\(\int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx\) [363]

Optimal result

Integrand size = 22, antiderivative size = 277 \[ \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx=\frac {d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) \sqrt {a+b x}}{12 a^2 c^3 (b c-a d) (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}+\frac {d \left (9 b^3 c^3+15 a b^2 c^2 d-145 a^2 b c d^2+105 a^3 d^3\right ) \sqrt {a+b x}}{12 a^2 c^4 (b c-a d)^2 \sqrt {c+d x}}-\frac {\left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{9/2}} \] Output:

1/12*d*(-35*a^2*d^2+18*a*b*c*d+9*b^2*c^2)*(b*x+a)^(1/2)/a^2/c^3/(-a*d+b*c) 
/(d*x+c)^(3/2)-1/2*(b*x+a)^(1/2)/a/c/x^2/(d*x+c)^(3/2)+1/4*(7*a*d+3*b*c)*( 
b*x+a)^(1/2)/a^2/c^2/x/(d*x+c)^(3/2)+1/12*d*(105*a^3*d^3-145*a^2*b*c*d^2+1 
5*a*b^2*c^2*d+9*b^3*c^3)*(b*x+a)^(1/2)/a^2/c^4/(-a*d+b*c)^2/(d*x+c)^(1/2)- 
1/4*(35*a^2*d^2+10*a*b*c*d+3*b^2*c^2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2 
)/(d*x+c)^(1/2))/a^(5/2)/c^(9/2)
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx=\frac {\sqrt {a+b x} \left (9 b^3 c^3 x (c+d x)^2-3 a b^2 c^2 (2 c-5 d x) (c+d x)^2+a^2 b c d \left (12 c^3-33 c^2 d x-198 c d^2 x^2-145 d^3 x^3\right )+a^3 d^2 \left (-6 c^3+21 c^2 d x+140 c d^2 x^2+105 d^3 x^3\right )\right )}{12 a^2 c^4 (b c-a d)^2 x^2 (c+d x)^{3/2}}-\frac {\left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{9/2}} \] Input:

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

Output:

(Sqrt[a + b*x]*(9*b^3*c^3*x*(c + d*x)^2 - 3*a*b^2*c^2*(2*c - 5*d*x)*(c + d 
*x)^2 + a^2*b*c*d*(12*c^3 - 33*c^2*d*x - 198*c*d^2*x^2 - 145*d^3*x^3) + a^ 
3*d^2*(-6*c^3 + 21*c^2*d*x + 140*c*d^2*x^2 + 105*d^3*x^3)))/(12*a^2*c^4*(b 
*c - a*d)^2*x^2*(c + d*x)^(3/2)) - ((3*b^2*c^2 + 10*a*b*c*d + 35*a^2*d^2)* 
ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(5/2)*c^(9/ 
2))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {114, 27, 168, 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.

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 168

\(\displaystyle -\frac {-\frac {\int \frac {3 b^2 c^2+10 a b d c+35 a^2 d^2+4 b d (3 b c+7 a d) x}{2 x \sqrt {a+b x} (c+d x)^{5/2}}dx}{a c}-\frac {\sqrt {a+b x} (7 a d+3 b c)}{a c x (c+d x)^{3/2}}}{4 a c}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\int \frac {3 b^2 c^2+10 a b d c+35 a^2 d^2+4 b d (3 b c+7 a d) x}{x \sqrt {a+b x} (c+d x)^{5/2}}dx}{2 a c}-\frac {\sqrt {a+b x} (7 a d+3 b c)}{a c x (c+d x)^{3/2}}}{4 a c}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}\)

