Integrand size = 22, antiderivative size = 292 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^4} \, dx=\frac {d \left (b^2 c^2+26 a b c d+5 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 a c}-\frac {\left (\frac {3 b^2 c}{a}+40 b d+\frac {5 a d^2}{c}\right ) \sqrt {a+b x} (c+d x)^{3/2}}{24 x}-\frac {(3 b c+5 a d) \sqrt {a+b x} (c+d x)^{5/2}}{12 c x^2}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{3 x^3}+\frac {\left (b^3 c^3-15 a b^2 c^2 d-45 a^2 b c d^2-5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{3/2} \sqrt {c}}+\sqrt {b} d^{3/2} (5 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right ) \]
-1/3*(b*x+a)^(3/2)*(d*x+c)^(5/2)/x^3+d^(3/2)*(3*a*d+5*b*c)*arctanh(d^(1/2) *(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))*b^(1/2)+1/8*(-5*a^3*d^3-45*a^2*b*c*d ^2-15*a*b^2*c^2*d+b^3*c^3)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^( 1/2))/a^(3/2)/c^(1/2)-1/24*(3*b^2*c/a+40*b*d+5*a*d^2/c)*(d*x+c)^(3/2)*(b*x +a)^(1/2)/x-1/12*(5*a*d+3*b*c)*(d*x+c)^(5/2)*(b*x+a)^(1/2)/c/x^2+1/8*d*(5* a^2*d^2+26*a*b*c*d+b^2*c^2)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a/c
Time = 0.78 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^4} \, dx=-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (3 b^2 c^2 x^2+2 a b x \left (7 c^2+34 c d x-12 d^2 x^2\right )+a^2 \left (8 c^2+26 c d x+33 d^2 x^2\right )\right )}{24 a x^3}-\frac {\left (-b^3 c^3+15 a b^2 c^2 d+45 a^2 b c d^2+5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{8 a^{3/2} \sqrt {c}}+\sqrt {b} d^{3/2} (5 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right ) \]
-1/24*(Sqrt[a + b*x]*Sqrt[c + d*x]*(3*b^2*c^2*x^2 + 2*a*b*x*(7*c^2 + 34*c* d*x - 12*d^2*x^2) + a^2*(8*c^2 + 26*c*d*x + 33*d^2*x^2)))/(a*x^3) - ((-(b^ 3*c^3) + 15*a*b^2*c^2*d + 45*a^2*b*c*d^2 + 5*a^3*d^3)*ArcTanh[(Sqrt[a]*Sqr t[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/(8*a^(3/2)*Sqrt[c]) + Sqrt[b]*d^(3/2 )*(5*b*c + 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])]
Time = 0.47 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.06, 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, 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.
| \(\displaystyle \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^4} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {1}{3} \int \frac {\sqrt {a+b x} (c+d x)^{3/2} (3 b c+5 a d+8 b d x)}{2 x^3}dx-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{3 x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \int \frac {\sqrt {a+b x} (c+d x)^{3/2} (3 b c+5 a d+8 b d x)}{x^3}dx-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{3 x^3}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {1}{6} \left (\frac {\int \frac {(c+d x)^{3/2} \left (3 b^2 c^2+40 a b d c+5 a^2 d^2+2 b d (19 b c+5 a d) x\right )}{2 x^2 \sqrt {a+b x}}dx}{2 c}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (5 a d+3 b c)}{2 c x^2}\right )-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{3 x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \left (\frac {\int \frac {(c+d x)^{3/2} \left (3 b^2 c^2+40 a b d c+5 a^2 d^2+2 b d (19 b c+5 a d) x\right )}{x^2 \sqrt {a+b x}}dx}{4 c}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (5 a d+3 b c)}{2 c x^2}\right )-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{3 x^3}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {1}{6} \left (\frac {\frac {\int -\frac {3 \sqrt {c+d x} \left (b^3 c^3-15 a b^2 d c^2-45 a^2 b d^2 c-5 a^3 d^3-2 b d \left (b^2 c^2+26 a b d c+5 a^2 d^2\right ) x\right )}{2 x \sqrt {a+b x}}dx}{a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (5 a^2 d^2+40 a b c d+3 b^2 c^2\right )}{a x}}{4 c}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (5 a d+3 b c)}{2 c x^2}\right )-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{3 x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \left (\frac {-\frac {3 \int \frac {\sqrt {c+d x} \left (b^3 c^3-15 a b^2 d c^2-45 a^2 b d^2 c-5 a^3 d^3-2 b d \left (b^2 c^2+26 a b d c+5 a^2 d^2\right ) x\right )}{x \sqrt {a+b x}}dx}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (5 a^2 d^2+40 a b c d+3 b^2 c^2\right )}{a x}}{4 c}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (5 a d+3 b c)}{2 c x^2}\right )-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{3 