Integrand size = 22, antiderivative size = 304 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x} \, dx=-\frac {\left (5 b^3 c^3-55 a b^2 c^2 d-17 a^2 b c d^2+3 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^2 d}+\frac {1}{96} \left (50 a c-\frac {5 b c^2}{d}+\frac {3 a^2 d}{b}\right ) \sqrt {a+b x} (c+d x)^{3/2}+\frac {(5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{24 d}+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}-2 a^{3/2} c^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\frac {\left (5 b^4 c^4-60 a b^3 c^3 d-90 a^2 b^2 c^2 d^2+20 a^3 b c d^3-3 a^4 d^4\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{5/2} d^{3/2}} \]
1/4*(b*x+a)^(3/2)*(d*x+c)^(5/2)-2*a^(3/2)*c^(5/2)*arctanh(c^(1/2)*(b*x+a)^ (1/2)/a^(1/2)/(d*x+c)^(1/2))-1/64*(-3*a^4*d^4+20*a^3*b*c*d^3-90*a^2*b^2*c^ 2*d^2-60*a*b^3*c^3*d+5*b^4*c^4)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x +c)^(1/2))/b^(5/2)/d^(3/2)+1/96*(50*a*c-5*b*c^2/d+3*a^2*d/b)*(d*x+c)^(3/2) *(b*x+a)^(1/2)+1/24*(3*a*d+5*b*c)*(d*x+c)^(5/2)*(b*x+a)^(1/2)/d-1/64*(3*a^ 3*d^3-17*a^2*b*c*d^2-55*a*b^2*c^2*d+5*b^3*c^3)*(b*x+a)^(1/2)*(d*x+c)^(1/2) /b^2/d
Time = 0.66 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.83 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-9 a^3 d^3+3 a^2 b d^2 (19 c+2 d x)+a b^2 d \left (337 c^2+244 c d x+72 d^2 x^2\right )+b^3 \left (15 c^3+118 c^2 d x+136 c d^2 x^2+48 d^3 x^3\right )\right )}{192 b^2 d}-2 a^{3/2} c^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )-\frac {\left (5 b^4 c^4-60 a b^3 c^3 d-90 a^2 b^2 c^2 d^2+20 a^3 b c d^3-3 a^4 d^4\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{64 b^{5/2} d^{3/2}} \]
(Sqrt[a + b*x]*Sqrt[c + d*x]*(-9*a^3*d^3 + 3*a^2*b*d^2*(19*c + 2*d*x) + a* b^2*d*(337*c^2 + 244*c*d*x + 72*d^2*x^2) + b^3*(15*c^3 + 118*c^2*d*x + 136 *c*d^2*x^2 + 48*d^3*x^3)))/(192*b^2*d) - 2*a^(3/2)*c^(5/2)*ArcTanh[(Sqrt[a ]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])] - ((5*b^4*c^4 - 60*a*b^3*c^3*d - 90*a^2*b^2*c^2*d^2 + 20*a^3*b*c*d^3 - 3*a^4*d^4)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(64*b^(5/2)*d^(3/2))
Time = 0.47 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.10, 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.
