Integrand size = 22, antiderivative size = 218 \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x} \, dx=-\frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 d^2}+\frac {(b c+5 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d}+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}-2 a^{5/2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{5/2}} \]
1/8*(5*a^3*d^3+15*a^2*b*c*d^2-5*a*b^2*c^2*d+b^3*c^3)*arctanh(d^(1/2)*(b*x+ a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/d^(5/2)/b^(1/2)-2*a^(5/2)*arctanh(c^(1/2)* (b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))*c^(1/2)+1/12*(5*a*d+b*c)*(b*x+a)^(3/2 )*(d*x+c)^(1/2)/d+1/3*(b*x+a)^(5/2)*(d*x+c)^(1/2)-1/8*(-5*a*d+b*c)*(a*d+b* c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^2
Time = 0.56 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (33 a^2 d^2+2 a b d (7 c+13 d x)+b^2 \left (-3 c^2+2 c d x+8 d^2 x^2\right )\right )}{24 d^2}-2 a^{5/2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )+\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 \sqrt {b} d^{5/2}} \]
(Sqrt[a + b*x]*Sqrt[c + d*x]*(33*a^2*d^2 + 2*a*b*d*(7*c + 13*d*x) + b^2*(- 3*c^2 + 2*c*d*x + 8*d^2*x^2)))/(24*d^2) - 2*a^(5/2)*Sqrt[c]*ArcTanh[(Sqrt[ a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])] + ((b^3*c^3 - 5*a*b^2*c^2*d + 1 5*a^2*b*c*d^2 + 5*a^3*d^3)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(8*Sqrt[b]*d^(5/2))
Time = 0.37 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {112, 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.
| \(\displaystyle \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x} \, dx\) |
| \(\Big \downarrow \) 112 |
| \(\displaystyle \frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}-\frac {1}{3} \int -\frac {(a+b x)^{3/2} (6 a c+(b c+5 a d) x)}{2 x \sqrt {c+d x}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \int \frac {(a+b x)^{3/2} (6 a c+(b c+5 a d) x)}{x \sqrt {c+d x}}dx+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{6} \left (\frac {\int \frac {3 \sqrt {a+b x} \left (8 a^2 c d-(b c-5 a d) (b c+a d) x\right )}{2 x \sqrt {c+d x}}dx}{2 d}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 a d+b c)}{2 d}\right )+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \left (\frac {3 \int \frac {\sqrt {a+b x} \left (8 a^2 c d-(b c-5 a d) (b c+a d) x\right )}{x \sqrt {c+d x}}dx}{4 d}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 a d+b c)}{2 d}\right )+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{6} \left (\frac {3 \left (\frac {\int \frac {16 c d^2 a^3+\left (b^3 c^3-5 a b^2 d c^2+15 a^2 b d^2 c+5 a^3 d^3\right ) x}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{d}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{d}\right )}{4 d}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 a d+b c)}{2 d}\right )+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \left (\frac {3 \left (\frac {\int \frac {16 c d^2 a^3+\left (b^3 c^3-5 a b^2 d c^2+15 a^2 b d^2 c+5 a^3 d^3\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{d}\right )}{4 d}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 a d+b c)}{2 d}\right )+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {1}{6} \left (\frac {3 \left (\frac {16 a^3 c d^2 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+\left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{d}\right )}{4 d}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 a d+b c)}{2 d}\right )+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {1}{6} \left (\frac {3 \left (\frac {16 a^3 c d^2 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+2 \left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 d}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{d}\right )}{4 d}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 a d+b c)}{2 d}\right )+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{6} \left (\frac {3 \left (\frac {32 a^3 c d^2 \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+2 \left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 d}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{d}\right )}{4 d}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 a d+b c)}{2 d}\right )+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{6} \left (\frac {3 \left (\frac {\frac {2 \left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {d}}-32 a^{5/2} \sqrt {c} d^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{2 d}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{d}\right )}{4 d}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 a d+b c)}{2 d}\right )+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}\) |
((a + b*x)^(5/2)*Sqrt[c + d*x])/3 + (((b*c + 5*a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(2*d) + (3*(-(((b*c - 5*a*d)*(b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d *x])/d) + (-32*a^(5/2)*Sqrt[c]*d^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a ]*Sqrt[c + d*x])] + (2*(b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d ^3)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*Sqr t[d]))/(2*d)))/(4*d))/6
3.7.50.3.1 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. \(498\) vs. \(2(174)=348\).
Time = 2.04 (sec) , antiderivative size = 499, normalized size of antiderivative = 2.29
| method | result | size |
| default | \(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (16 b^{2} d^{2} x^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a^{3} d^{3}+45 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a^{2} b c \,d^{2}-15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a \,b^{2} c^{2} d +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, b^{3} c^{3}-48 \sqrt {b d}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{3} c \,d^{2}+52 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,d^{2} x +4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c d x +66 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2}+28 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d -6 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, d^{2} \sqrt {b d}\, \sqrt {a c}}\) | \(499\) |
1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(16*b^2*d^2*x^2*(a*c)^(1/2)*((b*x+a)*(d*x +c))^(1/2)*(b*d)^(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^3*d^3+45*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*c*d ^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*b^2*c^2*d+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*c)^(1/2)*b^3*c^3-48*(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*c*d^2+52* (b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*d^2*x+4*(b*d)^(1/2)*(a *c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c*d*x+66*(b*d)^(1/2)*(a*c)^(1/2)*((b *x+a)*(d*x+c))^(1/2)*a^2*d^2+28*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^ (1/2)*a*b*c*d-6*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c^2)/( (b*x+a)*(d*x+c))^(1/2)/d^2/(b*d)^(1/2)/(a*c)^(1/2)
Time = 3.07 (sec) , antiderivative size = 1197, normalized size of antiderivative = 5.49 \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x} \, dx=\text {Too large to display} \]
[1/96*(48*sqrt(a*c)*a^2*b*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*(b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c *d^2 + 5*a^3*d^3)*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*(8*b^3*d^3*x^2 - 3*b^3*c^2*d + 14*a*b^2*c*d^2 + 3 3*a^2*b*d^3 + 2*(b^3*c*d^2 + 13*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c)) /(b*d^3), 1/48*(24*sqrt(a*c)*a^2*b*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*(b^3*c^3 - 5*a*b^2*c^2*d + 1 5*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqr t(-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*(8*b^3*d^3*x^2 - 3*b^3*c^2*d + 14*a*b^2*c*d^2 + 33*a^2*b*d^ 3 + 2*(b^3*c*d^2 + 13*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^3), 1/96*(96*sqrt(-a*c)*a^2*b*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*(b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*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*(8* b^3*d^3*x^2 - 3*b^3*c^2*d + 14*a*b^2*c*d^2 + 33*a^2*b*d^3 + 2*(b^3*c*d^...
\[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x} \, dx=\int \frac {\left (a + b x\right )^{\frac {5}{2}} \sqrt {c + d x}}{x}\, dx \]
Exception generated. \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x} \, dx=\text {Exception raised: ValueError} \]
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. \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/2}\,\sqrt {c+d\,x}}{x} \,d x \]