x^3}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{6} \left (\frac {-\frac {3 \left (\frac {\int \frac {b c \left (b^3 c^3-15 a b^2 d c^2-45 a^2 b d^2 c-5 a^3 d^3-8 a b d^2 (5 b c+3 a d) x\right )}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{b}-2 d \sqrt {a+b x} \sqrt {c+d x} \left (5 a^2 d^2+26 a b c d+b^2 c^2\right )\right )}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (5 a^2 d^2+40 a b c d+3 b^2 c^2\right )}{a x}}{4 c}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (5 a d+3 b c)}{2 c x^2}\right )-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{3 x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \left (\frac {-\frac {3 \left (c \int \frac {b^3 c^3-15 a b^2 d c^2-45 a^2 b d^2 c-5 a^3 d^3-8 a b d^2 (5 b c+3 a d) x}{x \sqrt {a+b x} \sqrt {c+d x}}dx-2 d \sqrt {a+b x} \sqrt {c+d x} \left (5 a^2 d^2+26 a b c d+b^2 c^2\right )\right )}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (5 a^2 d^2+40 a b c d+3 b^2 c^2\right )}{a x}}{4 c}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (5 a d+3 b c)}{2 c x^2}\right )-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{3 x^3}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {1}{6} \left (\frac {-\frac {3 \left (c \left (\left (-5 a^3 d^3-45 a^2 b c d^2-15 a b^2 c^2 d+b^3 c^3\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx-8 a b d^2 (3 a d+5 b c) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx\right )-2 d \sqrt {a+b x} \sqrt {c+d x} \left (5 a^2 d^2+26 a b c d+b^2 c^2\right )\right )}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (5 a^2 d^2+40 a b c d+3 b^2 c^2\right )}{a x}}{4 c}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (5 a d+3 b c)}{2 c x^2}\right )-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{3 x^3}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {1}{6} \left (\frac {-\frac {3 \left (c \left (\left (-5 a^3 d^3-45 a^2 b c d^2-15 a b^2 c^2 d+b^3 c^3\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx-16 a b d^2 (3 a d+5 b c) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )-2 d \sqrt {a+b x} \sqrt {c+d x} \left (5 a^2 d^2+26 a b c d+b^2 c^2\right )\right )}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (5 a^2 d^2+40 a b c d+3 b^2 c^2\right )}{a x}}{4 c}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (5 a d+3 b c)}{2 c x^2}\right )-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{3 x^3}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{6} \left (\frac {-\frac {3 \left (c \left (2 \left (-5 a^3 d^3-45 a^2 b c d^2-15 a b^2 c^2 d+b^3 c^3\right ) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}-16 a b d^2 (3 a d+5 b c) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )-2 d \sqrt {a+b x} \sqrt {c+d x} \left (5 a^2 d^2+26 a b c d+b^2 c^2\right )\right )}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (5 a^2 d^2+40 a b c d+3 b^2 c^2\right )}{a x}}{4 c}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (5 a d+3 b c)}{2 c x^2}\right )-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{3 x^3}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{6} \left (\frac {-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (5 a^2 d^2+40 a b c d+3 b^2 c^2\right )}{a x}-\frac {3 \left (c \left (-\frac {2 \left (-5 a^3 d^3-45 a^2 b c d^2-15 a b^2 c^2 d+b^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} \sqrt {c}}-16 a \sqrt {b} d^{3/2} (3 a d+5 b c) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )\right )-2 d \sqrt {a+b x} \sqrt {c+d x} \left (5 a^2 d^2+26 a b c d+b^2 c^2\right )\right )}{2 a}}{4 c}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (5 a d+3 b c)}{2 c x^2}\right )-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{3 x^3}\) |
-1/3*((a + b*x)^(3/2)*(c + d*x)^(5/2))/x^3 + (-1/2*((3*b*c + 5*a*d)*Sqrt[a + b*x]*(c + d*x)^(5/2))/(c*x^2) + (-(((3*b^2*c^2 + 40*a*b*c*d + 5*a^2*d^2 )*Sqrt[a + b*x]*(c + d*x)^(3/2))/(a*x)) - (3*(-2*d*(b^2*c^2 + 26*a*b*c*d + 5*a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x] + c*((-2*(b^3*c^3 - 15*a*b^2*c^2*d - 45*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sq rt[c + d*x])])/(Sqrt[a]*Sqrt[c]) - 16*a*Sqrt[b]*d^(3/2)*(5*b*c + 3*a*d)*Ar cTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])))/(2*a))/(4*c))/6
3.7.22.3.1 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. \(607\) vs. \(2(242)=484\).