| \(\displaystyle \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x} \, dx\) |
| \(\Big \downarrow \) 112 |
| \(\displaystyle \frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}-\frac {1}{4} \int -\frac {\sqrt {a+b x} (c+d x)^{3/2} (8 a c+(5 b c+3 a d) x)}{2 x}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \int \frac {\sqrt {a+b x} (c+d x)^{3/2} (8 a c+(5 b c+3 a d) x)}{x}dx+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{8} \left (\frac {\int \frac {(c+d x)^{3/2} \left (48 a^2 c d-\left (5 b^2 c^2-50 a b d c-3 a^2 d^2\right ) x\right )}{2 x \sqrt {a+b x}}dx}{3 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \left (\frac {\int \frac {(c+d x)^{3/2} \left (48 a^2 c d-\left (5 b^2 c^2-50 a b d c-3 a^2 d^2\right ) x\right )}{x \sqrt {a+b x}}dx}{6 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {\int \frac {3 \sqrt {c+d x} \left (64 a^2 b c^2 d-\left (5 b^3 c^3-55 a b^2 d c^2-17 a^2 b d^2 c+3 a^3 d^3\right ) x\right )}{2 x \sqrt {a+b x}}dx}{2 b}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (-3 a^2 d^2-50 a b c d+5 b^2 c^2\right )}{2 b}}{6 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {3 \int \frac {\sqrt {c+d x} \left (64 a^2 b c^2 d-\left (5 b^3 c^3-55 a b^2 d c^2-17 a^2 b d^2 c+3 a^3 d^3\right ) x\right )}{x \sqrt {a+b x}}dx}{4 b}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (-3 a^2 d^2-50 a b c d+5 b^2 c^2\right )}{2 b}}{6 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {3 \left (\frac {\int \frac {128 a^2 b^2 c^3 d-\left (5 b^4 c^4-60 a b^3 d c^3-90 a^2 b^2 d^2 c^2+20 a^3 b d^3 c-3 a^4 d^4\right ) x}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{b}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (3 a^3 d^3-17 a^2 b c d^2-55 a b^2 c^2 d+5 b^3 c^3\right )}{b}\right )}{4 b}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (-3 a^2 d^2-50 a b c d+5 b^2 c^2\right )}{2 b}}{6 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {3 \left (\frac {\int \frac {128 a^2 b^2 c^3 d-\left (5 b^4 c^4-60 a b^3 d c^3-90 a^2 b^2 d^2 c^2+20 a^3 b d^3 c-3 a^4 d^4\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (3 a^3 d^3-17 a^2 b c d^2-55 a b^2 c^2 d+5 b^3 c^3\right )}{b}\right )}{4 b}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (-3 a^2 d^2-50 a b c d+5 b^2 c^2\right )}{2 b}}{6 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {3 \left (\frac {128 a^2 b^2 c^3 d \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx-\left (-3 a^4 d^4+20 a^3 b c d^3-90 a^2 b^2 c^2 d^2-60 a b^3 c^3 d+5 b^4 c^4\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (3 a^3 d^3-17 a^2 b c d^2-55 a b^2 c^2 d+5 b^3 c^3\right )}{b}\right )}{4 b}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (-3 a^2 d^2-50 a b c d+5 b^2 c^2\right )}{2 b}}{6 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {3 \left (\frac {128 a^2 b^2 c^3 d \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx-2 \left (-3 a^4 d^4+20 a^3 b c d^3-90 a^2 b^2 c^2 d^2-60 a b^3 c^3 d+5 b^4 c^4\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 b}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (3 a^3 d^3-17 a^2 b c d^2-55 a b^2 c^2 d+5 b^3 c^3\right )}{b}\right )}{4 b}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (-3 a^2 d^2-50 a b c d+5 b^2 c^2\right )}{2 b}}{6 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {3 \left (\frac {256 a^2 b^2 c^3 d \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}-2 \left (-3 a^4 d^4+20 a^3 b c d^3-90 a^2 b^2 c^2 d^2-60 a b^3 c^3 d+5 b^4 c^4\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 b}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (3 a^3 d^3-17 a^2 b c d^2-55 a b^2 c^2 d+5 b^3 c^3\right )}{b}\right )}{4 b}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (-3 a^2 d^2-50 a b c d+5 b^2 c^2\right )}{2 b}}{6 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{8} \left (\frac {\frac {3 \left (\frac {-256 a^{3/2} b^2 c^{5/2} d \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\frac {2 \left (-3 a^4 d^4+20 a^3 b c d^3-90 a^2 b^2 c^2 d^2-60 a b^3 c^3 d+5 b^4 c^4\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {d}}}{2 b}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (3 a^3 d^3-17 a^2 b c d^2-55 a b^2 c^2 d+5 b^3 c^3\right )}{b}\right )}{4 b}-\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (-3 a^2 d^2-50 a b c d+5 b^2 c^2\right )}{2 b}}{6 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{3 d}\right )+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}\) |
((a + b*x)^(3/2)*(c + d*x)^(5/2))/4 + (((5*b*c + 3*a*d)*Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*d) + (-1/2*((5*b^2*c^2 - 50*a*b*c*d - 3*a^2*d^2)*Sqrt[a + b*x]*(c + d*x)^(3/2))/b + (3*(-(((5*b^3*c^3 - 55*a*b^2*c^2*d - 17*a^2*b*c* d^2 + 3*a^3*d^3)*Sqrt[a + b*x]*Sqrt[c + d*x])/b) + (-256*a^(3/2)*b^2*c^(5/ 2)*d*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] - (2*(5*b^4* c^4 - 60*a*b^3*c^3*d - 90*a^2*b^2*c^2*d^2 + 20*a^3*b*c*d^3 - 3*a^4*d^4)*Ar cTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*Sqrt[d])) /(2*b)))/(4*b))/(6*d))/8
3.7.19.3.1 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. \(2(254)=508\).
