\(\int \frac {(a+b x)^{5/2}}{x^2 (c+d x)^{3/2}} \, dx\) [304]

Optimal result

Integrand size = 22, antiderivative size = 164 \[ \int \frac {(a+b x)^{5/2}}{x^2 (c+d x)^{3/2}} \, dx=-\frac {(2 b c-3 a d) (b c-a d) \sqrt {a+b x}}{c^2 d \sqrt {c+d x}}-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}}-\frac {a^{3/2} (5 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}+\frac {2 b^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{3/2}} \] Output:

-(-3*a*d+2*b*c)*(-a*d+b*c)*(b*x+a)^(1/2)/c^2/d/(d*x+c)^(1/2)-a*(b*x+a)^(3/ 
2)/c/x/(d*x+c)^(1/2)-a^(3/2)*(-3*a*d+5*b*c)*arctanh(c^(1/2)*(b*x+a)^(1/2)/ 
a^(1/2)/(d*x+c)^(1/2))/c^(5/2)+2*b^(5/2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^( 
1/2)/(d*x+c)^(1/2))/d^(3/2)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.91 \[ \int \frac {(a+b x)^{5/2}}{x^2 (c+d x)^{3/2}} \, dx=-\frac {\sqrt {a+b x} \left (2 b^2 c^2 x-4 a b c d x+a^2 d (c+3 d x)\right )}{c^2 d x \sqrt {c+d x}}+\frac {a^{3/2} (-5 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}+\frac {2 b^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{3/2}} \] Input:

Integrate[(a + b*x)^(5/2)/(x^2*(c + d*x)^(3/2)),x]
 

Output:

-((Sqrt[a + b*x]*(2*b^2*c^2*x - 4*a*b*c*d*x + a^2*d*(c + 3*d*x)))/(c^2*d*x 
*Sqrt[c + d*x])) + (a^(3/2)*(-5*b*c + 3*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x 
])/(Sqrt[a]*Sqrt[c + d*x])])/c^(5/2) + (2*b^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a 
+ b*x])/(Sqrt[b]*Sqrt[c + d*x])])/d^(3/2)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {109, 27, 167, 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.

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 167

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 175

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

\(\Big \downarrow \) 66

\(\displaystyle \frac {\frac {a^2 d (5 b c-3 a d) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+4 b^3 c^2 \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{c d}-\frac {2 \sqrt {a+b x} (2 b c-3 a d) (b c-a d)}{c d \sqrt {c+d x}}}{2 c}-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {\frac {2 a^2 d (5 b c-3 a d) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+4 b^3 c^2 \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{c d}-\frac {2 \sqrt {a+b x} (2 b c-3 a d) (b c-a d)}{c d \sqrt {c+d x}}}{2 c}-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {4 b^{5/2} c^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {d}}-\frac {2 a^{3/2} d (5 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {c}}}{c d}-\frac {2 \sqrt {a+b x} (2 b c-3 a d) (b c-a d)}{c d \sqrt {c+d x}}}{2 c}-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}}\)

Input:

Int[(a + b*x)^(5/2)/(x^2*(c + d*x)^(3/2)),x]
 

Output:

-((a*(a + b*x)^(3/2))/(c*x*Sqrt[c + d*x])) + ((-2*(2*b*c - 3*a*d)*(b*c - a 
*d)*Sqrt[a + b*x])/(c*d*Sqrt[c + d*x]) + ((-2*a^(3/2)*d*(5*b*c - 3*a*d)*Ar 
cTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[c] + (4*b^(5/ 
2)*c^2*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[d])/ 
(c*d))/(2*c)
 

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 167

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(501\) vs. \(2(132)=264\).

