Integrand size = 22, antiderivative size = 267 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}-\frac {4 c (4 b c-3 a d) (a+b x)^{5/2}}{3 d^2 (b c-a d)^2 \sqrt {c+d x}}-\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{4 d^4 (b c-a d)}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3 (b c-a d)^2}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{9/2}} \]
2/3*c^2*(b*x+a)^(5/2)/d^2/(-a*d+b*c)/(d*x+c)^(3/2)+1/4*(3*a^2*d^2-30*a*b*c *d+35*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/d^(9/2 )/b^(1/2)-4/3*c*(-3*a*d+4*b*c)*(b*x+a)^(5/2)/d^2/(-a*d+b*c)^2/(d*x+c)^(1/2 )+1/6*(3*a^2*d^2-30*a*b*c*d+35*b^2*c^2)*(b*x+a)^(3/2)*(d*x+c)^(1/2)/d^3/(- a*d+b*c)^2-1/4*(3*a^2*d^2-30*a*b*c*d+35*b^2*c^2)*(b*x+a)^(1/2)*(d*x+c)^(1/ 2)/d^4/(-a*d+b*c)
Time = 0.31 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.56 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\frac {\sqrt {a+b x} \left (a d \left (55 c^2+78 c d x+15 d^2 x^2\right )-b \left (105 c^3+140 c^2 d x+21 c d^2 x^2-6 d^3 x^3\right )\right )}{12 d^4 (c+d x)^{3/2}}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{4 \sqrt {b} d^{9/2}} \]
(Sqrt[a + b*x]*(a*d*(55*c^2 + 78*c*d*x + 15*d^2*x^2) - b*(105*c^3 + 140*c^ 2*d*x + 21*c*d^2*x^2 - 6*d^3*x^3)))/(12*d^4*(c + d*x)^(3/2)) + ((35*b^2*c^ 2 - 30*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[ a + b*x])])/(4*Sqrt[b]*d^(9/2))
Time = 0.35 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {100, 27, 87, 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.
\(\displaystyle \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {2 \int \frac {(a+b x)^{3/2} (c (5 b c-3 a d)-3 d (b c-a d) x)}{2 (c+d x)^{3/2}}dx}{3 d^2 (b c-a d)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\int \frac {(a+b x)^{3/2} (c (5 b c-3 a d)-3 d (b c-a d) x)}{(c+d x)^{3/2}}dx}{3 d^2 (b c-a d)}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{5/2} (4 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {\left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{b c-a d}}{3 d^2 (b c-a d)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{5/2} (4 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {\left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{b c-a d}}{3 d^2 (b c-a d)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{5/2} (4 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {\left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{b c-a d}}{3 d^2 (b c-a d)}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{5/2} (4 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {\left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{b c-a d}}{3 d^2 (b c-a d)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{5/2} (4 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {\left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{b c-a d}}{3 d^2 (b c-a d)}\) |
(2*c^2*(a + b*x)^(5/2))/(3*d^2*(b*c - a*d)*(c + d*x)^(3/2)) - ((4*c*(4*b*c - 3*a*d)*(a + b*x)^(5/2))/((b*c - a*d)*Sqrt[c + d*x]) - ((35*b^2*c^2 - 30 *a*b*c*d + 3*a^2*d^2)*(((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)))/(b*c - a*d ))/(3*d^2*(b*c - a*d))
3.7.41.3.1 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[((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*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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. \(675\) vs. \(2(229)=458\).
Time = 0.55 (sec) , antiderivative size = 676, normalized size of antiderivative = 2.53
method | result | size |
default | \(\frac {\sqrt {b x +a}\, \left (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^{2} d^{4} x^{2}-90 \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 c \,d^{3} x^{2}+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^{2} c^{2} d^{2} x^{2}+12 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,d^{3} x^{3}+18 \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} c \,d^{3} x -180 \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 \,c^{2} d^{2} x +210 \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^{2} c^{3} d x +30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,d^{3} x^{2}-42 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b c \,d^{2} x^{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^{2} c^{2} d^{2}-90 \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 \,c^{3} 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^{2} c^{4}+156 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a c \,d^{2} x -280 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,c^{2} d x +110 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,c^{2} d -210 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,c^{3}\right )}{24 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, d^{4} \left (d x +c \right )^{\frac {3}{2}}}\) | \(676\) |
