Integrand size = 22, antiderivative size = 299 \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx=\frac {\left (64 a^2 b c d^2+(b c-5 a d) (b c-a d) (3 b c+a d)\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b d^2}-\frac {(b c-5 a d) (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{32 d^2}+\frac {(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 d}+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}-2 a^{5/2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (3 b^4 c^4-20 a b^3 c^3 d+90 a^2 b^2 c^2 d^2+60 a^3 b c d^3-5 a^4 d^4\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{5/2}} \]
1/24*(5*a*d+3*b*c)*(b*x+a)^(3/2)*(d*x+c)^(3/2)/d+1/4*(b*x+a)^(5/2)*(d*x+c) ^(3/2)-2*a^(5/2)*c^(3/2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/ 2))+1/64*(-5*a^4*d^4+60*a^3*b*c*d^3+90*a^2*b^2*c^2*d^2-20*a*b^3*c^3*d+3*b^ 4*c^4)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(3/2)/d^(5/2 )-1/32*(-5*a*d+b*c)*(a*d+3*b*c)*(d*x+c)^(3/2)*(b*x+a)^(1/2)/d^2+1/64*(64*a ^2*b*c*d^2+(-5*a*d+b*c)*(-a*d+b*c)*(a*d+3*b*c))*(b*x+a)^(1/2)*(d*x+c)^(1/2 )/b/d^2
Time = 0.64 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^3 d^3+a^2 b d^2 (337 c+118 d x)+a b^2 d \left (57 c^2+244 c d x+136 d^2 x^2\right )+b^3 \left (-9 c^3+6 c^2 d x+72 c d^2 x^2+48 d^3 x^3\right )\right )}{192 b d^2}-2 a^{5/2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (3 b^4 c^4-20 a b^3 c^3 d+90 a^2 b^2 c^2 d^2+60 a^3 b c d^3-5 a^4 d^4\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{5/2}} \]
(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^3*d^3 + a^2*b*d^2*(337*c + 118*d*x) + a *b^2*d*(57*c^2 + 244*c*d*x + 136*d^2*x^2) + b^3*(-9*c^3 + 6*c^2*d*x + 72*c *d^2*x^2 + 48*d^3*x^3)))/(192*b*d^2) - 2*a^(5/2)*c^(3/2)*ArcTanh[(Sqrt[c]* Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] + ((3*b^4*c^4 - 20*a*b^3*c^3*d + 9 0*a^2*b^2*c^2*d^2 + 60*a^3*b*c*d^3 - 5*a^4*d^4)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(3/2)*d^(5/2))
Time = 0.47 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.09, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {112, 27, 171, 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.
| \(\displaystyle \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx\) |
| \(\Big \downarrow \) 112 |
| \(\displaystyle \frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}-\frac {1}{4} \int -\frac {(a+b x)^{3/2} \sqrt {c+d x} (8 a c+(3 b c+5 a d) x)}{2 x}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \int \frac {(a+b x)^{3/2} \sqrt {c+d x} (8 a c+(3 b c+5 a d) x)}{x}dx+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{8} \left (\frac {\int \frac {3 \sqrt {a+b x} \sqrt {c+d x} \left (16 a^2 c d-(b c-5 a d) (3 b c+a d) x\right )}{2 x}dx}{3 d}+\frac {(a+b x)^{3/2} (c+d x)^{3/2} (5 a d+3 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \left (\frac {\int \frac {\sqrt {a+b x} \sqrt {c+d x} \left (16 a^2 c d-(b c-5 a d) (3 b c+a d) x\right )}{x}dx}{2 d}+\frac {(a+b x)^{3/2} (c+d x)^{3/2} (5 a d+3 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {\int \frac {\sqrt {c+d x} \left (64 c d^2 a^3+\left (3 b^3 c^3-17 a b^2 d c^2+73 a^2 b d^2 c+5 a^3 d^3\right ) x\right )}{2 x \sqrt {a+b x}}dx}{2 d}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-5 a d) (a d+3 b c)}{2 d}}{2 d}+\frac {(a+b x)^{3/2} (c+d x)^{3/2} (5 a d+3 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {\int \frac {\sqrt {c+d x} \left (64 c d^2 a^3+\left (3 b^3 c^3-17 a b^2 d c^2+73 a^2 b d^2 c+5 a^3 d^3\right ) x\right )}{x \sqrt {a+b x}}dx}{4 d}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-5 a d) (a d+3 b c)}{2 d}}{2 d}+\frac {(a+b x)^{3/2} (c+d x)^{3/2} (5 a d+3 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {\frac {\int \frac {128 b c^2 