\(\int \frac {x^3}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx\) [394]

Optimal result

Integrand size = 22, antiderivative size = 198 \[ \int \frac {x^3}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {2 a^3}{b^3 (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}+\frac {2 \left (b^3 c^3+3 a^3 d^3\right ) \sqrt {a+b x}}{3 b^3 d^2 (b c-a d)^2 (c+d x)^{3/2}}-\frac {2 \left (4 b^3 c^3-9 a b^2 c^2 d-3 a^3 d^3\right ) \sqrt {a+b x}}{3 b^2 d^2 (b c-a d)^3 \sqrt {c+d x}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} d^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.76 \[ \int \frac {x^3}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=-\frac {2 (a+b x)^{3/2} \left (b c^3 d+\frac {3 b^2 c^3 (c+d x)}{a+b x}-\frac {9 a b c^2 d (c+d x)}{a+b x}-\frac {3 a^3 d^2 (c+d x)^2}{(a+b x)^2}\right )}{3 b d^2 (b c-a d)^3 (c+d x)^{3/2}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{3/2} d^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {109, 27, 162, 66, 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.

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 162

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(2*a*x^2)/(b*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) - ((2*c*Sqrt[a + b 
*x]*(c*(b*c - 3*a*d)*(3*b*c + a*d) + 2*d*(2*b^2*c^2 - 3*a*b*c*d - 3*a^2*d^ 
2)*x))/(3*d^2*(b*c - a*d)^2*(c + d*x)^(3/2)) - (2*(b*c - a*d)*ArcTanh[(Sqr 
t[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*d^(5/2)))/(b*(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 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f 
*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) 
+ c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) 
 + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || 
IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 162

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) 
)*((g_.) + (h_.)*(x_)), x_] :> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) 
 - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g + e*h) + d*e 
*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + 
e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)/(b^2*(b 
*c - a*d)^2*(m + 1)*(m + 2)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] + Sim 
p[(f*(h/b^2) - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d 
*(f*g + e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/( 
b^2*(b*c - a*d)^2*(m + 1)*(m + 2)))   Int[(a + b*x)^(m + 2)*(c + d*x)^n, x] 
, x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + 
 n + 3, 0] &&  !LtQ[n, -2]))
 

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. \(1288\) vs. \(2(170)=340\).

Time = 0.29 (sec) , antiderivative size = 1289, normalized size of antiderivative = 6.51

method	result	size
default	\(\text {Expression too large to display}\)	\(1289\)

Input:

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

Output:

1/3*(3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b 
)^(1/2))*a^4*d^5*x^2-3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/ 
2)+a*d+b*c)/(d*b)^(1/2))*b^4*c^5*x+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^( 
1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^4*c^2*d^3-3*ln(1/2*(2*b*d*x+2*((b 
*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^3*c^5+8*b^3*c^3 
*d*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-12*a^3*c*d^3*x*((b*x+a)*(d*x+c) 
)^(1/2)*(d*b)^(1/2)-16*a^2*b*c^3*d*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+3*l 
n(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2)) 
*a*b^3*c^4*d*x-18*a*b^2*c^2*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-18 
*a^2*b*c^2*d^2*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-8*a*b^2*c^3*d*x*((b*x 
+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-9*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2) 
*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b^2*c^2*d^3*x^2+15*ln(1/2*(2*b*d*x+ 
2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^3*c^3*d^2* 
x^2-15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b 
)^(1/2))*a^3*b*c^2*d^3*x+9*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b) 
^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b^2*c^3*d^2*x-6*ln(1/2*(2*b*d*x+2*((b*x+a 
)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*b^4*c^4*d*x^2+6*ln(1/2* 
(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^4*c 
*d^4*x-9*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d 
*b)^(1/2))*a^3*b*c^3*d^2-9*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 656 vs. \(2 (170) = 340\).

Time = 0.45 (sec) , antiderivative size = 1326, normalized size of antiderivative = 6.70 \[ \int \frac {x^3}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x^3/(b*x+a)^(3/2)/(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. 490 vs. \(2 (170) = 340\).

