3.8.74 \(\int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^{3/2}} \, dx\) [774]

3.8.74.1 Optimal result
3.8.74.2 Mathematica [A] (verified)
3.8.74.3 Rubi [A] (verified)
3.8.74.4 Maple [B] (verified)
3.8.74.5 Fricas [A] (verification not implemented)
3.8.74.6 Sympy [F]
3.8.74.7 Maxima [F(-2)]
3.8.74.8 Giac [B] (verification not implemented)
3.8.74.9 Mupad [F(-1)]

3.8.74.1 Optimal result

Integrand size = 22, antiderivative size = 230 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^{3/2}} \, dx=-\frac {5 (b c-a d)^2 (7 b c-a d) \sqrt {c+d x}}{8 a^4 c \sqrt {a+b x}}-\frac {5 (b c-a d) (7 b c-a d) (c+d x)^{3/2}}{24 a^3 c x \sqrt {a+b x}}+\frac {(7 b c-a d) (c+d x)^{5/2}}{12 a^2 c x^2 \sqrt {a+b x}}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}}+\frac {5 (b c-a d)^2 (7 b c-a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{9/2} \sqrt {c}} \]

output
5/8*(-a*d+b*c)^2*(-a*d+7*b*c)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c 
)^(1/2))/a^(9/2)/c^(1/2)-5/24*(-a*d+b*c)*(-a*d+7*b*c)*(d*x+c)^(3/2)/a^3/c/ 
x/(b*x+a)^(1/2)+1/12*(-a*d+7*b*c)*(d*x+c)^(5/2)/a^2/c/x^2/(b*x+a)^(1/2)-1/ 
3*(d*x+c)^(7/2)/a/c/x^3/(b*x+a)^(1/2)-5/8*(-a*d+b*c)^2*(-a*d+7*b*c)*(d*x+c 
)^(1/2)/a^4/c/(b*x+a)^(1/2)
 
3.8.74.2 Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.73 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^{3/2}} \, dx=-\frac {\sqrt {c+d x} \left (105 b^3 c^2 x^3+5 a b^2 c x^2 (7 c-38 d x)+a^3 \left (8 c^2+26 c d x+33 d^2 x^2\right )+a^2 b x \left (-14 c^2-68 c d x+81 d^2 x^2\right )\right )}{24 a^4 x^3 \sqrt {a+b x}}-\frac {5 (b c-a d)^2 (-7 b c+a d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{8 a^{9/2} \sqrt {c}} \]

input
Integrate[(c + d*x)^(5/2)/(x^4*(a + b*x)^(3/2)),x]
 
output
-1/24*(Sqrt[c + d*x]*(105*b^3*c^2*x^3 + 5*a*b^2*c*x^2*(7*c - 38*d*x) + a^3 
*(8*c^2 + 26*c*d*x + 33*d^2*x^2) + a^2*b*x*(-14*c^2 - 68*c*d*x + 81*d^2*x^ 
2)))/(a^4*x^3*Sqrt[a + b*x]) - (5*(b*c - a*d)^2*(-7*b*c + a*d)*ArcTanh[(Sq 
rt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/(8*a^(9/2)*Sqrt[c])
 
3.8.74.3 Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.88, 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.

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

\(\Big \downarrow \) 107

\(\displaystyle -\frac {(7 b c-a d) \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{3/2}}dx}{6 a c}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {(7 b c-a d) \left (-\frac {5 (b c-a d) \int \frac {(c+d x)^{3/2}}{x^2 (a+b x)^{3/2}}dx}{4 a}-\frac {(c+d x)^{5/2}}{2 a x^2 \sqrt {a+b x}}\right )}{6 a c}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {(7 b c-a d) \left (-\frac {5 (b c-a d) \left (-\frac {3 (b c-a d) \int \frac {\sqrt {c+d x}}{x (a+b x)^{3/2}}dx}{2 a}-\frac {(c+d x)^{3/2}}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {(c+d x)^{5/2}}{2 a x^2 \sqrt {a+b x}}\right )}{6 a c}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {(7 b c-a d) \left (-\frac {5 (b c-a d) \left (-\frac {3 (b c-a d) \left (\frac {c \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{a}+\frac {2 \sqrt {c+d x}}{a \sqrt {a+b x}}\right )}{2 a}-\frac {(c+d x)^{3/2}}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {(c+d x)^{5/2}}{2 a x^2 \sqrt {a+b x}}\right )}{6 a c}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {(7 b c-a d) \left (-\frac {5 (b c-a d) \left (-\frac {3 (b c-a d) \left (\frac {2 c \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{a}+\frac {2 \sqrt {c+d x}}{a \sqrt {a+b x}}\right )}{2 a}-\frac {(c+d x)^{3/2}}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {(c+d x)^{5/2}}{2 a x^2 \sqrt {a+b x}}\right )}{6 a c}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {(7 b c-a d) \left (-\frac {5 (b c-a d) \left (-\frac {3 (b c-a d) \left (\frac {2 \sqrt {c+d x}}{a \sqrt {a+b x}}-\frac {2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2}}\right )}{2 a}-\frac {(c+d x)^{3/2}}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {(c+d x)^{5/2}}{2 a x^2 \sqrt {a+b x}}\right )}{6 a c}-\frac {(c+d x)^{7/2}}{3 a c x^3 \sqrt {a+b x}}\)