1/24*(b*x+a)^(1/2)*(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^2*d^4*x^2-90*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*c*d^3*x^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))*b^2*c^2*d^2*x ^2+12*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b*d^3*x^3+18*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*c*d^3*x-180*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*c^2*d^2*x+210*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^2*c^3*d*x+30*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a* d^3*x^2-42*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b*c*d^2*x^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^2*c^2*d^2 -90*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*c^3*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^2*c^4+156*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*c* d^2*x-280*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b*c^2*d*x+110*((b*x+a)*(d*x+ c))^(1/2)*(b*d)^(1/2)*a*c^2*d-210*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b*c^ 3)/(b*d)^(1/2)/((b*x+a)*(d*x+c))^(1/2)/d^4/(d*x+c)^(3/2)
Time = 0.46 (sec) , antiderivative size = 594, normalized size of antiderivative = 2.22 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\left [\frac {3 \, {\left (35 \, b^{2} c^{4} - 30 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{2} d^{2} - 30 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{2} c^{3} d - 30 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (6 \, b^{2} d^{4} x^{3} - 105 \, b^{2} c^{3} d + 55 \, a b c^{2} d^{2} - 3 \, {\left (7 \, b^{2} c d^{3} - 5 \, a b d^{4}\right )} x^{2} - 2 \, {\left (70 \, b^{2} c^{2} d^{2} - 39 \, a b c d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (b d^{7} x^{2} + 2 \, b c d^{6} x + b c^{2} d^{5}\right )}}, -\frac {3 \, {\left (35 \, b^{2} c^{4} - 30 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{2} d^{2} - 30 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{2} c^{3} d - 30 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (6 \, b^{2} d^{4} x^{3} - 105 \, b^{2} c^{3} d + 55 \, a b c^{2} d^{2} - 3 \, {\left (7 \, b^{2} c d^{3} - 5 \, a b d^{4}\right )} x^{2} - 2 \, {\left (70 \, b^{2} c^{2} d^{2} - 39 \, a b c d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{24 \, {\left (b d^{7} x^{2} + 2 \, b c d^{6} x + b c^{2} d^{5}\right )}}\right ] \]
[1/48*(3*(35*b^2*c^4 - 30*a*b*c^3*d + 3*a^2*c^2*d^2 + (35*b^2*c^2*d^2 - 30 *a*b*c*d^3 + 3*a^2*d^4)*x^2 + 2*(35*b^2*c^3*d - 30*a*b*c^2*d^2 + 3*a^2*c*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)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a* b*d^2)*x) + 4*(6*b^2*d^4*x^3 - 105*b^2*c^3*d + 55*a*b*c^2*d^2 - 3*(7*b^2*c *d^3 - 5*a*b*d^4)*x^2 - 2*(70*b^2*c^2*d^2 - 39*a*b*c*d^3)*x)*sqrt(b*x + a) *sqrt(d*x + c))/(b*d^7*x^2 + 2*b*c*d^6*x + b*c^2*d^5), -1/24*(3*(35*b^2*c^ 4 - 30*a*b*c^3*d + 3*a^2*c^2*d^2 + (35*b^2*c^2*d^2 - 30*a*b*c*d^3 + 3*a^2* d^4)*x^2 + 2*(35*b^2*c^3*d - 30*a*b*c^2*d^2 + 3*a^2*c*d^3)*x)*sqrt(-b*d)*a rctan(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*(6*b^2*d^4*x^3 - 105*b^2 *c^3*d + 55*a*b*c^2*d^2 - 3*(7*b^2*c*d^3 - 5*a*b*d^4)*x^2 - 2*(70*b^2*c^2* d^2 - 39*a*b*c*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^7*x^2 + 2*b*c*d^6 *x + b*c^2*d^5)]
\[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\int \frac {x^{2} \left (a + b x\right )^{\frac {3}{2}}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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 = 0.38 (sec) , antiderivative size = 393, normalized size of antiderivative = 1.47 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\frac {{\left ({\left (3 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (b^{6} c d^{6} - a b^{5} d^{7}\right )} {\left (b x + a\right )}}{b^{4} c d^{7} {\left | b \right |} - a b^{3} d^{8} {\left | b \right |}} - \frac {7 \, b^{7} c^{2} d^{5} - 6 \, a b^{6} c d^{6} - a^{2} b^{5} d^{7}}{b^{4} c d^{7} {\left | b \right |} - a b^{3} d^{8} {\left | b \right |}}\right )} - \frac {4 \, {\left (35 \, b^{8} c^{3} d^{4} - 65 \, a b^{7} c^{2} d^{5} + 33 \, a^{2} b^{6} c d^{6} - 3 \, a^{3} b^{5} d^{7}\right )}}{b^{4} c d^{7} {\left | b \right |} - a b^{3} d^{8} {\left | b \right |}}\right )} {\left (b x + a\right )} - \frac {3 \, {\left (35 \, b^{9} c^{4} d^{3} - 100 \, a b^{8} c^{3} d^{4} + 98 \, a^{2} b^{7} c^{2} d^{5} - 36 \, a^{3} b^{6} c d^{6} + 3 \, a^{4} b^{5} d^{7}\right )}}{b^{4} c d^{7} {\left | b \right |} - a b^{3} d^{8} {\left | b \right |}}\right )} \sqrt {b x + a}}{12 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {{\left (35 \, b^{3} c^{2} - 30 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{4 \, \sqrt {b d} d^{4} {\left | b \right |}} \]
1/12*((3*(b*x + a)*(2*(b^6*c*d^6 - a*b^5*d^7)*(b*x + a)/(b^4*c*d^7*abs(b) - a*b^3*d^8*abs(b)) - (7*b^7*c^2*d^5 - 6*a*b^6*c*d^6 - a^2*b^5*d^7)/(b^4*c *d^7*abs(b) - a*b^3*d^8*abs(b))) - 4*(35*b^8*c^3*d^4 - 65*a*b^7*c^2*d^5 + 33*a^2*b^6*c*d^6 - 3*a^3*b^5*d^7)/(b^4*c*d^7*abs(b) - a*b^3*d^8*abs(b)))*( b*x + a) - 3*(35*b^9*c^4*d^3 - 100*a*b^8*c^3*d^4 + 98*a^2*b^7*c^2*d^5 - 36 *a^3*b^6*c*d^6 + 3*a^4*b^5*d^7)/(b^4*c*d^7*abs(b) - a*b^3*d^8*abs(b)))*sqr t(b*x + a)/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 1/4*(35*b^3*c^2 - 30*a* b^2*c*d + 3*a^2*b*d^2)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b* x + a)*b*d - a*b*d)))/(sqrt(b*d)*d^4*abs(b))
Timed out. \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\int \frac {x^2\,{\left (a+b\,x\right )}^{3/2}}{{\left (c+d\,x\right )}^{5/2}} \,d x \]