\(\int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx\) [317]

Optimal result

Integrand size = 22, antiderivative size = 189 \[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\frac {\left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^3 d^2}-\frac {(b c+3 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 b^2 d^2}+\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 b^2 d}+\frac {(b c-a d) \left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{5/2}} \] Output:

1/8*(5*a^2*d^2+2*a*b*c*d+b^2*c^2)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^3/d^2-1/4* 
(3*a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(3/2)/b^2/d^2+1/3*(b*x+a)^(3/2)*(d*x+c)^ 
(3/2)/b^2/d+1/8*(-a*d+b*c)*(5*a^2*d^2+2*a*b*c*d+b^2*c^2)*arctanh(d^(1/2)*( 
b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(7/2)/d^(5/2)
 

Mathematica [A] (verified)

Time = 10.23 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\frac {\sqrt {c+d x} \left (\sqrt {d} \sqrt {a+b x} \left (15 a^2 d^2-2 a b d (2 c+5 d x)+b^2 \left (-3 c^2+2 c d x+8 d^2 x^2\right )\right )+\frac {3 \sqrt {b c-a d} \left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {\frac {b (c+d x)}{b c-a d}}}\right )}{24 b^3 d^{5/2}} \] Input:

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

Output:

(Sqrt[c + d*x]*(Sqrt[d]*Sqrt[a + b*x]*(15*a^2*d^2 - 2*a*b*d*(2*c + 5*d*x) 
+ b^2*(-3*c^2 + 2*c*d*x + 8*d^2*x^2)) + (3*Sqrt[b*c - a*d]*(b^2*c^2 + 2*a* 
b*c*d + 5*a^2*d^2)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/Sqrt[ 
(b*(c + d*x))/(b*c - a*d)]))/(24*b^3*d^(5/2))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {101, 27, 90, 60, 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.

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

\(\Big \downarrow \) 101

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{3/2}}{3 b d}-\frac {\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+3 b c)}{2 b d}-\frac {3 \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b d}}{6 b d}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{3/2}}{3 b d}-\frac {\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+3 b c)}{2 b d}-\frac {3 \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \left (\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b d}}{6 b d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{3/2}}{3 b d}-\frac {\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+3 b c)}{2 b d}-\frac {3 \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b d}}{6 b d}\)

Input:

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

Output:

(x*Sqrt[a + b*x]*(c + d*x)^(3/2))/(3*b*d) - (((3*b*c + 5*a*d)*Sqrt[a + b*x 
]*(c + d*x)^(3/2))/(2*b*d) - (3*(b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*((Sqrt[a 
 + b*x]*Sqrt[c + d*x])/b + ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(S 
qrt[b]*Sqrt[c + d*x])])/(b^(3/2)*Sqrt[d])))/(4*b*d))/(6*b*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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, 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 90

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

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

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. \(394\) vs. \(2(157)=314\).

Time = 0.22 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.09

method	result	size
default	\(-\frac {\sqrt {x d +c}\, \sqrt {b x +a}\, \left (-16 b^{2} d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{3} d^{3}-9 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} b c \,d^{2}-3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a \,b^{2} c^{2} d -3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) b^{3} c^{3}+20 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a b \,d^{2} x -4 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, b^{2} c d x -30 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a^{2} d^{2}+8 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a b c d +6 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, b^{2} c^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, b^{3} d^{2} \sqrt {d b}}\)	\(395\)

Input:

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

Output:

-1/48*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(-16*b^2*d^2*x^2*((b*x+a)*(d*x+c))^(1/2) 
*(d*b)^(1/2)+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*d^3-9*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*c*d^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*b^2*c^2*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))*b^3*c^3+20*( 
d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*d^2*x-4*(d*b)^(1/2)*((b*x+a)*(d*x+c 
))^(1/2)*b^2*c*d*x-30*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*d^2+8*(d*b)^ 
(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c*d+6*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2 
)*b^2*c^2)/((b*x+a)*(d*x+c))^(1/2)/b^3/d^2/(d*b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.16 \[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\left [-\frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d - 4 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{4} d^{3}}, -\frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d - 4 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{4} d^{3}}\right ] \] Input:

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

Output:

[-1/96*(3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*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*(8*b^ 
3*d^3*x^2 - 3*b^3*c^2*d - 4*a*b^2*c*d^2 + 15*a^2*b*d^3 + 2*(b^3*c*d^2 - 5* 
a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*d^3), -1/48*(3*(b^3*c^3 + 
a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*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*(8*b^3*d^3*x^2 - 3*b^3*c^2*d - 4*a*b^2*c*d^2 
+ 15*a^2*b*d^3 + 2*(b^3*c*d^2 - 5*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c 
))/(b^4*d^3)]
 

Sympy [F]

\[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\int \frac {x^{2} \sqrt {c + d x}}{\sqrt {a + b x}}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

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

integrate(x^2*(d*x+c)^(1/2)/(b*x+a)^(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 [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.99 \[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, {\left (4 \, b x + 4 \, a + \frac {b c d^{3} - 13 \, a d^{4}}{d^{4}}\right )} {\left (b x + a\right )} - \frac {3 \, {\left (b^{2} c^{2} d^{2} + 2 \, a b c d^{3} - 11 \, a^{2} d^{4}\right )}}{d^{4}}\right )} \sqrt {b x + a} - \frac {3 \, {\left (b^{4} c^{3} + a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - 5 \, a^{3} b d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{2}}\right )} {\left | b \right |}}{24 \, b^{5}} \] Input:

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

Output:

1/24*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(4*b*x + 4*a + (b*c*d^3 - 13* 
a*d^4)/d^4)*(b*x + a) - 3*(b^2*c^2*d^2 + 2*a*b*c*d^3 - 11*a^2*d^4)/d^4)*sq 
rt(b*x + a) - 3*(b^4*c^3 + a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - 5*a^3*b*d^3)*lo 
g(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sq 
rt(b*d)*d^2))*abs(b)/b^5
 

Mupad [B] (verification not implemented)

Time = 29.23 (sec) , antiderivative size = 924, normalized size of antiderivative = 4.89 \[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx =\text {Too large to display} \] Input:

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

Output:

- ((((a + b*x)^(1/2) - a^(1/2))*((b^5*c^3)/4 - (5*a^3*b^2*d^3)/4 + (3*a^2* 
b^3*c*d^2)/4 + (a*b^4*c^2*d)/4))/(d^8*((c + d*x)^(1/2) - c^(1/2))) - (((a 
+ b*x)^(1/2) - a^(1/2))^5*((33*a^3*d^3)/2 + (19*b^3*c^3)/2 + (275*a*b^2*c^ 
2*d)/2 + (313*a^2*b*c*d^2)/2))/(d^6*((c + d*x)^(1/2) - c^(1/2))^5) - (((a 
+ b*x)^(1/2) - a^(1/2))^3*((17*b^4*c^3)/12 - (85*a^3*b*d^3)/12 + (17*a^2*b 
^2*c*d^2)/4 + (91*a*b^3*c^2*d)/4))/(d^7*((c + d*x)^(1/2) - c^(1/2))^3) + ( 
((a + b*x)^(1/2) - a^(1/2))^11*((b^3*c^3)/4 - (5*a^3*d^3)/4 + (a*b^2*c^2*d 
)/4 + (3*a^2*b*c*d^2)/4))/(b^3*d^3*((c + d*x)^(1/2) - c^(1/2))^11) - (((a 
+ b*x)^(1/2) - a^(1/2))^9*((17*b^3*c^3)/12 - (85*a^3*d^3)/12 + (91*a*b^2*c 
^2*d)/4 + (17*a^2*b*c*d^2)/4))/(b^2*d^4*((c + d*x)^(1/2) - c^(1/2))^9) - ( 
((a + b*x)^(1/2) - a^(1/2))^7*((33*a^3*d^3)/2 + (19*b^3*c^3)/2 + (275*a*b^ 
2*c^2*d)/2 + (313*a^2*b*c*d^2)/2))/(b*d^5*((c + d*x)^(1/2) - c^(1/2))^7) + 
 (a^(1/2)*c^(1/2)*((a + b*x)^(1/2) - a^(1/2))^6*(128*a^2*d^2 + 64*b^2*c^2 
+ (704*a*b*c*d)/3))/(d^5*((c + d*x)^(1/2) - c^(1/2))^6) + (a^(1/2)*c^(1/2) 
*(32*b*c^2 + 96*a*c*d)*((a + b*x)^(1/2) - a^(1/2))^8)/(d^4*((c + d*x)^(1/2 
) - c^(1/2))^8) + (a^(1/2)*c^(1/2)*(32*b^3*c^2 + 96*a*b^2*c*d)*((a + b*x)^ 
(1/2) - a^(1/2))^4)/(d^6*((c + d*x)^(1/2) - c^(1/2))^4))/(((a + b*x)^(1/2) 
 - a^(1/2))^12/((c + d*x)^(1/2) - c^(1/2))^12 + b^6/d^6 - (6*b^5*((a + b*x 
)^(1/2) - a^(1/2))^2)/(d^5*((c + d*x)^(1/2) - c^(1/2))^2) + (15*b^4*((a + 
b*x)^(1/2) - a^(1/2))^4)/(d^4*((c + d*x)^(1/2) - c^(1/2))^4) - (20*b^3*...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.69 \[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\frac {15 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b \,d^{3}-4 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} c \,d^{2}-10 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} d^{3} x -3 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} c^{2} d +2 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} c \,d^{2} x +8 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} d^{3} x^{2}-15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{3} d^{3}+9 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b c \,d^{2}+3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,b^{2} c^{2} d +3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{3} c^{3}}{24 b^{4} d^{3}} \] Input:

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

Output:

(15*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*d**3 - 4*sqrt(c + d*x)*sqrt(a + b*x 
)*a*b**2*c*d**2 - 10*sqrt(c + d*x)*sqrt(a + b*x)*a*b**2*d**3*x - 3*sqrt(c 
+ d*x)*sqrt(a + b*x)*b**3*c**2*d + 2*sqrt(c + d*x)*sqrt(a + b*x)*b**3*c*d* 
*2*x + 8*sqrt(c + d*x)*sqrt(a + b*x)*b**3*d**3*x**2 - 15*sqrt(d)*sqrt(b)*l 
og((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*d 
**3 + 9*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x) 
)/sqrt(a*d - b*c))*a**2*b*c*d**2 + 3*sqrt(d)*sqrt(b)*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*sqrt(d)* 
sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c 
))*b**3*c**3)/(24*b**4*d**3)