\(\int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx\) [250]

Optimal result

Integrand size = 22, antiderivative size = 248 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\frac {2 c^2 (a+b x)^{5/2}}{d^2 (b c-a d) \sqrt {c+d x}}+\frac {\left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b d^4}+\frac {\left (10 a c-\frac {35 b c^2}{d}+\frac {a^2 d}{b}\right ) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^2 (b c-a d)}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d^2}-\frac {(b c-a d) \left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{3/2} d^{9/2}} \] Output:

2*c^2*(b*x+a)^(5/2)/d^2/(-a*d+b*c)/(d*x+c)^(1/2)+1/8*(-a^2*d^2-10*a*b*c*d+ 
35*b^2*c^2)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b/d^4+1/12*(10*a*c-35*b*c^2/d+a^2* 
d/b)*(b*x+a)^(3/2)*(d*x+c)^(1/2)/d^2/(-a*d+b*c)+1/3*(b*x+a)^(5/2)*(d*x+c)^ 
(1/2)/b/d^2-1/8*(-a*d+b*c)*(-a^2*d^2-10*a*b*c*d+35*b^2*c^2)*arctanh(d^(1/2 
)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(3/2)/d^(9/2)
 

Mathematica [A] (verified)

Time = 10.31 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.79 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\frac {b \sqrt {d} \sqrt {a+b x} \left (3 a^2 d^2 (c+d x)+2 a b d \left (-50 c^2-19 c d x+7 d^2 x^2\right )+b^2 \left (105 c^3+35 c^2 d x-14 c d^2 x^2+8 d^3 x^3\right )\right )-3 (b c-a d)^{3/2} \left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \sqrt {\frac {b (c+d x)}{b c-a d}} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{24 b^2 d^{9/2} \sqrt {c+d x}} \] Input:

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

Output:

(b*Sqrt[d]*Sqrt[a + b*x]*(3*a^2*d^2*(c + d*x) + 2*a*b*d*(-50*c^2 - 19*c*d* 
x + 7*d^2*x^2) + b^2*(105*c^3 + 35*c^2*d*x - 14*c*d^2*x^2 + 8*d^3*x^3)) - 
3*(b*c - a*d)^(3/2)*(35*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*Sqrt[(b*(c + d*x)) 
/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(24*b^2*d^ 
(9/2)*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {100, 27, 90, 60, 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.

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{d^2 \sqrt {c+d x} (b c-a d)}-\frac {\frac {\left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{6 b}-\frac {(a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}{3 b}}{d^2 (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{d^2 \sqrt {c+d x} (b c-a d)}-\frac {\frac {\left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{6 b}-\frac {(a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}{3 b}}{d^2 (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{d^2 \sqrt {c+d x} (b c-a d)}-\frac {\frac {\left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{6 b}-\frac {(a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}{3 b}}{d^2 (b c-a d)}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{d^2 \sqrt {c+d x} (b c-a d)}-\frac {\frac {\left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{6 b}-\frac {(a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}{3 b}}{d^2 (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{d^2 \sqrt {c+d x} (b c-a d)}-\frac {\frac {\left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{6 b}-\frac {(a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}{3 b}}{d^2 (b c-a d)}\)

Input:

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

Output:

(2*c^2*(a + b*x)^(5/2))/(d^2*(b*c - a*d)*Sqrt[c + d*x]) - (-1/3*((b*c - a* 
d)*(a + b*x)^(5/2)*Sqrt[c + d*x])/b + ((35*b^2*c^2 - 10*a*b*c*d - a^2*d^2) 
*(((a + b*x)^(3/2)*Sqrt[c + d*x])/(2*d) - (3*(b*c - a*d)*((Sqrt[a + b*x]*S 
qrt[c + d*x])/d - ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sq 
rt[c + d*x])])/(Sqrt[b]*d^(3/2))))/(4*d)))/(6*b))/(d^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 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 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. \(691\) vs. \(2(212)=424\).

