Integrand size = 21, antiderivative size = 193 \[ \int \frac {x^2}{\sqrt {\frac {b+a x}{d+c x}}} \, dx=\frac {\sqrt {\frac {b+a x}{d+c x}} \left (15 b^2 c^2 d-4 a b c d^2-3 a^2 d^3+15 b^2 c^3 x-14 a b c^2 d x-a^2 c d^2 x-10 a b c^3 x^2+10 a^2 c^2 d x^2+8 a^2 c^3 x^3\right )}{24 a^3 c^2}+\frac {\left (-5 b^3 c^3+3 a b^2 c^2 d+a^2 b c d^2+a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {b+a x}{d+c x}}}{\sqrt {a}}\right )}{8 a^{7/2} c^{5/2}} \]
1/24*((a*x+b)/(c*x+d))^(1/2)*(8*a^2*c^3*x^3+10*a^2*c^2*d*x^2-10*a*b*c^3*x^ 2-a^2*c*d^2*x-14*a*b*c^2*d*x+15*b^2*c^3*x-3*a^2*d^3-4*a*b*c*d^2+15*b^2*c^2 *d)/a^3/c^2+1/8*(a^3*d^3+a^2*b*c*d^2+3*a*b^2*c^2*d-5*b^3*c^3)*arctanh(c^(1 /2)*((a*x+b)/(c*x+d))^(1/2)/a^(1/2))/a^(7/2)/c^(5/2)
Time = 10.30 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.04 \[ \int \frac {x^2}{\sqrt {\frac {b+a x}{d+c x}}} \, dx=\frac {\sqrt {c} (b+a x) \sqrt {\frac {a (d+c x)}{-b c+a d}} \left (15 b^2 c^2-2 a b c (2 d+5 c x)+a^2 \left (-3 d^2+2 c d x+8 c^2 x^2\right )\right )+3 \sqrt {-b c+a d} \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {b+a x} \text {arcsinh}\left (\frac {\sqrt {c} \sqrt {b+a x}}{\sqrt {-b c+a d}}\right )}{24 a^3 c^{5/2} \sqrt {\frac {b+a x}{d+c x}} \sqrt {\frac {a (d+c x)}{-b c+a d}}} \]
(Sqrt[c]*(b + a*x)*Sqrt[(a*(d + c*x))/(-(b*c) + a*d)]*(15*b^2*c^2 - 2*a*b* c*(2*d + 5*c*x) + a^2*(-3*d^2 + 2*c*d*x + 8*c^2*x^2)) + 3*Sqrt[-(b*c) + a* d]*(5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[b + a*x]*ArcSinh[(Sqrt[c]*Sqrt[b + a*x])/Sqrt[-(b*c) + a*d]])/(24*a^3*c^(5/2)*Sqrt[(b + a*x)/(d + c*x)]*Sq rt[(a*(d + c*x))/(-(b*c) + a*d)])
Time = 0.36 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.39, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2052, 315, 25, 298, 215, 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.