input
Int[(c + d*x)^(5/2)/(x^4*(a + b*x)^(3/2)),x]
 
output
-1/3*(c + d*x)^(7/2)/(a*c*x^3*Sqrt[a + b*x]) - ((7*b*c - a*d)*(-1/2*(c + d 
*x)^(5/2)/(a*x^2*Sqrt[a + b*x]) - (5*(b*c - a*d)*(-((c + d*x)^(3/2)/(a*x*S 
qrt[a + b*x])) - (3*(b*c - a*d)*((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)) 
)/(2*a)))/(4*a)))/(6*a*c)
 

3.8.74.3.1 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]
 
3.8.74.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(703\) vs. \(2(192)=384\).

Time = 0.59 (sec) , antiderivative size = 704, normalized size of antiderivative = 3.06

method result size
default \(-\frac {\sqrt {d x +c}\, \left (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} b \,d^{3} x^{4}-135 \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^{2} c \,d^{2} x^{4}+225 \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^{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 (d x +c \right )}+2 a c}{x}\right ) b^{4} c^{3} x^{4}+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^{4} d^{3} x^{3}-135 \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} b c \,d^{2} x^{3}+225 \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^{2} c^{2} d \,x^{3}-105 \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^{3} c^{3} x^{3}+162 a^{2} b \,d^{2} x^{3} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-380 a \,b^{2} c d \,x^{3} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+210 b^{3} c^{2} x^{3} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+66 a^{3} d^{2} x^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-136 a^{2} b c d \,x^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+70 a \,b^{2} c^{2} x^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+52 a^{3} c d x \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-28 a^{2} b \,c^{2} x \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+16 a^{3} c^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\right )}{48 a^{4} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, x^{3} \sqrt {a c}\, \sqrt {b x +a}}\) \(704\)

input
int((d*x+c)^(5/2)/x^4/(b*x+a)^(3/2),x,method=_RETURNVERBOSE)
 
output
-1/48*(d*x+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*b*d^3*x^4-135*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+225*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+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^4*d^3*x^3-135*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+225 
*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^3-105*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*b^3*c^3*x^3+162*a^2*b*d^2*x^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-380 
*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)+66*a^3*d^2*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^ 
(1/2)-136*a^2*b*c*d*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+70*a*b^2*c^2*x 
^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+52*a^3*c*d*x*(a*c)^(1/2)*((b*x+a)*( 
d*x+c))^(1/2)-28*a^2*b*c^2*x*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+16*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^3/(a* 
c)^(1/2)/(b*x+a)^(1/2)
 
3.8.74.5 Fricas [A] (verification not implemented)

Time = 1.02 (sec) , antiderivative size = 636, normalized size of antiderivative = 2.77 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^{3/2}} \, dx=\left [-\frac {15 \, {\left ({\left (7 \, b^{4} c^{3} - 15 \, a b^{3} c^{2} d + 9 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} x^{4} + {\left (7 \, a b^{3} c^{3} - 15 \, a^{2} b^{2} c^{2} d + 9 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} x^{3}\right )} \sqrt {a c} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (8 \, a^{4} c^{3} + {\left (105 \, a b^{3} c^{3} - 190 \, a^{2} b^{2} c^{2} d + 81 \, a^{3} b c d^{2}\right )} x^{3} + {\left (35 \, a^{2} b^{2} c^{3} - 68 \, a^{3} b c^{2} d + 33 \, a^{4} c d^{2}\right )} x^{2} - 2 \, {\left (7 \, a^{3} b c^{3} - 13 \, a^{4} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, {\left (a^{5} b c x^{4} + a^{6} c x^{3}\right )}}, -\frac {15 \, {\left ({\left (7 \, b^{4} c^{3} - 15 \, a b^{3} c^{2} d + 9 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} x^{4} + {\left (7 \, a b^{3} c^{3} - 15 \, a^{2} b^{2} c^{2} d + 9 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} x^{3}\right )} \sqrt {-a c} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) + 2 \, {\left (8 \, a^{4} c^{3} + {\left (105 \, a b^{3} c^{3} - 190 \, a^{2} b^{2} c^{2} d + 81 \, a^{3} b c d^{2}\right )} x^{3} + {\left (35 \, a^{2} b^{2} c^{3} - 68 \, a^{3} b c^{2} d + 33 \, a^{4} c d^{2}\right )} x^{2} - 2 \, {\left (7 \, a^{3} b c^{3} - 13 \, a^{4} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (a^{5} b c x^{4} + a^{6} c x^{3}\right )}}\right ] \]

