Integrand size = 20, antiderivative size = 268 \[ \int x (a+b x)^{5/2} \sqrt {c+d x} \, dx=-\frac {(b c-a d)^3 (7 b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{128 b^2 d^4}+\frac {(b c-a d)^2 (7 b c+3 a d) (a+b x)^{3/2} \sqrt {c+d x}}{192 b^2 d^3}-\frac {(b c-a d) (7 b c+3 a d) (a+b x)^{5/2} \sqrt {c+d x}}{240 b^2 d^2}-\frac {(7 b c+3 a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b^2 d}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b d}+\frac {(b c-a d)^4 (7 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{5/2} d^{9/2}} \]
1/5*(b*x+a)^(7/2)*(d*x+c)^(3/2)/b/d+1/128*(-a*d+b*c)^4*(3*a*d+7*b*c)*arcta nh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(5/2)/d^(9/2)+1/192*(-a* d+b*c)^2*(3*a*d+7*b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/2)/b^2/d^3-1/240*(-a*d+b*c )*(3*a*d+7*b*c)*(b*x+a)^(5/2)*(d*x+c)^(1/2)/b^2/d^2-1/40*(3*a*d+7*b*c)*(b* x+a)^(7/2)*(d*x+c)^(1/2)/b^2/d-1/128*(-a*d+b*c)^3*(3*a*d+7*b*c)*(b*x+a)^(1 /2)*(d*x+c)^(1/2)/b^2/d^4
Time = 0.53 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.85 \[ \int x (a+b x)^{5/2} \sqrt {c+d x} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-45 a^4 d^4+30 a^3 b d^3 (2 c+d x)+2 a^2 b^2 d^2 \left (-173 c^2+109 c d x+372 d^2 x^2\right )+2 a b^3 d \left (170 c^3-111 c^2 d x+88 c d^2 x^2+504 d^3 x^3\right )+b^4 \left (-105 c^4+70 c^3 d x-56 c^2 d^2 x^2+48 c d^3 x^3+384 d^4 x^4\right )\right )}{1920 b^2 d^4}+\frac {(b c-a d)^4 (7 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{128 b^{5/2} d^{9/2}} \]
(Sqrt[a + b*x]*Sqrt[c + d*x]*(-45*a^4*d^4 + 30*a^3*b*d^3*(2*c + d*x) + 2*a ^2*b^2*d^2*(-173*c^2 + 109*c*d*x + 372*d^2*x^2) + 2*a*b^3*d*(170*c^3 - 111 *c^2*d*x + 88*c*d^2*x^2 + 504*d^3*x^3) + b^4*(-105*c^4 + 70*c^3*d*x - 56*c ^2*d^2*x^2 + 48*c*d^3*x^3 + 384*d^4*x^4)))/(1920*b^2*d^4) + ((b*c - a*d)^4 *(7*b*c + 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])]) /(128*b^(5/2)*d^(9/2))
Time = 0.28 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.91, 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.
| \(\displaystyle \int x (a+b x)^{5/2} \sqrt {c+d x} \, dx\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b d}-\frac {(3 a d+7 b c) \int (a+b x)^{5/2} \sqrt {c+d x}dx}{10 b d}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b d}-\frac {(3 a d+7 b c) \left (\frac {(b c-a d) \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}}dx}{8 b}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}\right )}{10 b d}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b d}-\frac {(3 a d+7 b c) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{6 d}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}\right )}{10 b d}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b d}-\frac {(3 a d+7 b c) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{6 d}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}\right )}{10 b d}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b d}-\frac {(3 a d+7 b c) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{6 d}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}\right )}{10 b d}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b d}-\frac {(3 a d+7 b c) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{6 d}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}\right )}{10 b d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b d}-\frac {(3 a d+7 b c) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{6 d}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}\right )}{10 b d}\) |
((a + b*x)^(7/2)*(c + d*x)^(3/2))/(5*b*d) - ((7*b*c + 3*a*d)*(((a + b*x)^( 7/2)*Sqrt[c + d*x])/(4*b) + ((b*c - a*d)*(((a + b*x)^(5/2)*Sqrt[c + d*x])/ (3*d) - (5*(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*d)))/( 8*b)))/(10*b*d)
3.7.48.3.1 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. \(2(224)=448\).