| \(\displaystyle \int \frac {x^2}{\sqrt {\frac {a x+b}{c x+d}}} \, dx\) |
| \(\Big \downarrow \) 2052 |
| \(\displaystyle -2 (b c-a d) \int \frac {\left (b-\frac {d (b+a x)}{d+c x}\right )^2}{\left (a-\frac {c (b+a x)}{d+c x}\right )^4}d\sqrt {\frac {b+a x}{d+c x}}\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle -2 (b c-a d) \left (\frac {(b c-a d) \sqrt {\frac {a x+b}{c x+d}} \left (b-\frac {d (a x+b)}{c x+d}\right )}{6 a c \left (a-\frac {c (a x+b)}{c x+d}\right )^3}-\frac {\int -\frac {b (5 b c+a d)-\frac {3 d (b c+a d) (b+a x)}{d+c x}}{\left (a-\frac {c (b+a x)}{d+c x}\right )^3}d\sqrt {\frac {b+a x}{d+c x}}}{6 a c}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 (b c-a d) \left (\frac {\int \frac {b (5 b c+a d)-\frac {3 d (b c+a d) (b+a x)}{d+c x}}{\left (a-\frac {c (b+a x)}{d+c x}\right )^3}d\sqrt {\frac {b+a x}{d+c x}}}{6 a c}+\frac {(b c-a d) \sqrt {\frac {a x+b}{c x+d}} \left (b-\frac {d (a x+b)}{c x+d}\right )}{6 a c \left (a-\frac {c (a x+b)}{c x+d}\right )^3}\right )\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle -2 (b c-a d) \left (\frac {\frac {3}{4} \left (\frac {5 b^2 c}{a}+\frac {a d^2}{c}+2 b d\right ) \int \frac {1}{\left (a-\frac {c (b+a x)}{d+c x}\right )^2}d\sqrt {\frac {b+a x}{d+c x}}+\frac {(b c-a d) (3 a d+5 b c) \sqrt {\frac {a x+b}{c x+d}}}{4 a c \left (a-\frac {c (a x+b)}{c x+d}\right )^2}}{6 a c}+\frac {(b c-a d) \sqrt {\frac {a x+b}{c x+d}} \left (b-\frac {d (a x+b)}{c x+d}\right )}{6 a c \left (a-\frac {c (a x+b)}{c x+d}\right )^3}\right )\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle -2 (b c-a d) \left (\frac {\frac {3}{4} \left (\frac {5 b^2 c}{a}+\frac {a d^2}{c}+2 b d\right ) \left (\frac {\int \frac {1}{a-\frac {c (b+a x)}{d+c x}}d\sqrt {\frac {b+a x}{d+c x}}}{2 a}+\frac {\sqrt {\frac {a x+b}{c x+d}}}{2 a \left (a-\frac {c (a x+b)}{c x+d}\right )}\right )+\frac {(b c-a d) (3 a d+5 b c) \sqrt {\frac {a x+b}{c x+d}}}{4 a c \left (a-\frac {c (a x+b)}{c x+d}\right )^2}}{6 a c}+\frac {(b c-a d) \sqrt {\frac {a x+b}{c x+d}} \left (b-\frac {d (a x+b)}{c x+d}\right )}{6 a c \left (a-\frac {c (a x+b)}{c x+d}\right )^3}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -2 (b c-a d) \left (\frac {\frac {3}{4} \left (\frac {5 b^2 c}{a}+\frac {a d^2}{c}+2 b d\right ) \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {a x+b}{c x+d}}}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {c}}+\frac {\sqrt {\frac {a x+b}{c x+d}}}{2 a \left (a-\frac {c (a x+b)}{c x+d}\right )}\right )+\frac {(b c-a d) (3 a d+5 b c) \sqrt {\frac {a x+b}{c x+d}}}{4 a c \left (a-\frac {c (a x+b)}{c x+d}\right )^2}}{6 a c}+\frac {(b c-a d) \sqrt {\frac {a x+b}{c x+d}} \left (b-\frac {d (a x+b)}{c x+d}\right )}{6 a c \left (a-\frac {c (a x+b)}{c x+d}\right )^3}\right )\) |
-2*(b*c - a*d)*(((b*c - a*d)*Sqrt[(b + a*x)/(d + c*x)]*(b - (d*(b + a*x))/ (d + c*x)))/(6*a*c*(a - (c*(b + a*x))/(d + c*x))^3) + (((b*c - a*d)*(5*b*c + 3*a*d)*Sqrt[(b + a*x)/(d + c*x)])/(4*a*c*(a - (c*(b + a*x))/(d + c*x))^ 2) + (3*((5*b^2*c)/a + 2*b*d + (a*d^2)/c)*(Sqrt[(b + a*x)/(d + c*x)]/(2*a* (a - (c*(b + a*x))/(d + c*x))) + ArcTanh[(Sqrt[c]*Sqrt[(b + a*x)/(d + c*x) ])/Sqrt[a]]/(2*a^(3/2)*Sqrt[c])))/4)/(6*a*c))
3.24.96.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
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]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_S ymbol] :> With[{q = Denominator[p]}, Simp[q*e*(b*c - a*d) Subst[Int[x^(q* (p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a + b* x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(587\) vs. \(2(177)=354\).