input
integrate((d*x+c)^(5/2)/x^4/(b*x+a)^(3/2),x, algorithm="fricas")
 
output
[-1/96*(15*((7*b^4*c^3 - 15*a*b^3*c^2*d + 9*a^2*b^2*c*d^2 - a^3*b*d^3)*x^4 
 + (7*a*b^3*c^3 - 15*a^2*b^2*c^2*d + 9*a^3*b*c*d^2 - a^4*d^3)*x^3)*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 + (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^4*c^3 + (105*a*b^3*c^3 - 190*a^2*b^2*c^2*d + 81*a^3*b*c*d^2 
)*x^3 + (35*a^2*b^2*c^3 - 68*a^3*b*c^2*d + 33*a^4*c*d^2)*x^2 - 2*(7*a^3*b* 
c^3 - 13*a^4*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^5*b*c*x^4 + a^6*c*x 
^3), -1/48*(15*((7*b^4*c^3 - 15*a*b^3*c^2*d + 9*a^2*b^2*c*d^2 - a^3*b*d^3) 
*x^4 + (7*a*b^3*c^3 - 15*a^2*b^2*c^2*d + 9*a^3*b*c*d^2 - a^4*d^3)*x^3)*sqr 
t(-a*c)*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^4*c^3 + 
(105*a*b^3*c^3 - 190*a^2*b^2*c^2*d + 81*a^3*b*c*d^2)*x^3 + (35*a^2*b^2*c^3 
 - 68*a^3*b*c^2*d + 33*a^4*c*d^2)*x^2 - 2*(7*a^3*b*c^3 - 13*a^4*c^2*d)*x)* 
sqrt(b*x + a)*sqrt(d*x + c))/(a^5*b*c*x^4 + a^6*c*x^3)]
 
3.8.74.6 Sympy [F]

\[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^{3/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{x^{4} \left (a + b x\right )^{\frac {3}{2}}}\, dx \]

input
integrate((d*x+c)**(5/2)/x**4/(b*x+a)**(3/2),x)
 
output
Integral((c + d*x)**(5/2)/(x**4*(a + b*x)**(3/2)), x)
 
3.8.74.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]

input
integrate((d*x+c)^(5/2)/x^4/(b*x+a)^(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
 
3.8.74.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2324 vs. \(2 (192) = 384\).

Time = 5.07 (sec) , antiderivative size = 2324, normalized size of antiderivative = 10.10 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^{3/2}} \, dx=\text {Too large to display} \]

input
integrate((d*x+c)^(5/2)/x^4/(b*x+a)^(3/2),x, algorithm="giac")
 
output
5/8*(7*sqrt(b*d)*b^3*c^3*abs(b) - 15*sqrt(b*d)*a*b^2*c^2*d*abs(b) + 9*sqrt 
(b*d)*a^2*b*c*d^2*abs(b) - sqrt(b*d)*a^3*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^4*b) - 4*(sqrt(b*d)*b^3*c^3*abs(b) 
- 3*sqrt(b*d)*a*b^2*c^2*d*abs(b) + 3*sqrt(b*d)*a^2*b*c*d^2*abs(b) - sqrt(b 
*d)*a^3*d^3*abs(b))/((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)*a^4) - 1/12*(57*sqrt(b*d)*b^13*c^8*abs(b) - 
 436*sqrt(b*d)*a*b^12*c^7*d*abs(b) + 1452*sqrt(b*d)*a^2*b^11*c^6*d^2*abs(b 
) - 2748*sqrt(b*d)*a^3*b^10*c^5*d^3*abs(b) + 3230*sqrt(b*d)*a^4*b^9*c^4*d^ 
4*abs(b) - 2412*sqrt(b*d)*a^5*b^8*c^3*d^5*abs(b) + 1116*sqrt(b*d)*a^6*b^7* 
c^2*d^6*abs(b) - 292*sqrt(b*d)*a^7*b^6*c*d^7*abs(b) + 33*sqrt(b*d)*a^8*b^5 
*d^8*abs(b) - 285*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + 
 a)*b*d - a*b*d))^2*b^11*c^7*abs(b) + 1281*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^10*c^6*d*abs(b) - 1917*sq 
rt(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^9*c^5*d^2*abs(b) + 393*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^8*c^4*d^3*abs(b) + 1953*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^4*b^7*c^3 
*d^4*abs(b) - 2277*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^6*c^2*d^5*abs(b) + 1017*sqrt(b*d)*(sqrt(b*d)...
 
3.8.74.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^{3/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/2}}{x^4\,{\left (a+b\,x\right )}^{3/2}} \,d x \]

input
int((c + d*x)^(5/2)/(x^4*(a + b*x)^(3/2)),x)
 
output
int((c + d*x)^(5/2)/(x^4*(a + b*x)^(3/2)), x)