Time = 0.24 (sec) , antiderivative size = 692, normalized size of antiderivative = 2.79

method	result	size
default	\(-\frac {\sqrt {b x +a}\, \left (-16 b^{2} d^{3} x^{3} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+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^{3} d^{4} x +27 \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^{3} x -135 \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^{2} x +105 \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} d x -28 a b \,d^{3} x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+28 b^{2} c \,d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+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^{3} c \,d^{3}+27 \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^{2} d^{2}-135 \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^{3} d +105 \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^{4}-6 a^{2} d^{3} x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+76 a b c \,d^{2} x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}-70 b^{2} c^{2} d x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}-6 a^{2} c \,d^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+200 a b \,c^{2} d \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}-210 b^{2} c^{3} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\right )}{48 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {x d +c}\, d^{4} b}\)	\(692\)

Input:

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

Output:

-1/48*(b*x+a)^(1/2)*(-16*b^2*d^3*x^3*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(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^3*d^4*x+27*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^3*x-135*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^2*x+105*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*d*x-28 
*a*b*d^3*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+28*b^2*c*d^2*x^2*((b*x+a) 
*(d*x+c))^(1/2)*(d*b)^(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^3*c*d^3+27*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^2*d^2-135*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 
^3*d+105*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^4-6*a^2*d^3*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+76*a*b* 
c*d^2*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-70*b^2*c^2*d*x*((b*x+a)*(d*x+c 
))^(1/2)*(d*b)^(1/2)-6*a^2*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+200*a 
*b*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-210*b^2*c^3*((b*x+a)*(d*x+c)) 
^(1/2)*(d*b)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(d*b)^(1/2)/(d*x+c)^(1/2)/d^4/ 
b
 

Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 600, normalized size of antiderivative = 2.42 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\left [\frac {3 \, {\left (35 \, b^{3} c^{4} - 45 \, a b^{2} c^{3} d + 9 \, a^{2} b c^{2} d^{2} + a^{3} c d^{3} + {\left (35 \, b^{3} c^{3} d - 45 \, a b^{2} c^{2} d^{2} + 9 \, a^{2} b c d^{3} + a^{3} d^{4}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (8 \, b^{3} d^{4} x^{3} + 105 \, b^{3} c^{3} d - 100 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - 14 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{2} + {\left (35 \, b^{3} c^{2} d^{2} - 38 \, a b^{2} c d^{3} + 3 \, a^{2} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, {\left (b^{2} d^{6} x + b^{2} c d^{5}\right )}}, \frac {3 \, {\left (35 \, b^{3} c^{4} - 45 \, a b^{2} c^{3} d + 9 \, a^{2} b c^{2} d^{2} + a^{3} c d^{3} + {\left (35 \, b^{3} c^{3} d - 45 \, a b^{2} c^{2} d^{2} + 9 \, a^{2} b c d^{3} + a^{3} d^{4}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{4} x^{3} + 105 \, b^{3} c^{3} d - 100 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - 14 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{2} + {\left (35 \, b^{3} c^{2} d^{2} - 38 \, a b^{2} c d^{3} + 3 \, a^{2} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (b^{2} d^{6} x + b^{2} c d^{5}\right )}}\right ] \] Input:

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

Output:

