\(\int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{5/2}} \, dx\) [407]

Optimal result

Integrand size = 22, antiderivative size = 220 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{5/2}} \, dx=\frac {5 (7 b c-3 a d) (b c-a d) \sqrt {c+d x}}{4 a^4 \sqrt {a+b x}}+\frac {5 (7 b c-3 a d) (b c-a d) (c+d x)^{3/2}}{12 a^3 c (a+b x)^{3/2}}+\frac {(7 b c-3 a d) (c+d x)^{5/2}}{4 a^2 c x (a+b x)^{3/2}}-\frac {(c+d x)^{7/2}}{2 a c x^2 (a+b x)^{3/2}}-\frac {5 \sqrt {c} (7 b c-3 a d) (b c-a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{9/2}} \] Output:

5/4*(-3*a*d+7*b*c)*(-a*d+b*c)*(d*x+c)^(1/2)/a^4/(b*x+a)^(1/2)+5/12*(-3*a*d 
+7*b*c)*(-a*d+b*c)*(d*x+c)^(3/2)/a^3/c/(b*x+a)^(3/2)+1/4*(-3*a*d+7*b*c)*(d 
*x+c)^(5/2)/a^2/c/x/(b*x+a)^(3/2)-1/2*(d*x+c)^(7/2)/a/c/x^2/(b*x+a)^(3/2)- 
5/4*c^(1/2)*(-3*a*d+7*b*c)*(-a*d+b*c)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2 
)/(d*x+c)^(1/2))/a^(9/2)
 

Mathematica [A] (verified)

Time = 10.18 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.72 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{5/2}} \, dx=\frac {-3 a^{7/2} (c+d x)^{7/2}+\frac {1}{2} (7 b c-3 a d) x \left (3 a^{5/2} (c+d x)^{5/2}+5 (b c-a d) x \left (\sqrt {a} \sqrt {c+d x} (4 a c+3 b c x+a d x)-3 c^{3/2} (a+b x)^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )\right )\right )}{6 a^{9/2} c x^2 (a+b x)^{3/2}} \] Input:

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

Output:

(-3*a^(7/2)*(c + d*x)^(7/2) + ((7*b*c - 3*a*d)*x*(3*a^(5/2)*(c + d*x)^(5/2 
) + 5*(b*c - a*d)*x*(Sqrt[a]*Sqrt[c + d*x]*(4*a*c + 3*b*c*x + a*d*x) - 3*c 
^(3/2)*(a + b*x)^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d 
*x])])))/2)/(6*a^(9/2)*c*x^2*(a + b*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {107, 105, 105, 105, 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.