Time = 1.56 (sec) , antiderivative size = 608, normalized size of antiderivative = 2.08
| method | result | size |
| default | \(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (72 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b \,d^{3} x^{3} \sqrt {a c}+120 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c \,d^{2} x^{3} \sqrt {a c}-15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{3} d^{3} x^{3} \sqrt {b d}-135 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} b c \,d^{2} x^{3} \sqrt {b d}-45 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{2} d \,x^{3} \sqrt {b d}+3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) b^{3} c^{3} x^{3} \sqrt {b d}+48 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,d^{2} x^{3}-66 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2} x^{2}-136 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d \,x^{2}-6 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2} x^{2}-52 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c d x -28 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,c^{2} x -16 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c^{2}\right )}{48 a \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, x^{3} \sqrt {b d}\, \sqrt {a c}}\) | \(608\) |
1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a*(72*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^3*x^3*(a*c)^(1/2)+120*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^2*x^3*(a*c)^(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^3*d^3*x^3*(b*d)^(1/2)-135*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^2*x^3*(b*d)^(1/2)-45*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*d*x^3 *(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^3*c^3*x^3*(b*d)^(1/2)+48*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^ (1/2)*a*b*d^2*x^3-66*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*d ^2*x^2-136*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c*d*x^2-6*( b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c^2*x^2-52*(b*d)^(1/2)* (a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*c*d*x-28*(b*d)^(1/2)*(a*c)^(1/2)*( (b*x+a)*(d*x+c))^(1/2)*a*b*c^2*x-16*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+ c))^(1/2)*a^2*c^2)/((b*x+a)*(d*x+c))^(1/2)/x^3/(b*d)^(1/2)/(a*c)^(1/2)
Time = 1.91 (sec) , antiderivative size = 1353, normalized size of antiderivative = 4.63 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^4} \, dx=\text {Too large to display} \]
[1/96*(24*(5*a^2*b*c^2*d + 3*a^3*c*d^2)*sqrt(b*d)*x^3*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^3*c^3 - 15*a*b^2*c^2 *d - 45*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(a*c)*x^3*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*(24*a^2*b*c*d^2*x^3 - 8*a^3*c^3 - (3*a*b^2*c^3 + 68*a^2*b*c^2*d + 33*a^3*c*d^2)*x^2 - 2*(7*a^2 *b*c^3 + 13*a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c*x^3), -1/96* (48*(5*a^2*b*c^2*d + 3*a^3*c*d^2)*sqrt(-b*d)*x^3*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^3*c^3 - 15*a*b^2*c^2*d - 45*a^2*b*c*d^2 - 5*a ^3*d^3)*sqrt(a*c)*x^3*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*(24*a^2*b*c*d^2*x^3 - 8*a^3*c^3 - (3*a*b^2*c^ 3 + 68*a^2*b*c^2*d + 33*a^3*c*d^2)*x^2 - 2*(7*a^2*b*c^3 + 13*a^3*c^2*d)*x) *sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c*x^3), -1/48*(3*(b^3*c^3 - 15*a*b^2*c^ 2*d - 45*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(-a*c)*x^3*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)) - 12*(5*a^2*b*c^2*d + 3*a^3*c*d^2)*sqrt(b*d)*x^3*l og(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...
\[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^4} \, dx=\int \frac {\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{2}}}{x^{4}}\, dx \]
Exception generated. \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^4} \, dx=\text {Exception raised: ValueError} \]
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. 2316 vs. \(2 (242) = 484\).
Time = 3.75 (sec) , antiderivative size = 2316, normalized size of antiderivative = 7.93 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^4} \, dx=\text {Too large to display} \]
1/24*(24*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*d^2*abs(b) - 12 *(5*sqrt(b*d)*b*c*d*abs(b) + 3*sqrt(b*d)*a*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) + 3*(sqrt(b*d)*b^4*c^3 *abs(b) - 15*sqrt(b*d)*a*b^3*c^2*d*abs(b) - 45*sqrt(b*d)*a^2*b^2*c*d^2*abs (b) - 5*sqrt(b*d)*a^3*b*d^3*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)*a*b) - 2*(3*sqrt(b*d)*b^14*c^8*abs(b) + 50*sqrt(b*d)*a *b^13*c^7*d*abs(b) - 330*sqrt(b*d)*a^2*b^12*c^6*d^2*abs(b) + 762*sqrt(b*d) *a^3*b^11*c^5*d^3*abs(b) - 820*sqrt(b*d)*a^4*b^10*c^4*d^4*abs(b) + 342*sqr t(b*d)*a^5*b^9*c^3*d^5*abs(b) + 90*sqrt(b*d)*a^6*b^8*c^2*d^6*abs(b) - 130* sqrt(b*d)*a^7*b^7*c*d^7*abs(b) + 33*sqrt(b*d)*a^8*b^6*d^8*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^ 12*c^7*abs(b) - 267*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^11*c^6*d*abs(b) + 765*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^10*c^5*d^2*abs(b) - 255*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^9*c^4*d^3*abs(b) - 765*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^8*c^3*d^4*abs(b) + 495*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^7*c^2*d^5*abs(b) + 207*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*...
Timed out. \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^4} \, dx=\int \frac {{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{5/2}}{x^4} \,d x \]