\(\int \frac {(a+b x)^{5/2}}{x (c+d x)^{3/2}} \, dx\) [303]

Optimal result

Integrand size = 22, antiderivative size = 163 \[ \int \frac {(a+b x)^{5/2}}{x (c+d x)^{3/2}} \, dx=-\frac {2 (b c-a d) (a+b x)^{3/2}}{c d \sqrt {c+d x}}+\frac {b (3 b c-2 a d) \sqrt {a+b x} \sqrt {c+d x}}{c d^2}-\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}}-\frac {b^{3/2} (3 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x)^{5/2}}{x (c+d x)^{3/2}} \, dx=\frac {\sqrt {a+b x} \left (-4 a b c d+2 a^2 d^2+b^2 c (3 c+d x)\right )}{c d^2 \sqrt {c+d x}}-\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}}-\frac {b^{3/2} (3 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {109, 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.

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 171

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 175

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(491\) vs. \(2(131)=262\).

Time = 0.22 (sec) , antiderivative size = 492, normalized size of antiderivative = 3.02

method	result	size
default	\(\frac {\sqrt {b x +a}\, \left (5 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a \,b^{2} c \,d^{2} x \sqrt {a c}-3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) b^{3} c^{2} d x \sqrt {a c}-2 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{3} d^{3} x \sqrt {d b}+5 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) \sqrt {a c}\, a \,b^{2} c^{2} d -3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) \sqrt {a c}\, b^{3} c^{3}-2 \sqrt {d b}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{3} c \,d^{2}+2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, b^{2} c d x +4 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, a^{2} d^{2}-8 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, a b c d +6 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, b^{2} c^{2}\right )}{2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, \sqrt {x d +c}\, c \,d^{2}}\)	\(492\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (131) = 262\).

Time = 1.09 (sec) , antiderivative size = 1181, normalized size of antiderivative = 7.25 \[ \int \frac {(a+b x)^{5/2}}{x (c+d x)^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((b*x+a)^(5/2)/x/(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 [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.57 \[ \int \frac {(a+b x)^{5/2}}{x (c+d x)^{3/2}} \, dx=-\frac {2 \, \sqrt {b d} a^{3} b \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} c {\left | b \right |}} + \frac {\sqrt {b x + a} {\left (\frac {{\left (b x + a\right )} b^{3}}{d {\left | b \right |}} + \frac {3 \, b^{6} c^{2} d - 5 \, a b^{5} c d^{2} + 2 \, a^{2} b^{4} d^{3}}{b^{2} c d^{3} {\left | b \right |}}\right )}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {{\left (3 \, \sqrt {b d} b^{3} c - 5 \, \sqrt {b d} a b^{2} d\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{2 \, d^{3} {\left | b \right |}} \] Input:

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

Output:

-2*sqrt(b*d)*a^3*b*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)*c*abs(b)) + sqrt(b*x + a)*((b*x + a)*b^3/(d*abs(b)) + (3*b^6*c^2*d - 5* 
a*b^5*c*d^2 + 2*a^2*b^4*d^3)/(b^2*c*d^3*abs(b)))/sqrt(b^2*c + (b*x + a)*b* 
d - a*b*d) + 1/2*(3*sqrt(b*d)*b^3*c - 5*sqrt(b*d)*a*b^2*d)*log((sqrt(b*d)* 
sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(d^3*abs(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 671, normalized size of antiderivative = 4.12 \[ \int \frac {(a+b x)^{5/2}}{x (c+d x)^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(8*sqrt(c + d*x)*sqrt(a + b*x)*a**2*c*d**3 - 16*sqrt(c + d*x)*sqrt(a + b*x 
)*a*b*c**2*d**2 + 12*sqrt(c + d*x)*sqrt(a + b*x)*b**2*c**3*d + 4*sqrt(c + 
d*x)*sqrt(a + b*x)*b**2*c**2*d**2*x + 4*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*c*d**3 + 4*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*d**4*x + 4*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*c*d**3 + 4*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*d**4*x - 
4*sqrt(c)*sqrt(a)*log(2*sqrt(d)*sqrt(b)*sqrt(c + d*x)*sqrt(a + b*x) + 2*sq 
rt(d)*sqrt(c)*sqrt(b)*sqrt(a) + 2*b*d*x)*a**2*c*d**3 - 4*sqrt(c)*sqrt(a)*l 
og(2*sqrt(d)*sqrt(b)*sqrt(c + d*x)*sqrt(a + b*x) + 2*sqrt(d)*sqrt(c)*sqrt( 
b)*sqrt(a) + 2*b*d*x)*a**2*d**4*x + 20*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a 
 + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b*c**3*d + 20*sqrt(d)* 
sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c 
))*a*b*c**2*d**2*x - 12*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt( 
b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**2*c**4 - 12*sqrt(d)*sqrt(b)*log((sqr 
t(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**2*c**3*d*x 
 + 8*sqrt(d)*sqrt(b)*a**2*c**2*d**2 + 8*sqrt(d)*sqrt(b)*a**2*c*d**3*x -...