Time = 1.56 (sec) , antiderivative size = 709, normalized size of antiderivative = 2.33
| 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}+144 a \,b^{2} d^{3} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}+272 b^{3} c \,d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}+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}\, a^{4} d^{4}-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^{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}+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 \,b^{3} c^{3} d -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}\, 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^{2} b^{2} c^{3} d +12 \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 +236 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, b^{3} c^{2} d x -18 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a^{3} d^{3}+114 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} b c \,d^{2}+674 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a \,b^{2} c^{2} d +30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, b^{3} c^{3}\right )}{384 b^{2} d \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)+144*a*b^2*d^3*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1 /2)*(a*c)^(1/2)+272*b^3*c*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c )^(1/2)+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)*a^4*d^4-60*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^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+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*b^3*c^3*d-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)*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^2*b^2*c^3*d+12*((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+236*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*b^3*c^2*d*x-18*((b* x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^3*d^3+114*((b*x+a)*(d*x+c))^ (1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*b*c*d^2+674*((b*x+a)*(d*x+c))^(1/2)*(b*d )^(1/2)*(a*c)^(1/2)*a*b^2*c^2*d+30*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a* c)^(1/2)*b^3*c^3)/b^2/d/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(a*c)^(1/2)
Time = 8.34 (sec) , antiderivative size = 1481, normalized size of antiderivative = 4.87 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x} \, dx=\text {Too large to display} \]
[1/768*(384*sqrt(a*c)*a*b^3*c^2*d^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) - 3*(5*b^4*c^4 - 60*a*b^3*c^3*d - 9 0*a^2*b^2*c^2*d^2 + 20*a^3*b*c*d^3 - 3*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 + 1 5*b^4*c^3*d + 337*a*b^3*c^2*d^2 + 57*a^2*b^2*c*d^3 - 9*a^3*b*d^4 + 8*(17*b ^4*c*d^3 + 9*a*b^3*d^4)*x^2 + 2*(59*b^4*c^2*d^2 + 122*a*b^3*c*d^3 + 3*a^2* b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^2), 1/384*(192*sqrt(a*c)*a *b^3*c^2*d^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) + 3*(5*b^4*c^4 - 60*a*b^3*c^3*d - 90*a^2*b^2*c^2*d^2 + 20* a^3*b*c*d^3 - 3*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 + 15*b^4*c^3*d + 337*a*b^3*c^2*d^2 + 57*a^2*b ^2*c*d^3 - 9*a^3*b*d^4 + 8*(17*b^4*c*d^3 + 9*a*b^3*d^4)*x^2 + 2*(59*b^4*c^ 2*d^2 + 122*a*b^3*c*d^3 + 3*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/( b^3*d^2), 1/768*(768*sqrt(-a*c)*a*b^3*c^2*d^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)) - 3*(5*b^4*c^4 - 60*a*b^3*c^3*d - 90*a^2*b^2*c^2*d...
\[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x} \, dx=\int \frac {\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{2}}}{x}\, dx \]
Exception generated. \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x} \, dx=\text {Exception raised: ValueError} \]
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. \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x} \, dx=\int \frac {{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{5/2}}{x} \,d x \]