Integrand size = 22, antiderivative size = 171 \[ \int \frac {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx=\frac {d (7 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}-\frac {2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {\sqrt {d} \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2}} \]
-2*c^(5/2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(1/2)+1/ 4*(3*a^2*d^2-10*a*b*c*d+15*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/ (d*x+c)^(1/2))*d^(1/2)/b^(5/2)+1/2*d*(d*x+c)^(3/2)*(b*x+a)^(1/2)/b+1/4*d*( -3*a*d+7*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^2
Time = 0.58 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.87 \[ \int \frac {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx=\frac {1}{4} \left (\frac {d \sqrt {a+b x} \sqrt {c+d x} (9 b c-3 a d+2 b d x)}{b^2}-\frac {8 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {\sqrt {d} \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2}}\right ) \]
((d*Sqrt[a + b*x]*Sqrt[c + d*x]*(9*b*c - 3*a*d + 2*b*d*x))/b^2 - (8*c^(5/2 )*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[a] + (Sqr t[d]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x]) /(Sqrt[b]*Sqrt[c + d*x])])/b^(5/2))/4
Time = 0.30 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.08, 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.
| \(\displaystyle \int \frac {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx\) |
| \(\Big \downarrow \) 113 |
| \(\displaystyle \frac {\int \frac {\sqrt {c+d x} \left (4 b c^2+d (7 b c-3 a d) x\right )}{2 x \sqrt {a+b x}}dx}{2 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {c+d x} \left (4 b c^2+d (7 b c-3 a d) x\right )}{x \sqrt {a+b x}}dx}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {\frac {\int \frac {8 b^2 c^3+d \left (15 b^2 c^2-10 a b d c+3 a^2 d^2\right ) x}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{b}}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {8 b^2 c^3+d \left (15 b^2 c^2-10 a b d c+3 a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{b}}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {\frac {d \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx+8 b^2 c^3 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{b}}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {\frac {2 d \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+8 b^2 c^3 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{b}}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {\frac {2 d \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+16 b^2 c^3 \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{b}}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {2 \sqrt {d} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}}-\frac {16 b^2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{b}}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\) |
(d*Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*b) + ((d*(7*b*c - 3*a*d)*Sqrt[a + b*x ]*Sqrt[c + d*x])/b + ((-16*b^2*c^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sq rt[a]*Sqrt[c + d*x])])/Sqrt[a] + (2*Sqrt[d]*(15*b^2*c^2 - 10*a*b*c*d + 3*a ^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[b]) /(2*b))/(4*b)
3.8.17.3.1 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 = 0.57 (sec) , antiderivative size = 342, normalized size of antiderivative = 2.00
| method | result | size |
| default | \(-\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (8 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) b^{2} c^{3} \sqrt {b d}-4 b \,d^{2} x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}-3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} d^{3} \sqrt {a c}+10 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c \,d^{2} \sqrt {a c}-15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2} d \sqrt {a c}+6 a \,d^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}-18 b c d \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} \sqrt {b d}\, \sqrt {a c}}\) | \(342\) |
-1/8*(d*x+c)^(1/2)*(b*x+a)^(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)*b^2*c^3*(b*d)^(1/2)-4*b*d^2*x*((b*x+a)*(d*x+c))^ (1/2)*(b*d)^(1/2)*(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))*a^2*d^3*(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*c*d^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))*b^2*c^2*d*(a*c)^(1/2)+6*a*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1 /2)*(a*c)^(1/2)-18*b*c*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2))/ ((b*x+a)*(d*x+c))^(1/2)/b^2/(b*d)^(1/2)/(a*c)^(1/2)
Time = 1.53 (sec) , antiderivative size = 987, normalized size of antiderivative = 5.77 \[ \int \frac {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx=\left [\frac {8 \, b^{2} c^{2} \sqrt {\frac {c}{a}} \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^{2} c + {\left (a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {c}{a}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + {\left (15 \, b^{2} c^{2} - 10 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {\frac {d}{b}} \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^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b d^{2} x + 9 \, b c d - 3 \, a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b^{2}}, \frac {4 \, b^{2} c^{2} \sqrt {\frac {c}{a}} \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^{2} c + {\left (a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {c}{a}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - {\left (15 \, b^{2} c^{2} - 10 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) + 2 \, {\left (2 \, b d^{2} x + 9 \, b c d - 3 \, a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b^{2}}, \frac {16 \, b^{2} c^{2} \sqrt {-\frac {c}{a}} \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 {c}{a}}}{2 \, {\left (b c d x^{2} + a c^{2} + {\left (b c^{2} + a c d\right )} x\right )}}\right ) + {\left (15 \, b^{2} c^{2} - 10 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {\frac {d}{b}} \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^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b d^{2} x + 9 \, b c d - 3 \, a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b^{2}}, \frac {8 \, b^{2} c^{2} \sqrt {-\frac {c}{a}} \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 {c}{a}}}{2 \, {\left (b c d x^{2} + a c^{2} + {\left (b c^{2} + a c d\right )} x\right )}}\right ) - {\left (15 \, b^{2} c^{2} - 10 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) + 2 \, {\left (2 \, b d^{2} x + 9 \, b c d - 3 \, a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b^{2}}\right ] \]
[1/16*(8*b^2*c^2*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2 )*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c /a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2 )*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b^2*d *x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a *b*d^2)*x) + 4*(2*b*d^2*x + 9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*sqrt(d*x + c) )/b^2, 1/8*(4*b^2*c^2*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^ 2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*s qrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - (15*b^2*c^2 - 10*a*b*c*d + 3*a^ 2*d^2)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) + 2*(2*b*d^2*x + 9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/b^2, 1/16*(16*b^2*c^2*sqrt (-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt (-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) + (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8* (b^2*c*d + a*b*d^2)*x) + 4*(2*b*d^2*x + 9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*s qrt(d*x + c))/b^2, 1/8*(8*b^2*c^2*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a* d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) - (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*sqrt(-d/b)*arctan(1...
\[ \int \frac {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{x \sqrt {a + b x}}\, dx \]
Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx=\text {Exception raised: ValueError} \]
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. \[ \int \frac {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/2}}{x\,\sqrt {a+b\,x}} \,d x \]