Integrand size = 22, antiderivative size = 171 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^3} \, dx=-\frac {(3 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{4 c x}-\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 x^2}-\frac {\left (3 b^2 c^2+6 a b c d-a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a} c^{3/2}}+2 b^{3/2} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right ) \]
-1/4*(-a^2*d^2+6*a*b*c*d+3*b^2*c^2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/ (d*x+c)^(1/2))/c^(3/2)/a^(1/2)+2*b^(3/2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^( 1/2)/(d*x+c)^(1/2))*d^(1/2)-1/2*(b*x+a)^(3/2)*(d*x+c)^(1/2)/x^2-1/4*(a*d+3 *b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c/x
Time = 0.49 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^3} \, dx=\frac {1}{4} \left (-\frac {\sqrt {a+b x} \sqrt {c+d x} (2 a c+5 b c x+a d x)}{c x^2}+\frac {\left (-3 b^2 c^2-6 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{\sqrt {a} c^{3/2}}+8 b^{3/2} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )\right ) \]
(-((Sqrt[a + b*x]*Sqrt[c + d*x]*(2*a*c + 5*b*c*x + a*d*x))/(c*x^2)) + ((-3 *b^2*c^2 - 6*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*S qrt[a + b*x])])/(Sqrt[a]*c^(3/2)) + 8*b^(3/2)*Sqrt[d]*ArcTanh[(Sqrt[b]*Sqr t[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/4
Time = 0.28 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {108, 27, 166, 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.
| \(\displaystyle \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^3} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt {a+b x} (3 b c+a d+4 b d x)}{2 x^2 \sqrt {c+d x}}dx-\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \frac {\sqrt {a+b x} (3 b c+a d+4 b d x)}{x^2 \sqrt {c+d x}}dx-\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {1}{4} \left (\frac {\int \frac {3 c^2 b^2+8 c d x b^2+6 a c d b-a^2 d^2}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{c}-\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{c x}\right )-\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {\int \frac {3 c^2 b^2+8 c d x b^2+6 a c d b-a^2 d^2}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 c}-\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{c x}\right )-\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {1}{4} \left (\frac {\left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+8 b^2 c d \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 c}-\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{c x}\right )-\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {1}{4} \left (\frac {\left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+16 b^2 c d \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 c}-\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{c x}\right )-\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{4} \left (\frac {2 \left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+16 b^2 c d \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 c}-\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{c x}\right )-\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{4} \left (\frac {16 b^{3/2} c \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )-\frac {2 \left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} \sqrt {c}}}{2 c}-\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{c x}\right )-\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 x^2}\) |
-1/2*((a + b*x)^(3/2)*Sqrt[c + d*x])/x^2 + (-(((3*b*c + a*d)*Sqrt[a + b*x] *Sqrt[c + d*x])/(c*x)) + ((-2*(3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*ArcTanh[(S qrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(Sqrt[a]*Sqrt[c]) + 16*b^( 3/2)*c*Sqrt[d]*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/( 2*c))/4
3.7.3.3.1 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[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h )*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
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. \(343\) vs. \(2(133)=266\).
Time = 0.54 (sec) , antiderivative size = 344, normalized size of antiderivative = 2.01
| method | result | size |
| default | \(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (\ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} d^{2} x^{2} \sqrt {b d}-6 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a b c d \,x^{2} \sqrt {b d}-3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) b^{2} c^{2} x^{2} \sqrt {b d}+8 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c d \,x^{2} \sqrt {a c}-2 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a d x -10 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b c x -4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a c \sqrt {b d}\, \sqrt {a c}\right )}{8 c \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, x^{2} \sqrt {b d}\, \sqrt {a c}}\) | \(344\) |
1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c*(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^2*d^2*x^2*(b*d)^(1/2)-6*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*d*x^2*(b*d)^(1/2)-3*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^2*x^2*(b*d) ^(1/2)+8*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*d*x^2*(a*c)^(1/2)-2*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x +c))^(1/2)*a*d*x-10*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b*c*x- 4*((b*x+a)*(d*x+c))^(1/2)*a*c*(b*d)^(1/2)*(a*c)^(1/2))/((b*x+a)*(d*x+c))^( 1/2)/x^2/(b*d)^(1/2)/(a*c)^(1/2)