[1/96*(3*(35*b^3*c^4 - 45*a*b^2*c^3*d + 9*a^2*b*c^2*d^2 + a^3*c*d^3 + (35* 
b^3*c^3*d - 45*a*b^2*c^2*d^2 + 9*a^2*b*c*d^3 + a^3*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)*sqr 
t(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 
^4*x^3 + 105*b^3*c^3*d - 100*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - 14*(b^3*c*d^3 
 - a*b^2*d^4)*x^2 + (35*b^3*c^2*d^2 - 38*a*b^2*c*d^3 + 3*a^2*b*d^4)*x)*sqr 
t(b*x + a)*sqrt(d*x + c))/(b^2*d^6*x + b^2*c*d^5), 1/48*(3*(35*b^3*c^4 - 4 
5*a*b^2*c^3*d + 9*a^2*b*c^2*d^2 + a^3*c*d^3 + (35*b^3*c^3*d - 45*a*b^2*c^2 
*d^2 + 9*a^2*b*c*d^3 + a^3*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*(8*b^3*d^4*x^3 + 105*b^3*c^3*d - 100*a*b^2*c^2*d^2 
+ 3*a^2*b*c*d^3 - 14*(b^3*c*d^3 - a*b^2*d^4)*x^2 + (35*b^3*c^2*d^2 - 38*a* 
b^2*c*d^3 + 3*a^2*b*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^6*x + b^2* 
c*d^5)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x^2*(b*x+a)^(3/2)/(d*x+c)^(3/2),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 = 0.18 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.14 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\frac {{\left ({\left (b x + a\right )} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{d {\left | b \right |}} - \frac {7 \, b^{3} c d^{5} + 5 \, a b^{2} d^{6}}{b^{2} d^{7} {\left | b \right |}}\right )} + \frac {35 \, b^{4} c^{2} d^{4} - 10 \, a b^{3} c d^{5} - a^{2} b^{2} d^{6}}{b^{2} d^{7} {\left | b \right |}}\right )} + \frac {3 \, {\left (35 \, b^{5} c^{3} d^{3} - 45 \, a b^{4} c^{2} d^{4} + 9 \, a^{2} b^{3} c d^{5} + a^{3} b^{2} d^{6}\right )}}{b^{2} d^{7} {\left | b \right |}}\right )} \sqrt {b x + a}}{24 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {{\left (35 \, b^{3} c^{3} - 45 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{8 \, \sqrt {b d} d^{4} {\left | b \right |}} \] Input:

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

Output:

1/24*((b*x + a)*(2*(b*x + a)*(4*(b*x + a)/(d*abs(b)) - (7*b^3*c*d^5 + 5*a* 
b^2*d^6)/(b^2*d^7*abs(b))) + (35*b^4*c^2*d^4 - 10*a*b^3*c*d^5 - a^2*b^2*d^ 
6)/(b^2*d^7*abs(b))) + 3*(35*b^5*c^3*d^3 - 45*a*b^4*c^2*d^4 + 9*a^2*b^3*c* 
d^5 + a^3*b^2*d^6)/(b^2*d^7*abs(b)))*sqrt(b*x + a)/sqrt(b^2*c + (b*x + a)* 
b*d - a*b*d) + 1/8*(35*b^3*c^3 - 45*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3) 
*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/ 
(sqrt(b*d)*d^4*abs(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 1.90 (sec) , antiderivative size = 707, normalized size of antiderivative = 2.85 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(24*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*c*d**3 + 24*sqrt(c + d*x)*sqrt(a + 
b*x)*a**2*b*d**4*x - 800*sqrt(c + d*x)*sqrt(a + b*x)*a*b**2*c**2*d**2 - 30 
4*sqrt(c + d*x)*sqrt(a + b*x)*a*b**2*c*d**3*x + 112*sqrt(c + d*x)*sqrt(a + 
 b*x)*a*b**2*d**4*x**2 + 840*sqrt(c + d*x)*sqrt(a + b*x)*b**3*c**3*d + 280 
*sqrt(c + d*x)*sqrt(a + b*x)*b**3*c**2*d**2*x - 112*sqrt(c + d*x)*sqrt(a + 
 b*x)*b**3*c*d**3*x**2 + 64*sqrt(c + d*x)*sqrt(a + b*x)*b**3*d**4*x**3 - 2 
4*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt 
(a*d - b*c))*a**3*c*d**3 - 24*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + 
 sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*d**4*x - 216*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**2*d**2 - 216*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sq 
rt(c + d*x))/sqrt(a*d - b*c))*a**2*b*c*d**3*x + 1080*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** 
3*d + 1080*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**2*x - 840*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**4 - 840* 
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*d*x + 3*sqrt(d)*sqrt(b)*a**3*c*d**3 + 3*sqrt(d)*sqrt( 
b)*a**3*d**4*x + 87*sqrt(d)*sqrt(b)*a**2*b*c**2*d**2 + 87*sqrt(d)*sqrt(b)* 
a**2*b*c*d**3*x - 615*sqrt(d)*sqrt(b)*a*b**2*c**3*d - 615*sqrt(d)*sqrt(...