d^2 a^3+\left (3 b^4 c^4-20 a b^3 d c^3+90 a^2 b^2 d^2 c^2+60 a^3 b d^3 c-5 a^4 d^4\right ) x}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (5 a^3 d^3+73 a^2 b c d^2-17 a b^2 c^2 d+3 b^3 c^3\right )}{b}}{4 d}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-5 a d) (a d+3 b c)}{2 d}}{2 d}+\frac {(a+b x)^{3/2} (c+d x)^{3/2} (5 a d+3 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {\frac {\int \frac {128 b c^2 d^2 a^3+\left (3 b^4 c^4-20 a b^3 d c^3+90 a^2 b^2 d^2 c^2+60 a^3 b d^3 c-5 a^4 d^4\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (5 a^3 d^3+73 a^2 b c d^2-17 a b^2 c^2 d+3 b^3 c^3\right )}{b}}{4 d}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-5 a d) (a d+3 b c)}{2 d}}{2 d}+\frac {(a+b x)^{3/2} (c+d x)^{3/2} (5 a d+3 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {\frac {128 a^3 b c^2 d^2 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+\left (-5 a^4 d^4+60 a^3 b c d^3+90 a^2 b^2 c^2 d^2-20 a b^3 c^3 d+3 b^4 c^4\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (5 a^3 d^3+73 a^2 b c d^2-17 a b^2 c^2 d+3 b^3 c^3\right )}{b}}{4 d}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-5 a d) (a d+3 b c)}{2 d}}{2 d}+\frac {(a+b x)^{3/2} (c+d x)^{3/2} (5 a d+3 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {\frac {128 a^3 b c^2 d^2 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+2 \left (-5 a^4 d^4+60 a^3 b c d^3+90 a^2 b^2 c^2 d^2-20 a b^3 c^3 d+3 b^4 c^4\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (5 a^3 d^3+73 a^2 b c d^2-17 a b^2 c^2 d+3 b^3 c^3\right )}{b}}{4 d}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-5 a d) (a d+3 b c)}{2 d}}{2 d}+\frac {(a+b x)^{3/2} (c+d x)^{3/2} (5 a d+3 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {\frac {256 a^3 b c^2 d^2 \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+2 \left (-5 a^4 d^4+60 a^3 b c d^3+90 a^2 b^2 c^2 d^2-20 a b^3 c^3 d+3 b^4 c^4\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (5 a^3 d^3+73 a^2 b c d^2-17 a b^2 c^2 d+3 b^3 c^3\right )}{b}}{4 d}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-5 a d) (a d+3 b c)}{2 d}}{2 d}+\frac {(a+b x)^{3/2} (c+d x)^{3/2} (5 a d+3 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {\frac {\sqrt {a+b x} \sqrt {c+d x} \left (5 a^3 d^3+73 a^2 b c d^2-17 a b^2 c^2 d+3 b^3 c^3\right )}{b}+\frac {\frac {2 \left (-5 a^4 d^4+60 a^3 b c d^3+90 a^2 b^2 c^2 d^2-20 a b^3 c^3 d+3 b^4 c^4\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {d}}-256 a^{5/2} b c^{3/2} d^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{2 b}}{4 d}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-5 a d) (a d+3 b c)}{2 d}}{2 d}+\frac {(a+b x)^{3/2} (c+d x)^{3/2} (5 a d+3 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}\) |
((a + b*x)^(5/2)*(c + d*x)^(3/2))/4 + (((3*b*c + 5*a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2))/(3*d) + (-1/2*((b*c - 5*a*d)*(3*b*c + a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2))/d + (((3*b^3*c^3 - 17*a*b^2*c^2*d + 73*a^2*b*c*d^2 + 5*a^3* d^3)*Sqrt[a + b*x]*Sqrt[c + d*x])/b + (-256*a^(5/2)*b*c^(3/2)*d^2*ArcTanh[ (Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] + (2*(3*b^4*c^4 - 20*a*b^ 3*c^3*d + 90*a^2*b^2*c^2*d^2 + 60*a^3*b*c*d^3 - 5*a^4*d^4)*ArcTanh[(Sqrt[d ]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*Sqrt[d]))/(2*b))/(4*d) )/(2*d))/8
3.7.59.3.1 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*(c + d*x)^n*((e + f*x)^(p + 1)/(f*(m + n + p + 1))), x] - Simp[1/(f*(m + n + p + 1)) Int[(a + b*x)^(m - 1)*(c + d*x) ^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a *f) + b*n*(d*e - c*f))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 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. \(708\) vs. \(2(249)=498\).
