Integrand size = 22, antiderivative size = 198 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^2} \, dx=\frac {d (11 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b}+\frac {3}{2} d \sqrt {a+b x} (c+d x)^{3/2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{x}-\frac {c^{3/2} (b c+5 a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {\sqrt {d} \left (15 b^2 c^2+10 a b c d-a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{3/2}} \]
-c^(3/2)*(5*a*d+b*c)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/ a^(1/2)+1/4*(-a^2*d^2+10*a*b*c*d+15*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2) /b^(1/2)/(d*x+c)^(1/2))*d^(1/2)/b^(3/2)+3/2*d*(d*x+c)^(3/2)*(b*x+a)^(1/2)- (d*x+c)^(5/2)*(b*x+a)^(1/2)/x+1/4*d*(a*d+11*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/ 2)/b
Time = 0.78 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^2} \, dx=\frac {1}{4} \left (\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-4 b c^2+9 b c d x+a d^2 x+2 b d^2 x^2\right )}{b x}-\frac {4 c^{3/2} (b c+5 a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {\sqrt {d} \left (15 b^2 c^2+10 a b c d-a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}}\right ) \]
((Sqrt[a + b*x]*Sqrt[c + d*x]*(-4*b*c^2 + 9*b*c*d*x + a*d^2*x + 2*b*d^2*x^ 2))/(b*x) - (4*c^(3/2)*(b*c + 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt [a]*Sqrt[c + d*x])])/Sqrt[a] + (Sqrt[d]*(15*b^2*c^2 + 10*a*b*c*d - a^2*d^2 )*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/b^(3/2))/4
Time = 0.36 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {108, 27, 171, 27, 171, 27, 175, 66, 104, 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.
| \(\displaystyle \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^2} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \int \frac {(c+d x)^{3/2} (b c+5 a d+6 b d x)}{2 x \sqrt {a+b x}}dx-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \frac {(c+d x)^{3/2} (b c+5 a d+6 b d x)}{x \sqrt {a+b x}}dx-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{x}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {b \sqrt {c+d x} (2 c (b c+5 a d)+d (11 b c+a d) x)}{x \sqrt {a+b x}}dx}{2 b}+3 d \sqrt {a+b x} (c+d x)^{3/2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {\sqrt {c+d x} (2 c (b c+5 a d)+d (11 b c+a d) x)}{x \sqrt {a+b x}}dx+3 d \sqrt {a+b x} (c+d x)^{3/2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{x}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {4 b (b c+5 a d) c^2+d \left (15 b^2 c^2+10 a b d c-a^2 d^2\right ) x}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{b}\right )+3 d \sqrt {a+b x} (c+d x)^{3/2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {4 b (b c+5 a d) c^2+d \left (15 b^2 c^2+10 a b d c-a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{b}\right )+3 d \sqrt {a+b x} (c+d x)^{3/2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{x}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {d \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx+4 b c^2 (5 a d+b c) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{b}\right )+3 d \sqrt {a+b x} (c+d x)^{3/2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{x}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {2 d \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+4 b c^2 (5 a d+b c) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{b}\right )+3 d \sqrt {a+b x} (c+d x)^{3/2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{x}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {2 d \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+8 b c^2 (5 a d+b c) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{b}\right )+3 d \sqrt {a+b x} (c+d x)^{3/2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{x}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\frac {2 \sqrt {d} \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}}-\frac {8 b c^{3/2} (5 a d+b c) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{b}\right )+3 d \sqrt {a+b x} (c+d x)^{3/2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{x}\) |
-((Sqrt[a + b*x]*(c + d*x)^(5/2))/x) + (3*d*Sqrt[a + b*x]*(c + d*x)^(3/2) + ((d*(11*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/b + ((-8*b*c^(3/2)*(b*c + 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[a] + (2*Sqrt[d]*(15*b^2*c^2 + 10*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[b])/(2*b))/2)/2
3.6.70.3.1 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 _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) , x] - Simp[1/(b*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c , d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (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_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2*n, 2*p]
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ )))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b Int[(c + d*x)^n*(e + f*x)^p, x] , x] + Simp[(b*g - a*h)/b Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
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. \(432\) vs. \(2(156)=312\).
