Integrand size = 22, antiderivative size = 212 \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^3} \, dx=\frac {b (11 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{4 c}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{4 c x}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}-\frac {\sqrt {a} \left (15 b^2 c^2+10 a b c d-a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 c^{3/2}}+\frac {b^{3/2} (b c+5 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {d}} \]
-1/4*(-a^2*d^2+10*a*b*c*d+15*b^2*c^2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2 )/(d*x+c)^(1/2))*a^(1/2)/c^(3/2)+b^(3/2)*(5*a*d+b*c)*arctanh(d^(1/2)*(b*x+ a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/d^(1/2)-1/4*(a*d+5*b*c)*(b*x+a)^(3/2)*(d*x +c)^(1/2)/c/x-1/2*(b*x+a)^(5/2)*(d*x+c)^(1/2)/x^2+1/4*b*(a*d+11*b*c)*(b*x+ a)^(1/2)*(d*x+c)^(1/2)/c
Time = 0.77 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.83 \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^3} \, dx=\frac {\sqrt {a} \left (-15 b^2 c^2-10 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )+\sqrt {c} \left (-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (9 a b c x-4 b^2 c x^2+a^2 (2 c+d x)\right )}{x^2}+\frac {4 b^{3/2} c (b c+5 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{\sqrt {d}}\right )}{4 c^{3/2}} \]
(Sqrt[a]*(-15*b^2*c^2 - 10*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[a]*Sqrt[c + d* x])/(Sqrt[c]*Sqrt[a + b*x])] + Sqrt[c]*(-((Sqrt[a + b*x]*Sqrt[c + d*x]*(9* a*b*c*x - 4*b^2*c*x^2 + a^2*(2*c + d*x)))/x^2) + (4*b^(3/2)*c*(b*c + 5*a*d )*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/Sqrt[d]))/(4*c ^(3/2))
Time = 0.35 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {108, 27, 166, 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.
| \(\displaystyle \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^3} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {1}{2} \int \frac {(a+b x)^{3/2} (5 b c+a d+6 b d x)}{2 x^2 \sqrt {c+d x}}dx-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \frac {(a+b x)^{3/2} (5 b c+a d+6 b d x)}{x^2 \sqrt {c+d x}}dx-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {1}{4} \left (\frac {\int \frac {\sqrt {a+b x} \left (15 b^2 c^2+10 a b d c-a^2 d^2+2 b d (11 b c+a d) x\right )}{2 x \sqrt {c+d x}}dx}{c}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {\int \frac {\sqrt {a+b x} \left (15 b^2 c^2+10 a b d c-a^2 d^2+2 b d (11 b c+a d) x\right )}{x \sqrt {c+d x}}dx}{2 c}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{4} \left (\frac {\frac {\int \frac {d \left (4 c (b c+5 a d) x b^2+a \left (15 b^2 c^2+10 a b d c-a^2 d^2\right )\right )}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{d}+2 b \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{2 c}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {\int \frac {4 c (b c+5 a d) x b^2+a \left (15 b^2 c^2+10 a b d c-a^2 d^2\right )}{x \sqrt {a+b x} \sqrt {c+d x}}dx+2 b \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{2 c}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {1}{4} \left (\frac {a \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+4 b^2 c (5 a d+b c) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx+2 b \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{2 c}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {1}{4} \left (\frac {a \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+8 b^2 c (5 a d+b c) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+2 b \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{2 c}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{4} \left (\frac {2 a \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+8 b^2 c (5 a d+b c) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+2 b \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{2 c}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{4} \left (\frac {-\frac {2 \sqrt {a} \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {c}}+\frac {8 b^{3/2} c (5 a d+b c) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {d}}+2 b \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{2 c}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}\) |
-1/2*((a + b*x)^(5/2)*Sqrt[c + d*x])/x^2 + (-(((5*b*c + a*d)*(a + b*x)^(3/ 2)*Sqrt[c + d*x])/(c*x)) + (2*b*(11*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x] - (2*Sqrt[a]*(15*b^2*c^2 + 10*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[c] + (8*b^(3/2)*c*(b*c + 5*a*d)*Arc Tanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[d])/(2*c))/4
3.7.52.3.1 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[((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. \(440\) vs. \(2(168)=336\).