Time = 1.59 (sec) , antiderivative size = 709, normalized size of antiderivative = 2.37
| method | result | size |
| default | \(-\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-96 b^{3} d^{3} x^{3} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}-272 a \,b^{2} d^{3} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}-144 b^{3} c \,d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \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 ) \sqrt {a c}\, a^{4} d^{4}-180 \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 ) \sqrt {a c}\, a^{3} b c \,d^{3}-270 \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 ) \sqrt {a c}\, a^{2} b^{2} c^{2} d^{2}+60 \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 ) \sqrt {a c}\, a \,b^{3} c^{3} d -9 \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 ) \sqrt {a c}\, b^{4} c^{4}+384 \sqrt {b d}\, \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} b \,c^{2} d^{2}-236 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} b \,d^{3} x -488 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a \,b^{2} c \,d^{2} x -12 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, b^{3} c^{2} d x -30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a^{3} d^{3}-674 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} b c \,d^{2}-114 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a \,b^{2} c^{2} d +18 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, b^{3} c^{3}\right )}{384 b \,d^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}}\) | \(709\) |
-1/384*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-96*b^3*d^3*x^3*((b*x+a)*(d*x+c))^(1/2 )*(b*d)^(1/2)*(a*c)^(1/2)-272*a*b^2*d^3*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^ (1/2)*(a*c)^(1/2)-144*b^3*c*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(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))*(a*c)^(1/2)*a^4*d^4-180*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*c)^(1/2)*a^3*b*c*d^3-270*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* c)^(1/2)*a^2*b^2*c^2*d^2+60*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*c)^(1/2)*a*b^3*c^3*d-9*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*c)^(1/2)*b^4* c^4+384*(b*d)^(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*b*c^2*d^2-236*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2 )*a^2*b*d^3*x-488*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a*b^2*c* d^2*x-12*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*b^3*c^2*d*x-30*(( b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^3*d^3-674*((b*x+a)*(d*x+c) )^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*b*c*d^2-114*((b*x+a)*(d*x+c))^(1/2)*(b *d)^(1/2)*(a*c)^(1/2)*a*b^2*c^2*d+18*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*( a*c)^(1/2)*b^3*c^3)/b/d^2/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(a*c)^(1/2)
Time = 8.55 (sec) , antiderivative size = 1481, normalized size of antiderivative = 4.95 \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx=\text {Too large to display} \]
[1/768*(384*sqrt(a*c)*a^2*b^2*c*d^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) - 3*(3*b^4*c^4 - 20*a*b^3*c^3*d + 9 0*a^2*b^2*c^2*d^2 + 60*a^3*b*c*d^3 - 5*a^4*d^4)*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*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) + 4*(48*b^4*d^4*x^3 - 9 *b^4*c^3*d + 57*a*b^3*c^2*d^2 + 337*a^2*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(9*b^ 4*c*d^3 + 17*a*b^3*d^4)*x^2 + 2*(3*b^4*c^2*d^2 + 122*a*b^3*c*d^3 + 59*a^2* b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^3), 1/384*(192*sqrt(a*c)*a ^2*b^2*c*d^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) - 3*(3*b^4*c^4 - 20*a*b^3*c^3*d + 90*a^2*b^2*c^2*d^2 + 60* a^3*b*c*d^3 - 5*a^4*d^4)*sqrt(-b*d)*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)) + 2*(48*b^4*d^4*x^3 - 9*b^4*c^3*d + 57*a*b^3*c^2*d^2 + 337*a^2*b^ 2*c*d^3 + 15*a^3*b*d^4 + 8*(9*b^4*c*d^3 + 17*a*b^3*d^4)*x^2 + 2*(3*b^4*c^2 *d^2 + 122*a*b^3*c*d^3 + 59*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/( b^2*d^3), 1/768*(768*sqrt(-a*c)*a^2*b^2*c*d^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)) - 3*(3*b^4*c^4 - 20*a*b^3*c^3*d + 90*a^2*b^2*c^2*d...
\[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx=\int \frac {\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {3}{2}}}{x}\, dx \]
Exception generated. \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx=\text {Exception raised: ValueError} \]
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
Exception generated. \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Timed out. \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{3/2}}{x} \,d x \]