3.6.72 \(\int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx\) [572]

3.6.72.1 Optimal result

Integrand size = 22, antiderivative size = 227 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=\frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 a^2 x}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 a x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}-\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} \sqrt {c}}+2 \sqrt {b} d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right ) \]

output

2*d^(5/2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))*b^(1/2)-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(c^(1/2)*(b*x+a)^ 
(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(5/2)/c^(1/2)-1/12*(5*a*d+b*c)*(d*x+c)^(3/2 
)*(b*x+a)^(1/2)/a/x^2-1/3*(d*x+c)^(5/2)*(b*x+a)^(1/2)/x^3+1/8*(-5*a*d+b*c) 
*(a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2/x
 

3.6.72.2 Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-3 b^2 c^2 x^2+2 a b c x (c+7 d x)+a^2 \left (8 c^2+26 c d x+33 d^2 x^2\right )\right )}{24 a^2 x^3}-\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} \sqrt {c}}+2 \sqrt {b} d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right ) \]

input

Integrate[(Sqrt[a + b*x]*(c + d*x)^(5/2))/x^4,x]
 

output

-1/24*(Sqrt[a + b*x]*Sqrt[c + d*x]*(-3*b^2*c^2*x^2 + 2*a*b*c*x*(c + 7*d*x) 
 + a^2*(8*c^2 + 26*c*d*x + 33*d^2*x^2)))/(a^2*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)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a 
]*Sqrt[c + d*x])])/(8*a^(5/2)*Sqrt[c]) + 2*Sqrt[b]*d^(5/2)*ArcTanh[(Sqrt[d 
]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])]
 

3.6.72.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 247, 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 = {108, 27, 166, 27, 166, 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.

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

\(\Big \downarrow \) 108

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 166

\(\displaystyle \frac {1}{6} \left (\frac {\int -\frac {3 \sqrt {c+d x} \left ((b c-5 a d) (b c+a d)-8 a b d^2 x\right )}{2 x^2 \sqrt {a+b x}}dx}{2 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{2 a x^2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (-\frac {3 \int \frac {\sqrt {c+d x} \left ((b c-5 a d) (b c+a d)-8 a b d^2 x\right )}{x^2 \sqrt {a+b x}}dx}{4 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{2 a x^2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 166

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\int -\frac {b^3 c^3-5 a b^2 d c^2+15 a^2 b d^2 c+5 a^3 d^3+16 a^2 b d^3 x}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{a}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{a x}\right )}{4 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{2 a x^2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (-\frac {\int \frac {b^3 c^3-5 a b^2 d c^2+15 a^2 b d^2 c+5 a^3 d^3+16 a^2 b d^3 x}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 a}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{a x}\right )}{4 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{2 a x^2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (-\frac {16 a^2 b d^3 \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx+\left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 a}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{a x}\right )}{4 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{2 a x^2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (-\frac {32 a^2 b d^3 \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+\left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 a}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{a x}\right )}{4 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{2 a x^2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (-\frac {32 a^2 b d^3 \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+2 \left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 a}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{a x}\right )}{4 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{2 a x^2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (-\frac {32 a^2 \sqrt {b} d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )-\frac {2 \left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} \sqrt {c}}}{2 a}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{a x}\right )}{4 a}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{2 a x^2}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}\)

input

Int[(Sqrt[a + b*x]*(c + d*x)^(5/2))/x^4,x]
 

output

-1/3*(Sqrt[a + b*x]*(c + d*x)^(5/2))/x^3 + (-1/2*((b*c + 5*a*d)*Sqrt[a + b 
*x]*(c + d*x)^(3/2))/(a*x^2) - (3*(-(((b*c - 5*a*d)*(b*c + a*d)*Sqrt[a + b 
*x]*Sqrt[c + d*x])/(a*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[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(S 
qrt[a]*Sqrt[c]) + 32*a^2*Sqrt[b]*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/( 
Sqrt[b]*Sqrt[c + d*x])])/(2*a)))/(4*a))/6
 

3.6.72.3.1 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 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 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]
 

3.6.72.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(516\) vs. \(2(183)=366\).

