Integrand size = 20, antiderivative size = 258 \[ \int x (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\frac {3 (b c-a d)^3 (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{128 b^3 d^3}-\frac {(b c-a d)^2 (b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{64 b^3 d^2}-\frac {(b c-a d) (b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{16 b^3 d}-\frac {(b c+a d) (a+b x)^{5/2} (c+d x)^{3/2}}{8 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {3 (b c-a d)^4 (b c+a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{7/2} d^{7/2}} \]
-1/8*(a*d+b*c)*(b*x+a)^(5/2)*(d*x+c)^(3/2)/b^2/d+1/5*(b*x+a)^(5/2)*(d*x+c) ^(5/2)/b/d-3/128*(-a*d+b*c)^4*(a*d+b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1 /2)/(d*x+c)^(1/2))/b^(7/2)/d^(7/2)-1/64*(-a*d+b*c)^2*(a*d+b*c)*(b*x+a)^(3/ 2)*(d*x+c)^(1/2)/b^3/d^2-1/16*(-a*d+b*c)*(a*d+b*c)*(b*x+a)^(5/2)*(d*x+c)^( 1/2)/b^3/d+3/128*(-a*d+b*c)^3*(a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^3/d^ 3
Time = 0.57 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.88 \[ \int x (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^4 d^4-10 a^3 b d^3 (4 c+d x)+2 a^2 b^2 d^2 \left (9 c^2+13 c d x+4 d^2 x^2\right )+2 a b^3 d \left (-20 c^3+13 c^2 d x+136 c d^2 x^2+88 d^3 x^3\right )+b^4 \left (15 c^4-10 c^3 d x+8 c^2 d^2 x^2+176 c d^3 x^3+128 d^4 x^4\right )\right )}{640 b^3 d^3}-\frac {3 (b c-a d)^4 (b c+a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{128 b^{7/2} d^{7/2}} \]
(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^4*d^4 - 10*a^3*b*d^3*(4*c + d*x) + 2*a^ 2*b^2*d^2*(9*c^2 + 13*c*d*x + 4*d^2*x^2) + 2*a*b^3*d*(-20*c^3 + 13*c^2*d*x + 136*c*d^2*x^2 + 88*d^3*x^3) + b^4*(15*c^4 - 10*c^3*d*x + 8*c^2*d^2*x^2 + 176*c*d^3*x^3 + 128*d^4*x^4)))/(640*b^3*d^3) - (3*(b*c - a*d)^4*(b*c + a *d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(128*b^(7/2) *d^(7/2))
Time = 0.28 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {90, 60, 60, 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.
| \(\displaystyle \int x (a+b x)^{3/2} (c+d x)^{3/2} \, dx\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {(a d+b c) \int (a+b x)^{3/2} (c+d x)^{3/2}dx}{2 b d}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {(a d+b c) \left (\frac {3 (b c-a d) \int (a+b x)^{3/2} \sqrt {c+d x}dx}{8 b}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b}\right )}{2 b d}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {(a d+b c) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\right )}{8 b}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b}\right )}{2 b d}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {(a d+b c) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\right )}{8 b}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b}\right )}{2 b d}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {(a d+b c) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\right )}{8 b}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b}\right )}{2 b d}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {(a d+b c) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\right )}{8 b}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b}\right )}{2 b d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b d}-\frac {(a d+b c) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\right )}{8 b}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b}\right )}{2 b d}\) |
((a + b*x)^(5/2)*(c + d*x)^(5/2))/(5*b*d) - ((b*c + a*d)*(((a + b*x)^(5/2) *(c + d*x)^(3/2))/(4*b) + (3*(b*c - a*d)*(((a + b*x)^(5/2)*Sqrt[c + d*x])/ (3*b) + ((b*c - a*d)*(((a + b*x)^(3/2)*Sqrt[c + d*x])/(2*d) - (3*(b*c - a* d)*((Sqrt[a + b*x]*Sqrt[c + d*x])/d - ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*d^(3/2))))/(4*d)))/(6*b)))/(8* b)))/(2*b*d)
3.7.8.3.1 Defintions of rubi rules used
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)^(-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. \(787\) vs. \(2(214)=428\).