\(\Big \downarrow \) 169

\(\displaystyle -\frac {-\frac {\frac {2 d \sqrt {a+b x} \left (-35 a^2 d^2+18 a b c d+9 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}-\frac {2 \int -\frac {3 (b c-a d) \left (3 b^2 c^2+10 a b d c+35 a^2 d^2\right )+2 b d \left (9 b^2 c^2+18 a b d c-35 a^2 d^2\right ) x}{2 x \sqrt {a+b x} (c+d x)^{3/2}}dx}{3 c (b c-a d)}}{2 a c}-\frac {\sqrt {a+b x} (7 a d+3 b c)}{a c x (c+d x)^{3/2}}}{4 a c}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\frac {\int \frac {3 (b c-a d) \left (3 b^2 c^2+10 a b d c+35 a^2 d^2\right )+2 b d \left (9 b^2 c^2+18 a b d c-35 a^2 d^2\right ) x}{x \sqrt {a+b x} (c+d x)^{3/2}}dx}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} \left (-35 a^2 d^2+18 a b c d+9 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}}{2 a c}-\frac {\sqrt {a+b x} (7 a d+3 b c)}{a c x (c+d x)^{3/2}}}{4 a c}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}\)

\(\Big \downarrow \) 169

\(\displaystyle -\frac {-\frac {\frac {\frac {2 d \sqrt {a+b x} \left (105 a^3 d^3-145 a^2 b c d^2+15 a b^2 c^2 d+9 b^3 c^3\right )}{c \sqrt {c+d x} (b c-a d)}-\frac {2 \int -\frac {3 (b c-a d)^2 \left (3 b^2 c^2+10 a b d c+35 a^2 d^2\right )}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{c (b c-a d)}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} \left (-35 a^2 d^2+18 a b c d+9 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}}{2 a c}-\frac {\sqrt {a+b x} (7 a d+3 b c)}{a c x (c+d x)^{3/2}}}{4 a c}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\frac {\frac {3 (b c-a d) \left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{c}+\frac {2 d \sqrt {a+b x} \left (105 a^3 d^3-145 a^2 b c d^2+15 a b^2 c^2 d+9 b^3 c^3\right )}{c \sqrt {c+d x} (b c-a d)}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} \left (-35 a^2 d^2+18 a b c d+9 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}}{2 a c}-\frac {\sqrt {a+b x} (7 a d+3 b c)}{a c x (c+d x)^{3/2}}}{4 a c}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {-\frac {\frac {\frac {6 (b c-a d) \left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{c}+\frac {2 d \sqrt {a+b x} \left (105 a^3 d^3-145 a^2 b c d^2+15 a b^2 c^2 d+9 b^3 c^3\right )}{c \sqrt {c+d x} (b c-a d)}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} \left (-35 a^2 d^2+18 a b c d+9 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}}{2 a c}-\frac {\sqrt {a+b x} (7 a d+3 b c)}{a c x (c+d x)^{3/2}}}{4 a c}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {-\frac {\frac {2 d \sqrt {a+b x} \left (-35 a^2 d^2+18 a b c d+9 b^2 c^2\right )}{3 c (c+d x)^{3/2} (b c-a d)}+\frac {\frac {2 d \sqrt {a+b x} \left (105 a^3 d^3-145 a^2 b c d^2+15 a b^2 c^2 d+9 b^3 c^3\right )}{c \sqrt {c+d x} (b c-a d)}-\frac {6 (b c-a d) \left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{3/2}}}{3 c (b c-a d)}}{2 a c}-\frac {\sqrt {a+b x} (7 a d+3 b c)}{a c x (c+d x)^{3/2}}}{4 a c}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}\)

Input:

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

Output:

-1/2*Sqrt[a + b*x]/(a*c*x^2*(c + d*x)^(3/2)) - (-(((3*b*c + 7*a*d)*Sqrt[a 
+ b*x])/(a*c*x*(c + d*x)^(3/2))) - ((2*d*(9*b^2*c^2 + 18*a*b*c*d - 35*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 + 
 15*a*b^2*c^2*d - 145*a^2*b*c*d^2 + 105*a^3*d^3)*Sqrt[a + b*x])/(c*(b*c - 
a*d)*Sqrt[c + d*x]) - (6*(b*c - a*d)*(3*b^2*c^2 + 10*a*b*c*d + 35*a^2*d^2) 
*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(Sqrt[a]*c^(3/2 
)))/(3*c*(b*c - a*d)))/(2*a*c))/(4*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 168

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] && ILtQ[m, -1]
 

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. \(1287\) vs. \(2(239)=478\).

