\(\int \frac {1}{x (a+b x)^{5/2} (c+d x)^{5/2}} \, dx\) [418]

Optimal result

Integrand size = 22, antiderivative size = 252 \[ \int \frac {1}{x (a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\frac {2 b}{3 a (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}+\frac {2 b (b c-3 a d)}{a^2 (b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/2}}+\frac {2 d \left (3 b^2 c^2-10 a b c d-a^2 d^2\right ) \sqrt {a+b x}}{3 a^2 c (b c-a d)^3 (c+d x)^{3/2}}+\frac {2 d (b c+a d) \left (3 b^2 c^2-14 a b c d+3 a^2 d^2\right ) \sqrt {a+b x}}{3 a^2 c^2 (b c-a d)^4 \sqrt {c+d x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{5/2} c^{5/2}} \] Output:

2/3*b/a/(-a*d+b*c)/(b*x+a)^(3/2)/(d*x+c)^(3/2)+2*b*(-3*a*d+b*c)/a^2/(-a*d+ 
b*c)^2/(b*x+a)^(1/2)/(d*x+c)^(3/2)+2/3*d*(-a^2*d^2-10*a*b*c*d+3*b^2*c^2)*( 
b*x+a)^(1/2)/a^2/c/(-a*d+b*c)^3/(d*x+c)^(3/2)+2/3*d*(a*d+b*c)*(3*a^2*d^2-1 
4*a*b*c*d+3*b^2*c^2)*(b*x+a)^(1/2)/a^2/c^2/(-a*d+b*c)^4/(d*x+c)^(1/2)-2*ar 
ctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(5/2)/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x (a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\frac {2 (a+b x)^{3/2} \left (a^2 c d^4-\frac {12 a^2 b c d^3 (c+d x)}{a+b x}+\frac {3 a^3 d^4 (c+d x)}{a+b x}+\frac {3 b^4 c^3 (c+d x)^2}{(a+b x)^2}-\frac {12 a b^3 c^2 d (c+d x)^2}{(a+b x)^2}+\frac {a b^4 c^2 (c+d x)^3}{(a+b x)^3}\right )}{3 a^2 c^2 (-b c+a d)^4 (c+d x)^{3/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{a^{5/2} c^{5/2}} \] Input:

Integrate[1/(x*(a + b*x)^(5/2)*(c + d*x)^(5/2)),x]
 

Output:

(2*(a + b*x)^(3/2)*(a^2*c*d^4 - (12*a^2*b*c*d^3*(c + d*x))/(a + b*x) + (3* 
a^3*d^4*(c + d*x))/(a + b*x) + (3*b^4*c^3*(c + d*x)^2)/(a + b*x)^2 - (12*a 
*b^3*c^2*d*(c + d*x)^2)/(a + b*x)^2 + (a*b^4*c^2*(c + d*x)^3)/(a + b*x)^3) 
)/(3*a^2*c^2*(-(b*c) + a*d)^4*(c + d*x)^(3/2)) - (2*ArcTanh[(Sqrt[a]*Sqrt[ 
c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/(a^(5/2)*c^(5/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.20, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {115, 27, 169, 27, 169, 27, 169, 27, 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.

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

\(\Big \downarrow \) 115

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 169

\(\displaystyle \frac {\frac {\frac {2 d \sqrt {a+b x} \left (-a^2 d^2-10 a b c d+3 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}-\frac {2 \int -\frac {3 (b c-a d)^3+2 b d \left (3 b^2 c^2-10 a b d c-a^2 d^2\right ) x}{2 x \sqrt {a+b x} (c+d x)^{3/2}}dx}{3 c (b c-a d)}}{a (b c-a d)}+\frac {2 b (b c-3 a d)}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b}{3 a (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {3 (b c-a d)^3+2 b d \left (3 b^2 c^2-10 a b d c-a^2 d^2\right ) x}{x \sqrt {a+b x} (c+d x)^{3/2}}dx}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} \left (-a^2 d^2-10 a b c d+3 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b (b c-3 a d)}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b}{3 a (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 169

