Integrand size = 22, antiderivative size = 275 \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x^3} \, dx=\frac {3 \left (b^2 c^2+6 a b c d+a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{4 c}+\frac {3 b (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(5 b c+3 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}-\frac {3 \sqrt {a} \left (5 b^2 c^2+10 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {c}}+\frac {3 \sqrt {b} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {d}} \]
-1/4*(3*a*d+5*b*c)*(b*x+a)^(3/2)*(d*x+c)^(3/2)/c/x-1/2*(b*x+a)^(5/2)*(d*x+ c)^(3/2)/x^2-3/4*(a^2*d^2+10*a*b*c*d+5*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^(1/2)+3/4*(5*a^2*d^2+10*a*b*c*d+b^2*c ^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))*b^(1/2)/d^(1/2)+3 /4*b*(a*d+3*b*c)*(d*x+c)^(3/2)*(b*x+a)^(1/2)/c+3/4*(a^2*d^2+6*a*b*c*d+b^2* c^2)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c
Time = 0.82 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.71 \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x^3} \, dx=\frac {1}{4} \left (\frac {\sqrt {a+b x} \sqrt {c+d x} \left (9 a b x (-c+d x)+b^2 x^2 (5 c+2 d x)-a^2 (2 c+5 d x)\right )}{x^2}-\frac {3 \sqrt {a} \left (5 b^2 c^2+10 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {c}}+\frac {3 \sqrt {b} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {d}}\right ) \]
((Sqrt[a + b*x]*Sqrt[c + d*x]*(9*a*b*x*(-c + d*x) + b^2*x^2*(5*c + 2*d*x) - a^2*(2*c + 5*d*x)))/x^2 - (3*Sqrt[a]*(5*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] + (3*Sqr t[b]*(b^2*c^2 + 10*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(S qrt[b]*Sqrt[c + d*x])])/Sqrt[d])/4
Time = 0.42 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {108, 27, 166, 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.
| \(\displaystyle \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x^3} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {1}{2} \int \frac {(a+b x)^{3/2} \sqrt {c+d x} (5 b c+3 a d+8 b d x)}{2 x^2}dx-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \frac {(a+b x)^{3/2} \sqrt {c+d x} (5 b c+3 a d+8 b d x)}{x^2}dx-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {1}{4} \left (\frac {\int \frac {3 \sqrt {a+b x} \sqrt {c+d x} \left (5 b^2 c^2+10 a b d c+a^2 d^2+4 b d (3 b c+a d) x\right )}{2 x}dx}{c}-\frac {(a+b x)^{3/2} (c+d x)^{3/2} (3 a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \int \frac {\sqrt {a+b x} \sqrt {c+d x} \left (5 b^2 c^2+10 a b d c+a^2 d^2+4 b d (3 b c+a d) x\right )}{x}dx}{2 c}-\frac {(a+b x)^{3/2} (c+d x)^{3/2} (3 a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {\int \frac {2 d \sqrt {c+d x} \left (a \left (5 b^2 c^2+10 a b d c+a^2 d^2\right )+2 b \left (b^2 c^2+6 a b d c+a^2 d^2\right ) x\right )}{x \sqrt {a+b x}}dx}{2 d}+2 b \sqrt {a+b x} (c+d x)^{3/2} (a d+3 b c)\right )}{2 c}-\frac {(a+b x)^{3/2} (c+d x)^{3/2} (3 a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (\int \frac {\sqrt {c+d x} \left (a \left (5 b^2 c^2+10 a b d c+a^2 d^2\right )+2 b \left (b^2 c^2+6 a b d c+a^2 d^2\right ) x\right )}{x \sqrt {a+b x}}dx+2 b \sqrt {a+b x} (c+d x)^{3/2} (a d+3 b