3.8.18 \(\int \frac {(c+d x)^{5/2}}{x^2 \sqrt {a+b x}} \, dx\) [718]

3.8.18.1 Optimal result
3.8.18.2 Mathematica [A] (verified)
3.8.18.3 Rubi [A] (verified)
3.8.18.4 Maple [B] (verified)
3.8.18.5 Fricas [A] (verification not implemented)
3.8.18.6 Sympy [F]
3.8.18.7 Maxima [F(-2)]
3.8.18.8 Giac [B] (verification not implemented)
3.8.18.9 Mupad [F(-1)]

3.8.18.1 Optimal result

Integrand size = 22, antiderivative size = 160 \[ \int \frac {(c+d x)^{5/2}}{x^2 \sqrt {a+b x}} \, dx=\frac {d (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{a b}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}+\frac {c^{3/2} (b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2}}+\frac {d^{3/2} (5 b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}} \]

output
c^(3/2)*(-5*a*d+b*c)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/ 
a^(3/2)+d^(3/2)*(-a*d+5*b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c) 
^(1/2))/b^(3/2)-c*(d*x+c)^(3/2)*(b*x+a)^(1/2)/a/x+d*(a*d+b*c)*(b*x+a)^(1/2 
)*(d*x+c)^(1/2)/a/b
 
3.8.18.2 Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.88 \[ \int \frac {(c+d x)^{5/2}}{x^2 \sqrt {a+b x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-b c^2+a d^2 x\right )}{a b x}+\frac {c^{3/2} (b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2}}+\frac {d^{3/2} (5 b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}} \]

input
Integrate[(c + d*x)^(5/2)/(x^2*Sqrt[a + b*x]),x]
 
output
(Sqrt[a + b*x]*Sqrt[c + d*x]*(-(b*c^2) + a*d^2*x))/(a*b*x) + (c^(3/2)*(b*c 
 - 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/a^(3/2 
) + (d^(3/2)*(5*b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c 
 + d*x])])/b^(3/2)
 
3.8.18.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {109, 27, 171, 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.

\(\displaystyle \int \frac {(c+d x)^{5/2}}{x^2 \sqrt {a+b x}} \, dx\)

\(\Big \downarrow \) 109

\(\displaystyle -\frac {\int \frac {\sqrt {c+d x} (c (b c-5 a d)-2 d (b c+a d) x)}{2 x \sqrt {a+b x}}dx}{a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\sqrt {c+d x} (c (b c-5 a d)-2 d (b c+a d) x)}{x \sqrt {a+b x}}dx}{2 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}\)

\(\Big \downarrow \) 171

\(\displaystyle -\frac {\frac {\int \frac {b c^2 (b c-5 a d)-a d^2 (5 b c-a d) x}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{b}-\frac {2 d \sqrt {a+b x} \sqrt {c+d x} (a d+b c)}{b}}{2 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}\)

\(\Big \downarrow \) 175

\(\displaystyle -\frac {\frac {b c^2 (b c-5 a d) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx-a d^2 (5 b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{b}-\frac {2 d \sqrt {a+b x} \sqrt {c+d x} (a d+b c)}{b}}{2 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}\)

\(\Big \downarrow \) 66

\(\displaystyle -\frac {\frac {b c^2 (b c-5 a d) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx-2 a d^2 (5 b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}-\frac {2 d \sqrt {a+b x} \sqrt {c+d x} (a d+b c)}{b}}{2 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {\frac {2 b c^2 (b c-5 a d) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}-2 a d^2 (5 b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}-\frac {2 d \sqrt {a+b x} \sqrt {c+d x} (a d+b c)}{b}}{2 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\frac {-\frac {2 b c^{3/2} (b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}-\frac {2 a d^{3/2} (5 b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}}}{b}-\frac {2 d \sqrt {a+b x} \sqrt {c+d x} (a d+b c)}{b}}{2 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}\)

input
Int[(c + d*x)^(5/2)/(x^2*Sqrt[a + b*x]),x]
 
output
-((c*Sqrt[a + b*x]*(c + d*x)^(3/2))/(a*x)) - ((-2*d*(b*c + a*d)*Sqrt[a + b 
*x]*Sqrt[c + d*x])/b + ((-2*b*c^(3/2)*(b*c - 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[ 
a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[a] - (2*a*d^(3/2)*(5*b*c - a*d)*A 
rcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[b])/b)/(2*a)
 

3.8.18.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 66
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]
 

rule 104
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]
 

rule 109
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f 
*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) 
+ c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) 
 + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || 
IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 171
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]
 

rule 175
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]
 

rule 221
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]
 
3.8.18.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(319\) vs. \(2(128)=256\).

Time = 1.64 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.00

method result size
default \(-\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (\ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} d^{3} x \sqrt {a c}-5 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c \,d^{2} x \sqrt {a c}+5 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a b \,c^{2} d x \sqrt {b d}-\ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) b^{2} c^{3} x \sqrt {b d}-2 a \,d^{2} x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}+2 b \,c^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\right )}{2 a \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, x \sqrt {b d}\, \sqrt {a c}\, b}\) \(320\)

input
int((d*x+c)^(5/2)/x^2/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 
output
-1/2*(d*x+c)^(1/2)*(b*x+a)^(1/2)/a*(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*x*(a*c)^(1/2)-5*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* 
x*(a*c)^(1/2)+5*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*b*c^2*d*x*(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)*b^2*c^3*x*(b*d)^(1/2)-2*a*d^2*x*((b*x+a)*(d*x+c))^(1/2 
)*(b*d)^(1/2)*(a*c)^(1/2)+2*b*c^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c 
)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/x/(b*d)^(1/2)/(a*c)^(1/2)/b
 