\(\displaystyle \frac {\frac {\frac {\frac {2 d \sqrt {a+b x} (a d+b c) \left (3 a^2 d^2-14 a b c d+3 b^2 c^2\right )}{c \sqrt {c+d x} (b c-a d)}-\frac {2 \int -\frac {3 (b c-a d)^4}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{c (b c-a d)}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} \left (-a^2 d^2-10 a b c d+3 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b (b c-3 a d)}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b}{3 a (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\frac {3 (b c-a d)^3 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{c}+\frac {2 d \sqrt {a+b x} (a d+b c) \left (3 a^2 d^2-14 a b c d+3 b^2 c^2\right )}{c \sqrt {c+d x} (b c-a d)}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} \left (-a^2 d^2-10 a b c d+3 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b (b c-3 a d)}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b}{3 a (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {\frac {\frac {\frac {6 (b c-a d)^3 \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{c}+\frac {2 d \sqrt {a+b x} (a d+b c) \left (3 a^2 d^2-14 a b c d+3 b^2 c^2\right )}{c \sqrt {c+d x} (b c-a d)}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} \left (-a^2 d^2-10 a b c d+3 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b (b c-3 a d)}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b}{3 a (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {\frac {2 d \sqrt {a+b x} (a d+b c) \left (3 a^2 d^2-14 a b c d+3 b^2 c^2\right )}{c \sqrt {c+d x} (b c-a d)}-\frac {6 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{3/2}}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} \left (-a^2 d^2-10 a b c d+3 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b (b c-3 a d)}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b}{3 a (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\)

Input:

Int[1/(x*(a + b*x)^(5/2)*(c + d*x)^(5/2)),x]
 

Output:

(2*b)/(3*a*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2)) + ((2*b*(b*c - 3*a 
*d))/(a*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) + ((2*d*(3*b^2*c^2 - 10 
*a*b*c*d - a^2*d^2)*Sqrt[a + b*x])/(3*c*(b*c - a*d)*(c + d*x)^(3/2)) + ((2 
*d*(b*c + a*d)*(3*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*Sqrt[a + b*x])/(c*(b*c 
 - a*d)*Sqrt[c + d*x]) - (6*(b*c - a*d)^3*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/ 
(Sqrt[a]*Sqrt[c + d*x])])/(Sqrt[a]*c^(3/2)))/(3*c*(b*c - a*d)))/(a*(b*c - 
a*d)))/(a*(b*c - a*d))
 

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 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 115

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[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*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2 
*n, 2*p]
 

rule 169

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 + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[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*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n, 2*p]
 

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. \(2032\) vs. \(2(218)=436\).

Time = 0.31 (sec) , antiderivative size = 2033, normalized size of antiderivative = 8.07

method	result	size
default	\(\text {Expression too large to display}\)	\(2033\)

Input:

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

Output:

-1/3*(-12*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^5*b*c^3*d^3+18*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*b^2*c^4*d^2-12*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^3*c^5*d+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^4*b^2*d^6*x^4+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^6*c^4*d^2*x^4+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^6*d^6*x^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^6*c^6*x^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^6*c^2*d^4+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^2*b^4*c^6+6*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^5*b*d^6*x^3+6*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^6*c^5*d*x^3+ 
6*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^6*c*d^ 
5*x+6*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^ 
5*c^6*x-6*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^5*d^5*x-6*(a*c)^(1/2)*((b* 
x+a)*(d*x+c))^(1/2)*b^5*c^5*x-8*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^5*c* 
d^4-8*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^4*c^5+18*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^4*c^2*d^4*x^4-12*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^5*c^3*d^3*x^ 
4-18*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^...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1029 vs. \(2 (218) = 436\).