Time = 0.29 (sec) , antiderivative size = 1288, normalized size of antiderivative = 4.65

method	result	size
default	\(\text {Expression too large to display}\)	\(1288\)

Input:

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

Output:

-1/24*(b*x+a)^(1/2)/a^2/c^4*(290*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b 
*c*d^4*x^3-30*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^2*c^2*d^3*x^3+396*(a 
*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b*c^2*d^3*x^2-48*(a*c)^(1/2)*((b*x+a 
)*(d*x+c))^(1/2)*a*b^2*c^3*d^2*x^2+66*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)* 
a^2*b*c^3*d^2*x-6*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^2*c^4*d*x+12*a^3 
*c^3*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+12*a*b^2*c^5*(a*c)^(1/2)*((b* 
x+a)*(d*x+c))^(1/2)+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^4*c^4*d^2*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)*b^4*c^5*d*x^3-360*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^4*x^3+108*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*d^3*x^3+24*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^4*d^2*x^3-18 
0*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^ 
3*d^3*x^2+54*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^4*d^2*x^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*b^3*c^5*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^4*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^4*c^6*x^2-42*(a*c)^(1/2)*((b*x+a)*(d*x+c)) 
^(1/2)*a^3*c^2*d^3*x-24*a^2*b*c^4*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-18 
0*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...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 583 vs. \(2 (239) = 478\).

Time = 1.54 (sec) , antiderivative size = 1186, normalized size of antiderivative = 4.28 \[ \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/48*(3*((3*b^4*c^4*d^2 + 4*a*b^3*c^3*d^3 + 18*a^2*b^2*c^2*d^4 - 60*a^3*b 
*c*d^5 + 35*a^4*d^6)*x^4 + 2*(3*b^4*c^5*d + 4*a*b^3*c^4*d^2 + 18*a^2*b^2*c 
^3*d^3 - 60*a^3*b*c^2*d^4 + 35*a^4*c*d^5)*x^3 + (3*b^4*c^6 + 4*a*b^3*c^5*d 
 + 18*a^2*b^2*c^4*d^2 - 60*a^3*b*c^3*d^3 + 35*a^4*c^2*d^4)*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 + (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*(6*a^2*b^2*c^6 - 12*a^3*b*c^5*d + 6*a^4*c^4*d^2 - (9*a*b^3*c^4*d^2 
+ 15*a^2*b^2*c^3*d^3 - 145*a^3*b*c^2*d^4 + 105*a^4*c*d^5)*x^3 - 2*(9*a*b^3 
*c^5*d + 12*a^2*b^2*c^4*d^2 - 99*a^3*b*c^3*d^3 + 70*a^4*c^2*d^4)*x^2 - 3*( 
3*a*b^3*c^6 + a^2*b^2*c^5*d - 11*a^3*b*c^4*d^2 + 7*a^4*c^3*d^3)*x)*sqrt(b* 
x + a)*sqrt(d*x + c))/((a^3*b^2*c^7*d^2 - 2*a^4*b*c^6*d^3 + a^5*c^5*d^4)*x 
^4 + 2*(a^3*b^2*c^8*d - 2*a^4*b*c^7*d^2 + a^5*c^6*d^3)*x^3 + (a^3*b^2*c^9 
- 2*a^4*b*c^8*d + a^5*c^7*d^2)*x^2), 1/24*(3*((3*b^4*c^4*d^2 + 4*a*b^3*c^3 
*d^3 + 18*a^2*b^2*c^2*d^4 - 60*a^3*b*c*d^5 + 35*a^4*d^6)*x^4 + 2*(3*b^4*c^ 
5*d + 4*a*b^3*c^4*d^2 + 18*a^2*b^2*c^3*d^3 - 60*a^3*b*c^2*d^4 + 35*a^4*c*d 
^5)*x^3 + (3*b^4*c^6 + 4*a*b^3*c^5*d + 18*a^2*b^2*c^4*d^2 - 60*a^3*b*c^3*d 
^3 + 35*a^4*c^2*d^4)*x^2)*sqrt(-a*c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sq 
rt(-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*(6*a^2*b^2*c^6 - 12*a^3*b*c^5*d + 6*a^4*c^4*d^2 - (9*a*b^3 
*c^4*d^2 + 15*a^2*b^2*c^3*d^3 - 145*a^3*b*c^2*d^4 + 105*a^4*c*d^5)*x^3 ...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1234 vs. \(2 (239) = 478\).