Time = 0.22 (sec) , antiderivative size = 502, normalized size of antiderivative = 3.06

method	result	size
default	\(\frac {\sqrt {b x +a}\, \left (3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{3} d^{3} x^{2} \sqrt {d b}-5 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{2} b c \,d^{2} x^{2} \sqrt {d b}+2 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) b^{3} c^{2} d \,x^{2} \sqrt {a c}+3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) \sqrt {d b}\, a^{3} c \,d^{2} x -5 \sqrt {d b}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{2} b \,c^{2} d x +2 \sqrt {a c}\, \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) b^{3} c^{3} x -6 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, a^{2} d^{2} x +8 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, a b c d x -4 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, b^{2} c^{2} x -2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, a^{2} c d \right )}{2 c^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, x \sqrt {d b}\, \sqrt {a c}\, \sqrt {x d +c}\, d}\)	\(502\)

Input:

int((b*x+a)^(5/2)/x^2/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/2*(b*x+a)^(1/2)*(3*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^3*x^2*(d*b)^(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^2*b*c*d^2*x^2*(d*b)^(1/2)+2*ln(1/2*(2*b*d*x+2 
*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*b^3*c^2*d*x^2*( 
a*c)^(1/2)+3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/ 
x)*(d*b)^(1/2)*a^3*c*d^2*x-5*(d*b)^(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*c^2*d*x+2*(a*c)^(1/2)*ln(1/2*(2*b*d*x 
+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*b^3*c^3*x-6*( 
(b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2)*a^2*d^2*x+8*((b*x+a)*(d*x+c 
))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2)*a*b*c*d*x-4*((b*x+a)*(d*x+c))^(1/2)*(d*b) 
^(1/2)*(a*c)^(1/2)*b^2*c^2*x-2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a*c)^( 
1/2)*a^2*c*d)/c^2/((b*x+a)*(d*x+c))^(1/2)/x/(d*b)^(1/2)/(a*c)^(1/2)/(d*x+c 
)^(1/2)/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (132) = 264\).

Time = 1.00 (sec) , antiderivative size = 1233, normalized size of antiderivative = 7.52 \[ \int \frac {(a+b x)^{5/2}}{x^2 (c+d x)^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((b*x+a)^(5/2)/x^2/(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

[1/4*(2*(b^2*c^2*d*x^2 + b^2*c^3*x)*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) - ((5*a*b*c*d^2 - 3*a^2*d^3)* 
x^2 + (5*a*b*c^2*d - 3*a^2*c*d^2)*x)*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) - 4*(a^2*c*d + (2* 
b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^2*d^2* 
x^2 + c^3*d*x), -1/4*(4*(b^2*c^2*d*x^2 + b^2*c^3*x)*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)) + ((5*a*b*c*d^2 - 3*a^2*d^3)*x^2 + (5*a*b*c^2* 
d - 3*a^2*c*d^2)*x)*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)*sqr 
t(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(a^2*c*d + (2*b^2*c^2 - 4*a*b*c 
*d + 3*a^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^2*d^2*x^2 + c^3*d*x), 1 
/2*(((5*a*b*c*d^2 - 3*a^2*d^3)*x^2 + (5*a*b*c^2*d - 3*a^2*c*d^2)*x)*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)) + (b^2*c^2*d*x^2 + b^2*c^3*x 
)*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) - 2*(a^2*c*d + (2*b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*x)*sqrt(b...
 

Sympy [F]

\[ \int \frac {(a+b x)^{5/2}}{x^2 (c+d x)^{3/2}} \, dx=\int \frac {\left (a + b x\right )^{\frac {5}{2}}}{x^{2} \left (c + d x\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((b*x+a)**(5/2)/x**2/(d*x+c)**(3/2),x)
 

Output:

Integral((a + b*x)**(5/2)/(x**2*(c + d*x)**(3/2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b x)^{5/2}}{x^2 (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x+a)^(5/2)/x^2/(d*x+c)^(3/2),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
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 560 vs. \(2 (132) = 264\).

