3.7.53 \(\int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^4} \, dx\) [653]

3.7.53.1 Optimal result

Integrand size = 22, antiderivative size = 228 \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^4} \, dx=-\frac {(5 b c-a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 c^2 x}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 c x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 x^3}-\frac {\left (5 b^3 c^3+15 a b^2 c^2 d-5 a^2 b c d^2+a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 \sqrt {a} c^{5/2}}+2 b^{5/2} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right ) \]

output

-1/8*(a^3*d^3-5*a^2*b*c*d^2+15*a*b^2*c^2*d+5*b^3*c^3)*arctanh(c^(1/2)*(b*x 
+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/c^(5/2)/a^(1/2)+2*b^(5/2)*arctanh(d^(1/2) 
*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))*d^(1/2)-1/12*(a*d+5*b*c)*(b*x+a)^(3/ 
2)*(d*x+c)^(1/2)/c/x^2-1/3*(b*x+a)^(5/2)*(d*x+c)^(1/2)/x^3-1/8*(-a*d+5*b*c 
)*(a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c^2/x
 

3.7.53.2 Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^4} \, dx=-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (33 b^2 c^2 x^2+2 a b c x (13 c+7 d x)+a^2 \left (8 c^2+2 c d x-3 d^2 x^2\right )\right )}{24 c^2 x^3}-\frac {\left (5 b^3 c^3+15 a b^2 c^2 d-5 a^2 b c d^2+a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{8 \sqrt {a} c^{5/2}}+2 b^{5/2} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right ) \]

input

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

output

-1/24*(Sqrt[a + b*x]*Sqrt[c + d*x]*(33*b^2*c^2*x^2 + 2*a*b*c*x*(13*c + 7*d 
*x) + a^2*(8*c^2 + 2*c*d*x - 3*d^2*x^2)))/(c^2*x^3) - ((5*b^3*c^3 + 15*a*b 
^2*c^2*d - 5*a^2*b*c*d^2 + a^3*d^3)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[ 
c]*Sqrt[a + b*x])])/(8*Sqrt[a]*c^(5/2)) + 2*b^(5/2)*Sqrt[d]*ArcTanh[(Sqrt[ 
b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])]
 

3.7.53.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 248, 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.

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

\(\Big \downarrow \) 108

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 175

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

input

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

output

-1/3*((a + b*x)^(5/2)*Sqrt[c + d*x])/x^3 + (-1/2*((5*b*c + a*d)*(a + b*x)^ 
(3/2)*Sqrt[c + d*x])/(c*x^2) + (3*(-(((5*b*c - a*d)*(b*c + a*d)*Sqrt[a + b 
*x]*Sqrt[c + d*x])/(c*x)) + ((-2*(5*b^3*c^3 + 15*a*b^2*c^2*d - 5*a^2*b*c*d 
^2 + a^3*d^3)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(S 
qrt[a]*Sqrt[c]) + 32*b^(5/2)*c^2*Sqrt[d]*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/( 
Sqrt[b]*Sqrt[c + d*x])])/(2*c)))/(4*c))/6
 

3.7.53.3.1 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]
 

3.7.53.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(516\) vs. \(2(184)=368\).

Time = 0.56 (sec) , antiderivative size = 517, normalized size of antiderivative = 2.27

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

input

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

output

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

3.7.53.5 Fricas [A] (verification not implemented)

Time = 1.46 (sec) , antiderivative size = 1241, normalized size of antiderivative = 5.44 \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^4} \, dx=\text {Too large to display} \]

input

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

output

[1/96*(48*sqrt(b*d)*a*b^2*c^3*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*(5*b^3*c^3 + 15*a*b^2*c^2*d - 5*a^2*b*c*d^2 
 + 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 + (33*a*b^2*c^3 + 14*a^2*b*c^2 
*d - 3*a^3*c*d^2)*x^2 + 2*(13*a^2*b*c^3 + a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt 
(d*x + c))/(a*c^3*x^3), -1/96*(96*sqrt(-b*d)*a*b^2*c^3*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*(5*b^3*c^3 + 15*a*b^2*c^2*d - 5*a^2*b* 
c*d^2 + 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 + (33*a*b^2*c^3 + 14*a^2* 
b*c^2*d - 3*a^3*c*d^2)*x^2 + 2*(13*a^2*b*c^3 + a^3*c^2*d)*x)*sqrt(b*x + a) 
*sqrt(d*x + c))/(a*c^3*x^3), 1/48*(24*sqrt(b*d)*a*b^2*c^3*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*(5*b^3*c^3 + 15 
*a*b^2*c^2*d - 5*a^2*b*c*d^2 + 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 + (33*a*b^2*c^3 + 14*a^2*b...
 

3.7.53.6 Sympy [F]

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

input

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

output

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

3.7.53.7 Maxima [F(-2)]

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

input

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

3.7.53.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2265 vs. \(2 (184) = 368\).

Time = 3.44 (sec) , antiderivative size = 2265, normalized size of antiderivative = 9.93 \[ \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^4} \, dx=\text {Too large to display} \]

input

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

output

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

3.7.53.9 Mupad [F(-1)]

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

input

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

output

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