Integrand size = 22, antiderivative size = 171 \[ \int \frac {(a+b x)^{5/2}}{x \sqrt {c+d x}} \, dx=-\frac {b (3 b c-7 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^2}+\frac {b (a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {c}}+\frac {\sqrt {b} \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{5/2}} \]
1/4*(15*a^2*d^2-10*a*b*c*d+3*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^(5/2)-2*a^(5/2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a ^(1/2)/(d*x+c)^(1/2))/c^(1/2)+1/2*b*(b*x+a)^(3/2)*(d*x+c)^(1/2)/d-1/4*b*(- 7*a*d+3*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^2
Time = 0.55 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b x)^{5/2}}{x \sqrt {c+d x}} \, dx=\frac {1}{4} \left (\frac {b \sqrt {a+b x} \sqrt {c+d x} (-3 b c+9 a d+2 b d x)}{d^2}-\frac {8 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{\sqrt {c}}+\frac {\sqrt {b} \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{d^{5/2}}\right ) \]
((b*Sqrt[a + b*x]*Sqrt[c + d*x]*(-3*b*c + 9*a*d + 2*b*d*x))/d^2 - (8*a^(5/ 2)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/Sqrt[c] + (Sq rt[b]*(3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x] )/(Sqrt[d]*Sqrt[a + b*x])])/d^(5/2))/4
Time = 0.31 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {113, 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.
| \(\displaystyle \int \frac {(a+b x)^{5/2}}{x \sqrt {c+d x}} \, dx\) |
| \(\Big \downarrow \) 113 |
| \(\displaystyle \frac {\int \frac {\sqrt {a+b x} \left (4 a^2 d-b (3 b c-7 a d) x\right )}{2 x \sqrt {c+d x}}dx}{2 d}+\frac {b (a+b x)^{3/2} \sqrt {c+d x}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {a+b x} \left (4 a^2 d-b (3 b c-7 a d) x\right )}{x \sqrt {c+d x}}dx}{4 d}+\frac {b (a+b x)^{3/2} \sqrt {c+d x}}{2 d}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {\frac {\int \frac {8 d^2 a^3+b \left (3 b^2 c^2-10 a b d c+15 a^2 d^2\right ) x}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{d}-\frac {b \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{d}}{4 d}+\frac {b (a+b x)^{3/2} \sqrt {c+d x}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {8 d^2 a^3+b \left (3 b^2 c^2-10 a b d c+15 a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}-\frac {b \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{d}}{4 d}+\frac {b (a+b x)^{3/2} \sqrt {c+d x}}{2 d}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {\frac {8 a^3 d^2 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+b \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}-\frac {b \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{d}}{4 d}+\frac {b (a+b x)^{3/2} \sqrt {c+d x}}{2 d}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {\frac {8 a^3 d^2 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+2 b \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 d}-\frac {b \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{d}}{4 d}+\frac {b (a+b x)^{3/2} \sqrt {c+d x}}{2 d}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {\frac {16 a^3 d^2 \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+2 b \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 d}-\frac {b \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{d}}{4 d}+\frac {b (a+b x)^{3/2} \sqrt {c+d x}}{2 d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {2 \sqrt {b} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {d}}-\frac {16 a^{5/2} d^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {c}}}{2 d}-\frac {b \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{d}}{4 d}+\frac {b (a+b x)^{3/2} \sqrt {c+d x}}{2 d}\) |
(b*(a + b*x)^(3/2)*Sqrt[c + d*x])/(2*d) + (-((b*(3*b*c - 7*a*d)*Sqrt[a + b *x]*Sqrt[c + d*x])/d) + ((-16*a^(5/2)*d^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/ (Sqrt[a]*Sqrt[c + d*x])])/Sqrt[c] + (2*Sqrt[b]*(3*b^2*c^2 - 10*a*b*c*d + 1 5*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[ d])/(2*d))/(4*d)
3.7.77.3.1 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[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]
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. \(341\) vs. \(2(133)=266\).
Time = 0.56 (sec) , antiderivative size = 342, normalized size of antiderivative = 2.00
| method | result | size |
| default | \(-\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (8 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{3} d^{2} \sqrt {b d}-4 b^{2} d x \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b \,d^{2} \sqrt {a c}+10 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c d \sqrt {a c}-3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{2} \sqrt {a c}-18 a b d \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+6 b^{2} c \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, d^{2} \sqrt {b d}\, \sqrt {a c}}\) | \(342\) |
-1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(8*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*(b*d)^(1/2)-4*b^2*d*x*(b*d)^(1/2)*(a*c)^ (1/2)*((b*x+a)*(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*(a*c)^(1/2)+10*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*(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*(a*c)^(1/2)-18*a*b*d*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a )*(d*x+c))^(1/2)+6*b^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)/d^2/(b*d)^(1/2)/(a*c)^(1/2)
Time = 1.54 (sec) , antiderivative size = 987, normalized size of antiderivative = 5.77 \[ \int \frac {(a+b x)^{5/2}}{x \sqrt {c+d x}} \, dx=\left [\frac {8 \, a^{2} d^{2} \sqrt {\frac {a}{c}} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c^{2} + {\left (b c^{2} + a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {a}{c}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + {\left (3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{2} d x - 3 \, b^{2} c + 9 \, a b d\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, d^{2}}, \frac {4 \, a^{2} d^{2} \sqrt {\frac {a}{c}} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c^{2} + {\left (b c^{2} + a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {a}{c}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - {\left (3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \, {\left (2 \, b^{2} d x - 3 \, b^{2} c + 9 \, a b d\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, d^{2}}, \frac {16 \, a^{2} d^{2} \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b d x^{2} + a^{2} c + {\left (a b c + a^{2} d\right )} x\right )}}\right ) + {\left (3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{2} d x - 3 \, b^{2} c + 9 \, a b d\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, d^{2}}, \frac {8 \, a^{2} d^{2} \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b d x^{2} + a^{2} c + {\left (a b c + a^{2} d\right )} x\right )}}\right ) - {\left (3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \, {\left (2 \, b^{2} d x - 3 \, b^{2} c + 9 \, a b d\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, d^{2}}\right ] \]
[1/16*(8*a^2*d^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) + (3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^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) + 4*(2*b^2*d*x - 3*b^2*c + 9*a*b*d)*sqrt(b*x + a)*sqrt(d*x + c) )/d^2, 1/8*(4*a^2*d^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)*s qrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - (3*b^2*c^2 - 10*a*b*c*d + 15*a^ 2*d^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)) + 2*(2*b^2*d*x - 3*b^2*c + 9*a*b*d)*sqrt(b*x + a)*sqrt(d*x + c))/d^2, 1/16*(16*a^2*d^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 + 15*a^2*d^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) + 4*(2*b^2*d*x - 3*b^2*c + 9*a*b*d)*sqrt(b*x + a)*s qrt(d*x + c))/d^2, 1/8*(8*a^2*d^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 + 15*a^2*d^2)*sqrt(-b/d)*arctan(1...
\[ \int \frac {(a+b x)^{5/2}}{x \sqrt {c+d x}} \, dx=\int \frac {\left (a + b x\right )^{\frac {5}{2}}}{x \sqrt {c + d x}}\, dx \]
Exception generated. \[ \int \frac {(a+b x)^{5/2}}{x \sqrt {c+d x}} \, dx=\text {Exception raised: ValueError} \]
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. \[ \int \frac {(a+b x)^{5/2}}{x \sqrt {c+d x}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {(a+b x)^{5/2}}{x \sqrt {c+d x}} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/2}}{x\,\sqrt {c+d\,x}} \,d x \]