default | \(\frac {\sqrt {b x +a}\, \left (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^{3} d^{4} 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 ) a^{2} b c \,d^{3} x +9 \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^{2} c^{2} d^{2} x -15 \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^{3} c^{3} d x +4 a b \,d^{3} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-4 b^{2} c \,d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b 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^{3} c \,d^{3}+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^{2} b \,c^{2} d^{2}+9 \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^{2} c^{3} d -15 \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^{3} c^{4}-6 a^{2} d^{3} x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-4 a b c \,d^{2} x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+10 b^{2} c^{2} d x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-6 a^{2} c \,d^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-8 a b \,c^{2} d \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+30 b^{2} c^{3} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\right )}{8 \left (a d -b c \right ) \sqrt {b d}\, b^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, d^{3} \sqrt {d x +c}}\) | \(673\) |
1/8*(b*x+a)^(1/2)*(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^3*d^4*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))*a^2*b*c*d^3*x+9*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^2*c^2*d^2*x-15* 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^3*c^3*d*x+4*a*b*d^3*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-4*b^2*c*d^ 2*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+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^3*c*d^3+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^2*b*c^2*d^2+ 9*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^2*c^3*d-15*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^3*c^4-6*a^2*d^3*x*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/ 2)-4*a*b*c*d^2*x*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+10*b^2*c^2*d*x*((b*x+ a)*(d*x+c))^(1/2)*(b*d)^(1/2)-6*a^2*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1 /2)-8*a*b*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+30*b^2*c^3*((b*x+a)*(d *x+c))^(1/2)*(b*d)^(1/2))/(a*d-b*c)/(b*d)^(1/2)/b^2/((b*x+a)*(d*x+c))^(1/2 )/d^3/(d*x+c)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (152) = 304\).
Time = 0.33 (sec) , antiderivative size = 630, normalized size of antiderivative = 3.60 \[ \int \frac {x^3}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\left [\frac {3 \, {\left (5 \, b^{3} c^{4} - 3 \, a b^{2} c^{3} d - a^{2} b c^{2} d^{2} - a^{3} c d^{3} + {\left (5 \, b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} - a^{2} b c d^{3} - a^{3} d^{4}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (15 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} - 3 \, a^{2} b c d^{3} - 2 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{2} + {\left (5 \, b^{3} c^{2} d^{2} - 2 \, a b^{2} c d^{3} - 3 \, a^{2} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (b^{4} c^{2} d^{4} - a b^{3} c d^{5} + {\left (b^{4} c d^{5} - a b^{3} d^{6}\right )} x\right )}}, -\frac {3 \, {\left (5 \, b^{3} c^{4} - 3 \, a b^{2} c^{3} d - a^{2} b c^{2} d^{2} - a^{3} c d^{3} + {\left (5 \, b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} - a^{2} b c d^{3} - a^{3} d^{4}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (15 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} - 3 \, a^{2} b c d^{3} - 2 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{2} + {\left (5 \, b^{3} c^{2} d^{2} - 2 \, a b^{2} c d^{3} - 3 \, a^{2} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (b^{4} c^{2} d^{4} - a b^{3} c d^{5} + {\left (b^{4} c d^{5} - a b^{3} d^{6}\right )} x\right )}}\right ] \]
[1/16*(3*(5*b^3*c^4 - 3*a*b^2*c^3*d - a^2*b*c^2*d^2 - a^3*c*d^3 + (5*b^3*c ^3*d - 3*a*b^2*c^2*d^2 - a^2*b*c*d^3 - a^3*d^4)*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)*sqrt(b*d)*s qrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(15*b^3*c^3*d - 4*a*b^2*c^2*d^2 - 3*a^2*b*c*d^3 - 2*(b^3*c*d^3 - a*b^2*d^4)*x^2 + (5*b^3*c ^2*d^2 - 2*a*b^2*c*d^3 - 3*a^2*b*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4 *c^2*d^4 - a*b^3*c*d^5 + (b^4*c*d^5 - a*b^3*d^6)*x), -1/8*(3*(5*b^3*c^4 - 3*a*b^2*c^3*d - a^2*b*c^2*d^2 - a^3*c*d^3 + (5*b^3*c^3*d - 3*a*b^2*c^2*d^2 - a^2*b*c*d^3 - a^3*d^4)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*s qrt(-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*(15*b^3*c^3*d - 4*a*b^2*c^2*d^2 - 3*a^2*b*c*d^3 - 2*(b^3* c*d^3 - a*b^2*d^4)*x^2 + (5*b^3*c^2*d^2 - 2*a*b^2*c*d^3 - 3*a^2*b*d^4)*x)* sqrt(b*x + a)*sqrt(d*x + c))/(b^4*c^2*d^4 - a*b^3*c*d^5 + (b^4*c*d^5 - a*b ^3*d^6)*x)]
\[ \int \frac {x^3}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\int \frac {x^{3}}{\sqrt {a + b x} \left (c + d x\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {x^3}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (152) = 304\).
Time = 0.34 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.74 \[ \int \frac {x^3}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\frac {{\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (b^{6} c d^{4} {\left | b \right |} - a b^{5} d^{5} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{9} c d^{5} - a b^{8} d^{6}} - \frac {5 \, b^{7} c^{2} d^{3} {\left | b \right |} + 2 \, a b^{6} c d^{4} {\left | b \right |} - 7 \, a^{2} b^{5} d^{5} {\left | b \right |}}{b^{9} c d^{5} - a b^{8} d^{6}}\right )} - \frac {15 \, b^{8} c^{3} d^{2} {\left | b \right |} - 9 \, a b^{7} c^{2} d^{3} {\left | b \right |} - 3 \, a^{2} b^{6} c d^{4} {\left | b \right |} + 5 \, a^{3} b^{5} d^{5} {\left | b \right |}}{b^{9} c d^{5} - a b^{8} d^{6}}\right )} \sqrt {b x + a}}{4 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} - \frac {3 \, {\left (5 \, b^{2} c^{2} {\left | b \right |} + 2 \, a b c d {\left | b \right |} + a^{2} d^{2} {\left | b \right |}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{4 \, \sqrt {b d} b^{3} d^{3}} \]
1/4*((b*x + a)*(2*(b^6*c*d^4*abs(b) - a*b^5*d^5*abs(b))*(b*x + a)/(b^9*c*d ^5 - a*b^8*d^6) - (5*b^7*c^2*d^3*abs(b) + 2*a*b^6*c*d^4*abs(b) - 7*a^2*b^5 *d^5*abs(b))/(b^9*c*d^5 - a*b^8*d^6)) - (15*b^8*c^3*d^2*abs(b) - 9*a*b^7*c ^2*d^3*abs(b) - 3*a^2*b^6*c*d^4*abs(b) + 5*a^3*b^5*d^5*abs(b))/(b^9*c*d^5 - a*b^8*d^6))*sqrt(b*x + a)/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) - 3/4*(5*b ^2*c^2*abs(b) + 2*a*b*c*d*abs(b) + a^2*d^2*abs(b))*log(abs(-sqrt(b*d)*sqrt (b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^3*d^3)
Timed out. \[ \int \frac {x^3}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\int \frac {x^3}{\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{3/2}} \,d x \]