\(\int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^3} \, dx\) [187]

Optimal result

Integrand size = 22, antiderivative size = 211 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^3} \, dx=\frac {d (b c+11 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 a}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 a x}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}+\frac {\sqrt {c} \left (b^2 c^2-10 a b c d-15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2}}+\frac {d^{3/2} (5 b c+a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}} \] Output:

1/4*d*(11*a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a-1/4*(5*a*d+b*c)*(b*x+a)^( 
1/2)*(d*x+c)^(3/2)/a/x-1/2*(b*x+a)^(1/2)*(d*x+c)^(5/2)/x^2+1/4*c^(1/2)*(-1 
5*a^2*d^2-10*a*b*c*d+b^2*c^2)*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^(1/2)
 

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^3} \, dx=\frac {\sqrt {c} \left (b^2 c^2-10 a b c d-15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\sqrt {a} \left (\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-2 a c^2-b c^2 x-9 a c d x+4 a d^2 x^2\right )}{x^2}+\frac {4 a d^{3/2} (5 b c+a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}}\right )}{4 a^{3/2}} \] Input:

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

Output:

(Sqrt[c]*(b^2*c^2 - 10*a*b*c*d - 15*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x 
])/(Sqrt[a]*Sqrt[c + d*x])] + Sqrt[a]*((Sqrt[a + b*x]*Sqrt[c + d*x]*(-2*a* 
c^2 - b*c^2*x - 9*a*c*d*x + 4*a*d^2*x^2))/x^2 + (4*a*d^(3/2)*(5*b*c + a*d) 
*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[b]))/(4*a^ 
(3/2))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {108, 27, 166, 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.

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

\(\Big \downarrow \) 108

\(\displaystyle \frac {1}{2} \int \frac {(c+d x)^{3/2} (b c+5 a d+6 b d x)}{2 x^2 \sqrt {a+b x}}dx-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \frac {(c+d x)^{3/2} (b c+5 a d+6 b d x)}{x^2 \sqrt {a+b x}}dx-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}\)

\(\Big \downarrow \) 166

\(\displaystyle \frac {1}{4} \left (\frac {\int -\frac {\sqrt {c+d x} \left (b^2 c^2-10 a b d c-15 a^2 d^2-2 b d (b c+11 a d) x\right )}{2 x \sqrt {a+b x}}dx}{a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{a x}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (-\frac {\int \frac {\sqrt {c+d x} \left (b^2 c^2-10 a b d c-15 a^2 d^2-2 b d (b c+11 a d) x\right )}{x \sqrt {a+b x}}dx}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{a x}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}\)

\(\Big \downarrow \) 171

\(\displaystyle \frac {1}{4} \left (-\frac {\frac {\int \frac {b \left (c \left (b^2 c^2-10 a b d c-15 a^2 d^2\right )-4 a d^2 (5 b c+a d) x\right )}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{b}-2 d \sqrt {a+b x} \sqrt {c+d x} (11 a d+b c)}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{a x}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (-\frac {\int \frac {c \left (b^2 c^2-10 a b d c-15 a^2 d^2\right )-4 a d^2 (5 b c+a d) x}{x \sqrt {a+b x} \sqrt {c+d x}}dx-2 d \sqrt {a+b x} \sqrt {c+d x} (11 a d+b c)}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{a x}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {1}{4} \left (-\frac {c \left (-15 a^2 d^2-10 a b c d+b^2 c^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx-4 a d^2 (a d+5 b c) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx-2 d \sqrt {a+b x} \sqrt {c+d x} (11 a d+b c)}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{a x}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {1}{4} \left (-\frac {c \left (-15 a^2 d^2-10 a b c d+b^2 c^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx-8 a d^2 (a d+5 b c) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}-2 d \sqrt {a+b x} \sqrt {c+d x} (11 a d+b c)}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{a x}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{4} \left (-\frac {2 c \left (-15 a^2 d^2-10 a b c d+b^2 c^2\right ) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}-8 a d^2 (a d+5 b c) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}-2 d \sqrt {a+b x} \sqrt {c+d x} (11 a d+b c)}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{a x}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{4} \left (-\frac {-\frac {2 \sqrt {c} \left (-15 a^2 d^2-10 a b c d+b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}-\frac {8 a d^{3/2} (a d+5 b c) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}}-2 d \sqrt {a+b x} \sqrt {c+d x} (11 a d+b c)}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{a x}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}\)