Time = 0.99 (sec) , antiderivative size = 1234, normalized size of antiderivative = 4.45 \[ \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-2/3*sqrt(b*x + a)*((11*b^4*c^5*d^5*abs(b) - 9*a*b^3*c^4*d^6*abs(b))*(b*x 
+ a)/(b^4*c^10*d - 2*a*b^3*c^9*d^2 + a^2*b^2*c^8*d^3) + 3*(4*b^5*c^6*d^4*a 
bs(b) - 7*a*b^4*c^5*d^5*abs(b) + 3*a^2*b^3*c^4*d^6*abs(b))/(b^4*c^10*d - 2 
*a*b^3*c^9*d^2 + a^2*b^2*c^8*d^3))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 
 1/4*(3*sqrt(b*d)*b^4*c^2 + 10*sqrt(b*d)*a*b^3*c*d + 35*sqrt(b*d)*a^2*b^2* 
d^2)*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^2*b*c^4*a 
bs(b)) + 1/2*(3*sqrt(b*d)*b^10*c^5 - sqrt(b*d)*a*b^9*c^4*d - 26*sqrt(b*d)* 
a^2*b^8*c^3*d^2 + 54*sqrt(b*d)*a^3*b^7*c^2*d^3 - 41*sqrt(b*d)*a^4*b^6*c*d^ 
4 + 11*sqrt(b*d)*a^5*b^5*d^5 - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt 
(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^8*c^4 - 28*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^7*c^3*d + 50*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 
^6*c^2*d^2 + 20*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^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^4*b^4*d^4 + 9*sqrt(b*d)*(sqrt(b*d 
)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^6*c^3 + 39*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^5*c^2*d + 31*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^4*c*d^2 + 33*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) ...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 1.99 (sec) , antiderivative size = 3963, normalized size of antiderivative = 14.31 \[ \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 84*sqrt(c + d*x)*sqrt(a + b*x)*a**5*c**4*d**3 + 294*sqrt(c + d*x)*sqrt 
(a + b*x)*a**5*c**3*d**4*x + 1960*sqrt(c + d*x)*sqrt(a + b*x)*a**5*c**2*d* 
*5*x**2 + 1470*sqrt(c + d*x)*sqrt(a + b*x)*a**5*c*d**6*x**3 + 156*sqrt(c + 
 d*x)*sqrt(a + b*x)*a**4*b*c**5*d**2 - 420*sqrt(c + d*x)*sqrt(a + b*x)*a** 
4*b*c**4*d**3*x - 2492*sqrt(c + d*x)*sqrt(a + b*x)*a**4*b*c**3*d**4*x**2 - 
 1820*sqrt(c + d*x)*sqrt(a + b*x)*a**4*b*c**2*d**5*x**3 - 60*sqrt(c + d*x) 
*sqrt(a + b*x)*a**3*b**2*c**6*d - 24*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b**2 
*c**5*d**2*x - 60*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b**2*c**4*d**3*x**2 - 8 
0*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b**2*c**3*d**4*x**3 - 12*sqrt(c + d*x)* 
sqrt(a + b*x)*a**2*b**3*c**7 + 132*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**3*c 
**6*d*x + 300*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**3*c**5*d**2*x**2 + 156*s 
qrt(c + d*x)*sqrt(a + b*x)*a**2*b**3*c**4*d**3*x**3 + 18*sqrt(c + d*x)*sqr 
t(a + b*x)*a*b**4*c**7*x + 36*sqrt(c + d*x)*sqrt(a + b*x)*a*b**4*c**6*d*x* 
*2 + 18*sqrt(c + d*x)*sqrt(a + b*x)*a*b**4*c**5*d**2*x**3 + 735*sqrt(c)*sq 
rt(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**5*c**2*d**5*x**2 + 1470*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**5*c*d**6*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**5*d**7*x**4 - 1155*sqrt(c)...