\(\int \frac {1}{x^2 (a+b x)^{3/2} (c+d x)^{5/2}} \, dx\) [399]

Optimal result

Integrand size = 22, antiderivative size = 264 \[ \int \frac {1}{x^2 (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=-\frac {b (3 b c-a d)}{a^2 c (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {1}{a c x \sqrt {a+b x} (c+d x)^{3/2}}-\frac {d \left (9 b^2 c^2-6 a b c d+5 a^2 d^2\right ) \sqrt {a+b x}}{3 a^2 c^2 (b c-a d)^2 (c+d x)^{3/2}}-\frac {d \left (9 b^3 c^3-9 a b^2 c^2 d+31 a^2 b c d^2-15 a^3 d^3\right ) \sqrt {a+b x}}{3 a^2 c^3 (b c-a d)^3 \sqrt {c+d x}}+\frac {(3 b c+5 a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{5/2} c^{7/2}} \] Output:

-b*(-a*d+3*b*c)/a^2/c/(-a*d+b*c)/(b*x+a)^(1/2)/(d*x+c)^(3/2)-1/a/c/x/(b*x+ 
a)^(1/2)/(d*x+c)^(3/2)-1/3*d*(5*a^2*d^2-6*a*b*c*d+9*b^2*c^2)*(b*x+a)^(1/2) 
/a^2/c^2/(-a*d+b*c)^2/(d*x+c)^(3/2)-1/3*d*(-15*a^3*d^3+31*a^2*b*c*d^2-9*a* 
b^2*c^2*d+9*b^3*c^3)*(b*x+a)^(1/2)/a^2/c^3/(-a*d+b*c)^3/(d*x+c)^(1/2)+(5*a 
*d+3*b*c)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(5/2)/c^( 
7/2)
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^2 (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {9 b^4 c^3 x (c+d x)^2+3 a b^3 c^2 (c-3 d x) (c+d x)^2-a^4 d^3 \left (3 c^2+20 c d x+15 d^2 x^2\right )+a^3 b d^2 \left (9 c^3+39 c^2 d x+11 c d^2 x^2-15 d^3 x^3\right )+a^2 b^2 c d \left (-9 c^3-9 c^2 d x+33 c d^2 x^2+31 d^3 x^3\right )}{3 a^2 c^3 (-b c+a d)^3 x \sqrt {a+b x} (c+d x)^{3/2}}+\frac {(3 b c+5 a d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{a^{5/2} c^{7/2}} \] Input:

Integrate[1/(x^2*(a + b*x)^(3/2)*(c + d*x)^(5/2)),x]
 

Output:

(9*b^4*c^3*x*(c + d*x)^2 + 3*a*b^3*c^2*(c - 3*d*x)*(c + d*x)^2 - a^4*d^3*( 
3*c^2 + 20*c*d*x + 15*d^2*x^2) + a^3*b*d^2*(9*c^3 + 39*c^2*d*x + 11*c*d^2* 
x^2 - 15*d^3*x^3) + a^2*b^2*c*d*(-9*c^3 - 9*c^2*d*x + 33*c*d^2*x^2 + 31*d^ 
3*x^3))/(3*a^2*c^3*(-(b*c) + a*d)^3*x*Sqrt[a + b*x]*(c + d*x)^(3/2)) + ((3 
*b*c + 5*a*d)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/(a 
^(5/2)*c^(7/2))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.17, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {114, 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.