Input:

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

Output:

-1/2*(Sqrt[a + b*x]*(c + d*x)^(5/2))/x^2 + (-(((b*c + 5*a*d)*Sqrt[a + b*x] 
*(c + d*x)^(3/2))/(a*x)) - (-2*d*(b*c + 11*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x 
] - (2*Sqrt[c]*(b^2*c^2 - 10*a*b*c*d - 15*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a 
 + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[a] - (8*a*d^(3/2)*(5*b*c + a*d)*Ar 
cTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[b])/(2*a))/4
 

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 108

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) 
, x] - Simp[1/(b*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* 
x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 
*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 166

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] && ILtQ[m, -1] && GtQ[n, 0]
 

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. \(440\) vs. \(2(167)=334\).

Time = 0.22 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.09

method	result	size
default	\(\frac {\sqrt {b x +a}\, \sqrt {x d +c}\, \left (4 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} d^{3} x^{2} \sqrt {a c}+20 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a b c \,d^{2} x^{2} \sqrt {a c}-15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{2} c \,d^{2} x^{2} \sqrt {d b}-10 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a b \,c^{2} d \,x^{2} \sqrt {d b}+\ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) b^{2} c^{3} x^{2} \sqrt {d b}+8 a \,d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}-18 a c d x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}-2 b \,c^{2} x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}-4 a \,c^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\right )}{8 a \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, x^{2} \sqrt {d b}\, \sqrt {a c}}\)	\(441\)

Input:

int((b*x+a)^(1/2)*(d*x+c)^(5/2)/x^3,x,method=_RETURNVERBOSE)
 

Output:

1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a*(4*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^2*d^3*x^2*(a*c)^(1/2)+20*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*c* 
d^2*x^2*(a*c)^(1/2)-15*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*c*d^2*x^2*(d*b)^(1/2)-10*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^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)*b^2*c^3*x^2*(d*b)^(1/2)+8* 
a*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2)-18*a*c*d*x*((b*x 
+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2)-2*b*c^2*x*((b*x+a)*(d*x+c))^(1/ 
2)*(d*b)^(1/2)*(a*c)^(1/2)-4*a*c^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a* 
c)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/x^2/(d*b)^(1/2)/(a*c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 1.34 (sec) , antiderivative size = 1094, normalized size of antiderivative = 5.18 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^3} \, dx=\text {Too large to display} \] Input:

integrate((b*x+a)^(1/2)*(d*x+c)^(5/2)/x^3,x, algorithm="fricas")
 

Output:

[1/16*(4*(5*a*b*c*d + a^2*d^2)*x^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) - (b^2*c^2 - 10*a*b*c*d - 15*a 
^2*d^2)*x^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) + 4*(4*a*d^2*x^2 - 2*a*c^2 - (b*c^2 + 9*a*c 
*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*x^2), -1/16*(8*(5*a*b*c*d + a^2*d^2 
)*x^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)) + (b^2*c^2 - 10*a* 
b*c*d - 15*a^2*d^2)*x^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) - 4*(4*a*d^2*x^2 - 2*a*c^2 - (b 
*c^2 + 9*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*x^2), -1/8*((b^2*c^2 - 
10*a*b*c*d - 15*a^2*d^2)*x^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)) - 2*(5*a*b*c*d + a^2*d^2)*x^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) - 2*(4*a*d^2*x^2 - 2*a*c^2 - 
 (b*c^2 + 9*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*x^2), -1/8*((b^2*c^2 
 - 10*a*b*c*d - 15*a^2*d^2)*x^2*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a...
 