Time = 0.66 (sec) , antiderivative size = 1027, normalized size of antiderivative = 6.01 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^3} \, dx=\left [\frac {8 \, \sqrt {b d} a b c^{2} x^{2} \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 ) - {\left (3 \, b^{2} c^{2} + 6 \, a b c d - a^{2} d^{2}\right )} \sqrt {a c} x^{2} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (2 \, a^{2} c^{2} + {\left (5 \, a b c^{2} + a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, a c^{2} x^{2}}, -\frac {16 \, \sqrt {-b d} a b c^{2} x^{2} \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 ) + {\left (3 \, b^{2} c^{2} + 6 \, a b c d - a^{2} d^{2}\right )} \sqrt {a c} x^{2} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (2 \, a^{2} c^{2} + {\left (5 \, a b c^{2} + a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, a c^{2} x^{2}}, \frac {4 \, \sqrt {b d} a b c^{2} x^{2} \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 ) + {\left (3 \, b^{2} c^{2} + 6 \, a b c d - a^{2} d^{2}\right )} \sqrt {-a c} x^{2} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (2 \, a^{2} c^{2} + {\left (5 \, a b c^{2} + a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, a c^{2} x^{2}}, -\frac {8 \, \sqrt {-b d} a b c^{2} x^{2} \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 ) - {\left (3 \, b^{2} c^{2} + 6 \, a b c d - a^{2} d^{2}\right )} \sqrt {-a c} x^{2} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) + 2 \, {\left (2 \, a^{2} c^{2} + {\left (5 \, a b c^{2} + a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, a c^{2} x^{2}}\right ] \]
[1/16*(8*sqrt(b*d)*a*b*c^2*x^2*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) - (3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*sqrt(a*c)*x^2* log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 + 4*(2*a*c + (b*c + a *d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^ 2) - 4*(2*a^2*c^2 + (5*a*b*c^2 + a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/ (a*c^2*x^2), -1/16*(16*sqrt(-b*d)*a*b*c^2*x^2*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)) + (3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*sqrt(a*c)*x^2*log(( 8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 + 4*(2*a*c + (b*c + a*d)*x )*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(2*a^2*c^2 + (5*a*b*c^2 + a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^ 2*x^2), 1/8*(4*sqrt(b*d)*a*b*c^2*x^2*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) + (3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*sqrt(-a* c)*x^2*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d* x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) - 2*(2*a^2*c^2 + ( 5*a*b*c^2 + a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^2*x^2), -1/8*(8* sqrt(-b*d)*a*b*c^2*x^2*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)) -...
\[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^3} \, dx=\int \frac {\left (a + b x\right )^{\frac {3}{2}} \sqrt {c + d x}}{x^{3}}\, dx \]
Exception generated. \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^3} \, dx=\text {Exception raised: ValueError} \]
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. 1127 vs. \(2 (133) = 266\).
Time = 0.80 (sec) , antiderivative size = 1127, normalized size of antiderivative = 6.59 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^3} \, dx=\text {Too large to display} \]
-1/4*(4*sqrt(b*d)*b*abs(b)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b* x + a)*b*d - a*b*d))^2) + (3*sqrt(b*d)*b^3*c^2*abs(b) + 6*sqrt(b*d)*a*b^2* c*d*abs(b) - sqrt(b*d)*a^2*b*d^2*abs(b))*arctan(-1/2*(b^2*c + a*b*d - (sqr t(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*c) + 2*(5*sqrt(b*d)*b^9*c^5*abs(b) - 19*sqrt(b* d)*a*b^8*c^4*d*abs(b) + 26*sqrt(b*d)*a^2*b^7*c^3*d^2*abs(b) - 14*sqrt(b*d) *a^3*b^6*c^2*d^3*abs(b) + sqrt(b*d)*a^4*b^5*c*d^4*abs(b) + sqrt(b*d)*a^5*b ^4*d^5*abs(b) - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^7*c^4*abs(b) + 16*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^6*c^3*d*abs(b) + 10*sqrt(b* d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b ^5*c^2*d^2*abs(b) - 8*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b *x + a)*b*d - a*b*d))^2*a^3*b^4*c*d^3*abs(b) - 3*sqrt(b*d)*(sqrt(b*d)*sqrt (b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^3*d^4*abs(b) + 15 *sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)) ^4*b^5*c^3*abs(b) + 13*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + ( b*x + a)*b*d - a*b*d))^4*a*b^4*c^2*d*abs(b) + 17*sqrt(b*d)*(sqrt(b*d)*sqrt (b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^3*c*d^2*abs(b) + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d) )^4*a^3*b^2*d^3*abs(b) - 5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^...
Timed out. \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^3} \, dx=\int \frac {{\left (a+b\,x\right )}^{3/2}\,\sqrt {c+d\,x}}{x^3} \,d x \]