\(\int \frac {x^2}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx\) [410]

Optimal result

Integrand size = 22, antiderivative size = 126 \[ \int \frac {x^2}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx=-\frac {2 a^2 \sqrt {c+d x}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {4 a (3 b c-2 a d) \sqrt {c+d x}}{3 b^2 (b c-a d)^2 \sqrt {a+b x}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2} \sqrt {d}} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.82 \[ \int \frac {x^2}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx=-\frac {2 a \sqrt {c+d x} \left (-6 b c+3 a d+\frac {a b (c+d x)}{a+b x}\right )}{3 b^2 (b c-a d)^2 \sqrt {a+b x}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{5/2} \sqrt {d}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {100, 27, 87, 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.

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 66

\(\displaystyle -\frac {-6 (b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}-\frac {4 a \sqrt {c+d x} (3 b c-2 a d)}{\sqrt {a+b x} (b c-a d)}}{3 b^2 (b c-a d)}-\frac {2 a^2 \sqrt {c+d x}}{3 b^2 (a+b x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 a^2 \sqrt {c+d x}}{3 b^2 (a+b x)^{3/2} (b c-a d)}-\frac {-\frac {6 (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {d}}-\frac {4 a \sqrt {c+d x} (3 b c-2 a d)}{\sqrt {a+b x} (b c-a d)}}{3 b^2 (b c-a d)}\)

Input:

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

Output:

(-2*a^2*Sqrt[c + d*x])/(3*b^2*(b*c - a*d)*(a + b*x)^(3/2)) - ((-4*a*(3*b*c 
 - 2*a*d)*Sqrt[c + d*x])/((b*c - a*d)*Sqrt[a + b*x]) - (6*(b*c - a*d)*ArcT 
anh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*Sqrt[d]))/( 
3*b^2*(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 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 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. \(603\) vs. \(2(102)=204\).

Time = 0.28 (sec) , antiderivative size = 604, normalized size of antiderivative = 4.79

method	result	size
default	\(\frac {\sqrt {x d +c}\, \left (3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} b^{2} d^{2} x^{2}-6 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a \,b^{3} c d \,x^{2}+3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) b^{4} c^{2} x^{2}+6 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{3} b \,d^{2} x -12 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} b^{2} c d x +6 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a \,b^{3} c^{2} x +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{4} d^{2}-6 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{3} b c d +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} b^{2} c^{2}-8 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a^{2} b d x +12 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a \,b^{2} c x -6 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a^{3} d +10 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a^{2} b c \right )}{3 \sqrt {d b}\, \left (a d -b c \right )^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, b^{2} \left (b x +a \right )^{\frac {3}{2}}}\)	\(604\)

Input:

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

Output:

1/3*(d*x+c)^(1/2)*(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))*a^2*b^2*d^2*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*b^3*c*d*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^2*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^3*b*d^2*x-12*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*d*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))*a*b^3*c^2*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*d^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^3*b*c* 
d+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^2*b^2*c^2-8*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b*d*x+12*(d*b) 
^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^2*c*x-6*(d*b)^(1/2)*((b*x+a)*(d*x+c))^( 
1/2)*a^3*d+10*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b*c)/(d*b)^(1/2)/(a* 
d-b*c)^2/((b*x+a)*(d*x+c))^(1/2)/b^2/(b*x+a)^(3/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (102) = 204\).

Time = 0.18 (sec) , antiderivative size = 670, normalized size of antiderivative = 5.32 \[ \int \frac {x^2}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx=\left [\frac {3 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{2} + 2 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (5 \, a^{2} b^{2} c d - 3 \, a^{3} b d^{2} + 2 \, {\left (3 \, a b^{3} c d - 2 \, a^{2} b^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (a^{2} b^{5} c^{2} d - 2 \, a^{3} b^{4} c d^{2} + a^{4} b^{3} d^{3} + {\left (b^{7} c^{2} d - 2 \, a b^{6} c d^{2} + a^{2} b^{5} d^{3}\right )} x^{2} + 2 \, {\left (a b^{6} c^{2} d - 2 \, a^{2} b^{5} c d^{2} + a^{3} b^{4} d^{3}\right )} x\right )}}, -\frac {3 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{2} + 2 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (5 \, a^{2} b^{2} c d - 3 \, a^{3} b d^{2} + 2 \, {\left (3 \, a b^{3} c d - 2 \, a^{2} b^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (a^{2} b^{5} c^{2} d - 2 \, a^{3} b^{4} c d^{2} + a^{4} b^{3} d^{3} + {\left (b^{7} c^{2} d - 2 \, a b^{6} c d^{2} + a^{2} b^{5} d^{3}\right )} x^{2} + 2 \, {\left (a b^{6} c^{2} d - 2 \, a^{2} b^{5} c d^{2} + a^{3} b^{4} d^{3}\right )} x\right )}}\right ] \] Input:

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

Output:

[1/6*(3*(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2 + (b^4*c^2 - 2*a*b^3*c*d + a^ 
2*b^2*d^2)*x^2 + 2*(a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x)*sqrt(b*d)*lo 
g(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*(5*a^ 
2*b^2*c*d - 3*a^3*b*d^2 + 2*(3*a*b^3*c*d - 2*a^2*b^2*d^2)*x)*sqrt(b*x + a) 
*sqrt(d*x + c))/(a^2*b^5*c^2*d - 2*a^3*b^4*c*d^2 + a^4*b^3*d^3 + (b^7*c^2* 
d - 2*a*b^6*c*d^2 + a^2*b^5*d^3)*x^2 + 2*(a*b^6*c^2*d - 2*a^2*b^5*c*d^2 + 
a^3*b^4*d^3)*x), -1/3*(3*(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2 + (b^4*c^2 - 
 2*a*b^3*c*d + a^2*b^2*d^2)*x^2 + 2*(a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2 
)*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*(5*a^2* 
b^2*c*d - 3*a^3*b*d^2 + 2*(3*a*b^3*c*d - 2*a^2*b^2*d^2)*x)*sqrt(b*x + a)*s 
qrt(d*x + c))/(a^2*b^5*c^2*d - 2*a^3*b^4*c*d^2 + a^4*b^3*d^3 + (b^7*c^2*d 
- 2*a*b^6*c*d^2 + a^2*b^5*d^3)*x^2 + 2*(a*b^6*c^2*d - 2*a^2*b^5*c*d^2 + a^ 
3*b^4*d^3)*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x^2/(b*x+a)^(5/2)/(d*x+c)^(1/2),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. 318 vs. \(2 (102) = 204\).

Time = 0.21 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.52 \[ \int \frac {x^2}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx=-\frac {\frac {3 \, \sqrt {b d} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{d {\left | b \right |}} - \frac {8 \, {\left (3 \, \sqrt {b d} a b^{5} c^{2} - 5 \, \sqrt {b d} a^{2} b^{4} c d + 2 \, \sqrt {b d} a^{3} b^{3} d^{2} - 6 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{3} c + 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{2} d + 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a b\right )}}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{3} {\left | b \right |}}}{3 \, b^{2}} \] Input:

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

Output:

-1/3*(3*sqrt(b*d)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b* 
d - a*b*d))^2)/(d*abs(b)) - 8*(3*sqrt(b*d)*a*b^5*c^2 - 5*sqrt(b*d)*a^2*b^4 
*c*d + 2*sqrt(b*d)*a^3*b^3*d^2 - 6*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sq 
rt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^3*c + 3*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^2*d + 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)/(( 
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*abs(b)))/b^2
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.77 \[ \int \frac {x^2}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx=\frac {2 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{3} d^{2}-4 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b c d +2 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b \,d^{2} x +2 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,b^{2} c^{2}-4 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,b^{2} c d x +2 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{3} c^{2} x -\frac {4 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a^{2} b c d}{3}-\frac {4 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a \,b^{2} c d x}{3}-2 \sqrt {d x +c}\, a^{3} b \,d^{2}+\frac {10 \sqrt {d x +c}\, a^{2} b^{2} c d}{3}-\frac {8 \sqrt {d x +c}\, a^{2} b^{2} d^{2} x}{3}+4 \sqrt {d x +c}\, a \,b^{3} c d x}{\sqrt {b x +a}\, b^{3} d \left (a^{2} b \,d^{2} x -2 a \,b^{2} c d x +b^{3} c^{2} x +a^{3} d^{2}-2 a^{2} b c d +a \,b^{2} c^{2}\right )} \] Input:

int(x^2/(b*x+a)^(5/2)/(d*x+c)^(1/2),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*d**2 - 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**2 
*b*c*d + 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**2*b*d**2*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*b**2*c**2 - 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*b**2*c*d*x + 3*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))*b**3*c**2*x - 2*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**2*b*c*d - 2 
*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b**2*c*d*x - 3*sqrt(c + d*x)*a**3*b*d**2 
+ 5*sqrt(c + d*x)*a**2*b**2*c*d - 4*sqrt(c + d*x)*a**2*b**2*d**2*x + 6*sqr 
t(c + d*x)*a*b**3*c*d*x))/(3*sqrt(a + b*x)*b**3*d*(a**3*d**2 - 2*a**2*b*c* 
d + a**2*b*d**2*x + a*b**2*c**2 - 2*a*b**2*c*d*x + b**3*c**2*x))