Sympy [F]

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

integrate((b*x+a)**(1/2)*(d*x+c)**(5/2)/x**3,x)
 

Output:

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

Maxima [F(-2)]

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

integrate((b*x+a)^(1/2)*(d*x+c)^(5/2)/x^3,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. 1200 vs. \(2 (167) = 334\).

Time = 0.73 (sec) , antiderivative size = 1200, normalized size of antiderivative = 5.69 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^3} \, dx=\text {Too large to display} \] Input:

integrate((b*x+a)^(1/2)*(d*x+c)^(5/2)/x^3,x, algorithm="giac")
 

Output:

1/4*(4*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 + (sqrt(b*d)*b^3*c^3*ab 
s(b) - 10*sqrt(b*d)*a*b^2*c^2*d*abs(b) - 15*sqrt(b*d)*a^2*b*c*d^2*abs(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)*a*b) - 2*(sqrt(b 
*d)*b^9*c^6*abs(b) + 5*sqrt(b*d)*a*b^8*c^5*d*abs(b) - 30*sqrt(b*d)*a^2*b^7 
*c^4*d^2*abs(b) + 50*sqrt(b*d)*a^3*b^6*c^3*d^3*abs(b) - 35*sqrt(b*d)*a^4*b 
^5*c^2*d^4*abs(b) + 9*sqrt(b*d)*a^5*b^4*c*d^5*abs(b) - 3*sqrt(b*d)*(sqrt(b 
*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^7*c^5*abs(b) 
- 20*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^6*c^4*d*abs(b) + 22*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^5*c^3*d^2*abs(b) + 28*sqrt(b*d)*(sq 
rt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^4*c^2 
*d^3*abs(b) - 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + 
a)*b*d - a*b*d))^2*a^4*b^3*c*d^4*abs(b) + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x 
+ a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^5*c^4*abs(b) + 29*sqrt(b*d 
)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^4* 
c^3*d*abs(b) + 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + 
 a)*b*d - a*b*d))^4*a^2*b^3*c^2*d^2*abs(b) + 27*sqrt(b*d)*(sqrt(b*d)*sq...
 

Mupad [F(-1)]

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

int(((a + b*x)^(1/2)*(c + d*x)^(5/2))/x^3,x)
 

Output:

int(((a + b*x)^(1/2)*(c + d*x)^(5/2))/x^3, x)
 

Reduce [B] (verification not implemented)

Time = 1.16 (sec) , antiderivative size = 1054, normalized size of antiderivative = 5.00 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^3} \, dx =\text {Too large to display} \] Input:

int((b*x+a)^(1/2)*(d*x+c)^(5/2)/x^3,x)
 

Output:

( - 8*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b*c**2*d - 36*sqrt(c + d*x)*sqrt(a 
+ b*x)*a**3*b*c*d**2*x + 16*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b*d**3*x**2 - 
 8*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**2*c**3 - 40*sqrt(c + d*x)*sqrt(a + 
b*x)*a**2*b**2*c**2*d*x + 16*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**2*c*d**2* 
x**2 - 4*sqrt(c + d*x)*sqrt(a + b*x)*a*b**3*c**3*x + 30*sqrt(c)*sqrt(a)*lo 
g( - 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*b*d**3*x**2 + 50*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**2*c*d**2*x**2 + 18*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*b**3*c**2*d*x**2 - 2*sqrt(c)*sqrt(a)*lo 
g( - 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))*b**4*c**3*x**2 + 30*sqrt(c)*sqrt(a)*log(sq 
rt(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*b*d**3*x**2 + 50*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) + sqr 
t(b)*sqrt(c + d*x))*a**2*b**2*c*d**2*x**2 + 18*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) + sqr 
t(b)*sqrt(c + d*x))*a*b**3*c**2*d*x**2 - 2*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...