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 169

\(\displaystyle -\frac {\frac {\frac {2 d \sqrt {a+b x} \left (5 a^2 d^2-6 a b c d+9 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}-\frac {2 \int -\frac {3 (3 b c+5 a d) (b c-a d)^2+2 b d \left (9 b^2 c^2-6 a b d c+5 a^2 d^2\right ) x}{2 x \sqrt {a+b x} (c+d x)^{3/2}}dx}{3 c (b c-a d)}}{a (b c-a d)}+\frac {2 b (3 b c-a d)}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{2 a c}-\frac {1}{a c x \sqrt {a+b x} (c+d x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {\int \frac {3 (3 b c+5 a d) (b c-a d)^2+2 b d \left (9 b^2 c^2-6 a b d c+5 a^2 d^2\right ) x}{x \sqrt {a+b x} (c+d x)^{3/2}}dx}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} \left (5 a^2 d^2-6 a b c d+9 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b (3 b c-a d)}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{2 a c}-\frac {1}{a c x \sqrt {a+b x} (c+d x)^{3/2}}\)

\(\Big \downarrow \) 169

\(\displaystyle -\frac {\frac {\frac {\frac {2 d \sqrt {a+b x} \left (-15 a^3 d^3+31 a^2 b c d^2-9 a b^2 c^2 d+9 b^3 c^3\right )}{c \sqrt {c+d x} (b c-a d)}-\frac {2 \int -\frac {3 (b c-a d)^3 (3 b c+5 a d)}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{c (b c-a d)}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} \left (5 a^2 d^2-6 a b c d+9 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b (3 b c-a d)}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{2 a c}-\frac {1}{a c x \sqrt {a+b x} (c+d x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {\frac {3 (5 a d+3 b c) (b c-a d)^2 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{c}+\frac {2 d \sqrt {a+b x} \left (-15 a^3 d^3+31 a^2 b c d^2-9 a b^2 c^2 d+9 b^3 c^3\right )}{c \sqrt {c+d x} (b c-a d)}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} \left (5 a^2 d^2-6 a b c d+9 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b (3 b c-a d)}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{2 a c}-\frac {1}{a c x \sqrt {a+b x} (c+d x)^{3/2}}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {\frac {\frac {\frac {6 (5 a d+3 b c) (b c-a d)^2 \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{c}+\frac {2 d \sqrt {a+b x} \left (-15 a^3 d^3+31 a^2 b c d^2-9 a b^2 c^2 d+9 b^3 c^3\right )}{c \sqrt {c+d x} (b c-a d)}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} \left (5 a^2 d^2-6 a b c d+9 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b (3 b c-a d)}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{2 a c}-\frac {1}{a c x \sqrt {a+b x} (c+d x)^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\frac {\frac {2 d \sqrt {a+b x} \left (5 a^2 d^2-6 a b c d+9 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}+\frac {\frac {2 d \sqrt {a+b x} \left (-15 a^3 d^3+31 a^2 b c d^2-9 a b^2 c^2 d+9 b^3 c^3\right )}{c \sqrt {c+d x} (b c-a d)}-\frac {6 (b c-a d)^2 (5 a d+3 b c) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{3/2}}}{3 c (b c-a d)}}{a (b c-a d)}+\frac {2 b (3 b c-a d)}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{2 a c}-\frac {1}{a c x \sqrt {a+b x} (c+d x)^{3/2}}\)

Input:

Int[1/(x^2*(a + b*x)^(3/2)*(c + d*x)^(5/2)),x]
 

Output:

-(1/(a*c*x*Sqrt[a + b*x]*(c + d*x)^(3/2))) - ((2*b*(3*b*c - a*d))/(a*(b*c 
- a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) + ((2*d*(9*b^2*c^2 - 6*a*b*c*d + 5*a 
^2*d^2)*Sqrt[a + b*x])/(3*c*(b*c - a*d)*(c + d*x)^(3/2)) + ((2*d*(9*b^3*c^ 
3 - 9*a*b^2*c^2*d + 31*a^2*b*c*d^2 - 15*a^3*d^3)*Sqrt[a + b*x])/(c*(b*c - 
a*d)*Sqrt[c + d*x]) - (6*(b*c - a*d)^2*(3*b*c + 5*a*d)*ArcTanh[(Sqrt[c]*Sq 
rt[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)))/(2*a*c)
 

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 114

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] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

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. \(1691\) vs. \(2(232)=464\).