Time = 1.58 (sec) , antiderivative size = 788, normalized size of antiderivative = 2.94
| method | result | size |
| default | \(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (768 b^{4} d^{4} x^{4} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+2016 a \,b^{3} d^{4} x^{3} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+96 b^{4} c \,d^{3} x^{3} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+1488 a^{2} b^{2} d^{4} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+352 a \,b^{3} c \,d^{3} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-112 b^{4} c^{2} d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+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^{5} d^{5}-75 \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}-150 \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}+450 \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}-375 \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 +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 ) b^{5} c^{5}+60 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{3} b \,d^{4} x +436 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b^{2} c \,d^{3} x -444 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{3} c^{2} d^{2} x +140 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{4} c^{3} d x -90 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{4} d^{4}+120 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{3} b c \,d^{3}-692 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b^{2} c^{2} d^{2}+680 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{3} c^{3} d -210 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{4} c^{4}\right )}{3840 b^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, d^{4} \sqrt {b d}}\) | \(788\) |
1/3840*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(768*b^4*d^4*x^4*((b*x+a)*(d*x+c))^(1/2 )*(b*d)^(1/2)+2016*a*b^3*d^4*x^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+96*b^ 4*c*d^3*x^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+1488*a^2*b^2*d^4*x^2*((b*x +a)*(d*x+c))^(1/2)*(b*d)^(1/2)+352*a*b^3*c*d^3*x^2*((b*x+a)*(d*x+c))^(1/2) *(b*d)^(1/2)-112*b^4*c^2*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+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^5*d^5-75*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^4-150*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+450*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^2-375*l n(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+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))*b^5*c^5+60*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^3*b*d^ 4*x+436*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^2*b^2*c*d^3*x-444*((b*x+a)*( d*x+c))^(1/2)*(b*d)^(1/2)*a*b^3*c^2*d^2*x+140*((b*x+a)*(d*x+c))^(1/2)*(b*d )^(1/2)*b^4*c^3*d*x-90*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^4*d^4+120*((b *x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^3*b*c*d^3-692*((b*x+a)*(d*x+c))^(1/2)*( b*d)^(1/2)*a^2*b^2*c^2*d^2+680*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*b^3*c ^3*d-210*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b^4*c^4)/b^2/((b*x+a)*(d*x+c) )^(1/2)/d^4/(b*d)^(1/2)