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

\(\Big \downarrow \) 107

\(\displaystyle -\frac {(7 b c-3 a d) \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{5/2}}dx}{4 a c}-\frac {(c+d x)^{7/2}}{2 a c x^2 (a+b x)^{3/2}}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {(7 b c-3 a d) \left (-\frac {5 (b c-a d) \int \frac {(c+d x)^{3/2}}{x (a+b x)^{5/2}}dx}{2 a}-\frac {(c+d x)^{5/2}}{a x (a+b x)^{3/2}}\right )}{4 a c}-\frac {(c+d x)^{7/2}}{2 a c x^2 (a+b x)^{3/2}}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {(7 b c-3 a d) \left (-\frac {5 (b c-a d) \left (\frac {c \int \frac {\sqrt {c+d x}}{x (a+b x)^{3/2}}dx}{a}+\frac {2 (c+d x)^{3/2}}{3 a (a+b x)^{3/2}}\right )}{2 a}-\frac {(c+d x)^{5/2}}{a x (a+b x)^{3/2}}\right )}{4 a c}-\frac {(c+d x)^{7/2}}{2 a c x^2 (a+b x)^{3/2}}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {(7 b c-3 a d) \left (-\frac {5 (b c-a d) \left (\frac {c \left (\frac {c \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{a}+\frac {2 \sqrt {c+d x}}{a \sqrt {a+b x}}\right )}{a}+\frac {2 (c+d x)^{3/2}}{3 a (a+b x)^{3/2}}\right )}{2 a}-\frac {(c+d x)^{5/2}}{a x (a+b x)^{3/2}}\right )}{4 a c}-\frac {(c+d x)^{7/2}}{2 a c x^2 (a+b x)^{3/2}}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {(7 b c-3 a d) \left (-\frac {5 (b c-a d) \left (\frac {c \left (\frac {2 c \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{a}+\frac {2 \sqrt {c+d x}}{a \sqrt {a+b x}}\right )}{a}+\frac {2 (c+d x)^{3/2}}{3 a (a+b x)^{3/2}}\right )}{2 a}-\frac {(c+d x)^{5/2}}{a x (a+b x)^{3/2}}\right )}{4 a c}-\frac {(c+d x)^{7/2}}{2 a c x^2 (a+b x)^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {(7 b c-3 a d) \left (-\frac {5 (b c-a d) \left (\frac {c \left (\frac {2 \sqrt {c+d x}}{a \sqrt {a+b x}}-\frac {2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2}}\right )}{a}+\frac {2 (c+d x)^{3/2}}{3 a (a+b x)^{3/2}}\right )}{2 a}-\frac {(c+d x)^{5/2}}{a x (a+b x)^{3/2}}\right )}{4 a c}-\frac {(c+d x)^{7/2}}{2 a c x^2 (a+b x)^{3/2}}\)

Input:

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

Output:

-1/2*(c + d*x)^(7/2)/(a*c*x^2*(a + b*x)^(3/2)) - ((7*b*c - 3*a*d)*(-((c + 
d*x)^(5/2)/(a*x*(a + b*x)^(3/2))) - (5*(b*c - a*d)*((2*(c + d*x)^(3/2))/(3 
*a*(a + b*x)^(3/2)) + (c*((2*Sqrt[c + d*x])/(a*Sqrt[a + b*x]) - (2*Sqrt[c] 
*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/a^(3/2)))/a))/( 
2*a)))/(4*a*c)
 

Defintions of rubi rules used

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 105

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 + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

rule 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 
 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

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. \(757\) vs. \(2(182)=364\).

Time = 0.26 (sec) , antiderivative size = 758, normalized size of antiderivative = 3.45

method	result	size
default	\(-\frac {\sqrt {x d +c}\, \left (45 \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^{2} c \,d^{2} x^{4}-150 \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^{3} c^{2} d \,x^{4}+105 \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^{4} c^{3} x^{4}+90 \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} b c \,d^{2} x^{3}-300 \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^{2} c^{2} d \,x^{3}+210 \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^{3} c^{3} x^{3}+45 \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^{4} c \,d^{2} x^{2}-150 \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} b \,c^{2} d \,x^{2}+105 \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^{2} c^{3} x^{2}-32 a^{2} b \,d^{2} x^{3} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+230 a \,b^{2} c d \,x^{3} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}-210 b^{3} c^{2} x^{3} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}-48 a^{3} d^{2} x^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+316 a^{2} b c d \,x^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}-280 a \,b^{2} c^{2} x^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+54 a^{3} c d x \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}-42 a^{2} b \,c^{2} x \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+12 a^{3} c^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\right )}{24 a^{4} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, x^{2} \sqrt {a c}\, \left (b x +a \right )^{\frac {3}{2}}}\)	\(758\)

Input:

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

Output:

-1/24*(d*x+c)^(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^2*c*d^2*x^4-150*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^3*c^2*d*x^4+105*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^4*c^3*x^4+90*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*b*c*d^2*x^3-300*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^2*c^2*d*x^3+210*l 
n((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^3*c^3*x 
^3+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^4* 
c*d^2*x^2-150*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*b*c^2*d*x^2+105*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^2*c^3*x^2-32*a^2*b*d^2*x^3*(a*c)^(1/2)*((b*x+a)*(d*x+c 
))^(1/2)+230*a*b^2*c*d*x^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-210*b^3*c^2 
*x^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-48*a^3*d^2*x^2*(a*c)^(1/2)*((b*x+ 
a)*(d*x+c))^(1/2)+316*a^2*b*c*d*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-28 
0*a*b^2*c^2*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+54*a^3*c*d*x*(a*c)^(1/ 
2)*((b*x+a)*(d*x+c))^(1/2)-42*a^2*b*c^2*x*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1 
/2)+12*a^3*c^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/a^4/((b*x+a)*(d*x+c))^ 
(1/2)/x^2/(a*c)^(1/2)/(b*x+a)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 659, normalized size of antiderivative = 3.00 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{5/2}} \, dx=\left [\frac {15 \, {\left ({\left (7 \, b^{4} c^{2} - 10 \, a b^{3} c d + 3 \, a^{2} b^{2} d^{2}\right )} x^{4} + 2 \, {\left (7 \, a b^{3} c^{2} - 10 \, a^{2} b^{2} c d + 3 \, a^{3} b d^{2}\right )} x^{3} + {\left (7 \, a^{2} b^{2} c^{2} - 10 \, a^{3} b c d + 3 \, a^{4} d^{2}\right )} x^{2}\right )} \sqrt {\frac {c}{a}} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a^{2} c + {\left (a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {c}{a}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (6 \, a^{3} c^{2} - {\left (105 \, b^{3} c^{2} - 115 \, a b^{2} c d + 16 \, a^{2} b d^{2}\right )} x^{3} - 2 \, {\left (70 \, a b^{2} c^{2} - 79 \, a^{2} b c d + 12 \, a^{3} d^{2}\right )} x^{2} - 3 \, {\left (7 \, a^{2} b c^{2} - 9 \, a^{3} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}, \frac {15 \, {\left ({\left (7 \, b^{4} c^{2} - 10 \, a b^{3} c d + 3 \, a^{2} b^{2} d^{2}\right )} x^{4} + 2 \, {\left (7 \, a b^{3} c^{2} - 10 \, a^{2} b^{2} c d + 3 \, a^{3} b d^{2}\right )} x^{3} + {\left (7 \, a^{2} b^{2} c^{2} - 10 \, a^{3} b c d + 3 \, a^{4} d^{2}\right )} x^{2}\right )} \sqrt {-\frac {c}{a}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {c}{a}}}{2 \, {\left (b c d x^{2} + a c^{2} + {\left (b c^{2} + a c d\right )} x\right )}}\right ) - 2 \, {\left (6 \, a^{3} c^{2} - {\left (105 \, b^{3} c^{2} - 115 \, a b^{2} c d + 16 \, a^{2} b d^{2}\right )} x^{3} - 2 \, {\left (70 \, a b^{2} c^{2} - 79 \, a^{2} b c d + 12 \, a^{3} d^{2}\right )} x^{2} - 3 \, {\left (7 \, a^{2} b c^{2} - 9 \, a^{3} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{24 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}\right ] \] Input:

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

Output:

[1/48*(15*((7*b^4*c^2 - 10*a*b^3*c*d + 3*a^2*b^2*d^2)*x^4 + 2*(7*a*b^3*c^2 
 - 10*a^2*b^2*c*d + 3*a^3*b*d^2)*x^3 + (7*a^2*b^2*c^2 - 10*a^3*b*c*d + 3*a 
^4*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*(6*a^3*c^2 - (105*b^3*c^2 - 115*a*b^2* 
c*d + 16*a^2*b*d^2)*x^3 - 2*(70*a*b^2*c^2 - 79*a^2*b*c*d + 12*a^3*d^2)*x^2 
 - 3*(7*a^2*b*c^2 - 9*a^3*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^4*b^2*x^ 
4 + 2*a^5*b*x^3 + a^6*x^2), 1/24*(15*((7*b^4*c^2 - 10*a*b^3*c*d + 3*a^2*b^ 
2*d^2)*x^4 + 2*(7*a*b^3*c^2 - 10*a^2*b^2*c*d + 3*a^3*b*d^2)*x^3 + (7*a^2*b 
^2*c^2 - 10*a^3*b*c*d + 3*a^4*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*(6*a^3*c^2 - (105*b^3*c^2 - 115*a*b^2*c*d + 16*a^2*b 
*d^2)*x^3 - 2*(70*a*b^2*c^2 - 79*a^2*b*c*d + 12*a^3*d^2)*x^2 - 3*(7*a^2*b* 
c^2 - 9*a^3*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^4*b^2*x^4 + 2*a^5*b*x^ 
3 + a^6*x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((d*x+c)^(5/2)/x^3/(b*x+a)^(5/2),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. 1694 vs. \(2 (182) = 364\).

Time = 1.36 (sec) , antiderivative size = 1694, normalized size of antiderivative = 7.70 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-5/4*(7*sqrt(b*d)*b^2*c^3*abs(b) - 10*sqrt(b*d)*a*b*c^2*d*abs(b) + 3*sqrt( 
b*d)*a^2*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^4*b) + 1/2*(11*sqrt(b*d)*b^8*c^6*abs(b) - 53*sqrt(b*d)*a*b^7*c^5* 
d*abs(b) + 102*sqrt(b*d)*a^2*b^6*c^4*d^2*abs(b) - 98*sqrt(b*d)*a^3*b^5*c^3 
*d^3*abs(b) + 47*sqrt(b*d)*a^4*b^4*c^2*d^4*abs(b) - 9*sqrt(b*d)*a^5*b^3*c* 
d^5*abs(b) - 33*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a 
)*b*d - a*b*d))^2*b^6*c^5*abs(b) + 56*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^5*c^4*d*abs(b) + 14*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^4* 
c^3*d^2*abs(b) - 64*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^3*c^2*d^3*abs(b) + 27*sqrt(b*d)*(sqrt(b*d)*sqr 
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^2*c*d^4*abs(b) + 
 33*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b* 
d))^4*b^4*c^4*abs(b) - 5*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^3*c^3*d*abs(b) + 15*sqrt(b*d)*(sqrt(b*d)*sq 
rt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^2*c^2*d^2*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))^4*a^3*b*c*d^3*abs(b) - 11*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt 
(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^2*c^3*abs(b) + 2*sqrt(b*d)*(sqrt(b...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 2.09 (sec) , antiderivative size = 1944, normalized size of antiderivative = 8.84 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(45*sqrt(c)*sqrt(a)*sqrt(a + b*x)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sq 
rt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**4*b 
*d**3*x**2 + 165*sqrt(c)*sqrt(a)*sqrt(a + b*x)*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*b**2*c*d**2*x**2 + 45*sqrt(c)*sqrt(a)*sqrt(a + b*x)*log( - sqr 
t(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**2*d**3*x**3 - 945*sqrt(c)*sqrt(a)*sqrt(a + 
 b*x)*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**3*c**2*d*x**2 + 165*sqrt(c 
)*sqrt(a)*sqrt(a + b*x)*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**3*c*d**2 
*x**3 + 735*sqrt(c)*sqrt(a)*sqrt(a + b*x)*log( - sqrt(2*sqrt(d)*sqrt(c)*sq 
rt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x) 
)*a*b**4*c**3*x**2 - 945*sqrt(c)*sqrt(a)*sqrt(a + b*x)*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**4*c**2*d*x**3 + 735*sqrt(c)*sqrt(a)*sqrt(a + b*x)*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))*b**5*c**3*x**3 + 45*sqrt(c)*sqrt(a)*sqrt(a + 
 b*x)*log(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sq 
rt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**4*b*d**3*x**2 + 165*sqrt(c)*sqr...