Time = 0.54 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.19
| method | result | size |
| default | \(-\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (\ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} d^{3} x \sqrt {a c}-10 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c \,d^{2} x \sqrt {a c}-15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2} d x \sqrt {a c}+20 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a b \,c^{2} d x \sqrt {b d}+4 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) b^{2} c^{3} x \sqrt {b d}-4 b \,d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}-2 a \,d^{2} x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}-18 b c d x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}+8 b \,c^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\right )}{8 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b x}\) | \(433\) |
-1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(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^3*x*(a*c)^(1/2)-10*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^2*x *(a*c)^(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))*b^2*c^2*d*x*(a*c)^(1/2)+20*ln((a*d*x+b*c*x+2*(a*c)^(1/2) *((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*a*b*c^2*d*x*(b*d)^(1/2)+4*ln((a*d*x+b*c *x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*b^2*c^3*x*(b*d)^(1/2)-4 *b*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-2*a*d^2*x*((b*x +a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-18*b*c*d*x*((b*x+a)*(d*x+c))^(1 /2)*(b*d)^(1/2)*(a*c)^(1/2)+8*b*c^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a *c)^(1/2))/(b*d)^(1/2)/(a*c)^(1/2)/((b*x+a)*(d*x+c))^(1/2)/b/x
Time = 1.55 (sec) , antiderivative size = 1079, normalized size of antiderivative = 5.45 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^2} \, dx=\text {Too large to display} \]
[-1/16*((15*b^2*c^2 + 10*a*b*c*d - a^2*d^2)*x*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(b^2*c^2 + 5*a* b*c*d)*x*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8* (a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*b*d^2*x^2 - 4*b*c^2 + (9*b*c*d + a*d^2) *x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*x), -1/8*((15*b^2*c^2 + 10*a*b*c*d - a ^2*d^2)*x*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d *x + c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) - 2*(b^2*c^2 + 5*a*b*c*d)*x*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x ^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 2*(2*b*d^2*x^2 - 4*b*c^2 + (9*b*c*d + a *d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*x), 1/16*(8*(b^2*c^2 + 5*a*b*c*d) *x*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) - (15*b^2*c^2 + 10* a*b*c*d - a^2*d^2)*x*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a ^2*d^2 - 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/ b) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(2*b*d^2*x^2 - 4*b*c^2 + (9*b*c*d + a*d^ 2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*x), 1/8*(4*(b^2*c^2 + 5*a*b*c*d)*x*s qrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c...
\[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^2} \, dx=\int \frac {\sqrt {a + b x} \left (c + d x\right )^{\frac {5}{2}}}{x^{2}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^2} \, dx=\text {Exception raised: ValueError} \]
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. 597 vs. \(2 (156) = 312\).
Time = 0.60 (sec) , antiderivative size = 597, normalized size of antiderivative = 3.02 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^2} \, dx=\frac {2 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} d^{2} {\left | b \right |}}{b^{2}} + \frac {9 \, b^{3} c d^{3} {\left | b \right |} - a b^{2} d^{4} {\left | b \right |}}{b^{4} d^{2}}\right )} - \frac {8 \, {\left (\sqrt {b d} b^{2} c^{3} {\left | b \right |} + 5 \, \sqrt {b d} a b c^{2} d {\left | b \right |}\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} b} - \frac {16 \, {\left (\sqrt {b d} b^{4} c^{4} {\left | b \right |} - 2 \, \sqrt {b d} a b^{3} c^{3} d {\left | b \right |} + \sqrt {b d} a^{2} b^{2} c^{2} d^{2} {\left | b \right |} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c^{3} {\left | b \right |} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b c^{2} d {\left | b \right |}\right )}}{b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4}} - \frac {{\left (15 \, \sqrt {b d} b^{2} c^{2} {\left | b \right |} + 10 \, \sqrt {b d} a b c d {\left | b \right |} - \sqrt {b d} a^{2} d^{2} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{b^{2}}}{8 \, b} \]
1/8*(2*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*d^2* abs(b)/b^2 + (9*b^3*c*d^3*abs(b) - a*b^2*d^4*abs(b))/(b^4*d^2)) - 8*(sqrt( b*d)*b^2*c^3*abs(b) + 5*sqrt(b*d)*a*b*c^2*d*abs(b))*arctan(-1/2*(b^2*c + a *b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/ (sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b) - 16*(sqrt(b*d)*b^4*c^4*abs(b) - 2* sqrt(b*d)*a*b^3*c^3*d*abs(b) + sqrt(b*d)*a^2*b^2*c^2*d^2*abs(b) - sqrt(b*d )*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c^ 3*abs(b) - sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*c^2*d*abs(b))/(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(s qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2* (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4) - (15*s qrt(b*d)*b^2*c^2*abs(b) + 10*sqrt(b*d)*a*b*c*d*abs(b) - sqrt(b*d)*a^2*d^2* abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d) )^2)/b^2)/b
Timed out. \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^2} \, dx=\int \frac {\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{5/2}}{x^2} \,d x \]