Time = 0.53 (sec) , antiderivative size = 517, normalized size of antiderivative = 2.28

method	result	size
default	\(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (48 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b \,d^{3} x^{3} \sqrt {a c}-15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{3} d^{3} x^{3} \sqrt {b d}-45 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} b c \,d^{2} x^{3} \sqrt {b d}+15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{2} d \,x^{3} \sqrt {b d}-3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) b^{3} c^{3} x^{3} \sqrt {b d}-66 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2} x^{2}-28 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d \,x^{2}+6 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2} x^{2}-52 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c d x -4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,c^{2} x -16 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c^{2}\right )}{48 a^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, x^{3} \sqrt {b d}\, \sqrt {a c}}\)	\(517\)

input

int((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^4,x,method=_RETURNVERBOSE)
 

output

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2*(48*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*b*d^3*x^3*(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^3*d^3*x^3* 
(b*d)^(1/2)-45*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^3*(b*d)^(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*b^2*c^2*d*x^3*(b*d)^(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)*b^3*c^3*x^3*(b*d)^(1/2)-66*( 
b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*d^2*x^2-28*(b*d)^(1/2)* 
(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c*d*x^2+6*(b*d)^(1/2)*(a*c)^(1/2)* 
((b*x+a)*(d*x+c))^(1/2)*b^2*c^2*x^2-52*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d 
*x+c))^(1/2)*a^2*c*d*x-4*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a 
*b*c^2*x-16*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*c^2)/((b*x 
+a)*(d*x+c))^(1/2)/x^3/(b*d)^(1/2)/(a*c)^(1/2)
 

3.6.72.5 Fricas [A] (verification not implemented)

Time = 1.40 (sec) , antiderivative size = 1245, normalized size of antiderivative = 5.48 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=\text {Too large to display} \]

input

integrate((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^4,x, algorithm="fricas")
 

output

[1/96*(48*sqrt(b*d)*a^3*c*d^2*x^3*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) + 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(a*c)*x^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) - 4*(8*a^3*c^3 - (3*a*b^2*c^3 - 14*a^2*b*c^2* 
d - 33*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + 13*a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt 
(d*x + c))/(a^3*c*x^3), -1/96*(96*sqrt(-b*d)*a^3*c*d^2*x^3*arctan(1/2*(2*b 
*d*x + b*c + a*d)*sqrt(-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)) - 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(a*c)*x^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) + 4*(8*a^3*c^3 - (3*a*b^2*c^3 - 14*a^2*b 
*c^2*d - 33*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + 13*a^3*c^2*d)*x)*sqrt(b*x + a) 
*sqrt(d*x + c))/(a^3*c*x^3), 1/48*(24*sqrt(b*d)*a^3*c*d^2*x^3*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) + 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(-a*c)*x^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)) - 2*(8*a^3*c^3 - (3*a*b^2*c^3 - 14*a^2*b*...
 

3.6.72.6 Sympy [F]

\[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=\int \frac {\sqrt {a + b x} \left (c + d x\right )^{\frac {5}{2}}}{x^{4}}\, dx \]

input

integrate((d*x+c)**(5/2)*(b*x+a)**(1/2)/x**4,x)
 

output

Integral(sqrt(a + b*x)*(c + d*x)**(5/2)/x**4, x)
 

3.6.72.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=\text {Exception raised: ValueError} \]

input

integrate((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^4,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
 

3.6.72.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2265 vs. \(2 (183) = 366\).

Time = 3.49 (sec) , antiderivative size = 2265, normalized size of antiderivative = 9.98 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=\text {Too large to display} \]

input

integrate((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^4,x, algorithm="giac")
 

output

-1/24*(24*sqrt(b*d)*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) + 3*(sqrt(b*d)*b^4*c^3*abs(b) - 5*sqrt(b*d)*a* 
b^3*c^2*d*abs(b) + 15*sqrt(b*d)*a^2*b^2*c*d^2*abs(b) + 5*sqrt(b*d)*a^3*b*d 
^3*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^2*b 
) - 2*(3*sqrt(b*d)*b^14*c^8*abs(b) - 32*sqrt(b*d)*a*b^13*c^7*d*abs(b) + 96 
*sqrt(b*d)*a^2*b^12*c^6*d^2*abs(b) - 72*sqrt(b*d)*a^3*b^11*c^5*d^3*abs(b) 
- 170*sqrt(b*d)*a^4*b^10*c^4*d^4*abs(b) + 432*sqrt(b*d)*a^5*b^9*c^3*d^5*ab 
s(b) - 408*sqrt(b*d)*a^6*b^8*c^2*d^6*abs(b) + 184*sqrt(b*d)*a^7*b^7*c*d^7* 
abs(b) - 33*sqrt(b*d)*a^8*b^6*d^8*abs(b) - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b* 
x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^12*c^7*abs(b) + 111*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^11*c^6*d*abs(b) - 111*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^10*c^5*d^2*abs(b) - 201*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^9*c^4*d^3* 
abs(b) + 219*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^8*c^3*d^4*abs(b) + 309*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x 
+ a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^5*b^7*c^2*d^5*abs(b) - 477 
*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)) 
^2*a^6*b^6*c*d^6*abs(b) + 165*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt...
 

3.6.72.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=\int \frac {\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{5/2}}{x^4} \,d x \]

input

int(((a + b*x)^(1/2)*(c + d*x)^(5/2))/x^4,x)
 

output

int(((a + b*x)^(1/2)*(c + d*x)^(5/2))/x^4, x)