Time = 0.32 (sec) , antiderivative size = 1692, normalized size of antiderivative = 6.41

method	result	size
default	\(\text {Expression too large to display}\)	\(1692\)

Input:

int(1/x^2/(b*x+a)^(3/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/6*(-30*a^4*d^5*x^2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+18*b^4*c^5*x*((b* 
x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-6*a^4*c^2*d^3*((b*x+a)*(d*x+c))^(1/2)*(a*c 
)^(1/2)+6*a*b^3*c^5*((b*x+a)*(d*x+c))^(1/2)*(a*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^4*b*d^6*x^4-9*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^5*c^4*d^2*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^5*d^6*x^ 
3-9*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^5*c^ 
6*x^2-18*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*b 
^5*c^5*d*x^3+30*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*c*d^5*x^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^5*c^2*d^4*x-9*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^4*c^6*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^4*c^5*d*x^2-36*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*c^3*d^3*x+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^3*b^2*c^4*d^2*x+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^2*b^3*c^5*d*x-30*a^3 
*b*d^5*x^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+18*b^4*c^3*d^2*x^3*((b*x+a) 
*(d*x+c))^(1/2)*(a*c)^(1/2)+36*b^4*c^4*d*x^2*((b*x+a)*(d*x+c))^(1/2)*(a*c) 
^(1/2)-40*a^4*c*d^4*x*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+18*a^3*b*c^3*d^2 
*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-18*a^2*b^2*c^4*d*((b*x+a)*(d*x+c))...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 819 vs. \(2 (232) = 464\).

Time = 2.20 (sec) , antiderivative size = 1658, normalized size of antiderivative = 6.28 \[ \int \frac {1}{x^2 (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/x^2/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

[1/12*(3*((3*b^5*c^4*d^2 - 4*a*b^4*c^3*d^3 - 6*a^2*b^3*c^2*d^4 + 12*a^3*b^ 
2*c*d^5 - 5*a^4*b*d^6)*x^4 + (6*b^5*c^5*d - 5*a*b^4*c^4*d^2 - 16*a^2*b^3*c 
^3*d^3 + 18*a^3*b^2*c^2*d^4 + 2*a^4*b*c*d^5 - 5*a^5*d^6)*x^3 + (3*b^5*c^6 
+ 2*a*b^4*c^5*d - 14*a^2*b^3*c^4*d^2 + 19*a^4*b*c^2*d^4 - 10*a^5*c*d^5)*x^ 
2 + (3*a*b^4*c^6 - 4*a^2*b^3*c^5*d - 6*a^3*b^2*c^4*d^2 + 12*a^4*b*c^3*d^3 
- 5*a^5*c^2*d^4)*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*(3*a^2*b^3*c^6 - 9*a^3*b^2*c^5*d + 9* 
a^4*b*c^4*d^2 - 3*a^5*c^3*d^3 + (9*a*b^4*c^4*d^2 - 9*a^2*b^3*c^3*d^3 + 31* 
a^3*b^2*c^2*d^4 - 15*a^4*b*c*d^5)*x^3 + (18*a*b^4*c^5*d - 15*a^2*b^3*c^4*d 
^2 + 33*a^3*b^2*c^3*d^3 + 11*a^4*b*c^2*d^4 - 15*a^5*c*d^5)*x^2 + (9*a*b^4* 
c^6 - 3*a^2*b^3*c^5*d - 9*a^3*b^2*c^4*d^2 + 39*a^4*b*c^3*d^3 - 20*a^5*c^2* 
d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/((a^3*b^4*c^7*d^2 - 3*a^4*b^3*c^6*d^3 
 + 3*a^5*b^2*c^5*d^4 - a^6*b*c^4*d^5)*x^4 + (2*a^3*b^4*c^8*d - 5*a^4*b^3*c 
^7*d^2 + 3*a^5*b^2*c^6*d^3 + a^6*b*c^5*d^4 - a^7*c^4*d^5)*x^3 + (a^3*b^4*c 
^9 - a^4*b^3*c^8*d - 3*a^5*b^2*c^7*d^2 + 5*a^6*b*c^6*d^3 - 2*a^7*c^5*d^4)* 
x^2 + (a^4*b^3*c^9 - 3*a^5*b^2*c^8*d + 3*a^6*b*c^7*d^2 - a^7*c^6*d^3)*x), 
-1/6*(3*((3*b^5*c^4*d^2 - 4*a*b^4*c^3*d^3 - 6*a^2*b^3*c^2*d^4 + 12*a^3*b^2 
*c*d^5 - 5*a^4*b*d^6)*x^4 + (6*b^5*c^5*d - 5*a*b^4*c^4*d^2 - 16*a^2*b^3*c^ 
3*d^3 + 18*a^3*b^2*c^2*d^4 + 2*a^4*b*c*d^5 - 5*a^5*d^6)*x^3 + (3*b^5*c^...
 