c)\right )}{2 c}-\frac {(a+b x)^{3/2} (c+d x)^{3/2} (3 a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {\int \frac {b c \left (a \left (5 b^2 c^2+10 a b d c+a^2 d^2\right )+b \left (b^2 c^2+10 a b d c+5 a^2 d^2\right ) x\right )}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{b}+2 \sqrt {a+b x} \sqrt {c+d x} \left (a^2 d^2+6 a b c d+b^2 c^2\right )+2 b \sqrt {a+b x} (c+d x)^{3/2} (a d+3 b c)\right )}{2 c}-\frac {(a+b x)^{3/2} (c+d x)^{3/2} (3 a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (c \int \frac {a \left (5 b^2 c^2+10 a b d c+a^2 d^2\right )+b \left (b^2 c^2+10 a b d c+5 a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}}dx+2 \sqrt {a+b x} \sqrt {c+d x} \left (a^2 d^2+6 a b c d+b^2 c^2\right )+2 b \sqrt {a+b x} (c+d x)^{3/2} (a d+3 b c)\right )}{2 c}-\frac {(a+b x)^{3/2} (c+d x)^{3/2} (3 a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (c \left (b \left (5 a^2 d^2+10 a b c d+b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx+a \left (a^2 d^2+10 a b c d+5 b^2 c^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx\right )+2 \sqrt {a+b x} \sqrt {c+d x} \left (a^2 d^2+6 a b c d+b^2 c^2\right )+2 b \sqrt {a+b x} (c+d x)^{3/2} (a d+3 b c)\right )}{2 c}-\frac {(a+b x)^{3/2} (c+d x)^{3/2} (3 a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (c \left (a \left (a^2 d^2+10 a b c d+5 b^2 c^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+2 b \left (5 a^2 d^2+10 a b c d+b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )+2 \sqrt {a+b x} \sqrt {c+d x} \left (a^2 d^2+6 a b c d+b^2 c^2\right )+2 b \sqrt {a+b x} (c+d x)^{3/2} (a d+3 b c)\right )}{2 c}-\frac {(a+b x)^{3/2} (c+d x)^{3/2} (3 a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (c \left (2 a \left (a^2 d^2+10 a b c d+5 b^2 c^2\right ) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+2 b \left (5 a^2 d^2+10 a b c d+b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )+2 \sqrt {a+b x} \sqrt {c+d x} \left (a^2 d^2+6 a b c d+b^2 c^2\right )+2 b \sqrt {a+b x} (c+d x)^{3/2} (a d+3 b c)\right )}{2 c}-\frac {(a+b x)^{3/2} (c+d x)^{3/2} (3 a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (c \left (\frac {2 \sqrt {b} \left (5 a^2 d^2+10 a b c d+b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {d}}-\frac {2 \sqrt {a} \left (a^2 d^2+10 a b c d+5 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {c}}\right )+2 \sqrt {a+b x} \sqrt {c+d x} \left (a^2 d^2+6 a b c d+b^2 c^2\right )+2 b \sqrt {a+b x} (c+d x)^{3/2} (a d+3 b c)\right )}{2 c}-\frac {(a+b x)^{3/2} (c+d x)^{3/2} (3 a d+5 b c)}{c x}\right )-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}\) |
-1/2*((a + b*x)^(5/2)*(c + d*x)^(3/2))/x^2 + (-(((5*b*c + 3*a*d)*(a + b*x) ^(3/2)*(c + d*x)^(3/2))/(c*x)) + (3*(2*(b^2*c^2 + 6*a*b*c*d + a^2*d^2)*Sqr t[a + b*x]*Sqrt[c + d*x] + 2*b*(3*b*c + a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2) + c*((-2*Sqrt[a]*(5*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] + (2*Sqrt[b]*(b^2*c^2 + 10*a* b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x]) ])/Sqrt[d])))/(2*c))/4
3.7.61.3.1 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. \(558\) vs. \(2(223)=446\).