Time = 1.99 (sec) , antiderivative size = 2078, normalized size of antiderivative = 8.25 \[ \int \frac {1}{x (a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/6*(3*(a^2*b^4*c^6 - 4*a^3*b^3*c^5*d + 6*a^4*b^2*c^4*d^2 - 4*a^5*b*c^3*d 
^3 + a^6*c^2*d^4 + (b^6*c^4*d^2 - 4*a*b^5*c^3*d^3 + 6*a^2*b^4*c^2*d^4 - 4* 
a^3*b^3*c*d^5 + a^4*b^2*d^6)*x^4 + 2*(b^6*c^5*d - 3*a*b^5*c^4*d^2 + 2*a^2* 
b^4*c^3*d^3 + 2*a^3*b^3*c^2*d^4 - 3*a^4*b^2*c*d^5 + a^5*b*d^6)*x^3 + (b^6* 
c^6 - 9*a^2*b^4*c^4*d^2 + 16*a^3*b^3*c^3*d^3 - 9*a^4*b^2*c^2*d^4 + a^6*d^6 
)*x^2 + 2*(a*b^5*c^6 - 3*a^2*b^4*c^5*d + 2*a^3*b^3*c^4*d^2 + 2*a^4*b^2*c^3 
*d^3 - 3*a^5*b*c^2*d^4 + a^6*c*d^5)*x)*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*(4*a^2*b^4*c^6 - 1 
2*a^3*b^3*c^5*d - 12*a^5*b*c^3*d^3 + 4*a^6*c^2*d^4 + (3*a*b^5*c^4*d^2 - 11 
*a^2*b^4*c^3*d^3 - 11*a^3*b^3*c^2*d^4 + 3*a^4*b^2*c*d^5)*x^3 + 6*(a*b^5*c^ 
5*d - 3*a^2*b^4*c^4*d^2 - 4*a^3*b^3*c^3*d^3 - 3*a^4*b^2*c^2*d^4 + a^5*b*c* 
d^5)*x^2 + 3*(a*b^5*c^6 - a^2*b^4*c^5*d - 8*a^3*b^3*c^4*d^2 - 8*a^4*b^2*c^ 
3*d^3 - a^5*b*c^2*d^4 + a^6*c*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^5*b^ 
4*c^9 - 4*a^6*b^3*c^8*d + 6*a^7*b^2*c^7*d^2 - 4*a^8*b*c^6*d^3 + a^9*c^5*d^ 
4 + (a^3*b^6*c^7*d^2 - 4*a^4*b^5*c^6*d^3 + 6*a^5*b^4*c^5*d^4 - 4*a^6*b^3*c 
^4*d^5 + a^7*b^2*c^3*d^6)*x^4 + 2*(a^3*b^6*c^8*d - 3*a^4*b^5*c^7*d^2 + 2*a 
^5*b^4*c^6*d^3 + 2*a^6*b^3*c^5*d^4 - 3*a^7*b^2*c^4*d^5 + a^8*b*c^3*d^6)*x^ 
3 + (a^3*b^6*c^9 - 9*a^5*b^4*c^7*d^2 + 16*a^6*b^3*c^6*d^3 - 9*a^7*b^2*c^5* 
d^4 + a^9*c^3*d^6)*x^2 + 2*(a^4*b^5*c^9 - 3*a^5*b^4*c^8*d + 2*a^6*b^3*c...
 

Sympy [F]

\[ \int \frac {1}{x (a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\int \frac {1}{x \left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{x (a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/x/(b*x+a)^(5/2)/(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. 942 vs. \(2 (218) = 436\).

Time = 0.84 (sec) , antiderivative size = 942, normalized size of antiderivative = 3.74 \[ \int \frac {1}{x (a+b x)^{5/2} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-2/3*sqrt(b*x + a)*((11*b^7*c^6*d^5*abs(b) - 36*a*b^6*c^5*d^6*abs(b) + 42* 
a^2*b^5*c^4*d^7*abs(b) - 20*a^3*b^4*c^3*d^8*abs(b) + 3*a^4*b^3*c^2*d^9*abs 
(b))*(b*x + a)/(b^9*c^11*d - 7*a*b^8*c^10*d^2 + 21*a^2*b^7*c^9*d^3 - 35*a^ 
3*b^6*c^8*d^4 + 35*a^4*b^5*c^7*d^5 - 21*a^5*b^4*c^6*d^6 + 7*a^6*b^3*c^5*d^ 
7 - a^7*b^2*c^4*d^8) + 3*(4*b^8*c^7*d^4*abs(b) - 17*a*b^7*c^6*d^5*abs(b) + 
 28*a^2*b^6*c^5*d^6*abs(b) - 22*a^3*b^5*c^4*d^7*abs(b) + 8*a^4*b^4*c^3*d^8 
*abs(b) - a^5*b^3*c^2*d^9*abs(b))/(b^9*c^11*d - 7*a*b^8*c^10*d^2 + 21*a^2* 
b^7*c^9*d^3 - 35*a^3*b^6*c^8*d^4 + 35*a^4*b^5*c^7*d^5 - 21*a^5*b^4*c^6*d^6 
 + 7*a^6*b^3*c^5*d^7 - a^7*b^2*c^4*d^8))/(b^2*c + (b*x + a)*b*d - a*b*d)^( 
3/2) + 4/3*(3*sqrt(b*d)*b^9*c^3 - 17*sqrt(b*d)*a*b^8*c^2*d + 25*sqrt(b*d)* 
a^2*b^7*c*d^2 - 11*sqrt(b*d)*a^3*b^6*d^3 - 6*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x 
 + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^7*c^2 + 30*sqrt(b*d)*(sqr 
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^6*c*d - 
24*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^5*d^2 + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b* 
x + a)*b*d - a*b*d))^4*b^5*c - 9*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^4*d)/((a^2*b^3*c^3*abs(b) - 3*a^3*b 
^2*c^2*d*abs(b) + 3*a^4*b*c*d^2*abs(b) - a^5*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)^3) - 2* 
sqrt(b*d)*b*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqr...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x (a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\int \frac {1}{x\,{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x (a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\int \frac {1}{x \left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {5}{2}}}d x \] Input:

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

Output:

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