Time = 0.27 (sec) , antiderivative size = 490, normalized size of antiderivative = 2.47 \[ \int \frac {x^3}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {4 \, a^{3} b d}{{\left (\sqrt {b d} b^{2} c^{2} {\left | b \right |} - 2 \, \sqrt {b d} a b c d {\left | b \right |} + \sqrt {b d} a^{2} d^{2} {\left | b \right |}\right )} {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}} - \frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (4 \, b^{8} c^{5} d^{2} - 17 \, a b^{7} c^{4} d^{3} + 22 \, a^{2} b^{6} c^{3} d^{4} - 9 \, a^{3} b^{5} c^{2} d^{5}\right )} {\left (b x + a\right )}}{b^{7} c^{5} d^{3} {\left | b \right |} - 5 \, a b^{6} c^{4} d^{4} {\left | b \right |} + 10 \, a^{2} b^{5} c^{3} d^{5} {\left | b \right |} - 10 \, a^{3} b^{4} c^{2} d^{6} {\left | b \right |} + 5 \, a^{4} b^{3} c d^{7} {\left | b \right |} - a^{5} b^{2} d^{8} {\left | b \right |}} + \frac {3 \, {\left (b^{9} c^{6} d - 6 \, a b^{8} c^{5} d^{2} + 12 \, a^{2} b^{7} c^{4} d^{3} - 10 \, a^{3} b^{6} c^{3} d^{4} + 3 \, a^{4} b^{5} c^{2} d^{5}\right )}}{b^{7} c^{5} d^{3} {\left | b \right |} - 5 \, a b^{6} c^{4} d^{4} {\left | b \right |} + 10 \, a^{2} b^{5} c^{3} d^{5} {\left | b \right |} - 10 \, a^{3} b^{4} c^{2} d^{6} {\left | b \right |} + 5 \, a^{4} b^{3} c d^{7} {\left | b \right |} - a^{5} b^{2} d^{8} {\left | b \right |}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {\log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{\sqrt {b d} d^{2} {\left | b \right |}} \] Input:

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

Output:

4*a^3*b*d/((sqrt(b*d)*b^2*c^2*abs(b) - 2*sqrt(b*d)*a*b*c*d*abs(b) + sqrt(b 
*d)*a^2*d^2*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)) - 2/3*sqrt(b*x + a)*((4*b^8*c^5*d^2 - 17*a* 
b^7*c^4*d^3 + 22*a^2*b^6*c^3*d^4 - 9*a^3*b^5*c^2*d^5)*(b*x + a)/(b^7*c^5*d 
^3*abs(b) - 5*a*b^6*c^4*d^4*abs(b) + 10*a^2*b^5*c^3*d^5*abs(b) - 10*a^3*b^ 
4*c^2*d^6*abs(b) + 5*a^4*b^3*c*d^7*abs(b) - a^5*b^2*d^8*abs(b)) + 3*(b^9*c 
^6*d - 6*a*b^8*c^5*d^2 + 12*a^2*b^7*c^4*d^3 - 10*a^3*b^6*c^3*d^4 + 3*a^4*b 
^5*c^2*d^5)/(b^7*c^5*d^3*abs(b) - 5*a*b^6*c^4*d^4*abs(b) + 10*a^2*b^5*c^3* 
d^5*abs(b) - 10*a^3*b^4*c^2*d^6*abs(b) + 5*a^4*b^3*c*d^7*abs(b) - a^5*b^2* 
d^8*abs(b)))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - log((sqrt(b*d)*sqrt(b 
*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(b*d)*d^2*abs(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 1112, normalized size of antiderivative = 5.62 \[ \int \frac {x^3}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*(3*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*s 
qrt(c + d*x))/sqrt(a*d - b*c))*a**3*c**2*d**3 + 6*sqrt(d)*sqrt(b)*sqrt(a + 
 b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c)) 
*a**3*c*d**4*x + 3*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x 
) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*d**5*x**2 - 9*sqrt(d)*sqr 
t(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqr 
t(a*d - b*c))*a**2*b*c**3*d**2 - 18*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqr 
t(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b*c**2*d 
**3*x - 9*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt( 
b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b*c*d**4*x**2 + 9*sqrt(d)*sqrt(b)* 
sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d 
 - b*c))*a*b**2*c**4*d + 18*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqr 
t(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b**2*c**3*d**2*x + 
9*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt( 
c + d*x))/sqrt(a*d - b*c))*a*b**2*c**2*d**3*x**2 - 3*sqrt(d)*sqrt(b)*sqrt( 
a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b* 
c))*b**3*c**5 - 6*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) 
 + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**3*c**4*d*x - 3*sqrt(d)*sqrt( 
b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt( 
a*d - b*c))*b**3*c**3*d**2*x**2 + 6*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**3*...