\(\int \frac {(a+b x)^{5/2}}{x^3 (c+d x)^{5/2}} \, dx\) [314]

Optimal result

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

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

Mathematica [A] (verified)

Time = 10.19 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.72 \[ \int \frac {(a+b x)^{5/2}}{x^3 (c+d x)^{5/2}} \, dx=\frac {-3 c^{7/2} (a+b x)^{7/2}-\frac {1}{2} (3 b c-7 a d) x \left (3 c^{5/2} (a+b x)^{5/2}-5 (b c-a d) x \left (\sqrt {c} \sqrt {a+b x} (4 a c+b c x+3 a d x)-3 a^{3/2} (c+d x)^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )\right )\right )}{6 a c^{9/2} x^2 (c+d x)^{3/2}} \] Input:

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

Output:

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

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

\(\Big \downarrow \) 107

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-1/2*(a + b*x)^(7/2)/(a*c*x^2*(c + d*x)^(3/2)) + ((3*b*c - 7*a*d)*(-((a + 
b*x)^(5/2)/(c*x*(c + d*x)^(3/2))) + (5*(b*c - a*d)*((2*(a + b*x)^(3/2))/(3 
*c*(c + d*x)^(3/2)) + (a*((2*Sqrt[a + b*x])/(c*Sqrt[c + d*x]) - (2*Sqrt[a] 
*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/c^(3/2)))/c))/( 
2*c)))/(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 = 0.24 (sec) , antiderivative size = 758, normalized size of antiderivative = 3.45

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

Input:

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

Output:

-1/24*(b*x+a)^(1/2)*(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^3*d^4*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^2*b*c*d^3*x^4+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^2*x^4+210*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*c*d^3*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*c^2*d^2*x^3+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*b^2*c^3*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^3 
*c^2*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^2*b*c^3*d*x^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^4*x^2-210*a^2*d^3*x^3*(a*c)^(1/2)*((b*x+a)*(d*x+c) 
)^(1/2)+230*a*b*c*d^2*x^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-32*b^2*c^2*d 
*x^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-280*a^2*c*d^2*x^2*(a*c)^(1/2)*((b 
*x+a)*(d*x+c))^(1/2)+316*a*b*c^2*d*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2) 
-48*b^2*c^3*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-42*a^2*c^2*d*x*(a*c)^( 
1/2)*((b*x+a)*(d*x+c))^(1/2)+54*a*b*c^3*x*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1 
/2)+12*a^2*c^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/c^4/((b*x+a)*(d*x+c))^ 
(1/2)/x^2/(a*c)^(1/2)/(d*x+c)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 1.70 (sec) , antiderivative size = 659, normalized size of antiderivative = 3.00 \[ \int \frac {(a+b x)^{5/2}}{x^3 (c+d x)^{5/2}} \, dx=\left [\frac {15 \, {\left ({\left (3 \, b^{2} c^{2} d^{2} - 10 \, a b c d^{3} + 7 \, a^{2} d^{4}\right )} x^{4} + 2 \, {\left (3 \, b^{2} c^{3} d - 10 \, a b c^{2} d^{2} + 7 \, a^{2} c d^{3}\right )} x^{3} + {\left (3 \, b^{2} c^{4} - 10 \, a b c^{3} d + 7 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {\frac {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^{2} + {\left (b c^{2} + a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {a}{c}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (6 \, a^{2} c^{3} - {\left (16 \, b^{2} c^{2} d - 115 \, a b c d^{2} + 105 \, a^{2} d^{3}\right )} x^{3} - 2 \, {\left (12 \, b^{2} c^{3} - 79 \, a b c^{2} d + 70 \, a^{2} c d^{2}\right )} x^{2} + 3 \, {\left (9 \, a b c^{3} - 7 \, a^{2} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (c^{4} d^{2} x^{4} + 2 \, c^{5} d x^{3} + c^{6} x^{2}\right )}}, \frac {15 \, {\left ({\left (3 \, b^{2} c^{2} d^{2} - 10 \, a b c d^{3} + 7 \, a^{2} d^{4}\right )} x^{4} + 2 \, {\left (3 \, b^{2} c^{3} d - 10 \, a b c^{2} d^{2} + 7 \, a^{2} c d^{3}\right )} x^{3} + {\left (3 \, b^{2} c^{4} - 10 \, a b c^{3} d + 7 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {-\frac {a}{c}} \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 {a}{c}}}{2 \, {\left (a b d x^{2} + a^{2} c + {\left (a b c + a^{2} d\right )} x\right )}}\right ) - 2 \, {\left (6 \, a^{2} c^{3} - {\left (16 \, b^{2} c^{2} d - 115 \, a b c d^{2} + 105 \, a^{2} d^{3}\right )} x^{3} - 2 \, {\left (12 \, b^{2} c^{3} - 79 \, a b c^{2} d + 70 \, a^{2} c d^{2}\right )} x^{2} + 3 \, {\left (9 \, a b c^{3} - 7 \, a^{2} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{24 \, {\left (c^{4} d^{2} x^{4} + 2 \, c^{5} d x^{3} + c^{6} x^{2}\right )}}\right ] \] Input:

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

Output:

[1/48*(15*((3*b^2*c^2*d^2 - 10*a*b*c*d^3 + 7*a^2*d^4)*x^4 + 2*(3*b^2*c^3*d 
 - 10*a*b*c^2*d^2 + 7*a^2*c*d^3)*x^3 + (3*b^2*c^4 - 10*a*b*c^3*d + 7*a^2*c 
^2*d^2)*x^2)*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^2 + (b*c^2 + a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(a/c) 
+ 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(6*a^2*c^3 - (16*b^2*c^2*d - 115*a*b*c 
*d^2 + 105*a^2*d^3)*x^3 - 2*(12*b^2*c^3 - 79*a*b*c^2*d + 70*a^2*c*d^2)*x^2 
 + 3*(9*a*b*c^3 - 7*a^2*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^4*d^2*x^ 
4 + 2*c^5*d*x^3 + c^6*x^2), 1/24*(15*((3*b^2*c^2*d^2 - 10*a*b*c*d^3 + 7*a^ 
2*d^4)*x^4 + 2*(3*b^2*c^3*d - 10*a*b*c^2*d^2 + 7*a^2*c*d^3)*x^3 + (3*b^2*c 
^4 - 10*a*b*c^3*d + 7*a^2*c^2*d^2)*x^2)*sqrt(-a/c)*arctan(1/2*(2*a*c + (b* 
c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-a/c)/(a*b*d*x^2 + a^2*c + (a 
*b*c + a^2*d)*x)) - 2*(6*a^2*c^3 - (16*b^2*c^2*d - 115*a*b*c*d^2 + 105*a^2 
*d^3)*x^3 - 2*(12*b^2*c^3 - 79*a*b*c^2*d + 70*a^2*c*d^2)*x^2 + 3*(9*a*b*c^ 
3 - 7*a^2*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^4*d^2*x^4 + 2*c^5*d*x^ 
3 + c^6*x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((b*x+a)^(5/2)/x^3/(d*x+c)^(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. 1278 vs. \(2 (182) = 364\).

Time = 1.59 (sec) , antiderivative size = 1278, normalized size of antiderivative = 5.81 \[ \int \frac {(a+b x)^{5/2}}{x^3 (c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

2/3*sqrt(b*x + a)*((2*b^6*c^7*d^2*abs(b) - 13*a*b^5*c^6*d^3*abs(b) + 20*a^ 
2*b^4*c^5*d^4*abs(b) - 9*a^3*b^3*c^4*d^5*abs(b))*(b*x + a)/(b^3*c^9*d - a* 
b^2*c^8*d^2) + 3*(b^7*c^8*d*abs(b) - 6*a*b^6*c^7*d^2*abs(b) + 12*a^2*b^5*c 
^6*d^3*abs(b) - 10*a^3*b^4*c^5*d^4*abs(b) + 3*a^4*b^3*c^4*d^5*abs(b))/(b^3 
*c^9*d - a*b^2*c^8*d^2))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 5/4*(3*sq 
rt(b*d)*a*b^4*c^2 - 10*sqrt(b*d)*a^2*b^3*c*d + 7*sqrt(b*d)*a^3*b^2*d^2)*ar 
ctan(-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^4*abs(b)) - 1/ 
2*(9*sqrt(b*d)*a*b^10*c^5 - 47*sqrt(b*d)*a^2*b^9*c^4*d + 98*sqrt(b*d)*a^3* 
b^8*c^3*d^2 - 102*sqrt(b*d)*a^4*b^7*c^2*d^3 + 53*sqrt(b*d)*a^5*b^6*c*d^4 - 
 11*sqrt(b*d)*a^6*b^5*d^5 - 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*b^8*c^4 + 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^2*b^7*c^3*d - 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^3 
*b^6*c^2*d^2 - 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^4*b^5*c*d^3 + 33*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^4*d^4 + 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*b^6*c^3 - 15 
*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^5*c^2*d + 5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + ...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 1.91 (sec) , antiderivative size = 2594, normalized size of antiderivative = 11.79 \[ \int \frac {(a+b x)^{5/2}}{x^3 (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 84*sqrt(c + d*x)*sqrt(a + b*x)*a**3*c**4*d**2 + 294*sqrt(c + d*x)*sqrt 
(a + b*x)*a**3*c**3*d**3*x + 1960*sqrt(c + d*x)*sqrt(a + b*x)*a**3*c**2*d* 
*4*x**2 + 1470*sqrt(c + d*x)*sqrt(a + b*x)*a**3*c*d**5*x**3 - 12*sqrt(c + 
d*x)*sqrt(a + b*x)*a**2*b*c**5*d - 336*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b* 
c**4*d**2*x - 1932*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*c**3*d**3*x**2 - 140 
0*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*c**2*d**4*x**3 - 54*sqrt(c + d*x)*sqr 
t(a + b*x)*a*b**2*c**5*d*x + 20*sqrt(c + d*x)*sqrt(a + b*x)*a*b**2*c**4*d* 
*2*x**2 - 6*sqrt(c + d*x)*sqrt(a + b*x)*a*b**2*c**3*d**3*x**3 + 48*sqrt(c 
+ d*x)*sqrt(a + b*x)*b**3*c**5*d*x**2 + 32*sqrt(c + d*x)*sqrt(a + b*x)*b** 
3*c**4*d**2*x**3 + 735*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 
**3*c**2*d**4*x**2 + 1470*sqrt(c)*sqrt(a)*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**3*c*d**5*x**3 + 735*sqrt(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqr 
t(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x)) 
*a**3*d**6*x**4 - 945*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*c**3*d**3*x**2 - 1890*sqrt(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*s 
qrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x 
))*a**2*b*c**2*d**4*x**3 - 945*sqrt(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sq...