Time = 0.09 (sec) , antiderivative size = 588, normalized size of antiderivative = 3.05
| method | result | size |
| default | \(\frac {\left (a x +b \right ) \left (-12 \sqrt {a c}\, \sqrt {a c \,x^{2}+a d x +c x b +b d}\, a^{2} c d x -36 \sqrt {a c}\, \sqrt {a c \,x^{2}+a d x +c x b +b d}\, a b \,c^{2} x +3 \ln \left (\frac {2 a c x +2 \sqrt {a c \,x^{2}+a d x +c x b +b d}\, \sqrt {a c}+a d +b c}{2 \sqrt {a c}}\right ) a^{3} d^{3}+3 \ln \left (\frac {2 a c x +2 \sqrt {a c \,x^{2}+a d x +c x b +b d}\, \sqrt {a c}+a d +b c}{2 \sqrt {a c}}\right ) a^{2} b c \,d^{2}-15 \ln \left (\frac {2 a c x +2 \sqrt {a c \,x^{2}+a d x +c x b +b d}\, \sqrt {a c}+a d +b c}{2 \sqrt {a c}}\right ) a \,b^{2} c^{2} d +9 \ln \left (\frac {2 a c x +2 \sqrt {a c \,x^{2}+a d x +c x b +b d}\, \sqrt {a c}+a d +b c}{2 \sqrt {a c}}\right ) b^{3} c^{3}+24 \ln \left (\frac {2 a c x +2 \sqrt {\left (c x +d \right ) \left (a x +b \right )}\, \sqrt {a c}+a d +b c}{2 \sqrt {a c}}\right ) a \,b^{2} c^{2} d -24 \ln \left (\frac {2 a c x +2 \sqrt {\left (c x +d \right ) \left (a x +b \right )}\, \sqrt {a c}+a d +b c}{2 \sqrt {a c}}\right ) b^{3} c^{3}+48 \sqrt {\left (c x +d \right ) \left (a x +b \right )}\, \sqrt {a c}\, b^{2} c^{2}+16 \left (a c \,x^{2}+a d x +c x b +b d \right )^{\frac {3}{2}} a c \sqrt {a c}-6 \sqrt {a c}\, \sqrt {a c \,x^{2}+a d x +c x b +b d}\, a^{2} d^{2}-24 \sqrt {a c}\, \sqrt {a c \,x^{2}+a d x +c x b +b d}\, a b c d -18 \sqrt {a c}\, \sqrt {a c \,x^{2}+a d x +c x b +b d}\, b^{2} c^{2}\right )}{48 a^{3} \sqrt {\frac {a x +b}{c x +d}}\, \sqrt {\left (c x +d \right ) \left (a x +b \right )}\, c^{2} \sqrt {a c}}\) | \(588\) |
1/48*(a*x+b)/a^3*(-12*(a*c)^(1/2)*(a*c*x^2+a*d*x+b*c*x+b*d)^(1/2)*a^2*c*d* x-36*(a*c)^(1/2)*(a*c*x^2+a*d*x+b*c*x+b*d)^(1/2)*a*b*c^2*x+3*ln(1/2*(2*a*c *x+2*(a*c*x^2+a*d*x+b*c*x+b*d)^(1/2)*(a*c)^(1/2)+a*d+b*c)/(a*c)^(1/2))*a^3 *d^3+3*ln(1/2*(2*a*c*x+2*(a*c*x^2+a*d*x+b*c*x+b*d)^(1/2)*(a*c)^(1/2)+a*d+b *c)/(a*c)^(1/2))*a^2*b*c*d^2-15*ln(1/2*(2*a*c*x+2*(a*c*x^2+a*d*x+b*c*x+b*d )^(1/2)*(a*c)^(1/2)+a*d+b*c)/(a*c)^(1/2))*a*b^2*c^2*d+9*ln(1/2*(2*a*c*x+2* (a*c*x^2+a*d*x+b*c*x+b*d)^(1/2)*(a*c)^(1/2)+a*d+b*c)/(a*c)^(1/2))*b^3*c^3+ 24*ln(1/2*(2*a*c*x+2*((c*x+d)*(a*x+b))^(1/2)*(a*c)^(1/2)+a*d+b*c)/(a*c)^(1 /2))*a*b^2*c^2*d-24*ln(1/2*(2*a*c*x+2*((c*x+d)*(a*x+b))^(1/2)*(a*c)^(1/2)+ a*d+b*c)/(a*c)^(1/2))*b^3*c^3+48*((c*x+d)*(a*x+b))^(1/2)*(a*c)^(1/2)*b^2*c ^2+16*(a*c*x^2+a*d*x+b*c*x+b*d)^(3/2)*a*c*(a*c)^(1/2)-6*(a*c)^(1/2)*(a*c*x ^2+a*d*x+b*c*x+b*d)^(1/2)*a^2*d^2-24*(a*c)^(1/2)*(a*c*x^2+a*d*x+b*c*x+b*d) ^(1/2)*a*b*c*d-18*(a*c)^(1/2)*(a*c*x^2+a*d*x+b*c*x+b*d)^(1/2)*b^2*c^2)/((a *x+b)/(c*x+d))^(1/2)/((c*x+d)*(a*x+b))^(1/2)/c^2/(a*c)^(1/2)
Time = 0.32 (sec) , antiderivative size = 419, normalized size of antiderivative = 2.17 \[ \int \frac {x^2}{\sqrt {\frac {b+a x}{d+c x}}} \, dx=\left [-\frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a c} \log \left (-2 \, a c x - b c - a d - 2 \, \sqrt {a c} {\left (c x + d\right )} \sqrt {\frac {a x + b}{c x + d}}\right ) - 2 \, {\left (8 \, a^{3} c^{4} x^{3} + 15 \, a b^{2} c^{3} d - 4 \, a^{2} b c^{2} d^{2} - 3 \, a^{3} c d^{3} - 10 \, {\left (a^{2} b c^{4} - a^{3} c^{3} d\right )} x^{2} + {\left (15 \, a b^{2} c^{4} - 14 \, a^{2} b c^{3} d - a^{3} c^{2} d^{2}\right )} x\right )} \sqrt {\frac {a x + b}{c x + d}}}{48 \, a^{4} c^{3}}, \frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} {\left (c x + d\right )} \sqrt {\frac {a x + b}{c x + d}}}{a c x + b c}\right ) + {\left (8 \, a^{3} c^{4} x^{3} + 15 \, a b^{2} c^{3} d - 4 \, a^{2} b c^{2} d^{2} - 3 \, a^{3} c d^{3} - 10 \, {\left (a^{2} b c^{4} - a^{3} c^{3} d\right )} x^{2} + {\left (15 \, a b^{2} c^{4} - 14 \, a^{2} b c^{3} d - a^{3} c^{2} d^{2}\right )} x\right )} \sqrt {\frac {a x + b}{c x + d}}}{24 \, a^{4} c^{3}}\right ] \]
[-1/48*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*sqrt(a*c)*lo g(-2*a*c*x - b*c - a*d - 2*sqrt(a*c)*(c*x + d)*sqrt((a*x + b)/(c*x + d))) - 2*(8*a^3*c^4*x^3 + 15*a*b^2*c^3*d - 4*a^2*b*c^2*d^2 - 3*a^3*c*d^3 - 10*( a^2*b*c^4 - a^3*c^3*d)*x^2 + (15*a*b^2*c^4 - 14*a^2*b*c^3*d - a^3*c^2*d^2) *x)*sqrt((a*x + b)/(c*x + d)))/(a^4*c^3), 1/24*(3*(5*b^3*c^3 - 3*a*b^2*c^2 *d - a^2*b*c*d^2 - a^3*d^3)*sqrt(-a*c)*arctan(sqrt(-a*c)*(c*x + d)*sqrt((a *x + b)/(c*x + d))/(a*c*x + b*c)) + (8*a^3*c^4*x^3 + 15*a*b^2*c^3*d - 4*a^ 2*b*c^2*d^2 - 3*a^3*c*d^3 - 10*(a^2*b*c^4 - a^3*c^3*d)*x^2 + (15*a*b^2*c^4 - 14*a^2*b*c^3*d - a^3*c^2*d^2)*x)*sqrt((a*x + b)/(c*x + d)))/(a^4*c^3)]