Sympy [F]

\[ \int \frac {1}{x^2 (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int \frac {1}{x^{2} \left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/x**2/(b*x+a)**(3/2)/(d*x+c)**(5/2),x)
 

Output:

Integral(1/(x**2*(a + b*x)**(3/2)*(c + d*x)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {1}{x^2 (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {5}{2}} x^{2}} \,d x } \] Input:

integrate(1/x^2/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + a)^(3/2)*(d*x + c)^(5/2)*x^2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1336 vs. \(2 (232) = 464\).

Time = 1.16 (sec) , antiderivative size = 1336, normalized size of antiderivative = 5.06 \[ \int \frac {1}{x^2 (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/x^2/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

-2/3*sqrt(b*x + a)*((11*b^6*c^6*d^5*abs(b) - 28*a*b^5*c^5*d^6*abs(b) + 23* 
a^2*b^4*c^4*d^7*abs(b) - 6*a^3*b^3*c^3*d^8*abs(b))*(b*x + a)/(b^7*c^11*d - 
 5*a*b^6*c^10*d^2 + 10*a^2*b^5*c^9*d^3 - 10*a^3*b^4*c^8*d^4 + 5*a^4*b^3*c^ 
7*d^5 - a^5*b^2*c^6*d^6) + 6*(2*b^7*c^7*d^4*abs(b) - 7*a*b^6*c^6*d^5*abs(b 
) + 9*a^2*b^5*c^5*d^6*abs(b) - 5*a^3*b^4*c^4*d^7*abs(b) + a^4*b^3*c^3*d^8* 
abs(b))/(b^7*c^11*d - 5*a*b^6*c^10*d^2 + 10*a^2*b^5*c^9*d^3 - 10*a^3*b^4*c 
^8*d^4 + 5*a^4*b^3*c^7*d^5 - a^5*b^2*c^6*d^6))/(b^2*c + (b*x + a)*b*d - a* 
b*d)^(3/2) - 2*(3*sqrt(b*d)*b^9*c^5 - 9*sqrt(b*d)*a*b^8*c^4*d + 12*sqrt(b* 
d)*a^2*b^7*c^3*d^2 - 10*sqrt(b*d)*a^3*b^6*c^2*d^3 + 5*sqrt(b*d)*a^4*b^5*c* 
d^4 - sqrt(b*d)*a^5*b^4*d^5 - 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^4 + 2*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^6*c^3*d - 6*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* 
c^2*d^2 + 2*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^4*c*d^3 + 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^3 - 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*c^2*d - sqrt(b*d)*(sqr 
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^3*c*d^ 
2 + 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^2*d^3)/((a^2*b^2*c^5*abs(b) - 2*a^3*b*c^4*d*abs(b) + a^4*c^...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int \frac {1}{x^2\,{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int(1/(x^2*(a + b*x)^(3/2)*(c + d*x)^(5/2)),x)
 

Output:

int(1/(x^2*(a + b*x)^(3/2)*(c + d*x)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^2 (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int \frac {1}{x^{2} \left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {5}{2}}}d x \] Input:

int(1/x^2/(b*x+a)^(3/2)/(d*x+c)^(5/2),x)
 

Output:

int(1/x^2/(b*x+a)^(3/2)/(d*x+c)^(5/2),x)