Time = 0.56 (sec) , antiderivative size = 559, normalized size of antiderivative = 2.03
| method | result | size |
| default | \(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b \,d^{2} x^{2} \sqrt {a c}+30 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c d \,x^{2} \sqrt {a c}+3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{2} x^{2} \sqrt {a c}-3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{3} d^{2} x^{2} \sqrt {b d}-30 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} b c d \,x^{2} \sqrt {b d}-15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{2} x^{2} \sqrt {b d}+4 b^{2} d \,x^{3} \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+18 a b d \,x^{2} \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+10 b^{2} c \,x^{2} \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-10 a^{2} d x \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-18 a b c x \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-4 a^{2} c \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, x^{2} \sqrt {b d}\, \sqrt {a c}}\) | \(559\) |
1/8*(b*x+a)^(1/2)*(d*x+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))*a^2*b*d^2*x^2*(a*c)^(1/2)+30*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)+3*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)-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)*a^3*d^2*x^2*(b*d)^(1/2)-30*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)+4*b^2*d*x^3*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)* (d*x+c))^(1/2)+18*a*b*d*x^2*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2 )+10*b^2*c*x^2*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-10*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 = 1.57 (sec) , antiderivative size = 1185, normalized size of antiderivative = 4.31 \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x^3} \, dx=\text {Too large to display} \]
[1/16*(3*(b^2*c^2 + 10*a*b*c*d + 5*a^2*d^2)*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) + 3*(5*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*(2*b^2*d*x^3 - 2*a^2*c + (5*b^2*c + 9*a*b*d)*x^2 - (9*a*b*c + 5*a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/x^2, -1/16*(6*(b^2*c^2 + 10*a*b*c*d + 5*a^2*d^2)*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)) - 3*(5*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*(2*b^2*d*x^3 - 2*a^2*c + (5*b^2*c + 9*a*b*d)*x^2 - (9*a*b*c + 5*a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/x^2, 1/16*(6*(5*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)) + 3*(b^2*c^2 + 10*a*b*c*d + 5*a^2*d^2)*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)*sqr t(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(2*b^2*d *x^3 - 2*a^2*c + (5*b^2*c + 9*a*b*d)*x^2 - (9*a*b*c + 5*a^2*d)*x)*sqrt(...
\[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x^3} \, dx=\int \frac {\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {3}{2}}}{x^{3}}\, dx \]
Exception generated. \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x^3} \, dx=\text {Exception raised: ValueError} \]
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. 1239 vs. \(2 (223) = 446\).
Time = 1.18 (sec) , antiderivative size = 1239, normalized size of antiderivative = 4.51 \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x^3} \, dx=\text {Too large to display} \]
1/8*(2*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*d*abs(b) + (5*b*c* d^2*abs(b) + 7*a*d^3*abs(b))/d^2)*sqrt(b*x + a) - 3*(sqrt(b*d)*b^2*c^2*abs (b) + 10*sqrt(b*d)*a*b*c*d*abs(b) + 5*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)/d - 6*(5*sqrt (b*d)*a*b^3*c^2*abs(b) + 10*sqrt(b*d)*a^2*b^2*c*d*abs(b) + sqrt(b*d)*a^3*b *d^2*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) - 4*(9*sqrt(b*d)*a*b^9*c^5*abs(b) - 31*sqrt(b*d)*a^2*b^8*c^4*d*abs(b) + 34 *sqrt(b*d)*a^3*b^7*c^3*d^2*abs(b) - 6*sqrt(b*d)*a^4*b^6*c^2*d^3*abs(b) - 1 1*sqrt(b*d)*a^5*b^5*c*d^4*abs(b) + 5*sqrt(b*d)*a^6*b^4*d^5*abs(b) - 27*sqr t(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) + 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^2*b^6*c^3*d*abs(b) + 34*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) - 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^4*b^4*c*d^3*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*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) + 33*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) + 37*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) ...
Timed out. \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x^3} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{3/2}}{x^3} \,d x \]