Time = 0.27 (sec) , antiderivative size = 704, normalized size of antiderivative = 2.63 \[ \int x (a+b x)^{5/2} \sqrt {c+d x} \, dx=\left [\frac {15 \, {\left (7 \, b^{5} c^{5} - 25 \, a b^{4} c^{4} d + 30 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} - 5 \, a^{4} b c d^{4} + 3 \, a^{5} d^{5}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (384 \, b^{5} d^{5} x^{4} - 105 \, b^{5} c^{4} d + 340 \, a b^{4} c^{3} d^{2} - 346 \, a^{2} b^{3} c^{2} d^{3} + 60 \, a^{3} b^{2} c d^{4} - 45 \, a^{4} b d^{5} + 48 \, {\left (b^{5} c d^{4} + 21 \, a b^{4} d^{5}\right )} x^{3} - 8 \, {\left (7 \, b^{5} c^{2} d^{3} - 22 \, a b^{4} c d^{4} - 93 \, a^{2} b^{3} d^{5}\right )} x^{2} + 2 \, {\left (35 \, b^{5} c^{3} d^{2} - 111 \, a b^{4} c^{2} d^{3} + 109 \, a^{2} b^{3} c d^{4} + 15 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{7680 \, b^{3} d^{5}}, -\frac {15 \, {\left (7 \, b^{5} c^{5} - 25 \, a b^{4} c^{4} d + 30 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} - 5 \, a^{4} b c d^{4} + 3 \, a^{5} d^{5}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (384 \, b^{5} d^{5} x^{4} - 105 \, b^{5} c^{4} d + 340 \, a b^{4} c^{3} d^{2} - 346 \, a^{2} b^{3} c^{2} d^{3} + 60 \, a^{3} b^{2} c d^{4} - 45 \, a^{4} b d^{5} + 48 \, {\left (b^{5} c d^{4} + 21 \, a b^{4} d^{5}\right )} x^{3} - 8 \, {\left (7 \, b^{5} c^{2} d^{3} - 22 \, a b^{4} c d^{4} - 93 \, a^{2} b^{3} d^{5}\right )} x^{2} + 2 \, {\left (35 \, b^{5} c^{3} d^{2} - 111 \, a b^{4} c^{2} d^{3} + 109 \, a^{2} b^{3} c d^{4} + 15 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3840 \, b^{3} d^{5}}\right ] \]
[1/7680*(15*(7*b^5*c^5 - 25*a*b^4*c^4*d + 30*a^2*b^3*c^3*d^2 - 10*a^3*b^2* c^2*d^3 - 5*a^4*b*c*d^4 + 3*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)*s qrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(384*b^5*d^5*x^4 - 105*b^5*c^4 *d + 340*a*b^4*c^3*d^2 - 346*a^2*b^3*c^2*d^3 + 60*a^3*b^2*c*d^4 - 45*a^4*b *d^5 + 48*(b^5*c*d^4 + 21*a*b^4*d^5)*x^3 - 8*(7*b^5*c^2*d^3 - 22*a*b^4*c*d ^4 - 93*a^2*b^3*d^5)*x^2 + 2*(35*b^5*c^3*d^2 - 111*a*b^4*c^2*d^3 + 109*a^2 *b^3*c*d^4 + 15*a^3*b^2*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^5), -1 /3840*(15*(7*b^5*c^5 - 25*a*b^4*c^4*d + 30*a^2*b^3*c^3*d^2 - 10*a^3*b^2*c^ 2*d^3 - 5*a^4*b*c*d^4 + 3*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*(384*b^5*d^5*x^4 - 105*b^5*c^4*d + 340*a*b^4*c^3*d^ 2 - 346*a^2*b^3*c^2*d^3 + 60*a^3*b^2*c*d^4 - 45*a^4*b*d^5 + 48*(b^5*c*d^4 + 21*a*b^4*d^5)*x^3 - 8*(7*b^5*c^2*d^3 - 22*a*b^4*c*d^4 - 93*a^2*b^3*d^5)* x^2 + 2*(35*b^5*c^3*d^2 - 111*a*b^4*c^2*d^3 + 109*a^2*b^3*c*d^4 + 15*a^3*b ^2*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^5)]
\[ \int x (a+b x)^{5/2} \sqrt {c+d x} \, dx=\int x \left (a + b x\right )^{\frac {5}{2}} \sqrt {c + d x}\, dx \]
Exception generated. \[ \int x (a+b x)^{5/2} \sqrt {c+d x} \, dx=\text {Exception raised: ValueError} \]
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. 1011 vs. \(2 (224) = 448\).
Time = 0.42 (sec) , antiderivative size = 1011, normalized size of antiderivative = 3.77 \[ \int x (a+b x)^{5/2} \sqrt {c+d x} \, dx=\text {Too large to display} \]
1/1920*(30*(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))*a*abs(b) + 240*(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^2*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))*b*abs(b) + ...
Timed out. \[ \int x (a+b x)^{5/2} \sqrt {c+d x} \, dx=\int x\,{\left (a+b\,x\right )}^{5/2}\,\sqrt {c+d\,x} \,d x \]