Time = 0.70 (sec) , antiderivative size = 560, normalized size of antiderivative = 3.41 \[ \int \frac {(a+b x)^{5/2}}{x^2 (c+d x)^{3/2}} \, dx=-\frac {\sqrt {b d} b^{3} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{d^{2} {\left | b \right |}} - \frac {{\left (5 \, \sqrt {b d} a^{2} b^{3} c - 3 \, \sqrt {b d} a^{3} b^{2} d\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} b c^{2} {\left | b \right |}} - \frac {2 \, {\left (\sqrt {b d} a^{2} b^{5} c^{2} - 2 \, \sqrt {b d} a^{3} b^{4} c d + \sqrt {b d} a^{4} b^{3} d^{2} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{3} c - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{3} b^{2} d\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4}\right )} c^{2} {\left | b \right |}} - \frac {2 \, {\left (b^{4} c^{2} {\left | b \right |} - 2 \, a b^{3} c d {\left | b \right |} + a^{2} b^{2} d^{2} {\left | b \right |}\right )} \sqrt {b x + a}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} b^{2} c^{2} d} \] Input:

integrate((b*x+a)^(5/2)/x^2/(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

-sqrt(b*d)*b^3*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - 
 a*b*d))^2)/(d^2*abs(b)) - (5*sqrt(b*d)*a^2*b^3*c - 3*sqrt(b*d)*a^3*b^2*d) 
*arctan(-1/2*(b^2*c + a*b*d - (sqrt(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)*b*c^2*abs(b)) - 
 2*(sqrt(b*d)*a^2*b^5*c^2 - 2*sqrt(b*d)*a^3*b^4*c*d + sqrt(b*d)*a^4*b^3*d^ 
2 - sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b* 
d))^2*a^2*b^3*c - sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + 
 a)*b*d - a*b*d))^2*a^3*b^2*d)/((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)*c^2*ab 
s(b)) - 2*(b^4*c^2*abs(b) - 2*a*b^3*c*d*abs(b) + a^2*b^2*d^2*abs(b))*sqrt( 
b*x + a)/(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*b^2*c^2*d)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{5/2}}{x^2 (c+d x)^{3/2}} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/2}}{x^2\,{\left (c+d\,x\right )}^{3/2}} \,d x \] Input:

int((a + b*x)^(5/2)/(x^2*(c + d*x)^(3/2)),x)
 

Output:

int((a + b*x)^(5/2)/(x^2*(c + d*x)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 1475, normalized size of antiderivative = 8.99 \[ \int \frac {(a+b x)^{5/2}}{x^2 (c+d x)^{3/2}} \, dx =\text {Too large to display} \] Input:

int((b*x+a)^(5/2)/x^2/(d*x+c)^(3/2),x)
 

Output:

( - 6*sqrt(c + d*x)*sqrt(a + b*x)*a**3*c**2*d**3 - 18*sqrt(c + d*x)*sqrt(a 
 + b*x)*a**3*c*d**4*x - 2*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*c**3*d**2 + 1 
8*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*c**2*d**3*x - 4*sqrt(c + d*x)*sqrt(a 
+ b*x)*a*b**2*c**3*d**2*x - 4*sqrt(c + d*x)*sqrt(a + b*x)*b**3*c**4*d*x - 
9*sqrt(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b* 
c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**3*c*d**4*x - 9*sqrt 
(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + s 
qrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**3*d**5*x**2 + 12*sqrt(c)* 
sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt( 
d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**2*b*c**2*d**3*x + 12*sqrt(c)* 
sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt( 
d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**2*b*c*d**4*x**2 + 5*sqrt(c)*s 
qrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d 
)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a*b**2*c**3*d**2*x + 5*sqrt(c)*sq 
rt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d) 
*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a*b**2*c**2*d**3*x**2 - 9*sqrt(c)* 
sqrt(a)*log(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)* 
sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**3*c*d**4*x - 9*sqrt(c)*sqrt(a)*l 
og(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + 
b*x) + sqrt(b)*sqrt(c + d*x))*a**3*d**5*x**2 + 12*sqrt(c)*sqrt(a)*log(s...