3.8.18.5 Fricas [A] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 991, normalized size of antiderivative = 6.19 \[ \int \frac {(c+d x)^{5/2}}{x^2 \sqrt {a+b x}} \, dx=\left [-\frac {{\left (5 \, a b c d - a^{2} d^{2}\right )} x \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 ) + {\left (b^{2} c^{2} - 5 \, a b c d\right )} x \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 ) - 4 \, {\left (a d^{2} x - b c^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, a b x}, -\frac {2 \, {\left (5 \, a b c d - a^{2} d^{2}\right )} x \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 ) + {\left (b^{2} c^{2} - 5 \, a b c d\right )} x \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 ) - 4 \, {\left (a d^{2} x - b c^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, a b x}, -\frac {2 \, {\left (b^{2} c^{2} - 5 \, a b c d\right )} x \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 (5 \, a b c d - a^{2} d^{2}\right )} x \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 (a d^{2} x - b c^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, a b x}, -\frac {{\left (b^{2} c^{2} - 5 \, a b c d\right )} x \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 (5 \, a b c d - a^{2} d^{2}\right )} x \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 (a d^{2} x - b c^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, a b x}\right ] \]

input
integrate((d*x+c)^(5/2)/x^2/(b*x+a)^(1/2),x, algorithm="fricas")
 
output
[-1/4*((5*a*b*c*d - a^2*d^2)*x*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) + (b^2*c^2 - 5*a*b*c*d)*x*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) - 4*(a*d^2*x - b*c^2)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b*x), -1/ 
4*(2*(5*a*b*c*d - a^2*d^2)*x*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*s 
qrt(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)) + (b^2*c^2 - 5*a*b*c*d)*x*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) - 4*(a*d^2*x - b*c^2)*sq 
rt(b*x + a)*sqrt(d*x + c))/(a*b*x), -1/4*(2*(b^2*c^2 - 5*a*b*c*d)*x*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)) + (5*a*b*c*d - a^2*d^2)*x*sq 
rt(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*(a*d^2*x - b*c^2)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b*x), -1/2*(( 
b^2*c^2 - 5*a*b*c*d)*x*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)) 
 + (5*a*b*c*d - a^2*d^2)*x*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*...
 
3.8.18.6 Sympy [F]

\[ \int \frac {(c+d x)^{5/2}}{x^2 \sqrt {a+b x}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{x^{2} \sqrt {a + b x}}\, dx \]

input
integrate((d*x+c)**(5/2)/x**2/(b*x+a)**(1/2),x)
 
output
Integral((c + d*x)**(5/2)/(x**2*sqrt(a + b*x)), x)
 
3.8.18.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^2 \sqrt {a+b x}} \, dx=\text {Exception raised: ValueError} \]

input
integrate((d*x+c)^(5/2)/x^2/(b*x+a)^(1/2),x, algorithm="maxima")
 
output
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
 
3.8.18.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 547 vs. \(2 (128) = 256\).

Time = 0.51 (sec) , antiderivative size = 547, normalized size of antiderivative = 3.42 \[ \int \frac {(c+d x)^{5/2}}{x^2 \sqrt {a+b x}} \, dx=\frac {\frac {2 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} d^{2} {\left | b \right |}}{b^{2}} - \frac {{\left (5 \, \sqrt {b d} b c d {\left | b \right |} - \sqrt {b d} a d^{2} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{b^{2}} + \frac {2 \, {\left (\sqrt {b d} b^{2} c^{3} {\left | b \right |} - 5 \, \sqrt {b d} a b c^{2} d {\left | b \right |}\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} a b} - \frac {4 \, {\left (\sqrt {b d} b^{4} c^{4} {\left | b \right |} - 2 \, \sqrt {b d} a b^{3} c^{3} d {\left | b \right |} + \sqrt {b d} a^{2} b^{2} c^{2} d^{2} {\left | b \right |} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c^{3} {\left | b \right |} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b c^{2} d {\left | b \right |}\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4}\right )} a}}{2 \, b} \]

input
integrate((d*x+c)^(5/2)/x^2/(b*x+a)^(1/2),x, algorithm="giac")
 
output
1/2*(2*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*d^2*abs(b)/b^2 - 
(5*sqrt(b*d)*b*c*d*abs(b) - sqrt(b*d)*a*d^2*abs(b))*log((sqrt(b*d)*sqrt(b* 
x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/b^2 + 2*(sqrt(b*d)*b^2*c^ 
3*abs(b) - 5*sqrt(b*d)*a*b*c^2*d*abs(b))*arctan(-1/2*(b^2*c + a*b*d - (sqr 
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b* 
c*d)*b))/(sqrt(-a*b*c*d)*a*b) - 4*(sqrt(b*d)*b^4*c^4*abs(b) - 2*sqrt(b*d)* 
a*b^3*c^3*d*abs(b) + sqrt(b*d)*a^2*b^2*c^2*d^2*abs(b) - sqrt(b*d)*(sqrt(b* 
d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c^3*abs(b) - 
 sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)) 
^2*a*b*c^2*d*abs(b))/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*d)* 
sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d 
)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d 
)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)*a))/b
 
3.8.18.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/2}}{x^2 \sqrt {a+b x}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/2}}{x^2\,\sqrt {a+b\,x}} \,d x \]

input
int((c + d*x)^(5/2)/(x^2*(a + b*x)^(1/2)),x)
 
output
int((c + d*x)^(5/2)/(x^2*(a + b*x)^(1/2)), x)