\[ \int \frac {x^2}{\sqrt {\frac {b+a x}{d+c x}}} \, dx=\int \frac {x^{2}}{\sqrt {\frac {a x + b}{c x + d}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.84 \[ \int \frac {x^2}{\sqrt {\frac {b+a x}{d+c x}}} \, dx=-\frac {3 \, {\left (5 \, b^{3} c^{5} - 3 \, a b^{2} c^{4} d - a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )} \left (\frac {a x + b}{c x + d}\right )^{\frac {5}{2}} - 8 \, {\left (5 \, a b^{3} c^{4} - 3 \, a^{2} b^{2} c^{3} d - 3 \, a^{3} b c^{2} d^{2} + a^{4} c d^{3}\right )} \left (\frac {a x + b}{c x + d}\right )^{\frac {3}{2}} + 3 \, {\left (11 \, a^{2} b^{3} c^{3} - 13 \, a^{3} b^{2} c^{2} d + a^{4} b c d^{2} + a^{5} d^{3}\right )} \sqrt {\frac {a x + b}{c x + d}}}{24 \, {\left (a^{6} c^{2} - \frac {3 \, {\left (a x + b\right )} a^{5} c^{3}}{c x + d} + \frac {3 \, {\left (a x + b\right )}^{2} a^{4} c^{4}}{{\left (c x + d\right )}^{2}} - \frac {{\left (a x + b\right )}^{3} a^{3} c^{5}}{{\left (c x + d\right )}^{3}}\right )}} + \frac {{\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (\frac {c \sqrt {\frac {a x + b}{c x + d}} - \sqrt {a c}}{c \sqrt {\frac {a x + b}{c x + d}} + \sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c^{2}} \]
-1/24*(3*(5*b^3*c^5 - 3*a*b^2*c^4*d - a^2*b*c^3*d^2 - a^3*c^2*d^3)*((a*x + b)/(c*x + d))^(5/2) - 8*(5*a*b^3*c^4 - 3*a^2*b^2*c^3*d - 3*a^3*b*c^2*d^2 + a^4*c*d^3)*((a*x + b)/(c*x + d))^(3/2) + 3*(11*a^2*b^3*c^3 - 13*a^3*b^2* c^2*d + a^4*b*c*d^2 + a^5*d^3)*sqrt((a*x + b)/(c*x + d)))/(a^6*c^2 - 3*(a* x + b)*a^5*c^3/(c*x + d) + 3*(a*x + b)^2*a^4*c^4/(c*x + d)^2 - (a*x + b)^3 *a^3*c^5/(c*x + d)^3) + 1/16*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^ 3*d^3)*log((c*sqrt((a*x + b)/(c*x + d)) - sqrt(a*c))/(c*sqrt((a*x + b)/(c* x + d)) + sqrt(a*c)))/(sqrt(a*c)*a^3*c^2)
Time = 0.34 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.20 \[ \int \frac {x^2}{\sqrt {\frac {b+a x}{d+c x}}} \, dx=\frac {1}{24} \, \sqrt {a c x^{2} + b c x + a d x + b d} {\left (2 \, x {\left (\frac {4 \, x}{a \mathrm {sgn}\left (c x + d\right )} - \frac {5 \, a b c^{2} \mathrm {sgn}\left (c x + d\right ) - a^{2} c d \mathrm {sgn}\left (c x + d\right )}{a^{3} c^{2}}\right )} + \frac {15 \, b^{2} c^{2} \mathrm {sgn}\left (c x + d\right ) - 4 \, a b c d \mathrm {sgn}\left (c x + d\right ) - 3 \, a^{2} d^{2} \mathrm {sgn}\left (c x + d\right )}{a^{3} c^{2}}\right )} + \frac {{\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left ({\left | -b c - a d - 2 \, \sqrt {a c} {\left (\sqrt {a c} x - \sqrt {a c x^{2} + b c x + a d x + b d}\right )} \right |}\right )}{16 \, \sqrt {a c} a^{3} c^{2} \mathrm {sgn}\left (c x + d\right )} \]