Time = 1.59 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.08
| method | result | size |
| default | \(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (20 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c d \,x^{2} \sqrt {a c}+4 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{2} x^{2} \sqrt {a c}+\ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{3} d^{2} x^{2} \sqrt {b d}-10 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} b c d \,x^{2} \sqrt {b d}-15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{2} x^{2} \sqrt {b d}+8 b^{2} c \,x^{2} \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-2 a^{2} d x \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-18 a b c x \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-4 a^{2} c \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\right )}{8 c \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, x^{2} \sqrt {b d}\, \sqrt {a c}}\) | \(441\) |
1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c*(20*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^2*c*d*x^2*(a*c)^(1/2)+4*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^3* c^2*x^2*(a*c)^(1/2)+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^3*d^2*x^2*(b*d)^(1/2)-10*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*b*c*d*x^2*(b*d)^(1/2)-15*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^2*c^2*x^2*(b*d)^(1/2)+8* b^2*c*x^2*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-2*a^2*d*x*(b*d)^ (1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-18*a*b*c*x*(b*d)^(1/2)*(a*c)^(1/ 2)*((b*x+a)*(d*x+c))^(1/2)-4*a^2*c*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c ))^(1/2))/((b*x+a)*(d*x+c))^(1/2)/x^2/(b*d)^(1/2)/(a*c)^(1/2)
Time = 1.25 (sec) , antiderivative size = 1098, normalized size of antiderivative = 5.18 \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^3} \, dx=\text {Too large to display} \]
[1/16*(4*(b^2*c^2 + 5*a*b*c*d)*x^2*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^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d* x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)*x) - (15*b^2*c^2 + 10*a*b*c*d - a ^2*d^2)*x^2*sqrt(a/c)*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^2 + (b*c^2 + a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(4*b^2*c*x^2 - 2*a^2*c - (9*a*b*c + a^2 *d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c*x^2), -1/16*(8*(b^2*c^2 + 5*a*b*c*d )*x^2*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-b/d)/(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x)) + (15*b^2*c^2 + 10 *a*b*c*d - a^2*d^2)*x^2*sqrt(a/c)*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^2 + (b*c^2 + a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c) *sqrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(4*b^2*c*x^2 - 2*a^2*c - (9 *a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c*x^2), 1/8*((15*b^2*c^2 + 10*a*b*c*d - a^2*d^2)*x^2*sqrt(-a/c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)* sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-a/c)/(a*b*d*x^2 + a^2*c + (a*b*c + a^2*d )*x)) + 2*(b^2*c^2 + 5*a*b*c*d)*x^2*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^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d *x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)*x) + 2*(4*b^2*c*x^2 - 2*a^2*c - (9*a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c*x^2), 1/8*((15*b^2*c^ 2 + 10*a*b*c*d - a^2*d^2)*x^2*sqrt(-a/c)*arctan(1/2*(2*a*c + (b*c + a*d...
\[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^3} \, dx=\int \frac {\left (a + b x\right )^{\frac {5}{2}} \sqrt {c + d x}}{x^{3}}\, dx \]
Exception generated. \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^3} \, dx=\text {Exception raised: ValueError} \]
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. 1197 vs. \(2 (168) = 336\).
Time = 0.91 (sec) , antiderivative size = 1197, normalized size of antiderivative = 5.65 \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^3} \, dx=\text {Too large to display} \]
1/4*(4*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*b*abs(b) - 2*(sqr t(b*d)*b^2*c*abs(b) + 5*sqrt(b*d)*a*b*d*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/d - (15*sqrt(b*d)*a*b^3*c^2*a bs(b) + 10*sqrt(b*d)*a^2*b^2*c*d*abs(b) - sqrt(b*d)*a^3*b*d^2*abs(b))*arct an(-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*c) - 2*(9*sqrt(b*d )*a*b^9*c^5*abs(b) - 35*sqrt(b*d)*a^2*b^8*c^4*d*abs(b) + 50*sqrt(b*d)*a^3* b^7*c^3*d^2*abs(b) - 30*sqrt(b*d)*a^4*b^6*c^2*d^3*abs(b) + 5*sqrt(b*d)*a^5 *b^5*c*d^4*abs(b) + sqrt(b*d)*a^6*b^4*d^5*abs(b) - 27*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^7*c^4*abs(b) + 28*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^6*c^3*d*abs(b) + 22*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^5*c^2*d^2*abs(b) - 20*sqrt(b*d)*(s qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*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^5*b^3*d^4*abs(b) + 27*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^5*c^3*abs(b) + 21*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^4 *c^2*d*abs(b) + 29*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^3*c*d^2*abs(b) + 3*sqrt(b*d)*(sqrt(b*d)*sqrt...
Timed out. \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^3} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/2}\,\sqrt {c+d\,x}}{x^3} \,d x \]