Time = 1.54 (sec) , antiderivative size = 788, normalized size of antiderivative = 3.05
| method | result | size |
| default | \(-\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-256 b^{4} d^{4} x^{4} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-352 a \,b^{3} d^{4} x^{3} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-352 b^{4} c \,d^{3} x^{3} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-16 a^{2} b^{2} d^{4} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-544 a \,b^{3} c \,d^{3} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-16 b^{4} c^{2} d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b 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 ) a^{5} d^{5}-45 \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} b c \,d^{4}+30 \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^{2} c^{2} d^{3}+30 \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^{3} c^{3} d^{2}-45 \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^{4} c^{4} 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^{5} c^{5}+20 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{3} b \,d^{4} x -52 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b^{2} c \,d^{3} x -52 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{3} c^{2} d^{2} x +20 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{4} c^{3} d x -30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{4} d^{4}+80 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{3} b c \,d^{3}-36 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b^{2} c^{2} d^{2}+80 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{3} c^{3} d -30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{4} c^{4}\right )}{1280 b^{3} d^{3} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}\) | \(788\) |
-1/1280*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-256*b^4*d^4*x^4*((b*x+a)*(d*x+c))^(1 /2)*(b*d)^(1/2)-352*a*b^3*d^4*x^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-352* b^4*c*d^3*x^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-16*a^2*b^2*d^4*x^2*((b*x +a)*(d*x+c))^(1/2)*(b*d)^(1/2)-544*a*b^3*c*d^3*x^2*((b*x+a)*(d*x+c))^(1/2) *(b*d)^(1/2)-16*b^4*c^2*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+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))*a ^5*d^5-45*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*b*c*d^4+30*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^2*c^2*d^3+30*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^3*c^3*d^2-45*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 ^4*c^4*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^5*c^5+20*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^3*b*d^4*x-5 2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^2*b^2*c*d^3*x-52*((b*x+a)*(d*x+c)) ^(1/2)*(b*d)^(1/2)*a*b^3*c^2*d^2*x+20*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)* b^4*c^3*d*x-30*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^4*d^4+80*((b*x+a)*(d* x+c))^(1/2)*(b*d)^(1/2)*a^3*b*c*d^3-36*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2) *a^2*b^2*c^2*d^2+80*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*b^3*c^3*d-30*((b *x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b^4*c^4)/b^3/d^3/((b*x+a)*(d*x+c))^(1/2)/ (b*d)^(1/2)
Time = 0.27 (sec) , antiderivative size = 694, normalized size of antiderivative = 2.69 \[ \int x (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\left [\frac {15 \, {\left (b^{5} c^{5} - 3 \, a b^{4} c^{4} d + 2 \, a^{2} b^{3} c^{3} d^{2} + 2 \, a^{3} b^{2} c^{2} d^{3} - 3 \, a^{4} b c d^{4} + a^{5} d^{5}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (128 \, b^{5} d^{5} x^{4} + 15 \, b^{5} c^{4} d - 40 \, a b^{4} c^{3} d^{2} + 18 \, a^{2} b^{3} c^{2} d^{3} - 40 \, a^{3} b^{2} c d^{4} + 15 \, a^{4} b d^{5} + 176 \, {\left (b^{5} c d^{4} + a b^{4} d^{5}\right )} x^{3} + 8 \, {\left (b^{5} c^{2} d^{3} + 34 \, a b^{4} c d^{4} + a^{2} b^{3} d^{5}\right )} x^{2} - 2 \, {\left (5 \, b^{5} c^{3} d^{2} - 13 \, a b^{4} c^{2} d^{3} - 13 \, a^{2} b^{3} c d^{4} + 5 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{2560 \, b^{4} d^{4}}, \frac {15 \, {\left (b^{5} c^{5} - 3 \, a b^{4} c^{4} d + 2 \, a^{2} b^{3} c^{3} d^{2} + 2 \, a^{3} b^{2} c^{2} d^{3} - 3 \, a^{4} b c d^{4} + a^{5} d^{5}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (128 \, b^{5} d^{5} x^{4} + 15 \, b^{5} c^{4} d - 40 \, a b^{4} c^{3} d^{2} + 18 \, a^{2} b^{3} c^{2} d^{3} - 40 \, a^{3} b^{2} c d^{4} + 15 \, a^{4} b d^{5} + 176 \, {\left (b^{5} c d^{4} + a b^{4} d^{5}\right )} x^{3} + 8 \, {\left (b^{5} c^{2} d^{3} + 34 \, a b^{4} c d^{4} + a^{2} b^{3} d^{5}\right )} x^{2} - 2 \, {\left (5 \, b^{5} c^{3} d^{2} - 13 \, a b^{4} c^{2} d^{3} - 13 \, a^{2} b^{3} c d^{4} + 5 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{1280 \, b^{4} d^{4}}\right ] \]
[1/2560*(15*(b^5*c^5 - 3*a*b^4*c^4*d + 2*a^2*b^3*c^3*d^2 + 2*a^3*b^2*c^2*d ^3 - 3*a^4*b*c*d^4 + a^5*d^5)*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)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(128*b^5*d^5*x^4 + 15*b^5*c^4*d - 40* a*b^4*c^3*d^2 + 18*a^2*b^3*c^2*d^3 - 40*a^3*b^2*c*d^4 + 15*a^4*b*d^5 + 176 *(b^5*c*d^4 + a*b^4*d^5)*x^3 + 8*(b^5*c^2*d^3 + 34*a*b^4*c*d^4 + a^2*b^3*d ^5)*x^2 - 2*(5*b^5*c^3*d^2 - 13*a*b^4*c^2*d^3 - 13*a^2*b^3*c*d^4 + 5*a^3*b ^2*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*d^4), 1/1280*(15*(b^5*c^5 - 3 *a*b^4*c^4*d + 2*a^2*b^3*c^3*d^2 + 2*a^3*b^2*c^2*d^3 - 3*a^4*b*c*d^4 + a^5 *d^5)*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*(128*b ^5*d^5*x^4 + 15*b^5*c^4*d - 40*a*b^4*c^3*d^2 + 18*a^2*b^3*c^2*d^3 - 40*a^3 *b^2*c*d^4 + 15*a^4*b*d^5 + 176*(b^5*c*d^4 + a*b^4*d^5)*x^3 + 8*(b^5*c^2*d ^3 + 34*a*b^4*c*d^4 + a^2*b^3*d^5)*x^2 - 2*(5*b^5*c^3*d^2 - 13*a*b^4*c^2*d ^3 - 13*a^2*b^3*c*d^4 + 5*a^3*b^2*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^ 4*d^4)]
\[ \int x (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\int x \left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {3}{2}}\, dx \]
Exception generated. \[ \int x (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\text {Exception raised: ValueError} \]
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. 1511 vs. \(2 (214) = 428\).
Time = 0.50 (sec) , antiderivative size = 1511, normalized size of antiderivative = 5.86 \[ \int x (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\text {Too large to display} \]
1/1920*(10*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)* (6*(b*x + a)/b^3 + (b^12*c*d^5 - 25*a*b^11*d^6)/(b^14*d^6)) - (5*b^13*c^2* d^4 + 14*a*b^12*c*d^5 - 163*a^2*b^11*d^6)/(b^14*d^6)) + 3*(5*b^14*c^3*d^3 + 9*a*b^13*c^2*d^4 + 15*a^2*b^12*c*d^5 - 93*a^3*b^11*d^6)/(b^14*d^6))*sqrt (b*x + a) + 3*(5*b^4*c^4 + 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 20*a^3*b*c* d^3 - 35*a^4*d^4)*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^2*d^3))*c*abs(b) + 160*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^2 + (b^6*c*d^3 - 13*a*b^5*d^4)/(b^7*d^4)) - 3*(b^7*c^2*d^2 + 2*a*b^6*c*d^3 - 11*a^2*b^5*d^ 4)/(b^7*d^4)) - 3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*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*d^2))*a*c*abs(b)/b + (sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(4*(b *x + a)*(6*(b*x + a)*(8*(b*x + a)/b^4 + (b^20*c*d^7 - 41*a*b^19*d^8)/(b^23 *d^8)) - (7*b^21*c^2*d^6 + 26*a*b^20*c*d^7 - 513*a^2*b^19*d^8)/(b^23*d^8)) + 5*(7*b^22*c^3*d^5 + 19*a*b^21*c^2*d^6 + 37*a^2*b^20*c*d^7 - 447*a^3*b^1 9*d^8)/(b^23*d^8))*(b*x + a) - 15*(7*b^23*c^4*d^4 + 12*a*b^22*c^3*d^5 + 18 *a^2*b^21*c^2*d^6 + 28*a^3*b^20*c*d^7 - 193*a^4*b^19*d^8)/(b^23*d^8))*sqrt (b*x + a) - 15*(7*b^5*c^5 + 5*a*b^4*c^4*d + 6*a^2*b^3*c^3*d^2 + 10*a^3*b^2 *c^2*d^3 + 35*a^4*b*c*d^4 - 63*a^5*d^5)*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^4))*d*abs(b) + ...
Timed out. \[ \int x (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\int x\,{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{3/2} \,d x \]