1/24*sqrt(a*c*x^2 + b*c*x + a*d*x + b*d)*(2*x*(4*x/(a*sgn(c*x + d)) - (5*a *b*c^2*sgn(c*x + d) - a^2*c*d*sgn(c*x + d))/(a^3*c^2)) + (15*b^2*c^2*sgn(c *x + d) - 4*a*b*c*d*sgn(c*x + d) - 3*a^2*d^2*sgn(c*x + d))/(a^3*c^2)) + 1/ 16*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*log(abs(-b*c - a*d - 2*sqrt(a*c)*(sqrt(a*c)*x - sqrt(a*c*x^2 + b*c*x + a*d*x + b*d))))/(sqrt( a*c)*a^3*c^2*sgn(c*x + d))
Time = 0.85 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.58 \[ \int \frac {x^2}{\sqrt {\frac {b+a x}{d+c x}}} \, dx=\frac {\frac {{\left (\frac {b+a\,x}{d+c\,x}\right )}^{5/2}\,\left (\frac {a^3\,d^3}{8}+\frac {a^2\,b\,c\,d^2}{8}+\frac {3\,a\,b^2\,c^2\,d}{8}-\frac {5\,b^3\,c^3}{8}\right )}{a^6}+\frac {{\left (\frac {b+a\,x}{d+c\,x}\right )}^{3/2}\,\left (\frac {a^3\,d^3}{3}-a^2\,b\,c\,d^2-a\,b^2\,c^2\,d+\frac {5\,b^3\,c^3}{3}\right )}{a^5\,c}-\frac {\sqrt {\frac {b+a\,x}{d+c\,x}}\,\left (\frac {a^3\,d^3}{8}+\frac {a^2\,b\,c\,d^2}{8}-\frac {13\,a\,b^2\,c^2\,d}{8}+\frac {11\,b^3\,c^3}{8}\right )}{a^4\,c^2}}{\frac {3\,c^2\,{\left (b+a\,x\right )}^2}{a^2\,{\left (d+c\,x\right )}^2}-\frac {c^3\,{\left (b+a\,x\right )}^3}{a^3\,{\left (d+c\,x\right )}^3}-\frac {3\,c\,\left (b+a\,x\right )}{a\,\left (d+c\,x\right )}+1}+\frac {\mathrm {atanh}\left (\frac {\sqrt {c}\,\sqrt {\frac {b+a\,x}{d+c\,x}}}{\sqrt {a}}\right )\,\left (a\,d-b\,c\right )\,\left (a^2\,d^2+2\,a\,b\,c\,d+5\,b^2\,c^2\right )}{8\,a^{7/2}\,c^{5/2}} \]
((((b + a*x)/(d + c*x))^(5/2)*((a^3*d^3)/8 - (5*b^3*c^3)/8 + (3*a*b^2*c^2* d)/8 + (a^2*b*c*d^2)/8))/a^6 + (((b + a*x)/(d + c*x))^(3/2)*((a^3*d^3)/3 + (5*b^3*c^3)/3 - a*b^2*c^2*d - a^2*b*c*d^2))/(a^5*c) - (((b + a*x)/(d + c* x))^(1/2)*((a^3*d^3)/8 + (11*b^3*c^3)/8 - (13*a*b^2*c^2*d)/8 + (a^2*b*c*d^ 2)/8))/(a^4*c^2))/((3*c^2*(b + a*x)^2)/(a^2*(d + c*x)^2) - (c^3*(b + a*x)^ 3)/(a^3*(d + c*x)^3) - (3*c*(b + a*x))/(a*(d + c*x)) + 1) + (atanh((c^(1/2 )*((b + a*x)/(d + c*x))^(1/2))/a^(1/2))*(a*d - b*c)*(a^2*d^2 + 5*b^2*c^2 + 2*a*b*c*d))/(8*a^(7/2)*c^(5/2))